foundations of algebraic geometry ... - stanford university

FOUNDATIONS OF ALGEBRAIC GEOMETRY:

Rough notes for 2007–08 course

December 10, 2007.

c© 2005, 2006, 2007 by Ravi Vakil.

Contents

Chapter 1. Strictly edited notes for myself 7

Chapter 2. Introduction 11

Part I. Preliminaries 27

Chapter 3. Some category theory 29

Chapter 4. Sheaves 67

Part II. Schemes and morphisms of schemes 91

Chapter 5. Toward affine schemes: the underlying set, and the underlyingtopological space 93

Chapter 6. The structure sheaf of an affine scheme, and the definition ofschemes in general 117

Chapter 7. Some properties of schemes 131

Part III. Morphisms of schemes 151

Chapter 8. Morphisms of schemes 153

Chapter 9. Some properties of morphisms of schemes 163

Chapter 10. Projective schemes 183

Chapter 11. Fibered products of schemes 195

Chapter 12. Separated morphisms, and varieties 213

Chapter 13. Rational maps 227

Chapter 14. Proper morphisms 239

Part IV. Harder properties of schemes 245

Chapter 15. Dimension 247

Chapter 16. Nonsingularity (or regularity or “Smoothness”) 275

Part V. Quasicoherent sheaves 299

3

Chapter 17. Quasicoherent and coherent sheaves 301

Chapter 18. Invertible sheaves and divisors 319

Chapter 19. Quasicoherent sheaves on projective A-schemes and gradedmodules 325

Chapter 20. Pushforwards and pullbacks of quasicoherent sheaves 333

Chapter 21. Constructing schemes by universal properties II: Sheafy spec andproj, and projective morphisms 343

Chapter 22. Differentials 353

Part VI. Cohomology 377

Chapter 23. Cech cohomology of quasicoherent sheaves 379

Chapter 24. ? Derived functor cohomology 407

Chapter 25. Running example: curves 411

Chapter 26. Base-point-free, ampleness, very ampleness 439

Part VII. Useful properties and constructions (not currentlyordered) 449

Chapter 27. ? Blowing up a scheme along a closed subscheme 451

Chapter 28. ? Varieties over non-algebraically closed fields 463

Chapter 29. Flatness 465

Chapter 30. Smooth, etale, unramified 495

Chapter 31. Serre duality 509

Chapter 32. Formal functions 523

Chapter 33. ? ? Comparing complex analytic and complex algebraic geometry:Serre’s “GAGA” Theorem 527

Part VIII. ? Further topics to explore 531

Chapter 34. ? Germs of topics 533

Chapter 35. ? Toric varieties 535

Chapter 36. ? Intersection theory 537

Chapter 37. ? Moduli spaces, moduli functors, and stacks 539

Chapter 38. Things still to place, and things not to include 545

Appendix A. Glossary 549

Appendix B. Crib notes of facts 553

Appendix. Bibliography 561

Appendix. Index 563

6

[forme:

CHAPTER 1

Strictly edited notes for myself

1.1 Parsimonious list of guidelines

Three editing environments:

• forreader [...]. These are comments to ask the (proof-)readers.• primordial = primordial ooze• forexperts (questions to ask; comments on controversial pedagogical choices)• forme (comparison to other results in the literature; how important exer-

cises are; links to elsewhere in the book, e.g. where results are used later;post mortem on how hard things are; things I still need to write or thinkabout writing TEND TO BE IN CAPS or STARRED***)

Sectioning:

• part (last)• chapter• vsection• subsection• Theorem Exercise (exercisedone) Definition Remark.

Other comment commands:

• notation• reference (also hartshorne)• lremind• cut (for things I don’t ever want to see, but want to keep in the file)

1.2 Tex tips

• varnothing ∅ not emptyset ∅• varprojlim lim←− not lim←, and varinjlim lim−→ not lim→.

hook, dotted, dashed arrows. a hi // b //___ c .

two arrows pointing to the right Aφ //ψ

// B

bowed arrows

U × Rn

π

U

f1,...,fn

VV

Boldface in index is index notation | textbf (no space)

7

8 Foundations of Algebraic Geometry

1.3 Conventions I will follow

Notation:

• Rings are A and B in the French style, not R and S. Scheme morphismis π : X → Y , although π may sometimes be f . Symbols that go together:X,A,m. Hence A is a B-algebra. I think we will have maps X → SpecBrather than X → SpecA, although I’m not sure about that. Ideals are I andJ. Modules are M and N . Schemes are X, Y , Z, and bases are often S, T .Sheaves are F , G.

• complexes/indices will be S•, where that symbol could be changed, e.g.possibly later to ∗. However, ∗ has too many meanings, and · is too small.But I may shrink the bullet. For graded rings, we have S0 = A. Sf is (S•)f .

• Γ(U,F), and later H0(U,F).• sheaves are underlined, e.g. Hom, not in calligraphic font.• Coordinates on affine space are xi. Projective coordinates x0/1.• Duals are ∨ not ∗.• FF stands for two things: fraction field (and total fraction ring, Eisenbud’s

total quotient ring), and function field. Defined for reduced and locallyNoetherian. Explain in the text why the usual notations of K(·), C(·), k(·)are not good.

• Blow-up is β : BlX Y → Y with exceptional divisor EXY .

Spellings:

• Leibniz (following Eisenbud). base-point-free. Unique factorization do-main, Krull’s Principal Ideal Theorem. Algebraic Hartogs’ Lemma (forNoetherian normal schemes). Cancellation Theorem for properties of schemes 12.1.19.

1.4 Various stages of notes (long-term planning)

• serious editing and writing “algebraic geometry done right” [and compre-hensively]

– teach class and order material– go through standard texts and make sure all important results are

here (see Checklist of things they should know); perhaps make a cribsheet

– then a write-through. In each section, have a forme of what the bigpicture is, and what they should know at the end.

• public posting

Extremely long-term issues:

• Names: FOAG is my working title. Schemes seems a good one. Kiran sug-gests Schemes and algebraic geometry, SAG or SAAG. Or Schemes and coho-

mology in algebraic geometry. Or Foundational/Fundamental Lessons/Lecturesin Algebraic Geometry. Possibly: “The Rising Sea: Foundations of Alge-braic Geometry”. Or “The Rising Tide”.

• commented out here: Tex for “Ravi Vakil, Palo Alto, today”• Some cover ideas:

– possibly: the most important diagrams from the text– the magic diagram Exercise 3.3.L– ellipse fact (see below); poncelet– quadric surface with 2 rulings, and also 2 lines meeting 4 lines– E6 singularity in the water

– Joe Harris’ picture: family of curves– Picture of schemes with associated points, maybe with map from Spec

of the dual numbers.– Serre duality statement

Ravi Vakil 9

– RR for curves

Fun ellipse fact in the style of Poncelet. Given an ellipse in the plane, suppose thereis a point such that when you do the following thing a finite number of times, youreturn to the starting point: take the other point on the ellipse with the same x-coordinate; then the same thing with the y-coordinate. Then show this is true nomatter where you start. Idea: degenerate elliptic curve on P1 × P1. To get concreteexamples: translate ellipse so it is centered at the origin, then scale so that it is ofthe form x2 + bxy + y2 = 1. Then

(x, y) 7→ (−by − x, y) 7→ (−by − x,−b(−by − x)− y) = (−by − x, bx+ (b2 − 1)y).

Then we get a matrix, whose eigenvalues satisfy

λ2 + λ(2− b2) + 1 = 0.

Here b ≤ 4, so we get a root of unity. We can get any root we want. A particularlyfun case is b =

√3. Then we get a 6th root of 1. Fun side fact: if we start over Q, then

the period can only a few finite things. Reason: roots of unity are not in quadraticextensions of Q.

1.4.1. 17 Sources. Hartshorne. Mumford Red book. Mumford Abelian varieties.

Complex projective varieties I. Shafarevich. Hindry-Silverman. Debarre; Smith et al.

Iitaka. Harris. Eisenbud-Harris. FAC, Dolgachev. Gathmann. Yves Andre. Milne.

Fundamentals of Algebraic Geometry (Barbara et al.) Not EGA. ]

CHAPTER 2

Introduction

[forexperts: The title “Foundations of Algebraic Geometry” was justa quickly chosen working title. What should this be called?]

[forme: Brief introduction goes here, and philosophy.]

2.1 Why algebraic geometry?

2.2 Background and conventions

1All rings are assumed to be commutative unless explicitly stated otherwise.All rings are assumed to have 1. Recall that maps of rings must send 1 to 1. Wedon’t require that 0 6= 1; in other words, the “0-ring” (with one element) is assumedto be the ring. (There is a ring map from any ring to the 0-ring; the 0-ring onlymaps to itself. The 0-ring is the final object in the category of rings.) [forme: 0-ring, zero ring

Many reasons for 0-ring, including: final object in rings. For all x ∈ A, Ax is a ring.]

We accept the axiom of choice. In particular, any ideal in a ring is containedin a maximal ideal. 2 AC

The reader should be familiar with some basic notions in commutative ringtheory, in particular the notion of ideals (including prime and maximal ideals)and localization. For example, the reader should be able to show that if S is amultiplicative set of a ring A, then the primes of S−1A are in natural bijectionwith those primes of A not meeting S (§5.3.E). The notion of tensor products andexact sequences of A-modules will be useful. [forme: Notions from commutative

algebra that we use: rings, modules, prime and maximal ideals, ideals and quotients,

localization, tensor product, exact sequences.]

We will not concern ourselves with set-theoretic issues of sets, classes, universes,small categories, etc. [forme: ADD MAX’s issues: About subtle foundational issues

(small categories, set-theoretic issues, etc.): It is true that some people should be careful

about these isues: but do you really want to be one of those people?]

Further background. It may be helpful to have books on other subjects handythat you can dip into for specific facts, rather than reading them in advance. In com-mutative algebra, Eisenbud [E] is good for this. Other popular choices are Atiyah-Macdonald [AM] and Matsumura [M-CRT]. For homological algebra, Weibel

1conventions

2assumeAC

11


[W] (homological algebra) has detailed and readable. Another good possibility isKashiwara-Schapira [KS].

Background from other parts of mathematics (topology, geometry, complexanalysis) will be helpful for developing intuition.

2.2.1. Acknowledgments. [primordial: In no particular order. Joe, Brendan,Johan, Mike Roth, Jason Starr, Brian Conrad, Brian Osserman, Kiran Kedlaya. Students (incl.216). Martin Olsson. Daniel Allcock. Justin Walker. Max Lieblich. Bierstone and Millman.Barbara Fantechi. David Eisenbud. Many students (my students, Jason Lo, Kirsten Wickel-

gren)! Greg Brumfiel. Many others who told me how to think about things will unfortunately gounacknowledged in the text.

][primordial:[to intro: Derived functors introduced, then to be read later. Needed only for Ext and Serre

duality. This is controversial: most would introduce derived functors early.]So: we work with Noetherian schemes in general, pointing out where things are nicer for finite

type schemes over k (and nicer still for k algebraically closed). We’re happy discussing things non-Noetherian, only where there is relatively little cost. (Coherence is the main issue.) But we won’tdo pages and pages of reduction to the Noetherian case. In this way, (i) non-Noetherian peoplewill clearly see where the issues are. (ii) Noetherian people will realize that non-Noetherian thingsare not so bad.

2.3 Area undergoing ongoing editing

2.3.1. Conventions.Pedagogy:

• Goal: complete first course. Everything developed from scratch, but in reasonable gen-erality. This may require leaving some things out of a course, but given for students toread; this way they have a reference. One year course (but very full year). Introductionto topics, all facts needed en route presented. Bonus topics. [primordial: We don’tmind proving things in generality. We won’t have harder proofs. ]

• A key point: there is a tendency I want to avoid: leaving to the reader to check thatsomething technical is true. Often this is easy for experts, but if you haven’t seensuch an argument before, you have no idea how to begin. One normally does this toavoid discussing something hard. A second tendency: doing something rather limitedin scope.

• I’m not going to cover topics with no pay-off (although even philosophical pay-off, orability to follow talks, counts as a pay-off). E.g. formal schemes are best done near theformal schemesstart of a course where they will be used.

• Not going for density of exposition; instead, pedagogically, schemes are obtained by glu-ing together affines. Instead, completeness of exposition, and as few “hard” argumentsas possible.

• Use as few as possible tools from elsewhere. Develop them from scratch (includingcommutative algebra).

• We try to make unnecessary hypotheses only when they substantially reduce the lengthor complexity of the exposition. Thus even though most readers will be Noetherian-minded (and indeed will prefer to work with finitely generated algebras over a field),they will have to deal with different variations of the notion of coherence; but theycan and should just merge these notions in their mind. On the other hand, we work

heavily with projective morphisms. The extension to proper morphisms sometimesinvolves more work.

• Some topics can be done naively, and then in a more sophisticated manner. Shouldthey then be done twice? It depends. If we can do it quickly, I want to do it. Hence:

Ravi Vakil 13

varieties: no. Rational maps: no (after separated morphisms! but we have rationalfunctions). Projective space: yes (gluing, then proj).

• ? (not “+”; in front or end of title?). ? = don’t read the first time. Possibly worthreading in a second reading. Instructors may wish (or be encouraged) to skip it. Notethat ? doesn’t necessarily mean “hard”, although it might. ??: handy as a reference,but definitely not to read. But sleep soundly at night knowing that you can read it ifyou wanted.

• Definition of unimportant: unimportant for the current exposition, not necessarilyunimportant in the larger scheme of things. Define “For pedants” (not intended asa put-down), “For Noetherian readers”. (Example: For pedants: we will ignore all“set/class” issues.)

• index: thorough; and bold-face is the definition, example . The “boldface” in the notationnotation indicates that the index entry is bold-faced.

• Exercises: be clear on how hard they are.– Fast answer v. find the reference v. long.– Easy v. hard.– Important result (used later) v. unimportant. [the meaning of “unimportant”.– Famous result, and hard, but we won’t use it in the exposition. (Shouldn’t be

exercise!)

2.3.2. Questions for experts and readers. Are there things from commutative algebraI used without proof or (sufficient) explanation? (I’d especially like to hear from non-experts.)

Notes to readers. [Later divide up into students and faculty.]

• Are there any epiphanies that should be included?• What are the sections where you can’t get a lot of insight out of them? Things that

seem like especially hard reading?• Are there some explanations that could be made easier, e.g. by rearranging the order

of various bits of material?

To experts: Here are some things I found troubling in a first course. How do you knowthat A3 is smooth (i) over the complex numbers, (ii) over an arbitrary field? A2 over an arbitraryfield?

“No offense, but I think that you make this argument too complicated, and do it the rightway. It is most easily seen in this way...”

Why is dimR[x] = dimR+1? I can show it if I know that the integral closure of a Noetherianlocal ring of dimension 1 is also Noetherian.

2.4 Introduction rough notes

[Note for people using this book as a text: you will need to supplement this to suit yourparticular needs. Also, following this book will force you to go through material that you mightprefer to excise by taking a more direct route to your goals.]

[For the most part, we will state results in sufficient generality that the proof justifies, but wewill not give a much harder proof where the generality of the result will not be used. (Two possibleexceptions: we prove a number theorems for proper morphisms, not just projective morphisms,and for coherent sheaves in general, not just for locally Noetherian schemes. However in bothcases, readers are invited to ignore the subtleties required for the greater generality.]

[Hensel’s Law / Newton’s method. Reasons why to think of things that generally. Everycommutative algebra is an algebra on a space.]

Warning: If you attempt to read this without working through a significant number of exer-cises, I will come to your house and pummel you with [SGA] until you grovel for mercy.

[fold in later: goal is to think topologically and categorically. Choices: don’t do things moregenerally when all we care about is a specific case, unless it makes the argument easier. Of course,

this depends on the generality we want — many readers will think we’re working too hard! Forexample, it will be weird to many readers that we will introduce varieties so late.]

[Disadvantage: In a good course, there should be some punchline, some interesting result ortopic at the very end. Because these notes are comprehensive, this isn’t the case here. But if you


are using this for a class, or to read on your own, you may (and should!) want to omit sectionsand jump ahead in order to give a good story. [Give example. For example, skip ample and veryample, and get to curves.] ]

Why algebraic geometry? A new type of “space”, allowing both singularities, and arithmeticapplications. Good way to speak about singularities. Analytic spaces. Want to distinguishsingularities, e.g. 4 lines through origin. Limits of schemes. Functorial point of view is the rightone: orbifolds. Grand unified theory of number theory and geometry. (Rings of integers in numberfields “behave like” smooth curves; in fact they “are”.) Lay out the program.

More generality lets one see commonality. Submersions: flat morphisms may be fine. Allowsingularities.

Old-fashioned varieties: algebraic sets in An. but we want to glue them together, e.g. Riemannsurfaces. But then they don’t sit in An at all (see theorem ???). We could stick them in projectivespace. How to show they are the same? Idea: abstract manifolds. Given a torus, we don’t ask“where does it sit? what’s inside?” We’ve learned that this isn’t the right question to ask; thoseare extrinsic issues.

Good handle on differential information.***Algebraic geometry finds itself repeatedly concerned with certain kinds of spaces (vari-

eties, schemes, algebraic spaces, stacks), morphisms between them, sheaves on them, cohomologiesthereof. How all of the above behave in families. Examples: curves, projective space, modifica-

tions.Why algebraic geometry (see introduction to notes; pull a fair bit out of there). Roughly:

varieties: things in projective space cut out by equations. Maps are by polynomials.This course will be hardcore. I want to cover a lot of materials. I’m going to hit the ground

running. You’ll have to do homework. On the other hand, I’m willing to put a lot of time intothis.

I’m doing this the wrong way. Usually start soft, with varieties. My apologies.Good for: algebraic geometry, number theory, symplectic geometry, certain kinds of topology.From start of 216. Algebraic geometry is a subject that somehow connects and unifies

several parts of mathematics, including obviously algebra and geometry, but also number theory,and depending on your point of view many other things, including topology, string theory, etc. Asa result, it can be a handy thing to know if you are in a variety of subjects, notably number theory,symplectic geometry, and certain kinds of topology. The power of the field arises from a point ofview that was developed in the 1960’s in Paris, by the group led by Alexandre Grothendieck. Thepower comes from rather heavy formal and technical machinery, in which it is easy to lose sightof the intuitive nature of the objects under consideration. This is one reason why it used to strikefear into the hearts of the uninitiated.

The rough edges have been softened over the ensuing decades, but there is an inescapableneed to understand the subject on its own terms.

This class is intended to be an experiment. I hope to try several things, which are mutuallyincompatible. Over the year, I want to cover the foundations of the subject fairly completely:the idea of varieties and schemes, the morphisms between them, their properties, cohomologytheories, and more. I would like to do this rigorously, while trying hard to keep track of thegeometric intuition behind it. This is the second time I will have taught such a class, and thefirst time I’m going to try to do this without working from a text. So in particular, I may findthat I talk myself into a corner, and may tell you about something, and then realize I’ll have togo backwards and say a little more about an earlier something.

Some of you have asked what background will be required, and how fast this class will move.In terms of background, I’m going to try to assume as little as possible, ideally just commutativering theory, and some comfort with things like prime ideals and localization. (All my rings willbe commutative, and have unit!) The more you know, the better, of course. But if I say thingsthat you don’t understand, please slow me down in class, and also talk to me after class. Giventhe amount of material that there is in the foundations of the subject, I’m afraid I’m going tomove faster than I would like, which means that for you it will be like drinking from a firehose,as one of you put it. If it helps, I’m very happy to do my part to make it easier for you, and I’mhappy to talk about things outside of class. I also intend to post notes for as many classes as Ican. They will usually appear before the next class, but not always.

Ravi Vakil 15

In particular, this will not be the type of class where you can sit back and hope to pickup things casually. The only way to avoid losing yourself in a sea of definitions is to becomecomfortable with the ideas by playing with examples.

To this end, I intend to give problem sets, to be handed in. They aren’t intended to beonerous, and if they become so, please tell me. But they are intended to force you to becomefamiliar with the ideas we’ll be using.

2.4.1. Why algebraic geometry?. It is hard to define algebraic geometry in its vast generalityin a couple of sentences. So I’ll talk around it a bit.

As a motivation, consider the study of manifolds. Real manifolds are things that locally looklike bits of real n-space, and they are glued together to make interesting shapes. There is alreadysome subtlety here — when you glue things together, you have to specify what kind of gluing isallowed. For example, if the transition functions are required to be differentiable, then you getthe notion of a differentiable manifold.

A great example of a manifold is a submanifold of Rn (consider a picture of a torus). In fact,any compact manifold can be described in such a way. You could even make this your definition,and not worry about gluing. This is a good way to think about manifolds, but not the best way.There is something arbitrary and inessential about defining manifolds in this way. Much cleaneris the notion of an abstract manifold, which is the current definition used by the mathematicalcommunity. Example: torus by gluing.

There is an even more sophisticated way of thinking about manifolds. A differentiable man-ifold is obviously a topological space, but it is a little bit more. There is a very clever way ofsummarizing what additional information is there, basically by declaring what functions on thistopological space are differentiable. The right notion is that of a sheaf, which is a simple idea,that I’ll soon define for you. It is true, but non-obvious, that this ring of functions that we aredeclaring to be differentiable determines the differentiable manifold structure.

Very roughly, algebraic geometry, at least in its geometric guise, is the kind of geometryyou can describe with polynomials. So you are allowed to talk about things like y2 = x3 + x,but not y = sinx. So some of the fundamental geometric objects under consideration are thingsin n-space cut out by polynomials. Depending on how you define them, they are called affinevarieties or affine schemes. They are the analogues of the patches on a manifold. Then you canglue these things together, using things that you can describe with polynomials, to obtain moregeneral varieties and schemes. So then we’ll have these algebraic objects, that we call varieties orschemes, and we can talk about maps between them, and things like that.

In comparison with manifold theory, we’ve really restricted ourselves by only letting ourselvesuse polynomials. But on the other hand, we have gained a huge amount too. First of all, we cannow talk about things that aren’t smooth (that are singular), and we can work with these things.Algebraic geometry provides particularly powerful tools for dealing with singular objects. (Onething we’ll have to do is to define what we mean by smooth and singular!) Also, we needn’t workover the real or complex numbers, so we can talk about arithmetic questions, such as: what are therational points on y2 = x3 + x2? (Here, we work over the field Q.) More generally, the recipe bywhich we make geometric objects out of things to do with polynomials can generalize drastically,and we can make a geometric object out of rings. This ends up being surprisingly useful — allsorts of old facts in algebra can be interpreted geometrically, and indeed progress in the field ofcommutative algebra these days usually requires a strong geometric background.

Let me give you some examples that will show you some surprising links between geometryand number theory. To the ring of integers Z, we will associate a smooth curve Spec Z. In fact, tothe ring of integers in a number field, there is always a smooth curve, and to its orders (subrings),we have singular = non-smooth curves.

An old flavor of Diophantine question is something like this. Given an equation in twovariables, y2 = x3 + x2, how many rational solutions are there? So we’re looking to solve thisequation over the field Q. Instead, let’s look at the equation over the field C. It turns out thatwe get a complex surface, perhaps singular, and certainly non-compact (draw picture). So let meseparate all the singular points, and compactify, by adding in points. The resulting thing turnsout to be a compact oriented surface, so (assuming it is connected) it has a genus g, which is thenumber of holes it has. For example, y2 = x3 + x2 turns out to have genus 0. Then Mordellconjectured that if the genus is at least 2, then there are at most a finite number of rationalsolutions. The set of complex solutions somehow tells you about the number of rational solutions!


Mordell’s conjecture was proved by Faltings, and earned him a Fields Medal in 1986. As anFaltings, Mordellapplication, consider Fermat’s Last Theorem. We’re looking for integer solutions to xn+yn = zn.If you think about it, we are basically looking for rational solutions to Xn + Y n = 1. Well, it

turns out that this has genus`n−1

2

´— we’ll verify something close to this at some point in the

future. Thus if n is at least 4, there are only a finite number of solutions. Thus Falting’s Theoremimplies that for each n ≥ 4, there are only a finite number of counterexamples to Fermat’s lasttheorem. Of course, we now know that Fermat is true — but Falting’s theorem applies muchmore widely — for example, in more variables. The equations x3 + y2 + z14 + xy + 17 = 0 and3x14 + x34y + · · · = 0, assuming their complex solutions form a surface of genus at least 2, whichthey probably do, have only a finite number of solutions.

So here is where we are going. Algebraic geometry involves a new kind of “space”, which willallow both singularities, and arithmetic interpretations. We are going to define these spaces, anddefine maps between them, and other geometric constructions such as vector bundles and sheaves,and pretty soon, cohomology groups.

Perhaps: discuss here how Spec Z is a geometric object. We’ll later see it as a “smoothcurve”. Points are primes. 27/4 example. Suppose we have a diophantine equation. We’d lookfor solutions mod n for each n. By the Chinese remainder theorem (which we’ll also see as analgebrogeometric fact), we need to check modulo pn. This is analoguos to power series — Zp.Solving differential equations term by term (Newton’s method?) is hensel’s lemma. Numbertheory ideas to geomtery: ideals matter. Geometric ideas to number theory: nonsingularity. It ishard to see where one begins and the other ends.

2.4.2. Philosophy. (This is a pedagogical section.) Hartshorne. Mumford (red book, cpv,abelian varieties [MAV], curves on an algebraic surface). Shafarevich. Eisenbud-Harris. FAC.EGA? Then more. Not Griffiths-Harris [GH].

The list of introductory books in algebraic geometry also includes: Debarre; Smith et al;Bump; Ueno triple; Iitaka; SGA; Harris; earlier Eisenbud-Harris. I’m not mentioning easier bookse.g. by Reid, and Fulton. [Q. Liu, Algebraic geometry and arithmetic, OUP.]

Competing priorities: examples. Get somewhere, quickly (ready to read more advancedtopics). Honest and rigorous. Leave out needless generality. Categorical pictures. Commonfoundations no matter what area of algebraic geometry you are heading in. It is too much to givea casual reader a good feeling. Accessible but still rigorous.

There are three sorts of arguments, that people should be exposed to: categorical arguments;Zariski topology; and hands-on algebra.

Regarding the use of advanced topics: I don’t mind introducing big machinery early when itwould help proving things. For example, sheaves; spectral sequences. However, I do not want tointroduce machinery that makes the basic results harder, just to get the reader to see the fancytopics. In particular, derived functor cohomology is left for much later (or may be ignored); propermorphisms may be held back; formal schemes are not discussed at all.

More on derived functors: it is certainly true that many topics in algebraic geometry are

best understood in the language of derived functors; but this is a view from the mountaintop,looking down, and not the best way to explore the forests. In order to appreciate derived functorsappropriately, it requires understanding the homological algebra behind it, and not just takingit as a black box. I want to emphasize that most foundational results can be understood in arelatively simple and straightforward manner using just Cech cohomology (using, for better or forworse, spectral sequences). The reader is then encouraged to go crazy and re-do everything usingderived functors, and then to see how much simpler things become.

Why is algebraic geometry hard? The traditional reason: you have to work hard to under-stand the basics. Really: Lots to know, but it is all accessible. Where do you start?

We’ll develop the algebra as we go, and see that it isn’t so bad.My philosophy toward pathologies (things that satisfy A and not B): metaphor with those

sticks on mountain highways. In winter, you can’t see so well. But they are warning signs:if you cross this line, you’re going to die! Examples for a similar reason: when someone tellsyou something (if you tell yourself something) with a straight face, these can be a reality check.Hypotheses: in general, I like having as few hypotheses as possible. Certainly a hypothesis thatisn’t necessary to the proof is a red herring. But if a reasonable hypothesis can make the proofmuch cleaner and memorable, I’m willing to do it. In particlar, Noetherian hypothesis are handywhen necessary. Finite type k-scheme occasionally — also especially when there are techniques

Ravi Vakil 17

unique to this situation that one should know. Restricting to k algebraically closed only as a lastresort.

“Collective wisdom, distilled through an imperfect filter”.Emphasize: this approach indicates how to define a geometric space in general. Manifolds of

various sorts (smooth, differentiable). Deals well with singular spaces. How to define morphisms,vector bundles and more generally sheaves. Examples of where this philosophy can work: algebraicspaces, orbifolds, stacks, analytic spaces, complex analytic spaces, rigid (p-adic) analytic spaces,formal schemes.

Central players: cohomology. Krull’s principal ideal theorem. Algebraic Hartogs’ Lemma(state clearly that this is the real reason we like normality).

• Noetherian people should just combine: coherent=finite type.• characteristic 0 people should just ignore the adjective “geometrically”, as in “geomet-

rically reduced”, “geometrically integral”, etc.

2.4.3. Assumptions.Background needed: I’ve tried to assume as little as possible. Comfort with rings (prime

ideals, tensor products, Noetherian rings, etc.) would be very helpful. (All my rings are com-mutative with unit. But unit could be 0.) Some vague familiarity with category theory (somecomfort with the notion of a functor for example) is also important. Other than that, of course itwould be helpful to have as much background as possible (a solid course in commutative algebra;experience with algebraic topology, and even if possible homological algebra; some experience withdifferential geometry and complex analysis; but I’m not expecting any of this, and I don’t thinkstudents should wait until they have this kind of background before starting to learn algebraicgeometry).

2.4.4. Some specific comments. Do sheaves without espace etale. Pullback more natural.Valuative criteria not so relevant early on, misleading. Cartier divisors done right, and early.Krull and algebraic Hartogs seem to be central.

Pleasant sheafy arguments: 4.7.1, 6.1.1, 7.3.2, 7.3.3, 17.1.1, 17.2.1.Functorial point of view: PnC . What right do we have to insist that it has this manifold

structure? We even made a choice of atlas. Answer: Shows how this behaves in families. Universalproperty. Hence we define the universal property.

Q: If I want a rough working knowledge of schemes, do I need to take a long course like thisin the subject?

A: Certainly not! But if that is your goal, you’ve come to the wrong place. This is for peoplewho want a very solid foundation in the subject.

Q: Why divisors so late? Why coherent sheaves and cohomology so early? Aren’t divisorson curve...

A: You’ll want to know cohomology anyway. It is important! You may as well elarn it early,because you can, and it isn’t so hard. (Plus, Cartier divisors aren’t easy! And they are lessfundamental than line bundles.) Also: how do you know that there is a curve not isomorphic toP1? 1) A1 6∼= P1. 2) Singular curves not P1: need differentials for this. But perhaps all nonsingularprojective curves are isomorphic to P1? Here’s an easy reason why not: if C is a nonsingular cubicplane curve, then χ(OP1 ) 6= χ(OC).

Q: Why don’t you work over an algebraically closed field? Doesn’t that make it easier?A: No.Q: I don’t like the way you treated [Topic X].A: I don’t like your FACE.pedagogical promises: No giving statements essential to the text without justification. Yes

giving as simple a proof as possible within the context of the book. No doing something in greatergenerality than needed. No ”By a standard fact”.

Large amount of background; yet attempt to be self-contained. feel free to use intuition fromother areas at occasional times.

Informally: fibers of pushforward correspond to global sections of fibers.Huge caution when thinking about quasicoherent sheaves in this [which?] way: any 2 line

bundles are the same on X: there exists an affine cover on which l and M are both trivial. A→ Af .hence they are isomorphic. On decisions of what to include: things that will allow the reader to


do interesting but not outrageously difficult exercises. Somewhat brutal decisions on what to cutdue to length.

“When I first started doing research in algebraic geometry, I thought the subject attractivefor two reasons: firstly, because it dealt with such down-to-earth and really concrete objects asprojective curves and surfaces; secondly, because it was a small, quiet field where a dozen peopledid not leap on each new idea the minute it became current. As it turned out, the field seemsto have acquired the reputation of being esoteric, exclusive and very abstract with adherents whoare secretly plotting to take over all the rest of mathematics! In one respect this last point isaccurate...” David Mumford (1975)

2.5 Politics

Philosophy: for the most part, the audience will be seeing quasiprojective schemes, so we oftenfirst prove things there, especially when arguments are much easier. A little bit of cohomologyfrees up a lot too.

Proper morphisms. Move it later; projective is easier. Don’t do valuative criterion, except inoptional section. In particular, don’t do what Hartshorne does: using valuative criteria to showproperties of properness (that are easier from first principles), to show properties of projectivemorphisms (again easier from first principles). Ditto for separatedness.

Cohomology: Do Cech cohomology first.Invertible sheaves (and maps to projective space) are more fundamental than divisors. Ef-

fective Cartier divisors should be introduced long before Cartier divisors.Inverse ideal sheaf is pretty useless; instead use scheme-theoretic preimage.We blow up closed subschemes, not quasicoherent ideal sheaves.Foreground etale and smooth morphisms more.Variety discussion: I define them very late. Perhaps define prevariety quite early. But I

needn’t work over an algebraically closed field. And things need not be irreducible. Varieties:sheaves just can have values in K(X); perhaps mention this.

Krull/Hartogs/associated points are key.

2.5.1. Old to-do list.Define dominant morphism.I need to sort out some definitions, of integral extensions and finite extensions. Here are

three possible conditions on A → B. (i) B a finitely generated A-module. (ii) All elements of Bare integral over A. (iii) A → B. (i) implies (ii) if A is Noetherian.

Behaves well with respect to exact sequences: rank, det, χ.More generally: the K-theory of coherent sheaves of Pn, and indeed on any projective scheme.

Well-defined, and well-behaved in exact sequences.

2.6 Checklist of things they should know

MMMMumford’s Red book (first LNM version) [M]

• I.1 Noether normalization ??15.4.5• I.1 Going-up theorem p. 4 15.2.2• I.6 Prop. 5 and corollary 12.1.14• I.7 Prop. 1 ??• I.7 Thm. 2 ??• I.7 alg version of Krull’s Principal Ideal Theorem ??• I.7 Def. 2 9.1.6

• I.7 Prop. 3.1 14.1.6• I.7 Prop 3.2 9.3.E(a)• I.7 Prop 3.3 finite f is surjective iff f∗ is injective 9.4.B• I.7 Geometric version of Noether normalization ??

Ravi Vakil 19

• I.7 Corollary 1 ??• I.7 Corollary 2 Krull dim = tr deg 15.4• I.7 Cor. 3 ??• I.7 Cor. 4 handy lemma for semicontinuity ??• I.7 Cor. 5 ??• I.9 Thm. 1 Pnk complete 21.4.2• I.10 p. 85–89, Chow’s lemma 21.6.2• II.2 Prop. 2 There exists a unique generic point of any closed subset of an affine scheme.

5.6.E• II.4 Cor.: picturing finite type k-schemes with k not algebraically closed 7.5.M• II.6 p. 160 Grothendieck’s existence problem 23.16• II.6 p. 160 faithfully flat descent, one version 37.0.7• II.6 Prop. 3 Hom(Spec(R,m)→ X) 8.4.I• II.6 Def. 2 locus where 2 morphisms X → Y agree 12.1.J• II.6 Prop. 4 locus where 2 morphisms agree is locally closed 12.1.J• II.6 Def. 3 of a scheme: he doesn’t use separated in the definition! Insetad he basically

uses the valuative criterion.• II.7 Def. 3 affine 9.1.3 finite 9.1.6,• II.7 Thm. 2 PnR → SpecR closed 21.4.2

• II.7 Prop. 4 14.1.6• II.7 Prop. 5 affineness and finiteness affine-local on the target 9.1.4 and 9.1.11 respec-

tively• III.2 Def. 1 Noetherian scheme, different definition 7.3.5• III.2 p. 212–3, 3 forms Nakayama A.2.6• III.4 Provisional Def. 1 16.1.1• III.4 Prop. 1 Jacobian def of tangent space 16.1.E• III.4 Final Def. 1 Zariski tangent space 16.1• III.5 Def. 1 (provisional definition of etale) 30.5.A• III.5 Theorem 2 (geometric variant of Hensel) 30.5.1• III.5 p. 247 Hensel’s lemma 30.5.2• III.5 Theorem 4 30.5.3• III.7 Thm. 1 regular local ring is UFD, if it comes from variety 16.4.2• III.8 Prop. 6 13.3.K• III.9 Mumford’s many versions of Zariski’s Main Theorem Connectedness version in

29.14.3• III.10 p. 295–7 flatness discussion 29.15.1• III.10 Theorem 3 and 3’ 30.2.H• III.10 Definition 3 (smoothness) 30.5.A

Mumford’s Abelian Varieties [MAV].

• p. 53 Cor. 3 29.8.A• p. 53 Cor. 4 29.8.B

HHHHartshorneComparisons to Hartshore.The exercises are too hard, and the reader doesn’t have the expertise to reasonably be able

to solve them.Many algebra facts are assumed.He doesn’t explain why OX -modules form an abelian category; similarly for quasicoherent

sheaves. Inverse image sheaf definition is unmotivated.Separatedness and properness are needlessly very hard sections — he proves everything

through the valuative criterion. For example, the proof that projective morphisms are properis just bizarre.

The “affine communication lemma” is buried.Pullback of quasicoherent sheaf is defined in an unusual way, making the connection to what

it is on rings obscure.He gives a misleading definition of coherent sheaf. Also quasifinite; for example, for him Spec

of a DVR over k is quasifinite over k.He has needless hypotheses, confusing the reader.


etale morphisms are buried, and are only defined for finite type k-schemes. Ditto for unram-ified.

Cohomology with compact supports is never used, and hence the big machinery makes thingsconfusing. Many of the cohomology facts could be proved easily and directly and instead rely onhard facts. Formal schemes are never used. Apparently derived functors are old-fashioned, andhe should use derived categories.

Kiran’s gripes:

• I don’t like having important stuff not just proved, but only stated in the exercises.E.g. dominant morhpism, reduced scheme, Zariski tangent space, Chow’s lemma, affinemorphisms.

• Defining schemes without stating the adjoint property [maps to affine schemes] makesthe definition look like Grothendieckian black magic; H belatedly gives a restrictedversion in Exercise II.2.4, but the point is that it’s true for all ringed spaces. To me,the point of the definition of scheme is given that Grothendieck decided that the “right”category to look for a geometric interpretation of commutative algebra is the categoryof locally ringed space, the definition of scheme is the onehe was forced to write down.If you make a different choice, you get algebraic spaces, or stacks, or whatever.

• I think my complaints about cohomology are the same as yours and you are alreadyplanning to address them. (Actually I think you feel more strongly about this than I do;I actually find the whole business about coeffaceability and such kind of convenient,because it means that a lot of comparison statements you make are automaticallytru without having to worry.) One point that I fundamentally fail to understandHartshorne’s proof of Serre duality; proving it for varieties using finite flat maps toprojective space is probably a much bettter idea.

• Definitely talk about flatness much sooner, and etaleness much more.• A good way to motivate the concept of scheme is to use it to prove one of these

statements you get in commutative algebra like, a module M is flat over a ring A iffthe same is true after localizaing at each maximal ideal. (Come to think of this, maybeyou can already use this as a sneaky way to smuggle the definition of scheme into thestudent’s heads, once they have the Zariski topology in hand but before they knowabout the sheaf of rings on it. I haven’t thought this through.)

• Somehow Hartshorne fails to even bother pooint out the following incredibly useful fact:if U , V are open affines in a scheme X, then any point x ∈ U ∩ V has a neighborhoodwhich is a basic open in both U and V . You use this a bazillion times when you’rechecking, say, locality on the base of your favorite property of morphisms. So this atleast needs to be stated as an exercise (which ends up being a good test of whetheryou’ve grokked the idea that functions on a scheme should be thought of as sections ofthe structure sheaf.)

H.I.1 (fully digested)

• Proposition 1.2 Zariski facts.• Theorem 1.3A Nullstellensatz A.2.11• Cor 1.4 inclusion-reversing correspondence.• 1.4.6 Noetherian topological space• 1.5 Irreducible components of a Noetherian topological space.• 1.6 Krull dimension• height of prime ideal• 1.7 dimension = Krull dimension ??• I1.8A (a) transcendence degree 15.4 (b) height plus codimension ??• I.1.9 15.4• 1.10 dimY = dimY• 1.11 Krull’s Hauptidealsatz ??• 1.12A N domain is UFD iff every prime ideal of height 1 is principal. [done]• 1.13 A variety in An has dimension n−1 iff it is zero set of single nonconstant irreducible

pol in k[x1, . . . , xn].

H.I.2 (digested except for exercises)

• Ex. I.2.1 10.5.A(a)• Ex. I.2.2 10.5.B

Ravi Vakil 21

H.I.3 (digested except for the following)

• p. 18 S(p). Exercise: Show that if Y is projective, then K(Y ) = · · · . Finiteness of

integral closure, see FiniteICGreg.pdf (Kirsten’s notes) in “other references” on Vaioin FOAG folder. Show that any nonsingular conic in P2

k, if k is algebraically closed, is

isomorphic to P1. Exercises.

H.I.5

• Thm. I.5.7A A.2.2

H.I.6

• Thm. I.6.1A A.2.3• Prop. I.6.8 25.1.2

H.II.1

• Prop. II.1.1 isomorphisms can be checked on stalks. 4.4.E• Ex. II.1.5 isomorphisms can be checked on stalks. 4.4.E

• Ex. II.1.7 4.5.E• Ex. II.1.8 left-exactness of the global section 4.5.C• Ex. II.1.13 espace etale 4.4.7, mentioned but not done.• Ex. II.1.14 support 4.5.I• Ex. II.1.15 Hom 4.3.B• Ex. II.1.16 flasque sheaves 4.5.J• Ex. II.1.17 skyscraper sheaves 4.2.10

H.II.2

• Lemma II.2.1 a and b 5.4.C• Ex. II.2.13(b) 5.5.D• Ex. II.3.21ad 15.6.4

H.II.3

• Prop. II.3.1 7.2.D• Ex. II.3.2 9.1.C• Ex. II.3.5(b) finite UC 14.1.6, make sure I have the right reference for finite UC• Ex. II.3.11 17.11• Ex. II.3.11(b) implied by 9.1.P• Ex. II.3.13 9.3.D• Ex. II.3.16 Noetherian induction 5.5.7• Ex. II.3.19 Chevalley 15.5.1• Ex. II.3.21 15.6.D??

H.II.4

• Prop. II.4.1 affine to affine is separated 12.1.C• Cor. II.4.2 comment after 12.1.4; 12.1.4• Thm. II.4.3 16.8.3• Lemma II.4.5 16.8.4• Cor. II.4.6(a) Exercise 12.1.B

• Cor. II.4.6(b) 12.1.13• Cor. II.4.6(c) separatedness preserved by base change 12.1.10• Cor. II.4.6(d) comment after Prop. 12.1.13; 12.1.15• Cor. II.4.6(e) Thm. 12.1.19(a)• Cor. II.4.6(f) 12.1.11• sep part of Thm. II.4.9 Cor. 12.1.14• Ex. II.4.1 21.4.A 14.1.2• Ex. II.4.3 12.1.8• Ex. II.4.6 23.17.1, except H uses something hard• Ex. II.4.8(d) 11.4.E• Ex. II.4.8(e) 12.1.19• Ex. II.4.8(f) 12.1.L• Ex. II.4.10 Chow’s lemma 21.6.2• Ex. II.4.11(b) 16.8.1, except he seems to do it a hard way

Finite morphisms are projective 21.4.


H.II.5

• Prop. II.5.2(d) 20.2.A• Thm. II.5.8 20.2.1• Ex. II.5.5b 9.1.N• Ex. II.5.7(c) 17.10.D• Ex. II.5.12a 26.0.B• Part of Ex. II.5.17(b), comment after Prop. 12.1.11

H.II.6

• Lem. II.6.1 16.3.G• Prop. II.6.2 [done]• Prop. II.6.3A 15.6.6• Example II.6.5.1 10.3.C• Ex. II.6.4 and 6.5a 7.4.I

H.II.7

• Thm. II.7.1 (maps to Pn) 20.4.1• Prop. II.7.3 21.4.3• Prop. II.7.5 26.0.5• Prop. II.7.13a 27.3.1• Prop. II.7.13b 27.2.2• Example II.7.13.3 27.4.I• Prop. II.7.14 (universal property of blowing up) 27.2 (Cor. 7.15 follows pretty imme-

diately)• Prop. II.7.16 27.3• Thm. II.7.17 27• Example II.7.17.3 27.4.I• Ex. II.7.5a 26.0.6• Ex. II.7.5c 26.0.6• Ex. II.7.5d 26.0.C

H.II.8

• Prop. II.8.1A which H refers to Matsumura p. 182; we prove it in section 22.5.4• Prop. II.8.2A (behaves well with respect to base change and localization) 22.2.C and

22.2.B respectively• Example II.8.2.1 22.2.4• Prop. II.8.3A relative cotangent exact sequence, 22.2.8• Prop. II.8.4A, conormal exact sequence, 22.2.9• Cor. II.8.5 22.2.F, roughly• Prop. II.8.7 22.2.12• Prop. II.8.10, 22.2.16• Prop. II.8.11, 22.2.15 relative cotangent sequence• Prop. II.8.12 22.2.15 conormal exact sequence• Theorem II.8.13 ??• Theorem II.8.14A localization of regular local ring is also regular 16.4 22.4.1

• Thm. 8.15 22.4.3• Cor. II.8.16 22.4.4; he has hypotheses k-variety, where k is algebraically closed. I may

go for more general hypotheses• Theorem II.8.17 conormal exact sequence in case it is nonsingular inside nonsingular,

22.4.15• Theorem II.8.18 Bertini’s theorem ??• Theorem II.8.21A(e) I/I2 for regular sequence in CM 16.7.4• Theorem II.8.22A (Serre R1+S2=normal) 16.6.B• part of Prop. II.8.23, CM implies normal iff R1 16.7.B

H.II.9

• Prop. II.9.1 (a) A.1.4• Prop. II.9.1 (b) A.1.4• Example II.9.1.2 A.1.5

Derived functor cohomology 24

Ravi Vakil 23

H.III.2

• Proposition III.2.1A 24 A-modules have enough injectives (I have no proof yet)• Prop. III.2.2 24• Theorem III.2.7 Grothendieck vanishing for abelian groups; I only do it for quasicoher-

ent sheaves, but mention this at 23.14.3 I discuss it a little at 24.1.1.

H.III.3 Cohomology of a Noetherian affine scheme

• Krull’s Thm, Prop. III.3.1A A.1.6• Thm. III.3.5 23.3.3 (higher coh of qcs on affines vanish; I do it w/o Noetherian hyp.,

but in Cech)

H.III.4 Cech cohomology

• 23.3• Cech cohomology 23.3.1• Thm. III.4.5 Cech is derived functor 24.2.3. He has just for schemes; he also says thatX is Noetherian.

H.III.5 ??

• Serre criterion for ampleness Prop. III.5.3 26.1.3• Exercise III.5.7(b) ampleness can be checked on the reduced locus 26.2.

H.III.6

• Props. 6.2, 6.7b, 6.8 for quasicoherent sheaves follow from an alternte definition of Ext.I haven’t needed them yet.

• Ex. III.6.10a-b 31.4.5 (I don’t do c or d.)

H.III.7 Serre duality

• Thm. III.7.1 duality for Pn 31.12.A (stated, but not proved; it needs Exts)• Prop. III.7.2 dualizing sheaf, if exists is unique 31.2.2• Lemmas III.7.3, III.7.4, Prop. III.7.5 31.13.2 (just stated)• Theorem III.7.6 duality for projective scheme 31.13.2• Cor. III.7.7 (Serre duality, cohomology version) 31.1.1• Prop. III.7.10A (Koszul complex fact) 16.7.7 (exercise with hint here)• Cor. III.7.13 (Serre symmetry of Hodge diamond) 31.7.A

H.III.8

• I can also do: Prop 8.1, Cor 8.2, Prop 8.5,• Cor. III.8.6 23.6.1 ([H] says Rif∗F is quasicoherent if X is Noetherian.)• Theorem III.8.8 (a–c) 23.8.B H requires Y Noetherian. Do I?• Ex. III.8.2 23.6.A; I suspect I do more.• Ex. III.8.3 ??, 23.7.E. Do we really need finite rank? Probably cut if we never use this.

H.III.9

• Prop. III.9.1A(b) 29.1.2• Prop. III.9.1A(c) 29.1.3• Prop. III.9.1A(d) 29.1.1• Prop. III.9.2(a) 29.1.6• Prop. III.9.2(b) 29.1.6• Prop. III.9.2(c) 29.1.6• Prop. III.9.3 cohomology and flat base change ?? See also 32.4.2• Remark III.9.3.1 cohomology and base change 32.4.2(a)• Prop. III.9.5 29.12.1• Cor. III.9.5 29.12.2• Thm. III.9.9 29.3.1 shows that flat implies constant euler characteristic (better than

H); I don’t show the converse (the result if target is reduced).• Exercise III.9.1 29.2.F• Exercise III.9.4 29.2.F locus where π is flat is open. I currently don’t prove that.

H.III.10 Smooth morphisms (everything is extracted.)

• Example III.10.0.1 AnY , PnY 30.1.3• Example 10.0.3 smooth over k 30.2.4• Prop. III.10.1(a) 30.1.3• Prop. III.10.1(b)(c) ?? 30.1.5 and 30.2.D


• Prop. III.10.1(d) 30.2.E• Theorem III.10.2 30.2.H• Lemma III.10.3A (local criterion of flatness) 29.13.2• Proposition III.10.4 30.2.A• Lemma III.10.5 generic smoothness on the source 30.3.1• Prop. III.10.6 30.3.4• Cor. III.10.7 generic smoothness on the target 30.3.3• Cor. III.10.9 30.3.8• Remark III.10.9.3 30.3.C• Ex. III.10.3 30.1.6 and ??• Ex. III.10.4 30.5.4• Ex. III.10.7 ??• Ex. III.10.8 ??• Ex. III.10.9 CM to regular is flat 16.7.3

H.III.11 Formal functions. (I think I have extracted everything I like. Maybe Ex. 11.3 too.)

• Thm. III.11.1 Formal functions 32.2.1• Remark III.11.1.1 (ref to EGA) 32.2.2• Cor. III.11.2 23.8.2; he has “of Noetherian schemes” and coherent sheaves.• Cor. III.11.3 (connectedness lemma) ??• Thm. III.11.4 Zariski’s Main Theorem 29.14.3• Thm. III.11.5 Stein factorization 29.14.2• Ex. III.11.1 23.8.2 [direct copy]• Ex. III.11.2 23.8.1• Ex. III.11.3 unramified bit 30.2.G• Ex. III.11.4 32.3.2• Ex. III.11.8 32.3.2

H.III.12

• Thm. III.12.8 Semicontinuity Theorem 29.6.4• Thm. III.12.9 Grauert’s theorem 29.8.1• Example III.12.9.2, on actual jumping. ??• Example III.12.9.3, on continuity of cohomology for smooth morphisms, is discussed

in 29.6.4• Prop. III.12.10 ??• Thm. III.12.11 Cohomology and base change. 29.8.2• Ex. III.12.1, dimension of Zariski tangent space is upper semicontinuous. 32.3.3• Ex. III.12.4, generalized in 29.9.1• Ex. III.12.5 29.9.D

H.IV.H.IV.1

• Prop. IV.2.1 22.4.8• Prop. IV.2.2 22.4.10• Prop. IV.2.3 22.4.11• Prop. IV.2.5.5 25.6.2

H.V.

• Lem. V.2.1 and Prop. V.2.2 ?? (more general thing discussed in 2 pars)

Uppersemicontinuity of dimension for projective morphisms.The nonsingular locus is open in nice situtations. Reason: Ω is coherent.Coherent sheaf: rank is upper semicontinuous.If X → Y is relative dimension 0, then it is finite on some dense open set of Y . (Even finite

flat if Y is generically reduced. Reason: rank f∗OX is constant on some large open set.)Codimension of intersection is at most sum of codimensions, in good situations. Caution:

Take 2 things of dimension n, and glue them together along a point. Intersection something on

the first, with something on the second. What hypotheses do we need? smooth? As long bothare codimension 1 and cut out by a single equation, using Krull.

Consequence: 2 things in Pn whose sums of dimensions is at least n must intersect. Reason:look at the affine cone over Pn, and use previous fact.

Ravi Vakil 25

If X → Y , both irreducible and finite type, and if there is some point where the relativedimension is expected, then f is dominant.

Hironaka’s example 29.3.7Serre duality 31.1.1Grothendieck topology and descent 37.0.5generic smoothness 30.3.1Bertini’s Theorem ?? [Dimension counts. Each smooth projective variety of dimension d can

be a closed immersion in P2d+1. Each curve can be a nodal curve. Remark: surfaces in P3 canbe used to prove facts.]

Kleiman-Bertini theorem 22.6.6etale unramified smoothProjective and quasifinite implies finite 23.8.3. Proper and quasifinite implies finite 23.17.D.

Projective and affine implies finite 23.8.1. Proper and affine implies finite 23.17.1.

2.6.1. Objects they should know. Toric varieties (weighted projective space). See 35.Quotients, e.g. by finite groups?Jacobian variety, abelian variety (in sections on functors, and group schemes)zeta functionK3 surface. Calabi-Yau variety. K3, Calabi-Yau

]

Part I

Preliminaries

CHAPTER 3

Some category theory

[forme: POSSIBLY ADD: Why say epi rather than surjective? because “surjec-

tive” will sometimes have an obvious meaning which may not agree with epi.]

That which does not kill me, makes me stronger. — Nietzsche [forme: 1844-

1900, Twilight of the Idols, 1888, or possibly Thus spake Zarathustra][forme: Outline of this chapter: Categories. Functors. Universal properties and

Yoneda’s lemma. Examples: limits. Adjoints. ? spectral sequences. [primordial:Place later: the category of vector bundles. What are maps between them? E.g.

×x : k[x]→ k[x]: injective, not surjective; cokernel. ]

Here is what they should know from this section: the definition of a category.

Functors (covariant and contravariant). Yoneda’s lemma. Universal properties, in

the important special example of colimits = direct limits. Direct limits and directed

systems. We need the adjoint version of Yoneda too, which is currently not in the

text below.]

3.1 Motivation

1Before we get to any interesting geometry, we need to develop the language todiscuss things cleanly and effectively. [forme: Counsel patience!] This is bestdone in the language of categories. There is not much to know about categories toget started; it is just a very useful language. Like all mathematical languages, cat-egory theory comes with an embedded logic, which allows us to abstract intuitionsin settings we know well to far more general situations.

Our motivation is as follows. We will be creating some new mathematicalobjects (such as schemes, and families of sheaves), and we expect them to act likeobjects we have seen before. We could try to nail down precisely what we meanby “act like”, and what minimal set of things we have to check in order to verifythat they act the way we expect. Fortunately, we don’t have to — other peoplehave done this before us, by defining key notions, such as abelian categories, whichbehave like modules over a ring.

For example, we will define the notion of product of the geometric spaces(schemes). We could just give a definition of product, but then you should want toknow why this precise definition deserves the name of “product”. As a motivation,we revisit the notion of product in a situation we know well: (the category of)sets. One way to define the product of sets U and V is as the set of ordered pairs(u, v) | u ∈ U, v ∈ V . But someone from a different mathematical culture might

reasonably define it as the set of symbols uv | u ∈ U, v ∈ V . These notions are“obviously the same”. Better: there is “an obvious bijection between the two”.

1catmot

29

30

This can be made precise by giving a better definition of product, in terms of auniversal property. Given two sets M and N , a product is a set P , along with mapsuniv. prop

µ : P →M and ν : P → N , such that for any other set P ′ with maps µ′ : P ′ →Mand ν′ : P ′ → N , these maps must factor uniquely through P :2

(1) P ′

∃!

ν′

((PPPPPPPPPPPPPPP

µ′

11111111111111

P ν//

µ

N

M

Thus a product is a diagram

Pν //

µ

N

M

and not just a set P , although the maps µ and ν are often left implicit.This definition agrees with the usual definition, with one twist: there isn’t just

a single product; but any two products come with a canonical isomorphism betweenthem. In other words, the product is unique up to unique isomorphism. Here iswhy: if you have a product

P1ν1 //

µ1

N

M

and I have a product

P2ν2 //

µ2

N

M

then by the universal property of my product (letting (P2, µ2, ν2) play the roleof (P, µ, ν), and (P1, µ1, ν1) play the role of (P ′, µ′, ν′) in (1)), there is a uniquemap f : P1 → P2 making the appropriate diagram commute (i.e. µ1 = µ2 f andν1 = ν2 f). Similarly by the universal property of your product, there is a uniquemap g : P2 → P1 making the appropriate diagram commute. Now consider theuniversal property of my product, this time letting (P2, µ2, ν2) play the role of both(P, µ, ν) and (P ′, µ′, ν′) in (1). There is a unique map h : P2 → P2 such that

P2

h

AAA

AAAA

Aν2

''PPPPPPPPPPPPPPP

µ2

000000000000000

P2 ν2

//

µ2

N

M

2setUP

31

commutes. However, I can name two such maps: the identity map idP2 , and g f .Thus g f = idP2 . Similarly, f g = idP1 . Thus the maps f and g arising fromthe universal property are bijections. In short, there is a unique bijection betweenP1 and P2 preserving the “product structure” (the maps to M and N). This givesus the right to name any such product M × N , since any two such products arecanonically identified.

This definition has the advantage that it works in many circumstances, andonce we define category, we will soon see that the above argument applies verbatimin any category to show that products, if they exist, are unique up to uniqueisomorphism. Even if you haven’t seen the definition of category before, you canverify that this agrees with your notion of product in some category that you haveseen before (such as the category of vector spaces, where the maps are taken tobe linear maps; or the category of real manifolds, where the maps are taken to besubmersions).

This is handy even in cases that you understand. For example, one way ofdefining the product of two manifolds M and N is to cut them both up into charts,then take products of charts, then glue them together. But if I cut up the manifoldsin one way, and you cut them up in another, how do we know our resulting manifoldsare the “same”? We could wave our hands, or make an annoying argument aboutrefining covers, but instead, we should just show that they are indeed products,and hence the “same” (i.e. isomorphic).

We will shortly formalize this argument in Section 3.3.Another set of notions we will abstract are categories that “behave like mod-

ules”. We will want to define kernels and cokernels for new notions, and we shouldmake sure that these notions behave the way we expect them to. This leads usto the definition of abelian categories, first defined by Grothendieck in his Tohoku abelian category

paper [Gr].In this chapter, we’ll give an informal introduction to these and related notions,

in the hope of giving just enough familiarity to comfortably use them in practice.[forme: The reader can also instead read [W, App. A].]

3.2 Categories and functors

We begin with an informal definition of categories and functors.

3.2.1. Categories. category

A category consists of a collection of objects, and for each pair of objects, aset of maps, or morphisms (or arrows), between them. The collection of objects of obj, mor, Mor, arrows

a category C are often denoted obj(C), but we will usually denote the collection Calso by C. If A, B ∈ C, then the morphisms from A to B are denoted Mor(A, B). Amorphism is often written f : A→ B, and A is said to be the source of f , and B thetarget of f . Morphisms compose as expected: there is a composition Mor(A, B) ×Mor(B, C) → Mor(A, C), and if f ∈ Mor(A, B) and g ∈ Mor(B, C), then theircomposition is denoted g f . Composition is associative: if f ∈ Mor(A, B), g ∈Mor(B, C), and h ∈ Mor(C, D), then h (g f) = (hg)f . For each object A ∈ C,there is always an identity morphism idA : A → A, such that when you (left- or

32

right-)compose a morphism with the identity, you get the same morphism. Moreprecisely, if f : A→ B is a morphism, then f idA = f = idB f .

If we have a category, then we have a notion of isomorphism between two objects(if we have two morphisms f : A→ B and g : B → A, both of whose compositionsare the identity on the appropriate object), and a notion of automorphism of anobject (an isomorphism of the object with itself).iso, aut

3.2.2. Example. The prototypical example to keep in mind is the category of sets,denoted Sets. The objects are sets, and the morphisms are maps of sets. (BecauseRussell’s paradox shows that there is no set of all sets, we did not say earlier thatthere is a set of all objects. But as stated in §2.2, we are deliberately omitting allset-theoretic issues.) 3

3.2.3. Example. Another good example is the category Veck of vector spacesover a given field k. The objects are k-vector spaces, and the morphisms are lineartransformations.4

3.2.A. Unimportant exercise. A category in which each morphism is an iso-morphism is called a groupoid. (This notion is not important in this book. Thegroupoid

point of this exercise is to give you some practice with categories, by relating themto an object you know well.)(a) A perverse definition of a group is: a groupoid with one element. Make senseof this.(b) Describe a groupoid that is not a group.(For readers with a topological background: if X is a topological space, then thefundamental groupoid is the category where the objects are points of x, and themorphisms from x → y are paths from x to y, up to homotopy. Then the auto-morphism group of x0 is the (pointed) fundamental group π1(X, x0). In the casewhere X is connected, and the π1(X) is not abelian, this illustrates the fact thatfor a connected groupoid — whose definition you can guess — the automorphismgroups of the objects are all isomorphic, but not canonically isomorphic.) [forme:

Cut this exercise most likely.]

3.2.B. Exercise. If A is an object in a category C, show that the Mor(A, A)forms a group (called the automorphism group of A, denoted Aut(A)). What areAut

the automorphism groups of the objects in Examples 3.2.2 and 3.2.3? Show thattwo isomorphic objects have isomorphic automorphism groups.

3.2.4. Example: abelian groups. The abelian groups, along with group homomor-phisms, form a category Ab.5

3.2.5. Example: modules over a ring. If A is a ring, then the A-modules forma category ModA. (This category has additional structure; it will be the proto-typical example of an abelian category, see Sect. 3.7.) Taking A = k, we obtainExample 3.2.3; taking A = Z, we obtain Example 3.2.4.

3Sets

4Veck

5Ab

33

3.2.6. Example: rings. There is a category Rings, where the objects are rings,and the morphisms are morphisms of rings (which I’ll assume send 1 to 1). [forme:

Chad remarks: the category Mod, (A,M,φ : A→ End(M)).]

3.2.7. Example: topological spaces. The topological spaces, along with continuousmaps, form a category Top. The isomorphisms are homeomorphisms.

3.2.8. Example: partially ordered sets. 6A partially ordered set, or poset, is a set(S,≥) along with a binary relation ≥ satisfying:

(i) x ≥ x,(ii) x ≥ y and y ≥ z imply x ≥ z (transitivity), and(iii) if x ≥ y and y ≥ x then x = y.

A partially ordered set (S,≥) can be interpreted as a category whose objects arethe elements of S, and with a single morphism from x to y if and only if x ≥ y (andno morphism otherwise).

A trivial example is (S,≥) where x ≥ y if and only if x = y. Another exampleis 7

(2) •

• // •Here there are three objects. The identity morphisms are omitted for convenience,and the three non-identity morphisms are depicted. A third example is

(3) •

// •

• // •Here the “obvious” morphisms are again omitted: the identity morphisms, and themorphism from the upper left to the lower right. Similarly,

· · · // • // • // •depicts a partially ordered set, where again, only the “generating morphisms” aredepicted.

3.2.9. Example: the category of subsets of a set, and the category of open sets ina topological space. If X is a set, then the subsets form a partially ordered set,where the order is given by inclusion. Similarly, if X is a topological space, thenthe open sets form a partially ordered set, where the order is given by inclusion.8

(What is the initial object? What is the final object?)

3.2.10. Example. A subcategory A of a category B has as its objects some of theobjects of B, and some of the morphisms, such that the morphisms of A includethe identity morphisms of the objects of A, and are closed under composition. (Forexample, (2) is in an obvious way a subcategory of (3).)

3.2.11. Functors.

6d:poset

7three3

8categoryofopensets

34

A covariant functor F from a category A to a category B, denoted F : A → B,is the following data. It is a map of objects F : obj(A) → obj(B), and for eacha1, a2 ∈ A a morphism m : a1 → a2, F (m) is a morphism from F (A1) → F (A2)in B. F preserves identity morphisms: for A ∈ A, F (idA) = idF (A). F preservescomposition: F (m1 m2) = F (m1) F (m2).covariant, functor

If F : A → B and G : B → C, then we may define a functor G F : A → C inthe obvious way. Composition of functors is associative.

3.2.12. Example: a forgetful functor. Consider the functor from the categoryof complex vector spaces Veck to Sets, that associates to each vector space itsunderlying set. The functor sends a linear transformation to its underlying map ofsets. This is an example of a forgetful functor, where some additional structure isforgetful functor

forgotten. Another example of a forgetful functor is ModA → Ab from A-modulesto abelian groups, remembering only the abelian group structure of the A-module.

3.2.13. Topological examples. Examples of covariant functors include the fun-damental group functor π1, which sends a topological space with X choice of apoint x0 ∈ X to a group π1(X, x0), and the ith homology functor Top → Ab,which sends a topological space X to its ith homology group Hi(X, Z). The co-variance corresponds to the fact that a (continuous) morphism of pointed topo-logical spaces f : X → Y with f(x0) = y0 induces a map of fundamental groupsπ1(X, x0)→ π1(Y, y0), and similarly for homology groups. 9

3.2.14. Example. Suppose A is an element of a category C. Then there is afunctor hA : C → Sets sending B ∈ C to Mor(A, B), and sending f : B1 → B2 toMor(A, B1)→ Mor(A, B2) described by10

[g : A→ B1] 7→ [f g : A→ B1 → B2].

3.2.15. Definition. A contravariant functor is defined in the same way asa covariant functor, except the arrows switch directions: in the above language,F (A1 → A2) is now an arrow from F (A2) to F (A1).

It is wise to always state whether a functor is covariant or contravariant. If itis not stated, the functor is often assumed to be covariant.

It can be more convenient to describe a contravariant functor C → D as acovariant functor Copp → D, where Copp is the same category as C except that thearrows go in the opposite direction.

3.2.16. Topological example (cf. Example 3.2.13). The the ith cohomology functorH i(·, Z) : Top→ Ab is a contravariant functor.

3.2.17. Example. If A is the category of complex vector spaces, then taking dualsgives a contravariant functor ∨ : A → A. Indeed, to each linear transformationf : V →W , we have a dual transformation f∨ : W∨ → V ∨, and (f g)∨ = g∨ f∨.

3.2.18. Example. There is a contravariant functor Top → Rings taking atopological space X to the continuous functions on X . A morphism of topological

9homologyexample, referred to in coho e.g.

10ha, referred to in contra e.g.

35

spaces X → Y (a continuous map) induces the pullback map from functions on Yto maps on X .

3.2.19. Example (cf. 3.2.14). Suppose A is an element of a category C. Thenthere is a contravariant functor hA : C → Sets sending B ∈ C to Mor(B, A), andsending f : B1 → B2 to Mor(B2, A)→ Mor(B1, A) described by11

[g : B2 → A] 7→ [g f : B2 → B1 → A].

This example initially looks weird and different, but the previous two examples arejust special cases of this; do you see how? What is A in each case?

3.2.20. Full and faithful functors. A covariant functor F : A → B is faithful iffor all A, A′ ∈ A, the map MorA(A, A′)→ MorB(F (A), F (A′)) is injective, and fullif it is surjective. A functor that is full and faithful is fully faithful. A subcategoryi : A → B is a full subcategory if i is full. [forme: Where will we use these notions? full, faithful functors

Full Yoneda, and Freyd-Mitchell]

3.3 Universal properties determine an object up to uniqueisomorphism

GET RID OF YONEDA!!!12 [forme: [H, p. 241] refers to the new edition of EGA I, Ch. 0, §1] Given

some category that we come up with, we often will have ways of producing newobjects from old. In good circumstances, such a definition can be made using thenotion of a universal property. Informally, we wish that there is an object withsome property. We first show that if it exists, then it is essentially unique, or moreprecisely, is unique up to unique isomorphism. Then we go about constructing anexample of such an object to show existence.

With a little practice, universal properties are useful in proving things quicklyslickly. However, explicit constructions are often intuitively easier to work with,and sometimes also lead to short proofs.

We have seen one important example of a universal property argument alreadyin Section 3.1: products. You should go back and verify that our discussion theregives a notion of product in category, and shows that products, if they exist, areunique up to canonical isomorphism. product

3.3.1. 13Another good example of a universal property construction is the notionof a tensor product of A-modules tensor product

⊗A : obj(ModA)× obj(ModA) // obj(ModA)

M ×N // M ⊗A N

The subscript A is often suppressed when it is clear from context. Tensor productis often defined as follows. Suppose you have two A-modules M and N . Then

11haha

12s:yoneda

13tpexists

36

elements of the tensor product M ⊗A N are of the form m ⊗ n (m ∈ M , n ∈ N),subject to relations (m1+m2)⊗n = m1⊗n+m2⊗n, m⊗(n1+n2) = m⊗n1+m⊗n2,a(m⊗ n) = (am)⊗ n = m⊗ (an) (where a ∈ A).

If A is a field k, we get the tensor product of vector spaces.

3.3.A. Exercise (if you haven’t seen tensor products before). CalculateZ/10⊗ZZ/12. (This exercise is intended to give some hands-on practice with tensorproducts.) 14 [forme: Have them generalize to Z/m ⊗Z Z/n?]

3.3.B. Exercise: right-exactness of · ⊗A N . Show that · ⊗A N gives acovariant functor ModA →ModA. Show that · ⊗A N is a right-exact functor, i.e.if

M ′ →M →M ′′ → 0

is an exact sequence of A-modules, then the induced sequence

M ′ ⊗A N →M ⊗A N →M ′′ ⊗A N → 0

is also exact. [forme: Tony Licata gave a universal property proof, at least partially.

By construction it is easy to check surjectivity, and that the composition is 0. So we

show that (M ⊗A N/M ′ ⊗A N) satisfies the universal property for M ′′ ⊗A N . Well, it

clearly has a map there, and then we win.] (You will be reminded of the definitionright exact

of right-exactness in §3.7.4.)15

ex.doneThis is a weird definition, and really the “wrong” definition. To motivate a

better one: notice that there is a natural A-bilinear map M × N → M ⊗A N .Any A-bilinear map M × N → C factors through the tensor product uniquely:M ×N →M ⊗A N → C. (Think this through!)tensor and ⊗ def

We can take this as the definition of the tensor product as follows. It is anA-module T along with an A-bilinear map t : M × N → T , such any other suchmap factors through t that given any other t′ : M ×N → T ′, there is a unique mapf : T → T ′ such that t′ = f t.

M ×Nt //

t′

##GGG

GGGG

GGT

∃!fT ′

3.3.C. Exercise. Show that (T, t : M × N → T ) is unique up to uniqueisomorphism. Hint: first figure out what “unique up to unique isomorphism” meansfor such pairs. Then follow the analogous argument for the product. [forme: Inclass: explain this. If I could create something satisfying this property, (M ⊗A N)′,and you were to create something else (M ⊗AN)′′, then by my universal property forC = (M ⊗A N)′′, there would be a unique map (M ⊗A N)′ → (M ⊗A N)′′ interpolatingM×N → (M⊗AN)′′, and similarly by your universal property there would be a uniqueuniversal map (M ⊗A N)′′ → (M ⊗A N)′. The composition of these two maps in oneorder

(M ⊗A N)′ → (M ⊗A N)′′ → (M ⊗A N)′

has to be the identity, by the universal property for C = (M ⊗A N)′, and similarly for

the other composition. Thus we have shown that these two maps are inverses, and

our two spaces are isomorphic. In short: our two definitions may not be the same,

14tensorbaby

15tensorrightexact

37

but there is a canonical isomorphism between them. Then the “usual” construction

works, but someone else may have another construction which works just as well.]

ex.done

In short: there is an A-bilinear map t : M×N →M⊗AN , unique up to uniqueisomorphism, defined by the following universal property: for any A-bilinear mapt′ : M ×N → T ′ there is a unique f : M ⊗A N → T ′ such that t′ = f t.

Note that this argument shows uniqueness assuming existence. We need tostill show the existence of such a tensor product. This forces us to do somethingconstructive.

3.3.D. Exercise. Show that the construction of §3.3.1 satisfies the universalproperty of tensor product.

ex.doneThe uniqueness of tensor product is our second example of the proof of unique-

ness (up to unique isomorphism) by a universal property. If you have never seenthis sort of argument before, then you might think you get it, but you don’t, so youshould think over it some more. We will be using such arguments repeatedly in thefuture. [forme: teaching: It’s hard, but easy. It is black, but white. This takes us

to a powerful fact, that is very zen: it is very deep, but also very shallow.]

Some stray remarks about tensor products that you should be familiar with:if S is a multiplicative set of A and M is an A-module, then there is a naturalisomorphism (S−1A) ⊗A B ∼= S−1M . If further A → B is a morphism of rings,then B ⊗A M naturally has the structure of a B-module. If further A → C is amorphism of rings, then B ⊗A C is a ring.

Here is another exercise involving a universal property.

3.3.2. Definition. An object of a category C is an initial object if it hasprecisely one map to every other object. It is a final object if it has precisely onemap from every other object. It is a zero-object if it is both an initial object anda final object. 16 initial, final, zero object

3.3.E. Exercise. Show that any two initial objects are canonically isomorphic.Show that any two final objects are canonically isomorphic.

This (partially) justifies the phrase “the initial object” rather than “an initialobject”, and similarly for “the final object” and “the zero object”.

3.3.F. Exercise. State what the initial and final objects are in Sets, Rings,and Top (if they exist).

[primordial:

3.3.3. Yoneda’s Lemma. 17 YonedaThis is an easy fact stated in a complicated way. Informally speaking, you can essentially

recover an obejct in a category by knowing the maps into it. For example, we’ve seen that thedata of maps to X×Y are naturally (canonically) the data of maps to X and to Y . Indeed, we’venow taken this as the definition of X × Y .

Recall Example 3.2.19. Suppose A is an object of category C. For any object C ∈ C, we havea set of morphisms Mor(C,A). If we have a morphism f : B → C, we get a map of sets 18

(4) Mor(C,A)→Mor(B,A),

16def0object

17yoneda

18yonedamap

38

by composition: given a map from C to A, we get a map from B to A by precomposing with f .Hence this gives a contravariant functor hA : C → Sets. Yoneda’s Lemma states that the functorhA determines A up to unique isomorphism. More precisely:

3.3.4. Yoneda’s lemma. — Given two objects A and A′,19 and bijections

(5) iC : Mor(C,A)→Mor(C,A′)

that commute with the maps (4), then the iC must be induced from a unique isomorphism A→ A′.

3.3.G. Important Exercise (that everyone should do once in their life). Prove this.(Hint: This sounds hard, but it really is not. This statement is so general that there are reallyonly a couple of things that you could possibly try. For example, if you’re hoping to find anisomorphism A → A′, where will you find it? Well, you’re looking for an element Mor(A,A′).So just plug in C = A to (5), and see where the identity goes. You’ll quickly find the desiredmorphism; show that it is an isomorphism, then show that it is unique.)

3.3.5. Remark. There is an analogous statement with the arrows reversed, where instead of mapsinto A, you think of maps from A. The role of the contravariant functor hA of Example 3.2.19 isplayed by the covariant functor of Example 3.2.14.

3.3.6. Remark: the full statement of Yoneda’s Lemma. [forme: teaching: don’t say thisin class] It won’t matter so much for us, but it is useful to know the full statement of Yoneda’sLemma. If C is a category, consider the contravariant functor

h : C → SetsC

where the category on the right is the “functor category” where the objects are contravariant

functors C → Sets. (What are the morphisms in this category? You will rediscover the notion

of natural transformation of functors.) [forme: Will we use the notion of natural trans-

formations? Will we use full and faithful? Yes, in Freyd-Mitchell. Will we use full

Yoneda? If so, then expand this section.] This functor h sends A to hA. Yoneda’sfull f, faithful f, functor

category, nat’l trans lemma states that this is a fully faithful functor, called the Yoneda embedding. (Fully faithful

Y embedding functors were defined in §3.2.20.) ][forme: This is the first of the Yoneda-related topics. Later topics include: group

schemes, and the existence of the fibered product. Brian Conrad has the followingsuggestions.

1. First get them comfortable with Yoneda’s Lemma. (My favorite examples fordoing that are: (i) functorial meaning of Pn and of its standard affine open coveringand the gluing among the pieces, [I should do this] and (ii) proving that a map

of group schemes which respects the multiplication law automatically respects theidentity and inversion – and have them unravel it to give a direct painful proof bydiagram-chasing of morphisms.) [I definitely want to do this.]

2. Define what it means to say a functor is a sheaf for the big Zariski siteover a scheme, have them work out examples (representable functors, Hom-functors,Hilbert or Quot functors, pullbacks of sheaves of O-modules, units, flags on a fixedvector bundle) and non-examples (line bundles up to isomorphism, presheaf quotientby group action).

3. Combining # 1 and # 2, deduce that representability of a functor on Sch/Xis equivalence to being a sheaf for big Zariski site over X and being representable”locally over X”.

4. In the case of constructing X ×Z Y , have them work out that it suffices toconsider a functor on Sch/Z, and then by # 3 reduce to the case when Z is affineopen. Then likewise with X and Y in turn.

5. Have them do the same in the category of smooth manifolds *provided* that

the maps to Z are submersions. ]

19yoneda2

39

3.3.7. Example: Fibered products. (This notion of fibered product will beimportant for us later.) Suppose we have morphisms X, Y → Z (in any category).Then the fibered product is an object X ×Z Y along with morphisms to X andY , where the two compositions X ×Z Y → Z agree, such that given any otherobject W with maps to X and Y (whose compositions to Z agree), these mapsfactor through some unique W → X ×Z Y :

W∃!

$$

666

6666

6666

6666

6

))TTTTTTTTTTTTTTTTTTT

X ×Z Y

πX

πY

// Y

g

X

f // Z

By the usual universal property argument, if it exists, it is unique up to uniqueisomorphism. (You should think this through until it is clear to you.) Thus the useof the phrase “the fibered product” (rather than “a fibered product”) is reasonable,and we should reasonably be allowed to give it the name X ×Z Y . We know whatmaps to it are: they are precisely maps to X and maps to Y that agree on mapsto Z.

Depending on your religion, the diagram

X ×Z Y

πX

πY

// Y

g

X

f // Z

is called a fibered diagram, Cartesian diagram, fibered square, or Cartesiansquare. Cartesian diagram, fibered square, and Cartesian square. 20 fibered/Cartesian

diagram/squareThe right way to interpret the notion of fibered product is first to think aboutwhat it means in the category of sets.

3.3.H. Exercise. Show that in Sets,

X ×Z Y = (x ∈ X, y ∈ Y ) | f(x) = g(y).More precisely, describe a natural isomorphism between the left and right sides.(This will help you build intuition for fibered products.)

3.3.I. Exercise. If X is a topological space, show that fibered products alwaysexist in the category of open sets of X , by describing what a fibered product is.(Hint: it has a one-word description.)

3.3.J. Exercise. If Z is the final object in a category C, and X, Y ∈ C, then“X ×Z Y = X × Y ”: “the” fibered product over Z is canonically isomorphic to“the” product. (This is an exercise about unwinding the definition.)21

3.3.K. Useful Exercise: towers of fiber diagrams are fiber diagrams.If the two squares in the following commutative diagram are fiber diagrams, show

20d:fibereddiagram

21absolutproduct

40

that the “outside rectangle” (involving U , V , Y , and Z) is also a fiber diagram.22

U //

V

W //

X

Y // Z

3.3.L. Useful exercise: The magic diagram. Suppose we are given mor-phisms W, X → Y and Y → Z. Describe the natural morphism W×Y X →W×ZX .Show that the following diagram is a commutative diagram.

W ×Y X //

W ×Z X

Y // Y ×Z Y

The following diagram is surprisingly incredibly useful — so useful that we will callit the magic diagram. 23

3.3.M. Exercise. Given X → Y → Z, show that there is a natural morphismX ×Y X → X ×Z X , assuming that both fibered products exist. (This is trivialonce you figure out what it is saying. The point of this exercise is to see why it istrivial.)

3.3.8. Monomorphisms and epimorphisms.

3.3.9. Definition. A morphism f : X → Y is a monomorphism if any twomonomorphism

morphisms g1, g2 : Z → X such that f g1 = f g2 must satisfy g1 = g2. This ageneralization of of an injection of sets. In other words, there is a unique way offilling in the dotted arrow so that the following diagram commutes.

Z

≤1

AAA

AAAA

Xf // Y.

Intuitively, it is the categorical version of an injective map, and indeed this notiongeneralizes the familiar notion of injective maps of sets. 24

3.3.N. Exercise. Show that the composition of two monomorphisms is amonomorphism. 25

The notion of an epimorphism is “dual” to this diagramatic definition, whereall the arrows are reversed. This concept will not be central for us, although it isnecessary for the definition of an abelian category. Intuitively, it is the categoricalversion of a surjective map. 26epimorphism

22towerofpower

23magicdiagram

24monodef

25monomorphismscompose

41

3.3.O. Exercise. Prove a morphism is a monomorphism if and only if thenatural morphism X → X ×Y X is an isomorphism. We may then take this asthe definition of monomorphism. (Monomorphisms aren’t very central to futurediscussions, although they will come up again. This exercise is just good practice.)

3.3.P. Exercise. Suppose X → Y is a monomorphism, and W, Z → X are twomorphisms. Show that W ×X Z and W ×Y Z are canonically isomorphic. We willuse this later when talking about fibered products. (Hint: for any object V , give anatural bijection between maps from V to the first and maps from V to the second.It is also possible to use the magic diagram, Exercise 3.3.L)

3.3.10. Coproducts.

3.3.Q. Exercise. Define coproduct in a category by reversing all the arrows inthe definition of product. Show that coproduct for Sets is disjoint union. [forme: coproduct

This definition is here because of the next exercise.]

3.3.R. Exercise. Suppose C → A, B are two ring morphisms, so in particular Aand B are C-modules. Define a ring structure A⊗C B with multiplication given by(a1⊗b1)(a2⊗b2) = (a1a2)⊗(b1b2). There is a natural morphism A→ A⊗C B givenby a 7→ (a, 1). (This is not necessarily an inclusion, see Exercise 3.3.A.) Similarly,there is a natural morphism B → A ⊗C B. Show that this gives a coproduct onrings, i.e. that

A⊗C B Boo

A

OO

Coo

OO

satisfies the universal property of coproduct. [forme: Not too hard.] 27

3.4 Limits and colimits

Limits and colimits provide two important examples defined by universal prop-erties. They generalize a number of familiar constructions. I’ll give the definitionfirst, and then show you why it is familiar. [forme: Ben’s analogy: kernel/cokernel,

product/coproduct, limit/colimit.] 28

3.4.1. Limits. We say that a category is an small category if the objects form index category

a set. (This is a technical condition intended only for experts.) Suppose I is anysmall category, and C is any category. Then a functor F : I → C (i.e. with an objectAi ∈ C for each element i ∈ I, and appropriate commuting morphisms dictated byI) is said to be a diagram indexed by I. We call I an index category. Our index index cat

categories will all be partially ordered sets (Example 3.2.8), in which in particularthere is at most one morphism between any two objects. (However, other examples

26epidef

27e:ringcoproduct

28d:limit

42

are sometimes useful.) [forme: e.g. equalizer] For example, if 2 is the category

•

// •

• // •and A is a category, then a functor 2 → A is precisely the data of a commutingsquare in A.

Then the limit is an object lim←−IAi of C along with morphisms fi : lim←−I

Ai suchinverse limit, limit, proj

lim that if m : i→ j is a morphism in I, then

lim←−IAi

fi

fj

""FFF

FFFF

FF

Ai

F (m) // Aj

commutes, and this object and maps to each Ai is universal (final) respect to thisproperty. [forme: Fix wording.] (The limit is sometimes called the inverse limitor projective limit.) By the usual universal property argument, if the limit exists,it is unique up to unique isomorphism.

3.4.2. Examples: products. For example, if I is the partially ordered set

•

• // •we obtain the fibered product.

If I is• •

we obtain the product.If I has an initial object e, then Ae is the limit, and in particular the limit

always exists.If I is a set (i.e. the only morphisms are the identity maps), then the limit is

called the product of the Ai, and is denoted∏

i Ai. The special case where I hastwo elements is the example of the previous paragraph.

3.4.3. Example: the p-adics. The p-adic numbers, Zp, are often described infor-mally (and somewhat unnaturally) as being of the form Zp =?+?p+?p2+?p3 + · · · .They are an example of a limit in the category of rings:

Zp

!!CCC

CCCC

C

((RRRRRRRRRRRRRRRRR

++VVVVVVVVVVVVVVVVVVVVVVVVVVV

· · · // Z/p3 // Z/p2 // Z/p

Limits do not always exist. For example, there is no limit of · · · → Z/p3 →Z/p2 → Z/p→ 0 in the category of finite rings.

However, you can often easily check that limits exist if the elements of yourcategory can be described as sets with additional structure, and arbitrary productsexist (respecting the set structure).

43

3.4.A. Exercise. Show that in the category Sets,

(ai)i∈I ∈∏

i

Ai : F (m)(ai) = aj for all [m : i→ j] ∈ Mor(I)

,

along with the projection maps to each Ai, is the limit lim←−IAi.

This clearly also works in the category ModA of A-modules, and its special-izations such as Veck and Ab.

From this point of view, 2 + 3p + 2p2 + · · · ∈ Zp can be understood as thesequence (2, 2 + 3p, 2 + 3p + 2p2, . . . ).

3.4.4. Colimits. 29 More immediately relevant for us will be the dual of thenotion of limit (or inverse limit). We just flip all the arrows in that definition,and get the notion of a colimit. Again, if it exists, it is unique up to unique direct limit

isomorphism. (The colimit is sometimes called the direct limit or injective limit.)A limit maps to all the objects in the big commutative diagram indexed by I.

A colimit has a map from all the objects.Even though we have just flipped the arrows, somehow colimits behave quite

differently from limits.

3.4.5. Example. The ring 5−∞Z of rational30 numbers whose denominators arepowers of 5 is a colimit lim−→ 5−iZ. More precisely, 5∞Z is the colimit of

Z // 5−1Z // 5−2Z // · · ·

The colimit over an index set I is called the coproduct, denoted∐

i Ai, and isthe dual notion to the product. sum

3.4.B. Exercise. (a) Interpret the statement “Q = lim−→1nZ”. (b) Interpret the

union of some subsets of a given set as a colimit. (Dually, the intersection can beinterpreted as a limit.)

Colimits always exist in the category of sets:

3.4.C. Exercise. Consider the set (i ∈ I, ai ∈ Ai) modulo the equivalencegenerated by: if m : i→ j is an arrow in I, then (i, ai) ∼ (j, F (m)(ai)). Show thatthis set, along with the obvious maps from each Ai, is the colimit.

Thus in Example 3.4.5, each element of the direct limit is an element of some-thing upstairs, but you can’t say in advance what it is an element of. For example,17/125 is an element of the 5−3Z (or 5−4Z, or later ones), but not 5−2Z.

3.4.6. Example: inverse limits of A-modules. A variant of this construction worksin a number of categories that can be interpreted as sets with additional structure(such as abelian groups, A-modules, groups, etc.). While in the case of sets, thedirect limit is a quotient object of the direct sum (= disjoint union) of the Ai, inthe case of A-modules (for example), the direct limit is a quotient object of thedirect sum. Thus the direct limit is ⊕Ai modulo aj −F (m)(ai) for every m : i→ jin I. 31

29directlimit

30five

31alsoused, used in definition of a stalk

44

[primordial: [forme: Why is this here? Possibly interpret how to understand

it as a group?] This is cleaner in the special case when I satisfies the cofinal condition: forcofinal

each i, j ∈ I, there is a k such that there are arrows i → k, j → k. For example, in the category

of A-modules ModA, each element a of the colimit lim−→Ai can be interpreted as elements of some

a ∈ Ai. The element a ∈ lim−→Ai is 0 if there is some m : i → j such that F (m)(a) = 0 (i.e.

if it becomes 0 “later in the diagram”). More generally, two elements interpreted as ai ∈ Ai

and aj ∈ Aj are the same if there are some arrows m : i → k and n : j → k such that

F (m)(ai) = F (n)(aj), i.e. if they become the same “later in the diagram”. To add two elements

interpreted as ai ∈ Ai and aj ∈ Aj , we choose arrows m : i→ k and n : j → k, and then interpret

their sum as F (m)(ai) + F (n)(aj ). ]

3.4.D. Exercise. Verify that the A-module described above is indeed the colimit.

3.4.7. Summary. One useful thing to informally keep in mind is the following. Ina category where the objects are “set-like”, an element of a colimit can be thoughtof (“has a representative that is”) an element of a single object in the diagram.And an element of a limit can be thought of as an element in each object in thediagram, that are “compatible”. Swap the order. Even though the definitions oflimit and colimit are the same, just with arrows reversed, these interpretations arequite different.

[forme: I should check out [AM, p. 32, Exercise 14].]

3.5 Adjoints

[forme: Say why we care?] Here is another example of a construction closelyrelated to universal properties.32 Just as a universal property “essentially” (up tounique isomorphism) determines an object in a category (assuming it exists), “ad-joints” essentially determine a functor (again, assuming it exists). [forme: Weibeladjoint functors

Defn 2.3.9] Two covariant functors F : A → B and G : B → A are adjoint if thereis a natural bijection for all A ∈ A and B ∈ B(6) τAB : MorB(F (A), B)→ MorA(A, G(B)).

We say that (F, G) form an adjoint pair, and that F is left-adjoint to G (and Gis right-adjoint to F ). By “natural” we mean the following. [forme: Make sure to

comment on the word ’natural’ the first time I use it.] For all f : A → A′ in A,we require33

(7) MorB(F (A′), B)Ff∗

//

τA′B

MorB(F (A), B)

τAB

MorA(A′, G(B))

f∗

// MorA(A, G(B))

to commute, and for all g : B → B′ in B we want a similar commutative diagramto commute. (Here f∗ is the map induced by f : A → A′, and Ff∗ is the mapinduced by Ff : L(A)→ L(A′).)

32adjoint

33adjointdiagram

45

3.5.A. Exercise. Write down what this diagram should be. (Hint: do it byextending diagram (7) above.)

[forme: We could figure out what this should mean if the functors were both

contravariant. I haven’t tried to see if this could make sense. Remark: unique up to

unique isomorphism, by universal properties.]

3.5.B. Exercise. Show that the map τAB (6) is given as follows. For each A thereis a map A→ GFA so that for any g : F (a)→ B, the corresponding f : A→ G(B)is given by the composition

A // GFAGg // GB.

Similarly, there is a map B → FGB for each B so that for any f : A→ G(B), thecorresponding map g : F (A)→ B is given by the composition

F (A)Ff // F (G(B)) // B.

[forexperts: Is there an easy way to check adjointness using these twomaps? Here is what I’ve figured out. In order to get naturality in thefirst term, we need the commutativity of

A

// A′

GFA // GFA′

The naturality in the second term is automatic. I think in order tomake these constructions opposite of each other, we need to have that(Gt)B tGB = idGB and tF!

FtA = idFA. I think it will suffice to have GtBand tGB to be inverse isomorphisms, and simlarly for F and A. This isnot necessary. ]

Here is an example.

3.5.C. Exercise. Suppose M , N , and P are A-modules. Describe a natural bijec-tion MorA(M ⊗A N, P ) = MorA(M, MorA(N, P )). (Hint: try to use the universalproperty.) If you want, you could check that · ⊗A N and MorA(N, ·) are adjointfunctors. (Checking adjointness is never any fun!) [forme: We may later see why

this implies left-exactness of tensor product; refer back to that, and forward.]

[forme: A good example is Frobenius reciprocity, because it is so useful!. MorG(V,WG) =

MorH(VH ,W ), and MorG(WG, V ) = MorH(W,VH). ·H and ·H are both left and right ex-

act.]

3.5.1. Example: groupification. 34Here is another motivating example: gettingan abelian group from an abelian semigroup. An abelian semigroup is just like agroup, except you don’t require an inverse. One example is the non-negative integers0, 1, 2, . . . under addition. Another is the positive integers under multiplication1, 2, . . . . From an abelian semigroup, you can create an abelian group, and thiscould be called groupification. Here is a formalization of that notion. If S is asemigroup, then its groupification is a map of semigroups π : S → G such that G

34groupification

46

is a group, and any other map of semigroups from S to a group G′ factors uniquelythrough G.

S //

π

@@@

@@@@

@ G

∃!

G′

3.5.D. Exercise. Define groupification H from the category of abelian semi-groups to the category of abelian groups. (One possibility of a construction: givenan abelian semigroup S, the elements of its groupification H(S) are (a, b), whichyou may think of as a− b, with the equivalence that (a, b) ∼ (c, d) if a + d = b + c.Describe addition in this group, and show that it satisfies the properties of anabelian group. Describe the semigroup map S → H(S).) Let F be the forget-ful morphism from the category of abelian groups Ab to the category of abeliansemigroups. Show that H is left-adjoint to F .

(Here is the general idea for experts: We have a full subcategory of a cat-egory. We want to “project” from the category to the subcategory. We haveMorcategory(S, H) = Morsubcategory(G, H) automatically; thus we are describing theleft adjoint to the forgetful functor. How the argument worked: we constructedsomething which was in the small category, which automatically satisfies the uni-versal property.)

3.5.E. Exercise. Show that if a semigroup is already a group then groupificationis the identity morphism, by the universal property.

3.5.F. Exercise. The purpose of this exercise is to give you some practice with“adjoints of forgetful functors”, the means by which we get groups from semigroups,and sheaves from presheaves. Suppose A is a ring, and S is a multiplicative subset.Then S−1A-modules are a fully faithful subcategory of the category of A-modules(meaning: the objects of the first category are a subset of the objects of the second;and the morphisms between any two objects of the second that are secretly objectsof the first are just the morphisms from the first). Then M → S−1M satisfies auniversal property. Figure out what the universal property is, and check that itholds. In other words, describe the universal property enjoyed by M → S−1M ,and prove that it holds.

(Here is the larger story. Let S−1A-Mod be the category of S−1A-modules, andA-Mod be the category of A-modules. Every S−1A-module is an A-module, andthis is an injective map, so we have a (covariant) forgetful functor F : S−1A-Mod→ A-Mod. In fact this is a fully faithful functor: it is injective on objects, andthe morphisms between any two S−1A-modules as A-modules are just the samewhen they are considered as S−1A-modules. Then there is a functor G : A-Mod→ S−1A-Mod, which might reasonably be called “localization with respect to S”,which is left-adjoint to the forgetful functor. Translation: If M is an A-module, andN is an S−1A-module, then Mor(GM, N) (morphisms as S−1A-modules, which isincidentally the same as morphisms as A-modules) are in natural bijection withMor(M, FN) (morphisms as A-modules).)

Here is a table of adjoints that will come up for us.

47

situation category category left-adjoint right-adjointA B F : A → B G : B → A

A-modules · ⊗A N HomA(N, ·)(pre)sheaves on a presheaves sheaves on X sheafification forgetfultopological space X on X(semi)groups semigroups groups groupification forgetfulsheaves, f : X → Y sheaves on Y sheaves on X f−1 f∗quasicoherent sheaves, quasicoherent quasicoherent f ∗ f∗f : X → Y sheaves on Y sheaves on X

3.5.2. Useful comment for experts. Here is one last useful comment intendedonly for people who have seen adjoints before. If (F, G) is an adjoint pair of functors,then F commutes with colimits, and G commutes with limits. We’ll prove thisin §3.7.6 when we’ll extend this result.35 Also, limits commute with limits andcolimits commute with colimits. (This isn’t hard once you figure out what it means.)[forme: Main idea: product index category I × I ′.]

[primordial:

3.6 ? Natural transformations of functors, and equivalencesof categories

define “essentially surjective”. We can’t say everything is hit; we can just say that everything ishit up to isomorphism.

Equivalence of categories. When two categories are are essentially the same. this is abstractnonsense. Never saying “the same as”. In a category: always saying “isomorphic to”. Betweencategories: there is something weaker, that still has the sense of “same”.

Example: vector spaces with basis, and kn. There are more objects in the first.Then: vector spaces, and kn.We want two functors, so that... their compositionsVector spaces and their double-duals.

Not needed: This is transitive. ]

3.7 Kernels, cokernels, and exact sequences: A briefintroduction to abelian categories

Since learning linear algebra, the reader has been familiar with the notions andbehaviors of kernels, cokernels, etc. Later in your life you saw them in the categoryof abelian groups, and later still in the category of A-modules. Each of these notionsgeneralizes the previous one. The notion of abelian category formalizes kernels etc.

[forme: THINK THROUGH PRECISELY WHAT I WANT TO HAVE HERE.HOW LITTLE CAN I INCLUDE? Why do I need to discuss abelian categories?

• We want to define notions which behave like modules over a ring. Ouronly examples are: Ab, ModA, sheaves of abelian groups, OX-modules,qcoh/ft/coherent sheaves.

35preadjointfun

48

• Motivate definitions of: 0-object, monomorphism (subobject) and epimor-phism, kernel and cokernel, image (not coimage). The reader should under-stand that they behave the way he expects. This could be done by defininga “good subcategory C′ of an abelian category C” (contains 0; given anymorphism f in the subcategory, its kernel and cokernel and image [in C]are actually in C′; then they are actually cokernels and images in the smallsubcategory; etc.), and in our cases they are all good subcategories of someModA.

• derived functors

Definitions of abelian catogories from various sources (2 from wikipedia, [KS],Gelfand-Manin, [W], Lang) are in the “primordial” bit below.

[primordial: Wiki1

Z. has 0-objectPu. has pullbacks and pushoutsN. monomorphisms and epimorphisms are normal

Wiki2

PAd. homsets are abeilan groups; composition of morphisms are bilinear. (tohere: pre-additive)

BiP. finite biproducts (?). This implies SP. (to here: additive)K. every morphism has a kernel and cokernelN. every mono and epi is normal

[KS, 1.2]

PAd+Z.SP. has finite direct sums and finite direct products (to here: additive)K.I. For all f , The canonical map Coim f → im f is an isomorphism.

[KS] says: finite products finite sums in an additive category.

Gelfand-Manin II.5.

PAd+Z.SP’. For any two objects X1, X2, there exist an object Y and morphisms

X1 Yp1oo p2 // X2

X1i1 // Y X2

i2

oo

such that p1i1 = idX1, p2i2 = idX2

, i1p1 + i2p2 = idY , p2i1 = p1i2 = 0. ImpliesSP, Lemma 5 in II.5.

K+I. Any morphism f : X → Y admits a sequence

Kk // X i // I

j // Y c // K′

where j i = f , k is the kernel,,c is the cokernel, I is both the kernel andthe cokernel. Equivalent to K+I.

[W, App. A.4]

PAd+Z.Pr. finite direct products. (Hence = finite direct sums). (to here: additive)K.

49

W. Every monic is the kernel of its cokernel, and every epi is the cokernel ofits kernel

Lang revised third edition, III, §3, p. 133.

PAd+Z.SP. (to here: additive, as in [KS])K.L. If f is a morphism with ker f = 0, then f is the kernel of its cokernel.

Similarly for cokernels.

What I’ll take: PAd+Z+(SP or Pr) = Additive. Additive + K + (I or W

or L) = Abelian. I may take L, because then we don’t need monomorphisms or

epimorphisms. ] ]

[forme: Weibel appendix. According to [MilneLET] p. 2, abelian categories were

defined by Buchsbaum (and he extended the Cartan-Eilenberg theory of derived

functors to such categories). The name is due to Grothendieck.]36We now briefly introduce a few notions about abelian categories. We will

soon define some new categories (certain sheaves) that will have familiar-lookingbehavior, reminiscent of that of modules over a ring. The notions of kernels, coker-nels, images, and more will make sense, and they will behave “the way we expect”from our experience with modules. This can be made precise through the notionof an abelian category. We will see enough to motivate the definitions that we willsee in general: monomorphism (and subobject), epimorphism, kernel, cokernel, andimage. But we will avoid having to show that they behave “the way we expect”in a general abelian category because the examples we will see will be directlyinterpretable in terms of modules over rings.

Abelian categories are the right general setting in which one can do “homolog-ical algebra”, in which notions of kernel, cokernel, and so on are used, and one canwork with complexes and exact sequences. 37

Two key examples of an abelian category are the category Ab of abelian groups,and the category ModA of A-modules. The first is a special case of the second (justtake A = Z). As we give the definitions, you should verify that ModA is an abeliancategory. zero object

additive categoryWe first define the notion of additive category. We will use it only as astepping stone to the notion of an abelian category.

3.7.1. Definition. A category C is said to be additive if it satisfies the followingproperties. 38

Ad1. For each A, B ∈ C, Mor(A, B) is an abelian group, such that compositionof morphisms distributes over addition. (You should think about whatthis means — it translates to two distinct statements).

Ad2. C has a zero-object, denoted 0. (This is an object that is simultaneouslyan initial object and a final object, Defn. 3.3.2.)

Ad3. It has products of two objects (a product A×B for any pair of objects),and hence by induction, products of any finite number of objects.

In an additive category, the morphisms are often called homomorphisms, andMor is denoted by Hom. In fact, this notation Hom is a good indication that you’re

36s:abcat

37aac

38d:ac

50

working in an additive category. A functor between additive categories preservingthe additive structure of Hom, is called an additive functor. [forme: Weibeladditive functor

doesn’t include the 0 to 0 requirement. Joe explains why this is a consequence.

(Lemma: a is a 0-object iff ida = 0a. Additive functors preserve both id and 0.) Also,

I haven’t checked that it preserves products.] (It is a consequence of the definitionthat additive functors send 0-objects to 0-objects, and preserve products.)

3.7.2. Remarks. It is a consequence of the definition of additive category thatfinite direct products are also finite direct sums=coproducts (the details don’t mat-ter to us). The symbol ⊕ is used for this notion. Also, it is quick to show that⊕additive functors send zero-objects to zero-objects (show that a is a 0-object if andonly if ida = 0a; additive functors preserve both id and 0).

One motivation for the name 0-object is that the 0-morphism in the abeliangroup Hom(A, B) is the composition A→ 0→ B.

Real (or complex) Banach spaces are an example of an additive category. Thecategory ModA of A-modules is another example, but it has even more structure,which we now formalize as an example of an abelian category.abelian category

3.7.3. Definition. Let C be an additive category. A kernel of a morphismkernel, cokernel, monic

morphism, epi mor-

phism

f : B → C is a map i : A → B such that f i = 0, and that is universal withrespect to this property. Diagramatically:

Z

@@@

@@@@

0

''OOOOOOOOOOOOOO

∃!

A

i //

0

77Bf // C

(Note that the kernel is not just an object; it is a morphism of an object to B.)Hence it is unique up to unique isomorphism by universal property nonsense. Acokernel is defined dually by reversing the arrows — do this yourself. Notice thatthe kernel of f : B → C is the limit (§3.4) of the diagram

0

B

f // C

and similarly the cokernel is a colimit. 39

A morphism i : A → B in C is monic if for all i g = 0, where the tail of g isA, implies g = 0. Diagramatically,

C

∴g=0

0

@@@

@@@@

Ai // B

(Once we know what an abelian category is, you may check that a monic morphismin an abelian category is a monomorphism.) [forme: Do we need ab cat for this?

Even if so, it is weird to refer forward only a couple of sentences.] If i : A → Bis monic, then we say that A is a subobject of B, where the map i is implicit.

39d:kernel

51

Dually, there is the notion of epi — reverse the arrows to find out what that is.The notion of quotient object is defined dually to subobject. sub and quotient obj

An abelian category is an additive category satisfying three additional prop-erties.

(1) Every map has a kernel and cokernel.(2) Every monic morphism is the kernel of its cokernel.(3) Every epi morphism is the cokernel of its kernel.

It is a non-obvious (and imprecisely stated) fact that every property you wantto be true about kernels, cokernels, etc. follows from these three.

The image of a morphism f : A → B is defined as im(f) = ker(coker f). It isthe unique factorization

Aepi // im(f)

monic // B

It is the cokernel of the kernel, and the kernel of the cokernel. The reader maywant to verify this as an exercise. It is unique up to unique isomorphism.

We will leave the foundations of abelian categories untouched. The key thingto remember is that if you understand kernels, cokernels, images and so on in thecategory of modules over a ring ModA, you can manipulate objects in any abeliancategory. This is made precise by Freyd-Mitchell Embedding Theorem. However, Freyd-Mitchell

the abelian categories we’ll come across will obviously be related to modules, andour intuition will clearly carry over. For example, we’ll show that sheaves of abeliangroups on a topological space X form an abelian category (§4.5), and the interpre-tation in terms of “compatible germs” will connect notions of kernels, cokernels etc.of sheaves of abelian groups to the corresponding notions of abelian groups.

[primordial: Freyd-Mitchell Embedding Theorem. If A is an abelian category such that

Hom(a, a′) is a set for all a, a′ ∈ A, then there is a ring A and an exact, full faithful functor

from A into ModA, which embeds A as a full subcategory. A proof is sketched in [W, §1.6], and

references to a complete proof are given there. We won’t use this fact, but we’ll take as a moral

that if we need to prove anything about a diagram in some abelian category (for example, that

it commutes), we are free to imagine that it is a diagram of modules over some ring, and we may

then “diagram-chase” elements. Moreover, any fact about kernels, cokernels, and so on that holds

in ModA holds in any abelian category. ]

3.7.4. Complexes, exactness, and homology.We say 4041

(8) Af // B

g // C

is a complex if g f = 0, and is exact if ker g = im f . If (8) is a complex,then its homology is ker g/ im f . We say that ker g are the cycles, and im f arethe boundaries. Homology (resp. cohomology) is denoted by H , often with asubscript (resp. superscript), and it should be clear from the context what thesubscript means (see for example the discussion below). complex exact homology

cycle boundaryAn42 exact sequence

(9) A• : · · · // Ai−1f i−1

// Aif i

// Ai+1f i+1

// · · ·

40ceh

41s:ceh

42infinitesequence2

52

can be “factored” into short exact sequences

0 // ker f i // Ai // ker f i+1 // 0

which is helpful in proving facts about long exact sequences by reducing them tofacts about short exact sequences.

More generally, if (9) is assumed only to be a complex, then it can be “factored”into short exact sequences

0 // ker f i // Ai // im f i // 0

0 // im f i−1 // ker f i // H i(A•) // 0

3.7.A. Exercise. Suppose

0d0

// A1 d1// · · · dn−1

// An dn//// 0

is a complex of k-vector spaces (often called A• for short). Show that∑

(−1)i dim Ai =∑

(−1)ihi(A•). (Recall that hi(A•) = dim ker(di)/ im(di−1).) In particular, if A•

is exact, then∑

(−1)i dim Ai = 0. (If you haven’t dealt much with cohomology,this will give you some practice.)

3.7.B. Important exercise. 43Suppose C is an abelian category. Define thecategory ComC as follows. The objects are infinite complexes

A• : · · · // Ai−1f i−1

// Aif i

// Ai+1f i+1

// · · ·

in C, and the morphisms A• → B• are commuting diagrams

A• :

· · · // Ai−1

f i−1

// Aif i

//

Ai+1f i+1

//

· · ·

B• : · · · // Bi−1f i−1

// Bif i

// Bi+1f i+1

// · · ·

43LES

53

Show that ComC is an abelian category. Show that a short exact sequence ofcomplexes

0 :

· · · // 0 //

0 //

0 //

· · ·

A• :

· · · // Ai−1

f i−1

// Aif i

//

Ai+1f i+1

//

· · ·

B• :

· · · // Bi−1

gi−1

// Bigi

//

Bi+1gi+1

//

· · ·

C• :

· · · // Ci−1hi−1

//

Cihi

//

Ci+1hi+1

//

· · ·

0 : · · · // 0 // 0 // 0 // · · ·induces a long exact sequence in cohomology LES

. . . // H i−1(C•) //

H i(A•) // H i(B•) // H i(C•) //

H i+1(A•) // · · ·

(This can be done with spectral sequences as well, see Exercise 3.8.E.)

3.7.5. Exactness of functors. If F : A → B is a covariant additive functor fromone abelian category to another, we say that F is right-exact if the exactness of

A′ // A // A′′ // 0,

in A implies that

F (A′) // F (A) // F (A′′) // 0

is also exact. [forme: Why is F (0) = 0?] Dually, we say that F is left-exact ifthe exactness of

0 // A′ // A // A′′ implies

0 // F (A′) // F (A) // F (A′′) is exact.

A contravariant functor is left-exact if the exactness of

A′ // A // A′′ // 0 implies

0 // F (A′′) // F (A) // F (A′) is exact.

54

The reader should be able to deduce what it means for a contravariant functor tobe right-exact.L- and R-exact

A covariant or contravariant functor is exact if it is both left-exact and right-exact.

When you see a left-exact functor, you should always dream that you are seeingthe end of a long exact sequence. If

0→M ′ →M →M ′′ → 0

is an exact sequence in abelian category A, and F : A → B is a left-exact functor,then

0→ FM ′ → FM → FM ′′

is exact, and you should always dream that it should continue in some natural way.For example, the next term should depend only on M ′, call it R1FM ′, and if it iszero, then FM → FM ′′ is an epimorphism. This remark holds true for left-exactand contravariant functors too. In good cases, such a continuation exists, and isincredibly useful. We’ll see this when we come to cohomology.

[primordial: [W, p. 429]. A-mod is a cocomplete abelian category, so filtered colimits

always exist. This is exact, and commutes with Tor. This gives an easy proof that S−1A is A-flat.

]

3.7.6. ? Interactions of adjoints, (co)limits, and (left- and right-) exact-ness. 44There are some useful properties of adjoints that make certain argumentsquite short. We present them as three facts. Suppose (F : C → D, G : D → C) is apair of adjoint functors.

Fact 1. F commutes with colimits, and G commutes with limits. [forme: [W,

2.6.10]]

This was promised in §3.5.2. We prove the second statement here. The first isthe same, “with the arrows reversed”. We begin with a useful fact. [forme: We

don’t seem to need this... And this isn’t a special case. Think about this more.]

3.7.C. Exercise: Mor(X, ·) commutes with limits. Suppose Ai (i ∈ I) is adiagram in D indexed by I, and lim←−Ai → Ai is its limit. Then for any X ∈ D,

Mor(X, lim←−Ai)→ Mor(X, Ai) is the limit lim←−Mor(X, Ai).45

ex.doneWe are now ready to prove (one direction of) Fact 1.

3.7.7. Proposition (right-adjoints commute with limits). — Suppose(F : C → D, G : D → C) is a pair of adjoint functors. If A = lim←−Ai is a limit

in D of a diagram indexed by I, then GA = lim←−GAi (with the corresponding maps

GA→ GAi) is a limit in C.

Proof. We must show that GA → GAi satisfies the universal property of limits.Suppose we have maps W → GAi commuting with the maps of I. We wish to showthat there exists a unique W → GA extending the W → GAi. By adjointness of Fand G, we can restate this as: Suppose we have maps FW → Ai commuting withthe maps of I. We wish to show that there exists a unique FW → A extending theFW → Ai. But this is precisely the universal property of the limit.

44adjointfun

45e:homlim

55

Suppose now further that C and D are abelian categories, and F and G areadditive functors. Kernels are limits and cokernels are colimits (§3.7.3), so we haveFact 2. F commutes with cokernels and G commutes with kernels.

Now suppose

M ′f // M // M ′′ // 0

is an exact sequence in C, so M ′′ = cokerf . [forme: Luis Diogo says: I’m using the

fact that the second map is the cokernel of the first, without explaining why. I can’t

see that I should comment about this.] Then by Fact 2, FM ′′ = cokerFf . Thus

FM ′ → FM → FM ′′ → 0

so we have: Fact 3. Left-adjoint additive functors are right-exact, and right-adjointadditive functors are left-exact. For example, the fact that (· ⊗A N, HomA(N, ·))are an adjoint pair (from the A-Mod to itself) imply that · ⊗A N is right-exact(Exercise 3.3.B) and Hom(N, ·) is left-exact. [forme: [W, 2.6.1], and Rotman p.

515, 7.105. [W, 2.3.10], with short proof: If an additive functor is right-adjoint to an

exact functor, then R preserves injectives. (Recall injective means Hom(·, I) is exact.)

Will this be handy? ]

[forme: Kirsten points out: Colimits taken over directed systems are in fact

exact. Reference: [E] Appendix A6, and colimits prop A6.4. Limits are left-exact.

Eisenbud uses the language of filtered category, and gives a careful definition with

two parts. Here’s one idea of why: colimits are left-adjoints, and hence are right-

exact. So we need only worry about left-exactness. Consequence: sheafification is

exact. Reason: exact sequence of presheaves gives exact sequence of stalks of those

presheaves, gives exact sequence of stalks of the sheaves, gives exact sequence of

sheaves. Further consequence: hence the forgetful map from sheaves to presheaves

sends injectives to injectives, from which I think global sections of injective sheaves

are injective.]

3.8 ? Spectral sequences

46Spectral sequences are a powerful book-keeping tool for proving things involv- spectral sequence

ing complicated commutative diagrams. [forme: They were introduced by Leray

in the 1940’s at the same time as he introduced sheaves. Reference from [KS, p. 138].

There is a “short history” as a section of [KS]. According to [MilneLET, p. 2] sheaves

were introduced in the 1940s by Leray. He also introduced sheaf cohomology and

spectral sequences. Spectral sequences were introduced at about the same time by

Roger Lyndon, who was trying to understand the relations between the cohomologies

of a group, a normal subgroup, and the quotient.] They have a reputation forbeing abstruse and difficult. It has been suggested that the name ‘spectral’ wasgiven because, like spectres, spectral sequences are terrifying, evil, and dangerous. Ihave heard no one disagree with this interpretation, which is perhaps not surprisingsince I just made it up.

Nonetheless, the goal of this section is to tell you enough that you can usespectral sequences without hesitation or fear, and why you shouldn’t be frightenedwhen they come up in a seminar. What is different in this presentation is that

46ss

56

we will use spectral sequence to prove things that you may have already seen, andthat you can prove easily in other ways. This will allow you to get some hands-onexperience for how to use them. We will also see them only in a “special case”of double complexes (which is the version by far the most often used in algebraicgeometry), and not in the general form usually presented (filtered complexes, exactcouples, etc.). See [W, Ch. 5] for more detailed information if you wish. [forme:

The exposition in [L] is just a few pages long. [MilneLET, p. 72] says: there is a three-

page explanation of spectral sequences in EC p. 307–309, a 14 page explanation in

Shatz, Profinite Groups, Arithemtic, and Geometry, Princeton 1972, II.4, and a 45-

page explanation in Weibel, chapter 5 including proof. Tim Chow’s Notices article

“You could have discovered...”.]

You should not read this section when you are reading the rest of Chapter 3.Instead, you should read it just before you need it for the first time. (You’ll bedirected back here at that time.) When you finally do read this section, you mustwork through the exercises.

[primordial: Discussion of proof: Possibly give a ref to the proof. Otherwise, give the

proof. In any case, tell them: Once you have read how to use spectral sequences, I strongly

encourage you to wait a healthy interval (at least a month) before reading the proof. The proof

will confuse you and fool you into believing that using spectral sequences is difficult. This is one

of the cases in mathematics where it is much more important to know how to use a machine than

to know why it works. ]We work in the category ModA of A-modules. However, everything we say

will apply in any abelian category.

3.8.1. Double complexes.A first-quadrant double complex (in ModA) is a collection of A-modulesdouble cx

Ep,q (p, q ∈ Z), which are zero unless p, q ≥ 0, and “rightward” morphisms dp,q> :

Ep,q → Ep+1,q and “upward” morphisms dp,q∧ : Ep,q → Ep,q+1. The subscript is

meant to suggest the direction of the arrows. We will always write these as d> andd∧ and ignore the superscripts. We require that d> and d∧ satisfying (a) d2

> = 0,(b) d2

∧ = 0, and one more condition: (c) either d>d∧ = d∧d> (all the squarescommute) or d>d∧ + d∧d> = 0 (they all anticommute). Both come up in nature,and you can switch from one to the other by replacing dp,q

∧ with dp∧(−1)q . So I’ll

hereafter assume that all the squares anticommute, but that you know how to turnthe commuting case into this one. (You will see that there is no difference in therecipe, basically because the image and kernel of a homomorphism f equal imageand kernel respectively of −f .)

Ep+1,qdp+1,q

> // Ep+1,q+1

anticommutes

Ep,q

dp,q∧

OO

dp,q> // Ep,q+1

dp,q+1∧

OO

There are variations on this definition, where for example the vertical arrowsgo downwards, or some different subset of the Ep,q are required to be zero, but I’llleave these straightforward variations to you.

57

From the double complex we construct a corresponding (single) complex E•

with Ek = ⊕iEi,k−i, with d = d> + d∧. In other words, when there is a single

superscript k, we mean a sum of the kth antidiagonal of the double complex. Thesingle complex is somtimes called the total complex. Note that d2 = (d>+d∧)2 =d2

> + (d>d∧ + d∧d>) + d2∧ = 0, so E• is indeed a complex.

The cohomology of the single complex is sometimes called the hypercoho-mology of the double complex. We will instead use the phrase “cohomology of thedouble complex”. hypercoh, coh of a dou-

ble cxOur initial goal will be to find the hypercohomology of the double complex.You will see later that we secretly have other goals.

A spectral sequence is a recipe for computing some information about thecohomology of the double complex. We won’t yet give the full recipe. Surprisingly,this fragmentary bit of information is sufficent to prove lots of things.

3.8.2. Approximate Definition. A spectral sequence with rightwardorientation is a sequence of tables or pages >Ep,q

0 , >Ep,q1 , >Ep,q

2 , . . . (p, q ∈ Z), page

where >Ep,q0 = Ep,q , along with a differential

>dp,qr : >Ep,q

r → >Ep+r,q−r+1

with >dp,qr >dp,q

r = 0, along with an isomorphism of the cohomology of >dr at

>Ep,q (i.e. ker >dp,qr / im >dp−r,q+r−1

r ) with >Ep,qr+1.

The orientation indicates that our 0th differential is the rightward one d0 = d>.The left subscript “>” is usually omitted. orientation

The order of the morphisms is depicted below.47

(10) •

•

•

•d0

//

d1

OOd2

WW//////////////

d3

ZZ5555

5555

5555

5555

5555

555

•

(although the morphisms each apply to different pages). Notice that the map alwaysis “degree 1” in the grading of the single complex E•.

The actual definition describes what E•,•r and d•,•

r actually are, in terms ofE•,•. We will describe d0, d1, and d2 below, and you should for now take on faiththat this sequence continues in some natural way.

Note that Ep,qr is always a subquotient of the corresponding term on the 0th

page Ep,q0 = Ep,q . In particular, Ep,q

r = 0 unless p, q ∈ Z≥0. Notice also that forany fixed p, q, once r is sufficiently large, Ep,q

r+1 is computed from (E•,•r , dr) using

the complex

0dp+r,q−r−1

r // Ep,qr

dp,qr // 0

47ssorder

58

and thus we have canonical isomorphisms

Ep,qr∼= Ep,q

r+1∼= Ep,q

r+2∼= · · ·

We denote this module Ep,q∞ .

We now describe the first few pages of the spectral sequence explicitly. Asstated above, the differential d0 on E•,•

0 = E•,• is defined to be d>. The rows arecomplexes:

• // • // •

The 0th page E0: • // • // •

• // • // •and so E1 is just the table of cohomologies of the rows. You should check thatthere are now vertical maps dp,q

1 : Ep,q1 → Ep+1,q

1 of the row cohomology groups,induced by d∧, and that these make the columns into complexes. (We have “usedup the horizontal morphisms”, but “the vertical differentials live on”.)

• • •

The 1st page E1: •

OO

•

OO

•

OO

•

OO

•

OO

•

OO

We take cohomology of d1 on E1, giving us a new table, Ep,q2 . It turns out that

there are natural morphisms from each entry to the entry two above and one to theleft, and that the composition of these two is 0. (It is a very worthwhile exercise towork out how this natural morphism d2 should be defined. Your argument may bereminiscent of the connecting homomorphism in the Snake Lemma 3.8.5 or in thelong exact sequence in cohomology, Exercise 3.7.B. This is no coincidence.)

• • •

The 2nd page E2: • • •

• •

WW..............

•

WW..............

This can go on until the cows come home.Then it is a theorem that there is a filtration of Hk(E•) by Ep,q

∞ where p+q = k.(We can’t yet state it as an official Theorem because we haven’t precisely definedthe pages and differentials in the spectral sequence.) More precisely, In other words,there is a filtration48

(11) Ek,0∞

Ek−1,1∞ // ?

Ek−2,2∞ // · · · E0,k

// Hk(E•)

48spectralfiltration

59

where the quotients are displayed above each inclusion. (I always forget which waythe quotients are supposed to go, i.e. whether Ek,0 or E0,k is the subobject. Oneway of remembering it is by having some idea of how the result is proved.)

We say that the spectral sequence >E•,•• converges to H•(E•). We often say

that >E•,•2 (or any other page) abuts to H•(E•). convergence

Although the filtration gives only partial information about H•(E•), sometimesone can find H•(E•) precisely. One example is if all Ei,k−i

∞ are zero, or if all butone of them are zero (e.g. if Ei,k−i

r has precisely one non-zero row or column, inwhich case one says that the spectral sequence collapses at the rth step, althoughwe will not use this term). Another example is in the category of vector spacesover a field, in which case we can find the dimension of Hk(E•). Also, in luckycircumstances, E2 (or some other small page) already equals E∞.

3.8.A. Exercise: information from the second page. Show that H0(E•) =

E0,0∞ = E0,0

2 and

0→ E1,02 → H1(E•)→ E0,1

2 → E2,02 → H2(E•).

[primordial:

3.8.B. Exercise. Suppose we are working in the category of vector spaces over a field, and

⊕p,qEp,q2 is a finite-dimensional vector space. Show that χ(H•(E•)) is well-defined, and equalsPp,q(−1)p+qEp,q2 . (Not infrequently, ⊕Ep,q0 will be an infinite-dimensional vector space, but that

⊕Ep,q2 will be finite-dimensional!) ]

3.8.3. The other orientation.The reader may have observed that we could as well have done everything in

the opposite direction, i.e. reversing the roles of horizontal and vertical morphisms.Then the sequences of arrows giving the spectral sequence would look like this(compare to (10)).49

(12) •

•

OO

//

''OOOOOOOOOOOOOO

%%JJJJJJJJJJJJJJJJJJJJJJJ •

•

•This spectral sequence is denoted ∧E•,•

• (“with the upwards orientation”). Thenwe would again get pieces of a filtration of H•(E•) (where we have to be a bitcareful with the order with which ∧Ep,q

∞ corresponds to the subquotients — it inthe opposite order to the >Ep,q

∞ ). Warning: in general there is no isomorphismbetween >Ep,q

∞ and ∧Ep,q∞ .

In fact, this observation was our secret goal all along. Both algorithms computeinformation about the same thing (H•(E•)), and usually we don’t care about thefinal answer — we often care about the answer we get in one way, and we get at itby doing the spectral sequence in the other way.

49ssorder2

60

3.8.4. Examples.We’re now ready to see how this is useful. The moral of these examples is the

following. In the past, you may have proved various facts involving various sortsof diagrams, which involved chasing elements around. Now, you’ll just plug theminto a spectral sequence, and let the spectral sequence machinery do your chasingfor you.

3.8.5. Example: Proving the Snake Lemma. Consider the diagram50

0 // D // E // F // 0

0 // A //

α

OO

B //

β

OO

C

γ

OO

// 0

where the rows are exact and the squares commute. (Normally the Snake Lemmais described with the vertical arrows pointing downwards, but I want to fit thisinto my spectral sequence conventions.) We wish to show that there is an exactsequence51

(13) 0→ kerα→ kerβ → kerγ → im α→ im β → im γ → 0.

We plug this into our spectral sequence machinery. We first compute the coho-mology using the rightwards orientation, i.e. using the order (10). Then because therows are exact, Ep,q

1 = 0, so the spectral sequence has already converged: Ep,q∞ = 0.

We next compute this “0” in another way, by computing the spectral sequenceusing the upwards orientation. Then ∧E•,•

1 (with its differentials) is:

0 // im α // im β // im γ // 0

0 // kerα // kerβ // ker γ // 0.

Then ∧E•,•2 is of the form:

0

''OOOOOOOOOOOOOOO ??

''OOOOOOOOOOOOOO ?

''OOOOOOOOOOOOOOO ? 0

0 ? ? ?? 0.

We see that after ∧E2, all the terms will stabilize except for the double-question-marks — all maps to and from the single question marks are to and from 0-entries.And after ∧E3, even these two double-quesion-mark terms will stabilize. But in theend our complex must be the 0 complex. This means that in ∧E2, all the entriesmust be zero, except for the two double-question-marks, and these two must bethe isormorphic. This means that 0 → kerα → kerβ → ker γ and im α → im β →im γ → 0 are both exact (that comes from the vanishing of the single-question-marks), and

coker(kerβ → ker γ) ∼= ker(im α→ im β)

is an isomorphism (that comes from the equality of the double-question-marks).Taken together, we have proved the exactness of (13), and hence the Snake Lemma!

50snakelemma

51e:snakelemma

61

Spectral sequences make it easy to see how to generalize results further. Forexample, if A→ B is no longer assumed to be injective, how would the conclusionchange? [forme: Should I have them prove a slightly weaker version of Snake

Lemma, where one of the short exact sequences isn’t injective? Used in proof of

chomology and base change, or possibly cohomology and flat base change. Give a

ref.]

3.8.6. Example: the Five Lemma. Suppose52

(14) F // G // H // I // J

A //

α

OO

B //

β

OO

C

γ

OO

// D //

δ

OO

E

ε

OO

where the rows are exact and the squares commute.Suppose α, β, δ, ε are isomorphisms. We’ll show that γ is an isomorphism.We first compute the cohomology of the total complex using the rightwards

orientation (10). We choose this because we see that we will get lots of zeros. Then

>E•,•1 looks like this:

? 0 0 0 ?

?

OO

0

OO

0

OO

0

OO

?

OO

Then >E2 looks similar, and the sequence will converge by E2 (as we will never getany arrows between two non-zero entries in a table thereafter). We can’t concludethat the cohomology of the total complex vanishes, but we can note that it vanishesin all but four degrees — and most important, in the two degrees corresponding tothe entries C and H (the source and target of γ).

We next compute this using the upwards orientation (12). Then ∧E1 looks likethis:

0 // 0 // ? // 0 // 0

0 // 0 // ? // 0 // 0

and the spectral sequence converges at this step. We wish to show that those twoquestion marks are zero. But they are precisely the cohomology groups of the totalcomplex that we just showed were zero — so we’re done!

3.8.C. Exercise: The subtle five lemma. By looking at this proof, provea subtler version of the Five Lemma, where one of the isomorphisms can insteadjust be required to be an injection, and another can instead just be required to bea surjection. (I am deliberately not telling you which ones, so you can see how thespectral sequence is telling you how to improve the result.) [forme: I’ve heard this

called the “Subtle Five Lemma”, but I like calling it the 4 12-lemma. Wait, is the next

one also called the subtle five lemma?] 53 5455

52fivehyp

53subtlefive

54fourandhalf

55fivelemma

62

3.8.D. Exercise. If β and δ (in (14)) are injective, and α is surjective, showthat γ is injective. State the dual statement. (Its proof is of course essentially thesame.) [forme: [KS, Ex. I.8, p. 72], [L, p. 169].]

3.8.E. Exercise. Use spectral sequences to show that a short exact sequence ofcomplexes gives a long exact sequence in cohomology (Exercise 3.7.B). 56

3.8.F. Exercise. Suppose µ : A• → B• is a morphism of complexes. SupposeC• is the single complex associated to the double complex A• → B•. (C• is calledthe mapping cone of µ.) Show that there is a long exact sequence of complexes:mapping cone

· · · → H i−1(C•)→ H i(A•)→ H i(B•)→ H i(C•)→ H i+1(A•)→ · · · .(There is a slight notational ambiguity here; depending on how you index your dou-ble complex, your long exact sequence might look slightly different.) In particular,we will use the fact that µ induces an isomorphism on cohomology if and only ifthe mapping cone is exact. [forme: Link to where we use it. Also refer to other

mentions of mapping cone, refmappingcone1, mappingcone3, mappingcone4.] 57

The Grothendieck spectral sequence [forme: ref!] will be an importantexample of a spectral sequence that specializes in a number of useful ways. [pri-mordial: GSS from [L]. Example: Leray. I give a Cech version in Section 23.9. Refer to

[L, p. 819] for the proof, or possible Weibel [W]. Example: if f : X → Y is affine and F is

a quasicoherent sheaf on X, then Hi(X,F) ∼= Hi(Y, f∗F). We’ve proved this in another way

(Exercise 23.6.A). ][forme: From [W, p. 11] the 3 × 3 lemma. If bottom two rows are exact, so is

the top. If the top and bottom rows are exact, and the middle is a complex, then

the middle row is also exact.] [forexperts: Do you have some classicalimportant fact that would be useful practice for people learning spectralsequences?]

You are now ready to go out into the world and use spectral sequences to yourhearts’ content!

3.8.7. ?? Complete definition of the spectral sequence, and proof.You should most definitely not read this section any time soon after reading

the introduction to spectral sequences above. Instead, flip quickly through it toconvince yourself that nothing fancy is involved. [forme: I’m trying to follow [L, p.

814].]

We consider the rightwards orientation. The upwards orientation is of coursea trivial variation of this.

3.8.8. Goals. We wish to describe the pages and differentials of the spectralsequence explicitly, and prove that they behave the way we said they did. Moreprecisely, we wish to:58

(a) describe Ep,qr ,

(b) verify that Hk(E•) is filtered by Ep,k−p∞ as in (11),

(c) describe dr and verify that d2r = 0, and

(d) verify that Ep,qr+1 is given by cohomology using dr.

56LES2

57mappingcone2

58ssgoals

63

Before tacking these goals, you can impress your friends by giving this shortdescription of the pages and differentials of the spectral sequence. We say that anelement of E•,• is a (p, q)-strip if it is an element of ⊕l≥0E

p+l,q+l (see Fig. 1). Itsnon-zero entries lie on a semi-infinite antidiagonal starting with position (p, q).59

We say that the (p, q)-entry (the projection to Ep,q) is the leading term of the

(p, q)-strip. Let Sp,q ⊂ E•,• be the submodule of all the (p, q)-strips. Clearly

Sp,q ⊂ Ep+q , and S0,k = Ek.

. . . 0 0 0 0

0 ∗p+2,q−2 0 0 0

0 0 ∗p+1,q−1 0 0

0 0 0 ∗p,q 0

0 0 0 0 0p−1,q+1

Figure 1. A (p, q)-strip (in Sp,q ⊂ Ep+q). Clearly S0,k = Ek.

Note that the differential d = d∧ + d> sends a (p, q)-strip x to a (p, q + 1)-stripdx. If dx is furthermore a (p + r, q + r + 1)-strip (r ∈ Z≥0), we say that x is an

r-closed (p, q)-strip. We denote the set of such Sp,qr (so for example Sp,q

0 = Sp,q,

and S0,k0 = Ek). An element of Sp,q

r may be depicted as:

. . .// ?

∗p+2,q−2

OO

// 0

∗p+1,q−1

OO

// 0

∗p,q //

OO

0

3.8.9. Preliminary definition of Ep,qr . 60We are now ready to give a first definition

of Ep,qr , which by construction should be a subquotient of Ep,q = Ep,q

0 . We describeit as such by describing two submodules Y p,q

r ⊂ Xp,qr ⊂ Ep,q, and defining Ep,q

r =

59pqstrip

60pd:epqr

64

Xp,qr /Y p,q

r . Let Xp,qr be those elements of Ep,q that are the leading terms of r-closed

(p, q)-strips. Note that by definition, d sends (r−1)-closed Sp−(r−1),q+(r−1)−1-stripsto (p, q)-strips. Let Y p,q

r be the leading ((p, q))-terms of the differential d of (r−1)-closed (p− (r − 1), q + (r − 1)− 1)-strips (where the differential is considered as a(p, q)-strip).

We next give the definition of the differential dr of such an element x ∈ Xp,qr .

We take any r-closed (p, q)-strip with leading term x. Its differential d is a (p+r, q−r + 1)-strip, and we take its leading term. The choice of the r-closed (p, q)-stripmeans that this is not a well-defined element of Ep,q. But it is well-defined modulothe (r − 1)-closed (p + 1, r + 1)-strips, and hence gives a map Ep,q

r → Ep+r,q−r+1r .

This definition is fairly short, but not much fun to work with, so we will forgetit, and instead dive into a snakes’ nest of subscripts and superscripts.

We begin with making some quick but important observations about (p, q)-strips.

3.8.G. Exercise. Verify the following. 61

(a) Sp,q = Sp+1,q−1 ⊕Ep,q .(b) (Any closed (p, q)-strip is r-closed for all r.) Any element x of Sp,q = Sp,q

0

that is a cycle (i.e. dx = 0) is automatically in Sp,qr for all r. For example,

this holds when x is a boundary (i.e. of the form dy). 62

(c) Show that for fixed p, q,

Sp,q0 ⊃ Sp,q

1 ⊃ · · · ⊃ Sp,qr ⊃ · · ·

stabilizes for r 0 (i.e. Sp,qr = Sp,q

r+1 = · · · ). Denote the stabilizedmodule Sp,q

∞ . Show Sp,q∞ is the set of closed (p, q)-strips (those (p, q)-strips

annihilated by d, i.e. the cycles). In particular, S0,kr is the set of cycles in

Ek.63

3.8.10. Defining Ep,qr .

Define Xp,qr := Sp,q

r /Sp+1,q−1r−1 and Y := dS

p−(r−1),q+(r−1)−1r−1 /Sp+1,q−1

r−1 .

Then Y p,qr ⊂ Xp,q

r by Exercise 3.8.G(b). We define 64

(15) Ep,qr =

Xp,qr

Y p,qr

=Sp,q

r

dSp−(r−1),q+(r−1)−1r−1 + Sp+1,q−1

r−1

We have completed Goal 3.8.8(a).You are welcome to verify that these definitions of Xp,q

r and Y p,qr and hence

Ep,qr agree with the earlier ones of §3.8.9 (and in particular Xp,q

r and Y p,qr are both

submodules of Ep,q), but we won’t need this fact.

3.8.H. Exercise: Ep,k−p∞ gives subquotients of Hk(E•). By Exercise 3.8.G(c),

Ep,qr stabilizes as r →∞. For r 0, interpret Sp,q

r /dSp−(r−1),q+(r−1)−1r−1 as the cy-

cles in Sp,q∞ ⊂ Ep+q modulo those boundary elements of dEp+q−1 contained in Sp,q

∞ .Finally, show that Hk(E•) is indeed filtered as described in (11).

ex.doneWe have completed Goal 3.8.8(b).

61e:pqstrip

62Zboundary

63ssstablize

64d:Epqr

65

3.8.11. Definition of dr.We shall see that the map dr : Ep,q

r → Ep+r,q−r+1 is just induced by ourdifferential d. Notice that d sends r-closed (p, q)-strips Sp,q

r to (p + r, q − r + 1)-strips Sp+r,q−r+1, by the definition “r-closed”. By Exercise 3.8.G(b), the imagelies in Sp+r,q−r+1

r .

3.8.I. Exercise. Verify that d sends

dSp−(r−1),q+(r−1)−1r−1 +Sp+1,q−1

r−1 → dS(p+r)−(r−1),(q−r+1)+(r−1)−1r−1 +S

(p+r)+1,(q−r+1)−1r−1 .

(The first term on the left goes to 0 from d2 = 0, and the second term on the leftgoes to the first term on the right.)

ex.doneThus we may define

dr : Ep,qr =

Sp,qr

dSp−(r−1),q+(r−1)−1r−1 + Sp+1,q−1

r−1

→

Sp+r,q−r+1r

dSp+1,q−1r−1 + Sp+r+1,q−r

r−1

= Ep+r,q−r+1r

and clearly d2r = 0 (as we may interpret it as taking an element of Sp,q

r and applyingd twice).

We have accomplished Goal 3.8.8(c).

3.8.12. Verifying that the cohomology of dr at Ep,qr is Ep,q

r+1. We are left with the

unpleasant job of verifying that the cohomology of 65

(16)

Sp−r,q+r−1r

dSp−2r+1,q−3r−1 +Sp−r+1,q+r−2

r−1

dr // Sp,qr

dSp−r+1,q+r−2r−1 +Sp+1,q−1

r−1

dr // Sp+r,q−r+1r

dSp+1,q−1r−1 +Sp+r+1,q−r

r−1

is naturally identified with

Sp,qr+1

dSp−r,q+r−1r + Sp+1,q−1

r

and this will conclude our final Goal 3.8.8(d).Let’s begin by understanding the kernel of the right map of (16). Suppose

a ∈ Sp,qr is mapped to 0. This means that da = db + c, where b ∈ Sp+1,q−1

r−1 .

If u = a − b, then u ∈ Sp,q , while du = c ∈ Sp+r+1,q−rr−1 ⊂ Sp+r+1,q−r, from

which u is r-closed, i.e. u ∈ Sp,qr+1. Hence a = b + u + x where dx = 0, from

which a − x = b + c ∈ Sp+1,q−1r−1 + Sp,q

r+1. However, x ∈ Sp,q , so x ∈ Sp,qr+1 by

Exercise 3.8.G(b). Thus a ∈ Sp+1,q−1r−1 +Sp,q

r+1. Conversely, any a ∈ Sp+1,q−1r−1 +Sp,q

r+1

satisfies

da ∈ dSp+r,q−r+1r−1 + dSp,q

r+1 ⊂ dSp+r,q−r+1r−1 + Sp+r+1,q−r

r−1

(using dSp,qr+1 ⊂ Sp+r+1,q−r

0 and Exercise 3.8.G(b)) so any such a is indeed in thekernel of

Sp,qr → Sp+r,q−r+1

r

dSp+1,q−1r−1 + Sp+r+1,q−r

r−1

.

65ssfinal

66

Hence the kernel of the right map of (16) is

ker =Sp+1,q−1

r−1 + Sp,qr+1

dSp−r+1,q+r−2r−1 + Sp+1,q−1

r−1

.

Next, the image of the left map of (16) is immediately

im =dSp−r,q+r−1

r + dSp−r+1,q+r−2r−1 + Sp+1,q−1

r−1

dSp−r+1,q+r−2r−1 + Sp+1,q−1

r−1

=dSp−r,q+r−1

r + Sp+1,q−1r−1

dSp−r+1,q+r−2r−1 + Sp+1,q−1

r−1

(as Sp−r,q−r+1r contains Sp−r+1,q+r−1

r−1 ).Thus the cohomology of (16) is

ker / im =Sp+1,q−1

r−1 + Sp,qr+1

dSp−r,q+r−1r + Sp+1,q−1

r−1

=Sp,q

r+1

Sp,qr+1 ∩ (dSp−r,q+r−1

r + Sp+1,q−1r−1 )

where the equality on the right uses the fact that dSp−r,q+r+1r ⊂ Sp,q

r+1 and anisomorphism theorem. We thus must show


r + Sp+1,q−1r−1 ) = dSp−r,q+r−1

r + Sp+1,q−1r .

However,


r + Sp+1,q−1r−1 ) = dSp−r,q+r−1

r + Sp,qr+1 ∩ Sp+1,q−1

r−1

and Sp,qr+1 ∩ Sp+1,q−1

r−1 consists of (p, q)-strips whose differential vanishes up to row

p + r, from which Sp,qr+1 ∩ Sp+1,q−1

r−1 = Sp,qr as desired.

This completes the explanation of how spectral sequences work for a first-quadrant double complex. The argument applies without significant change tomore general situations, including filtered complexes.

CHAPTER 4

Sheaves

[forme: New notation to add: AbX , OX-Mod]

[forme: Summary of exposition. Motivating example (including germs). Defini-

tions (presheaf with values in an abelian category, stalk, sheaf). Examples (holomor-

phic functions; sheaf of sections of a continuous map. sheaf of sets, abelian groups,

rings, OX-modules. pushforward/skyscraper etc.). Morphisms of (pre-)sheaves.

Presheaves form an abelian category. Morphisms and stalks; check mono/epi on

stalks. Sheafification. Sheaves form an abelian category. Left-exactness. ? Pull-

back sheaf = inverse image sheaf; exact functor. Sheaves on a base, 2 versions: one

intended for affine scheme, and one intended to be the “distinguished topology”. ]

It is perhaps suprising that geometric spaces are often best understood in termsof (nice) functions on them. For example, a differentiable manifold that is a subsetof Rn can be studied in terms of its differentiable functions. Because geometricspaces can have few functions, a more precise version of this insight is that thestructure of the space can be well understood by undestanding all functions on allopen subsets of the space. This information is encoded in something called a sheaf.We will define sheaves and describe many useful facts about them. Sheaves wereintroduced by Leray in the 1940s. [forme: [KS, p. 138], [MilneLET, p. 2]] ) Thereason for the name is from an earlier, different perspective on the definition, whichwe shall not discuss. [forme: [FAC]]

We will begin with a motivating example to convince you that the notion isnot so foreign.

One reason sheaves are often considered slippery to work with is that they keeptrack of a huge amount of information, and there are some subtle local-to-globalissues. There are also three different ways of getting a hold of them.

• in terms of open sets (the definition §4.2) — intuitive but in some waythe least helpful

• in terms of stalks (see §4.4)• in terms of a base of a topology (§4.7).

Knowing which idea to use requires experience, so it is essential to do a number ofexercises on different aspects of sheaves in order to truly understand the concept.

4.1 Motivating example: The sheaf of differentiablefunctions.

1We will consider differentiable functions on the topological X = Rn, althoughyou may consider a more general manifold X . The sheaf of differentiable functions

1s:ms

67

68

on X is the data of all differentiable functions on all open subsets on X ; we willsee how to manage this data, and observe some of its properties. To each open setU ⊂ X , we have a ring of differentiable functions. We denote this ring O(U).

Given a differentiable function on an open set, you can restrict it to a smalleropen set, obtaining a differentiable function there. In other words, if U ⊂ V is aninclusion of open sets, we have a map resV,U : O(V )→ O(U).

Take a differentiable function on a big open set, and restrict it to a mediumopen set, and then restrict that to a small open set. The result is the same as if yourestrict the differentiable function on the big open set directly to the small openset. In other words, if U → V →W , then the following diagram commutes:

O(W )resW,V //

resW,U ##HHH

HHHH

HHO(V )

resV,Uwwwww

wwww

O(U)

Next take two differentiable functions f1 and f2 on a big open set U , and anopen cover of U by some Ui. Suppose that f1 and f2 agree on each of these Ui. Thenthey must have been the same function to begin with. In other words, if Uii∈I

is a cover of U , and f1, f2 ∈ O(U), and resU,Ui f1 = resU,Ui f2, then f1 = f2. ThusI can identify functions on an open set by looking at them on a covering by smallopen sets.

Finally, given the same U and cover Ui, take a differentiable function on eachof the Ui — a function f1 on U1, a function f2 on U2, and so on — and theyagree on the pairwise overlaps. Then they can be “glued together” to make onedifferentiable function on all of U . In other words, given fi ∈ O(Ui) for all i, suchthat resUi,Ui∩Uj fi = resUj ,Ui∩Uj fj for all i, j, then there is some f ∈ O(U) suchthat resU,Ui f = fi for all i.

The entire example above would have worked just as well with continuousfunction, or smooth functions, or just functions. Thus all of these classes of “nice”functions share some common properties; we will soon formalize these propertiesin the notion of a sheaf.

4.1.1. Motivating example continued: the germ of a differentiable func-tion. 2Before we do, we first point out another definition, that of the germ of adifferentiable function at a point x ∈ X . Intuitively, it is a shred of a differentiablefunction at x. Germs are objects of the form (f, open U) | x ∈ U, f ∈ O(U) mod-ulo the relation that (f, U) ∼ (g, V ) if there is some open set W ⊂ U, V containingx where f |W = g|W (or in our earlier language, resU,W f = resV,W g). In otherwords, two functions that are the same in a neighborhood of x but (but may differelsewhere) have the same germ. We call this set of germs Ox. Notice that thisforms a ring: you can add two germs, and get another germ: if you have a functionf defined on U , and a function g defined on V , then f + g is defined on U ∩ V .Moreover, f + g is well-defined: if f ′ has the same germ as f , meaning that thereis some open set W containing x on which they agree, and g′ has the same germas g, meaning they agree on some open W ′ containing x, then f ′ + g′ is the samefunction as f + g on U ∩ V ∩W ∩W ′.

2germofdifffun

69

Notice also that if x ∈ U , you get a map O(U) → Ox. Experts may alreadysee that this is secretly a colimit.

We can see that Ox is a local ring as follows. Consider those germs vanishingat x, which we denote mx ⊂ Ox. They certainly form an ideal: mx is closed underaddition, and when you multiply something vanishing at x by any other function,the result also vanishes at x. Anything not in this ideal is invertible: given a germof a function f not vanishing at x, then f is non-zero near x by continuity, so 1/fis defined near x. We check that this ideal is maximal by showing that the quotientmap is a field:

0 // m := ideal of germs vanishing at x // Ox

f 7→f(x)// R // 0

4.1.A. Exercise (for those familiar with differentiable functions).Show that this is the only maximal ideal of Ox.

ex.done

Note that we can interpret the value of a function at a point, or the valueof a germ at a point, as an element of the local ring modulo the maximal ideal.(We will see that this doesn’t work for more general sheaves, but does work forthings behaving like sheaves of functions. This will be formalized in the notion ofa local-ringed space, which we will see (briefly) in §8.4.)

Side fact for those with more geometric experience. Notice that m/m2 is amodule over Ox/m ∼= R, i.e. it is a real vector space. It turns out to be naturally(whatever that means) the cotangent space to the manifold at x. This insight willprove handy later, when we define tangent and cotangent spaces of schemes.

[primordial: Left out: Something fancier: sheaf of differentiable sections of a vector

bundle. (Define what “sections of a vector bundle” means.) Here we get a “sheaf of groups”.

Again we have notion of “value at a point”. These are all modules over the rings of differentiable

functions. ]

4.2 Definition of sheaf and presheaf

We now formalize these notions, by defining presheaves and sheaves. Presheavesare simpler to define, and notions such as kernel and cokernel are straightforward— they are defined “open set by open set”. Sheaves are more complicated todefine, and some notions such as cokernel require more thought (and the notionof sheafification). But we like sheaves are useful because they are in some sensegeometric; you can get information about a sheaf locally. 3

4.2.1. Definition of sheaf and presheaf on a topological space X.To be concrete, we will define sheaves of sets. However, Sets can be replaced

by any category, and other important examples are abelian groups Ab, k-vectorspaces, rings, modules over a ring, and more. Sheaves (and presheaves) are oftenwritten in calligraphic font, or with an underline. The fact that F is a sheaf on a

3s:defsheaf

70

topological space X is often written as

F

X

4.2.2. Definition: Presheaf. A presheaf F on a topological space X is thepresheaf, sheaf, sections

over an open set following data.• To each open set U ⊂ X , we have a set F(U) (e.g. the set of differentiable

functions). (Notational warning: Several notations are in use, for various goodreasons: F(U) = Γ(U,F) = H0(U,F). We will use them all.) The elements ofF(U) are called sections of F over U .• For each inclusion U → V of open sets, we have a restriction map resV,U :

F(V )→ F(U) (just as we did for differentiable functions).• The map resU,U is the identity: resU,U = idF(U).• If U → V → W are inclusions of open sets, then the restriction maps

commute, i.e.

F(W )resW,V //

resW,U ##HHH

HHHH

HHF(V )

resV,Uwwwww

wwww

F(U)

commutes.

4.2.A. Interesting exercise for category-lovers: “A presheaf is thesame as a contravariant functor”. Given any topological space X , wecan get a category, called the “category of open sets” (Example 3.2.9), where theobjects are the open sets and the morphisms are inclusions. Verify that the dataof a presheaf is precisely the data of a contravariant functor from the category ofopen sets of X to the category of sets. (This interpretation is suprisingly useful.) 4category of open sets

4.2.3. Definition: Stalks and germs. We define the stalk of a sheaf at a pointin two different ways. One will be hands-on, and the other will be categorical usinguniversal properties (as a colimit).

4.2.4. 5We will define the stalk of F at x to be the set of germs of a presheaf F ata point x, Fx, as in the example of §4.1.1. Elements are (f, open U) | x ∈ U, f ∈O(U) modulo the relation that (f, U) ∼ (g, V ) if there is some open set W ⊂ U, Vwhere resU,W f = resV,W g. Elements of the stalk correspond to sections over someopen set containing x. Two of these sections are considered the same if they agreeon some smaller open set.stalk

germ

4.2.5. 6A useful (and better) equivalent definition of a stalk is as a colimit of allF(U) over all open sets U containing x:

Fx = lim−→F(U).

4presheaffunctor

5stalkexplicit

6stalkuniversal

71

(Those having thought about the category of open sets, §3.2.9, will have a warmfeeling in their stomachs.) The index category is a directed set (§3.4.6: givenany two such open sets, there is a third such set contained in both), so thesetwo definitions are the same by Remark/Exercise 3.4.6. [These refs to 3.4.6are bogus.] It would be good for you to think this through. Hence by thatRemark/Exercise, we can have stalks for sheaves of sets, groups, rings, and otherthings for which direct limits exist for directed sets.

Elements of the stalk Fx are called germs. If x ∈ U , and f ∈ F(U), then theimage of f in Fx is called the germ of f .

We repeat that it is useful to think of stalks in both ways, as colimits, and alsoexplicitly: a germ at p has as a representative a section over an open set near p.

If F is a sheaf of rings, then Fx is a ring, and ditto for rings replaced by abeliangroups (or indeed any category in which colimits exist).

(Warning: the value at a point of a section doesn’t make sense.)

4.2.6. Definition: Sheaf. A presheaf is a sheaf if it satisfies two more axioms,which will use the notion of when some open sets cover another.

Identity axiom. If Uii∈I is an open cover of U , and f1, f2 ∈ F(U), andresU,Ui f1 = resU,Ui f2, then f1 = f2. identity axiom

(A presheaf satisfying the identity axiom is sometimes called a separatedpresheaf, but we will not use that notation in any essential way.) [forme: separated presheaf

[Example of a presheaf not satisfying the identity criterion: vector bundles up to

isomorphism; this leads to stacks.]]

Gluability axiom. If Uii∈I is a open cover of U , then given fi ∈ F(Ui) gluability axiom

for all i, such that resUi,Ui∩Uj fi = resUj ,Ui∩Uj fj for all i, j, then there is somef ∈ F(U) such that resU,Ui f = fi for all i.

(For experts, and scholars of the empty set only: an additional axiom some-times included is that F (∅) is a one-element set, and in general, for a sheaf withvalues in a category, F (∅) is required to be the final object in the category. Thisactually follows from the above definitions, assuming that the empty product isappropriately defined as the final object.)

Example. If U and V are disjoint, then F(U ∪ V ) = F(U) × F(V ). Here weuse the fact that F (∅) is the final object.

[forexperts: Is this really common and/or necessary? If so, then thenotion of a sheaf of rings becomes problematic if you don’t have a 0-ring.] [forme: Weibel considers only sheaves with values in an abelian category, p. 26. zero ring

He then requires F (∅) = 0.]

The stalk of a sheaf at a point is just its stalk as a presheaf; the samedefinition applies.

Philosophical note. In mathematics, definitions often come paired: “at mostone” and “at least one”. In this case, identity means there is at most one wayto glue, and gluability means that there is at least one way to glue. [forme:

unramified, etale, and smooth.]

4.2.7. “Equalizer exact sequence” interpretation. The two axioms for a presheaf tobe a sheaf can be interpreted as “exactness” of the “equalizer exact sequence”. In-

formally · // F(U) // ∏F(Ui)////∏F(Ui ∩ Uj) . Identity is exactness

72

on the left part, and gluability is exactness on the right part. (Thus the sheafaxioms may be interpreted as saying that F(∪i∈IUi) is a certain limit.)7

We now give a number of examples of sheaves.

4.2.8. Example. (a) Verify that the examples of §4.1 are indeed sheaves (ofdifferentiable functions, or continuous functions, or smooth functions, or functionson a manifold or Rn).(b) Show that real-valued continuous functions on (open sets of) a topological spaceX form a sheaf.8

4.2.9. Important Example: Restriction of a sheaf. Suppose F is a sheaf on X ,and U ⊂ is an open set. Define the restriction of F to U , denoted F|U , to bethe collection F|U (V ) = F(V ) for all V ⊂ U . Clearly this is a sheaf on U . 9

4.2.10. Important Example: skyscraper sheaf. Suppose X is a topological space,with x ∈ X , and S is a set. Then Sx defined by F(U) = S if x ∈ U and F(U) = eif x /∈ U forms a sheaf. Here e is any one-element set. (Check this if it isn’t clearto you.) This is called a skyscraper sheaf, because the informal picture of it lookslike a skyscraper at x. 1011 There is an analogous definition for sheaves of abelianskyscraper sheaf

groups, except F(U) = 0 if x ∈ U ; and for sheaves with values in a categorymore generally, F(U) should be a final object. (Warning: the notation Sx is notideal, as the subscript of a point will also used to denote a stalk.) [forme: Perhaps

SkyS? Do I even need such notation?]

4.2.B. Important Exercise: constant presheaf and locally constantsheaf. (a) Let X be a topological space, and S a set with more than one element,and define F(U) = S for all open sets U . Show that this forms a presheaf (with theobvious restriction maps), and even satisfies the identity axiom. We denote thispresheaf Spre. Show that this needn’t form a sheaf. This is called the constantpresheaf with values in S.(b) Now let F(U) be the maps to S that are locally constant, i.e. for any pointx in U , there is a neighborhood of x where the function is constant. Show thatthis is a sheaf. (A better description is this: endow S with the discrete topology,and let F(U) be the continuous maps U → S. Using this description, this followsimmediately from Exercise 4.2.D below.) We will call this the locally constantsheaf. This is usually called the constant sheaf. [primordial: ***Mention later:

it is the sheafification of the constant presheaf. ] [forme: Jack and Kirsten: “locally

constant sheaf” should mean things like the sheaf of sections of a finite etale degree

d cover.] We denote this sheaf S. 12constant sheaf, locally

constant sheaf, Spre, S4.2.C. Unimportant exercise: more examples of presheaves that arenot sheaves. Show that the following are presheaves on C (with the usual topol-ogy), but not sheaves: (a) bounded functions, (b) holomorphic functions admittinga holomorphic square root.

7equalizerexactsequence

8sheafold

9sheafrestriction

10H.E.II.1.17, skyscraperdef

11skyscraperdef

12d:constantsheaf

73

4.2.D. Exercise (“morphisms glue”). 13Suppose Y is a topological space.Show that “continuous maps to Y ” form a sheaf of sets on X . More precisely, toeach open set U of X , we associate the set of continuous maps to Y . Show thatthis forms a sheaf. (Example 4.2.8(b), with Y = R, and Exercise 4.2.B(b), withY = S with the discrete topology, are both special cases.)

4.2.E. Exercise. This is a fancier example of the previous exercise.(a) Suppose we are given a continuous map f : Y → X . Show that “sections off” form a sheaf. More precisely, to each open set U of X , associate the set ofcontinuous maps s to Y such that f s = id|U . Show that this forms a sheaf.(For those who have heard of vector bundles, these are a good example.) This ismotivation for the phrase “section of a sheaf”.(b) (This exercise is for those who know what a topological group is. If you don’tknow what a topological group is, you might be able to guess.) Suppose that Y is atopological group. Show that maps to Y form a sheaf of groups. (Example 4.2.8(b),with Y = R, is a special case.) [primordial: Later: “group object in the category of

topological spaces”. ] 14

4.2.11. ? The espace etale of a (pre)sheaf. Depending on your background, youmay prefer the following perspective on sheaves, which we will not discuss further.Suppose F is a presheaf (e.g. a sheaf) on a topological space X . Construct atopological space Y along with a continuous map to X as follows: as a set, Y isthe disjoint union of all the stalks of X . This also describes a natural set mapY → X . We topologize Y as follows. Each section s of F over an open set Udetermines a section of Y → X over U , sending s to each of its germs for eachx ∈ U . The topology on Y is the weakest topology such that these sections arecontinuous. This is called the espace etale of the Then the reader may wish toshow that (a) if F is a sheaf, then the sheaf of sections of Y → X (see the previousexercise 4.2.E(a) can be naturally identified with the sheaf F itself. (b) Moreover,if F is a presheaf, the sheaf of sections of Y → X is the sheafification of F (to bedefined in Definition 4.4.4; see Remark 4.4.6).15

4.2.F. Important exercise: the direct image sheaf or pushforwardsheaf. Suppose f : X → Y is a continuous map, and F is a sheaf on X . direct image sh., push-

forwardThen define f∗F by f∗F(V ) = F(f−1(V )), where V is an open subset of Y . Showthat f∗F is a sheaf. This is called a direct image sheaf of pushforward sheaf. Moreprecisely, f∗F is called the pushforward of F by f . pushforward sheaf, f∗

ex.doneThe skyscraper sheaf (Exercise 4.2.10) can be interpreted as follows as the

pushforward of the constant sheaf S on a one-point space x, under the morphismf : x → X .

Once we realize that sheaves form a category, we will see that the pushforwardis a functor from sheaves on X to sheaves on Y (Exercise 4.3.A).

4.2.G. Exercise (pushforward induces maps of stalks). Suppose F is asheaf of sets (or rings or A-modules). If f(x) = y, describe the natural morphism

13e:morphismstsglue, mentioned above and in ringed space 8.3.A

14topsection

15espaceetale

74

of stalks (f∗F)y → Fx. (You can use the explicit definition of stalk using represen-tatives, §4.2.4, or the universal property, §4.2.5. If you prefer one way, you shouldtry the other.)

4.2.12. Important Example: Ringed spaces, and OX -modules.. SupposeOX is a sheaf of rings on a topological space X (i.e. a sheaf on X with values inthe category of Rings). Then (X,OX) is called a ringed space.16 The sheaf ofringed space

rings is often denoted by OX ; this is pronounced “oh-of-X”. This sheaf is calledthe structure sheaf of the ringed space. We now define the notion of an OX -module.str. sheaf

The notion is analagous to one we’ve seen before: just as we have modules over aring, we have OX -modules over the structure sheaf (of rings) OX .

There is only one possible definition that could go with this name. An OX -module is a sheaf of abelian groups F with the following additional structure. Foreach U , F(U) is a OX(U)-module. Furthermore, this structure should behave wellwith respect to restriction maps. This means the following. If U ⊂ V , then 17

(17) OX (V )×F(V )action //

resV,U × resV,U

F(V )

resV,U

OX(U)×F(U)

action // F(U)

commutes. (You should convince yourself that I haven’t forgotten anything.)Recall that the notion of A-module generalizes the notion of abelian group,

because an abelian group is the same thing as a Z-module. Similarly, the notionof OX -module generalizes the notion of sheaf of abelian groups, because the latteris the same thing as a Z-module, where Z is the locally constant sheaf with valuesin Z. Hence when we are proving things about OX -modules, we are also provingthings about sheaves of abelian groups.

4.2.13. For those who know about vector bundles. The motivating example ofOX -modules is the sheaf of sections of a vector bundle. If X is a differentiablemanifold, and π : V → X is a vector bundle over X , then the sheaf of differentiablesections φ : X → V is an OX -module. Indeed, given a section s of π over an opensubset U ⊂ X , and a function f on U , we can multiply s by f to get a new sectionfs of π over U . Moreover, if V is a smaller subset, then we could multiply f by sand then restrict to V , or we could restrict both f and s to V and then multiply,and we would get the same answer. That is precisely the commutativity of (17).

[primordial: This is useless: Someone pointed out that we can make the same notionof presheaf of OX -modules, where OX is a presheaf of rings. In this setting, presheaves of abeliangroups are the same as modules over the constant presheaf Zpre. I’m doubt we will use this [Wedon’t, except briefly, ese the interseting example of morphisms of presheaves of abelian groupsbelow], so feel free to ignore it.

16d:ringedspace

17OXmodulediagram

75

4.2.H. Less important exercise (or example). Suppose X is a topological space. U →continuousX → U up to isomorphism (say what that is). Show that this is a presheaf, but nota sheaf. (Notice that automorphism groups are the problem.) 18

4.2.I. Follow-up exercise. If we allow “no local automorphism group”, show that this indeeda sheaf.

4.2.J. Follow-up exercise. Show Isom(Y,Z → X) is a sheaf. (Hence Aut is a sheaf.)

4.2.K. +Ill-defined follow-up exercise. This exercise will help motivate “stacks” later.Think about what additional information you would need to patch this (Ex. 4.2.H), and allowthings with automorphisms.

]

4.3 Morphisms of presheaves and sheaves

Whenever one defines a new mathematical object, category theory has taught us totry to understand maps between them. We now define morphisms of presheaves,and similarly for sheaves. In other words, we will descibe the category of presheaves(of abelian groups, etc.) and the category of sheaves. morphism of

(pre)sheavesA morphism of presheaves of sets (or indeed with values in any category) f :F → G is the data of maps f(U) : F(U) → G(U) for all U behaving well withrespect to restriction: if U → V then

F(V )

resV,U

f(V ) // G(V )

resV,U

F(U)

f(U) // G(U)

commutes. (Notice: the underlying space remains X .)A morphism of sheaves is defined in the same way: the morphisms from a sheaf

F to a sheaf G are precisely the morphisms from F to G as presheaves. (Translation:The category of sheaves on X is a full subcategory of the category of presheaves onX .)

An example of a morphism of sheaves is the map from the sheaf of differentiablefunctions on R to the sheaf of continuous functions. This is a “forgetful map”: weare forgetting that these functions are differentiable, and remembering only thatthey are continuous.

4.3.1. Side-remarks for category-lovers. If you interpret a presheaf on X as a con-travariant functor (from the category of open sets), a morphism of presheaves on Xis a natural transformation of functors. We haven’t defined natural transformationof functors, but you might be able to guess the definition from this remark.

4.3.A. Exercise. Suppose f : X → Y is a continuous map of topological spaces(i.e. a morphism in the category of topological spaces). Show that pushforwardgives a functor from sheaves of sets on X to sheaves of sets on Y . Here“sets” can be replaced by any category.19 (Watch out for some possible confusion:

18stackyexercise

19pffunctor

76

a presheaf is a functor, and presheaves form a category. It may be best to forgetthat presheaves form a functor for the time being.) [forme: Other confusion:

Pushforward sends presheaves to presheaves. Later comment: I don’t see why this

is a confusion.]

4.3.B. Important exercise and definition: “Sheaf Hom”. Suppose F andG are two sheaves of abelian groups on X . (In fact, it will suffice that F is apresheaf.) Let Hom(F ,G) be the collection of data

Hom(F ,G)(U) := Hom(F|U ,G|U ).

(Recall the notation F|U , the restriction of the sheaf to the open set U , see lastday’s notes.) Show that this is a sheaf. This is called the “sheaf Hom”. Show thatif G is a sheaf of abelian groups, then Hom(F ,G) is a sheaf of abelian groups. 2021Hom, sheaf Hom, re-

striction of sheaf to open

set

(The same construction will obviously work for sheaves with values in any category.)[forme: Awkwardness: if we’re dealing with sheaves of sets, or more generally if we’re

not in an additive category, we shouldn’t use Hom, but Mor.]

4.3.2. Presheaves of abelian groups or OX-modules form an abelian cat-egory.

We can make module-like constructions using presheaves of abelian groups ona topological space X . (In this section, all (pre)sheaves are of abelian groups.) Forexample, we can clearly add maps of presheaves and get another map of presheaves:if f, g : F → G, then we define the map f + g by (f + g)(V ) = f(V )+ g(V ). (Thereis something small to check here: that the result is indeed a map of presheaves.) Inthis way, presheaves of abelian groups form an additive category (Defn. 3.7.1: themorphisms between any two presheaves of abelian groups form an abelian group;there is a 0-morphism; and one can take finite products.) For exactly the samereasons, sheaves of abelian groups also form an additive category.

If f : F → G is a morphism of presheaves, define the presheaf kernel kerpre fby (kerpre f)(U) = ker f(U).

4.3.C. Exercise. Show that kerpre f is a presheaf. (Hint: if U → V , thereis a natural map resV,U : G(V )/f(V )(F(V )) → G(U)/f(U)(F(U)) by chasing thefollowing diagram:

0 // kerpre f(V )

∃!

// F(V )

resV,U

// G(V )

resV,U

0 // kerpre f(U) // F(U) // G(U)

You should check that the restriction maps compose as desired.) [forme: Possible

hint: triple decker version of that diagram above.]ex.done

Define the presheaf cokernel cokerpre f similarly. It is a presheaf by essentiallythe same argument.

4.3.D. Exercise: the cokernel deserves its name. Show that the presheafcokernel satisfies the universal property of cokernels (Defn. 3.7.3) in the categoryof presheaves.

20H.E.II.1.15

21sheafHom

77

Similarly, kerpre f → F satisfies the unversal property for kernels (Defn. 3.7.3)in the category of presheaves.

It is not too tedious to verify that presheaves of abelian groups form an abeliancategory, and the reader is free to do so. (The key idea is that all abelian-categoricalnotions may be defined and verified open set by open set.) Hence we can defineterms such as subpresheaf, image presheaf, quotient presheaf, cokernel presheaf, andthey behave the way one expect. You construct kernels, quotients, cokernels, andimages open set by open set. Homological algebra (exact sequences etc.) works,and also “works open set by open set”. In particular:

4.3.E. Exercise. If 0 → F1 → F2 → · · · → Fn → 0 is an exact sequence ofpresheaves of abelian groups, then 0 → F1(U) → F2(U) → · · · → Fn(U) → 0 isalso an exact sequence for all U , and vice versa.

The above discussion carries over without any change to presheaves with valuesin any abelian category.

However, we are interested in more geometric objects, sheaves, where thingsare can be understood in terms of their local behavior, thanks to the identity andgluing axioms. We will soon see that sheaves of abelian groups also form an abeliancategory, but a complication will arise that will force the notion of sheafification on sheafification

us. Sheafification will be the answer to many of our prayers. We just don’t realizeit yet.

Kernels work just as with presheaves:

4.3.F. Important Exercise. Suppose f : F → G is a morphism of sheaves.Show that the presheaf kernel kerpre f is in fact a sheaf. Show that it satisfies theuniversal property of kernels (Defn. 3.7.3). (Hint: the second question follows im-mediately from the fact that kerpre f satisfies the universal property in the categoryof presheaves.)22

Thus if f is a morphism of sheaves, we define

ker f := kerpre f.

The problem arises with the cokernel.

4.3.G. Important Exercise. Let X be C with the classical topology, let Z be thelocally constant sheaf on X with group Z, OX the sheaf of holomorphic functions,and F the presheaf of functions admitting a holomorphic logarithm. (Why is F nota sheaf?) Consider

0 // Z // OXf 7→exp2πif // F // 0

where Z → OX is the natural inclusion. Show that this is an exact sequence ofpresheaves. Show that F is not a sheaf. (Hint: F does not satisfy the gluabilityaxiom. The problem is that there are functions that don’t have a logarithm thatlocally have a logarithm.) This will come up again in Example 4.4.9.

We will have to put our hopes for understanding cokernels of sheaves on holdfor a while. We will first take a look at how to understand sheaves using stalks.

22e:sheafkernel

78

4.4 Properties determined at the level of stalks

[forme: Make sure all things in [KS, §2.2] are here. Summary: Things determined

by stalks: sections, morphisms, isomorphisms. Last is the hardest of the three. Can’t

go open set by open set; can go stalk by stalk!] In this section, we’ll see that lotsof facts about sheaves can be checked “at the level of stalks”. This isn’t true forpresheaves, and reflects the local nature of sheaves.23 We will flag each case of aproperty determined by stalks.

4.4.A. Important exercise (sections are determined by stalks). 24 Provethat a section of a sheaf is determined by its germs, i.e. the natural map25

(18) F(U)→∏

x∈U

Fx

is injective. (Hint # 1: you won’t use the gluability axiom, so this is true for sepa-rated presheaves. Hint # 2: it is false for presheaves in general, see Exercise 4.4.F,so you will use the identity axiom.) [forme: Sol’n: Suppose f, g ∈ Γ(U,F), with

fx = gx in Fx for all x ∈ U . Say fx = (U, f) and gx = (U, g), and (U, f) = (U, g) means that

there is an open subset Ux of U , containing x, such that f |Ux = g|Ux . The Ux cover U ,

so by the identity axiom for this cover of U , f = g.]ex.done

This exercise suggests an important question: which elements of the right sideof (18) are in the image of the left side?

4.4.1. Important definition. We say that an element∏

x∈U sx of the rightside

∏

x∈U Fx of (18) consists of compatible germs if for all x ∈ U , there is somerepresentative (Ux, s′x ∈ Γ(Ux,F)) for sx (where x ∈ Ux ⊂ U) such that the germof s′x at all y ∈ Ux is sy. [forme: teaching: in english...] You’ll have to thinkabout this a little. Clearly any section s of F over U gives a choice of compatiblegerms for U — take (Ux, s′x) = (U, s).

4.4.B. Important exercise. Prove that any choice of compatible germs for Fover U is the image of a section of F over U . (Hint: you will use gluability.)

ex.doneWe have thus completely described the image of (18), in a way that we will

find useful.

4.4.2. Remark. This perspective is part of the motivation for the agriculturalterminology “sheaf”: it is the data of a bunch of stalks, bundled together appro-priately.

Now we throw morphisms into the mix.

4.4.C. Exercise. Show a morphism of (pre)sheaves (of sets, or rings, or abeliangroups, or OX -modules) induces a morphism of stalks. More precisely, if φ : F → Gis a morphism of (pre)sheaves on X , and x ∈ X , describe a natural map φx : Fx →Gx. 26

23s:morphismsstalks

24injectivesheaf

25sectiongerm

26morphismsofstalks

79

4.4.D. Exercise (morphisms are determined by stalks). Show that mor-phisms of sheaves are determined by morphisms of stalks. Hint: consider thefollowing diagram. 27 [forme: doesn’t use gluability, true for separated presheaves]28 29

(19) F(U) //_

G(U)_

∏

x∈U Fx //∏

x∈U Gx

4.4.E. Tricky Exercise (isomorphisms are determined by stalks). Showthat a morphism of sheaves is an isomorphism if and only if it induces an iso-morphism of all stalks. (Hint: Use (19). Injectivity uses the previous exercise4.4.D. Surjectivity will use gluability, and is more subtle.) [forme: Proof of

surjectivity: Say φ : F → G. Given g ∈ G(U), we get gx for all x ∈ U . Lift these

to fx = (Ux, hx ∈ F(Ux)) ∈ Fx, where x ∈ Ux ⊂ U . φ(fx) = gx, meaning that

(Ux, φ(hx)) ∼ (Ux, gx). Thus by the equivalence relation, there is a smaller Vx such

that φ(hx) = gx on Vx. Then we have some functions hx ∈ F(Vx). I claim we can glue

these into a single function f ∈ F(U). By gluability, we need only show that the agree

on the overlap. But they do, using injectivity F(Vx ∩ Vy) → G(Vx ∩ Vy). This will be

reminiscent of the compatible germs argument] 30 31 32

4.4.F. Exercise. (a) Show that Exercise 4.4.A is false for general presheaves.(b) Show that Exercise 4.4.D is false for general presheaves.(c) Show that Exercise 4.4.E is false for general presheaves.(General hint for finding counterexamples of this sort: consider a 2-point spacewith the discrete topology, i.e. every subset is open.) 33 discrete topology

4.4.3. Sheafification. sheafification

Every sheaf is a presheaf (and indeed by definition sheaves on X form a fullsubcategory of the category of presheaves on X). Just as groupification (§3.5.1)gives a group that best approximates a semigroup, sheafification gives the sheafthat best approximates a presheaf, with an analogous universal property.

4.4.4. Definition. 34If F is a presheaf on X , then a morphism of presheavessh : F → Fsh on X is a sheafification of F if Fsh is a sheaf, and for any othersheaf G, and any presheaf morphism g : F → G, there exists a unique morphism of

27injectivething

28morphismssheafstalk

29morphismdeterminedbystalks

30isoifonstalks

31H.P.II.1.1; make sure attached to right ex!

32H.E.II.1.5; make sure attached to right ex!

33presheaffalse

34d:sheafification

80

sheaves f : Fsh → G making the diagram

F sh //

g!!B

BBBB

BBB Fsh

f

G

commute.

4.4.G. Exercise. Show that sheafification is unique up to unique isomorphism.

Show that if F is a sheaf, then the sheafification is F id // F . (This should besecond nature by now.)

4.4.5. Construction. We next show that any presheaf has a sheafification.Suppose F is a presheaf. Define Fsh by defining Fsh(U) as the set of compatiblegerms of the presheaf F over U . Explicitly:

Fsh(U) := (fx ∈ Fx)x∈U : ∀x ∈ U, ∃x ∈ V ⊂ U, s ∈ F(V ) | sy = fy∀y ∈ V .(Those who want to worry about the empty set are welcome to.)

4.4.H. Easy Exercise. Show that Fsh (using the tautological restriction maps)forms a sheaf.

4.4.I. Easy Exercise. Describe a natural map sh : F → F sh.

4.4.J. Exercise. Show that the map sh satisfies the universal property 4.4.4 ofsheafification.

4.4.6. ? Remark. The espace etale construction (§4.2.11) yields a different de-scription of sheafification which may be preferred by some readers.35 The idea isidentical.

4.4.K. Exercise. Use the universal property to show that for any morphism ofpresheaves φ : F → G, we get a natural induced morphism of sheaves φsh : Fsh →Gsh.36 Show that sheafification is a functor from presheaves on X to sheaves on X .

4.4.L. Useful exercise for category-lovers. Show that the sheafificationfunctor is left-adjoint to the forgetful functor from sheaves on X to presheaves onX .

4.4.M. Exercise. Show F → Fsh induces an isomorphism of stalks. (Possiblehint: Use the concrete description of the stalks. Another possibility: judicious useof adjoints.) 37 [forme: Jarod’s solution: Let i : x → X. For any abelian group A,

HomAb(Fx, A) = Hom(F , i∗A) = Hom(Fsh, i∗A) (as i∗A is a sheaf) = HomAb(Fshx , A).]

4.4.7. Unimportant remark. Sheafification can be defined in a topological way,via the “espace etale” construction, see [H, II.1.13]. [forme: and I think Serre’s

FAC] This is essentially the same construction as the one given here. Another

35espaceetale2

36sheafifyingmorphisms

37shpreservesstalks

81

construction is described in [EH]. [forme: Give specific ref.] 38 [primordial:Sheafification via the + construction: Joe R says: This is Cech H0 of the presheaf. The +

construction is aside 7.19 in Milne’s online etale cohomology notes (“732”). ]

4.4.8. Subsheaves and quotient sheaves.

4.4.N. Exercise. Suppose φ : F → G is a morphism of sheaves (of sets) on attopological space X . Show that the following are equivalent.39

(a) φ is a monomorphism in the category of sheaves.(b) φ is injective on the level of stalks: φx : Fx → Gx injective for all x ∈ X .(c) φ is injective on the level of open sets: φ(U) : F(U) → G(U) is injective

for all open U ⊂ X .

(Possible hints: for (b) implies (a), recall that morphisms are determined by stalks,Exercise 4.4.D. For (a) implies (b), judiciously choose a skyscraper sheaf. For (a)implies (c), judiciously the “indicator sheaf” with one section over every open setcontained in U , and no section over any other open set.)

ex.doneIf these conditions hold, we say that F is a subsheaf of G (where the “inclusion”

φ is sometimes left implicit). subsheaf

4.4.O. Exercise. 40Continuing the notation of the previous exercise, show thatthe following are equivalent.41

(a) φ is a epimorphism in the category of sheaves.(b) φ is surjective on the level of stalks: φx : Fx → Gx surjective for all x ∈ X .

ex.doneIf these conditions hold, we say that G is a quotient sheaf of F . subsheaf

ex.doneThus monomorphisms and epimorphisms — subsheafiness and quo-

tient sheafiness — can be checked at the level of stalks.Both exercises generalize immediately to sheaves with values in any category,

where “injective” is replaced by “monomorphism” and “surjective” is replaced by“epimorphism”.

[forme: Solution: Suppose first that we have surjectivity on all stalks for amorphism φ : F → G. We want to check the definition of epimorphism. Suppose wehave α : F → H, and β, γ : G → H such that α = β φ = γ φ.

Fφ //

α @

@@@@

@@@ G

≤1?

H

Then by taking stalks at x, we have

Fxφx //

αx!!C

CCCC

CCC Gx

βx,γx

Hx

38H.E.II.1.13 mentioned but not done

39e:monostalk

40e:epistalk

41epilocal

82

By surjectivity (epimorphism-ness) of the morphisms of stalks, βx = γx. But asmorphisms are determined by morphisms at stalks (Exercise 4.4.D), we must haveβ = γ.

Next assume that φ : F → G is an epimorphism of sheaves, and x ∈ X. We willshow that φx : Fx → Gx is a epimorphism for any given x ∈ X. Choose for H anyskyscraper sheaf supported at x. (the stalk of a skyscraper sheaf at the skyscraperpoint is just the skyscraper set/group/ring). Then the maps α, β, γ factor throughthe stalk maps:

F //

G

i∗Fx //

##GGG

GGGG

GGi∗Gx

skyscrapers

Hand then we are basically done. ]

Notice that there was no part (c) to the previous exercise, and here is anexample showing why.

4.4.9. Example. 42Let X = C with the usual (analytic) topology, and defineOX to be the sheaf of holomorphic functions, and O∗

X to be the sheaf of invertible(nowhere zero) holomorphic functions. This is a sheaf of abelian groups undermultiplication. We have maps of sheaves43

(20) 0 // Z×2πi // OX

exp // O∗X

// 1

where Z is the locally constant sheaf associated to Z. (You can figure out what thesheaves 0 and 1 mean; they are isomorphic, and are written in this way for reasonsthat may be clear). We will soon interpret this as an exact sequence of sheavesof abelian groups (the exponential exact sequence), although we don’t yet have thelanguage to do so.exponential ES

4.4.P. Exercise. Show that OXexp // O∗

X describes O∗X as a quotient sheaf of

OX . Show that it is not surjective on all open sets. [forme: Remember to discuss

this further. [Another example: z → z2, given by O∗X → O∗X . Discuss cokernel and

image.] 44 ]

This is a great example to get a sense of what “surjectivity” means for sheaves.Nonzero holomorphic functions locally have logarithms, but they need not globally.

4.5 Sheaves of abelian groups, and OX-modules, form abeliancategories

45We are now ready to see that sheaves of abelian groups, and their cousins,OX -modules, form abelian categories. In other words, we may treat them in thesame way we treat vector spaces, and modules over a ring. In the process of doing

42e:exponentialES

43exponentialES

44weirdexample

45absheafabcat

83

this, we will see that this is much stronger than an analogy; kernels, cokernels,exactness, etc. can be understood at the level of germs (which are just abeliangroups), and the compatibility of the germs will come for free.

The category of sheaves of abelian groups is clearly an additive category (Def-inition 3.7.1). In order to show that it is an abelian category, we must show thatany morphism φ : F → G has a kernel and a cokernel. We have already seen thatφ has a kernel (Exercise 4.3.F): the presheaf kernel is a sheaf, and is a kernel.

4.5.A. Exercise. Show that the stalk of the kernel is the kernel of the stalks:there is a natural isomorphism

(ker(F → G))x∼= ker(Fx → Gx).

[forme: Watch out, you can’t get too categorical here. Limits and colimits tend not

to commute.]

So we next address the issue of the cokernel. Now φ : F → G has a cokernel inthe category of presheaves; call it Hpre (where the superscript is meant to remind

us that this is a presheaf). Let Hpre sh // H be its sheafification. Recall that thecokernel is defined using a universal property: it is the colimit of the diagram

F

φ // G

0

in the category of presheaves. We claim that H is the cokernel of φ in the categoryof sheaves, and show this by proving the universal property. Given any sheaf E anda commutative diagram

F

φ // G

0 // E

We construct

F

φ // G

0

**VVVVVVVVVVVVVVVVVVVVVVVV // Hpre sh // H

EWe show that there is a unique morphism H → E making the diagram commute.As Hpre is the cokernel in the category of presheaves, there is a unique morphismof presheaves Hpre → E making the diagram commute. But then by the universalproperty of sheafification (Defn. 4.4.4), there is a unique morphism of sheaves H →E making the diagram commute.

4.5.B. Exercise. Show that the stalk of the cokernel is naturally isomorphic tothe cokernel of the stalk. [forme: Fast categorical reason: colimits commute with

colimits.]ex.done

84

We have now defined the notions of kernel and cokernel, and verified that theymay be checked at the level of stalks. We have also verified that the qualities ofa morphism being monic or epi are also determined at the level of stalks (Exer-cises 4.4.N and 4.4.O). Hence sheaves of abelian groups on X form an abeliancategory.

We see more: all structures coming from the abelian nature of this categorymay be checked at the level of stalks. For example, exactness of a sequence ofsheaves may be checked at the level of stalks.46 A fancy-sounding conse-quence: taking stalks is an exact functor from sheaves of abelian groups on X toabelian groups.

4.5.C. Exercise (Left-exactness of the global section functor). 47

Suppose U ⊂ X is an open set, and 0 → F → G → H is an exact sequence ofsheaves of abelian groups. Show that

0→ F(U)→ G(U)→ H(U)

is exact. Give an example to show that the global section functor is not exact.(Hint: the exponential exact sequence (20).) 48 [forme: One-liner: forgetful mapleft-exactness of global

section functor to presheaves is right-adjoint hence left-exact.]

[forme: Hints of cohomology.]

4.5.D. Exercise: Left-exactness of pushforward. Suppose 0→ F → G →H is an exact sequence of sheaves of abelian groups on X . If f : X → Y is acontinuous map, show that

0→ f∗F → f∗G → f∗His exact. (The previous exercise, dealing with the left-exactness of the global sectionfunctor can be interpreted as a special case of this, in the case where Y is a point.)[forme: Do it for U first; direct limit is exact for rings. Also, do I define image

sheaf?]

4.5.E. Exercise. Suppose φ : F → G is a morphism of sheaves of abeliangroups. Show that the image sheaf im φ is the sheafification of the image presheaf.(You must use the definition of image in an abelian category. In fact, this gives theaccepted definition of image sheaf for a morphism of sheaves of sets.) [primordial:Suppose f : F → G is a morphism of sheaves of abelian groups or OX -modules. Let im f be the

sheafification of the “presheaf image”. Show that there are natural isomorphisms im f ∼= F/ ker f

and coker f ∼= G/ im f . (This problem shows that this construction deserves to be called the

“image”.) [forme: First one is easy: just sheafify the presheaf statement. For the

second, we need to show that (G/ im f)sh ∼= (G/(im f)sh)sh. This is immediate checking

at stalks. Chad also has a nice diagram chase.] ] 49

[primordial:

4.5.F. Exercise. (This philosophically is related to Exercises 4.4.E, ?? and 4.4.O — now a

couple of exercises in stalks section. [Later cross-reference all four of these.]) Show that taking

stalks at a point is an exact functor from the category of sheaves with values in some abelian

46exactnessstalklocal

47H.E.II.1.8

48s:leftexact

49H.E.II.1.7

85

category (with direct limits) to the abelian category. (Hint: direct limits are exact. [filtered direct

limits are exact, Weibel p. 429, put this in direct limit section.], cf. Section 3.4.4.) Perhaps I will

put this after the pullback section? It is a special case of exactness of pullback. ]

4.5.G. Exercise. Show that if (X,OX) is a ringed space, then OX -modules forman abelian category. (There isn’t much more to check!)

We end with a useful construction using some of the ideas in this section.

4.5.H. Important exercise: tensor products of OX -modules. (a) Sup-pose OX is a sheaf of rings on X . Define (categorically) what we should mean bytensor product of two OX -modules. Give an explicit construction, and show thatit satisfies your categorical definition. Hint: take the “presheaf tensor product” —which needs to be defined — and sheafify. Note: ⊗OX is often written ⊗ when thesubscript is clear from the context. (An example showing sheafification is necessarywill arise in Exercise 19.4.B. [forme: [EH] may have some examples in Ch. 1. What

about the Moebius line bundle?] ) 50 ⊗ of sheaves

(b) Show that the tensor product of stalks is the stalk of tensor product. [forme:

Reason: colimits commute with tensor products; I should explain why.]

[primordial:

4.5.I. Important exercise: support of a section. [H, Ex. II.1.14] Define support of asection of a sheaf of groups Supp. Define support of a sheaf of groups. Show that the support ofa section is closed. Show that the support of a sheaf need not be closed. 51 [not given in 216] support

4.5.J. Exercise: Flasque sheaves. [forme: This is annoying. Do I really need this?]A sheaf F is said to be flasque if for every U ⊂ V , the restriction map resV,U : F(V ) → F(U) issurjective. In other words, every section over U extends to a section over V . This is a very strongcondition, but it comes up surprisingly often.(a) Show that if 0→ F ′ → F → F ′′ → 0 is an exact sequence of sheaves of groups or OX -modules,and F ′ and F ′′ are flasque, then so is F . (Hint: 5-lemma, I think.) [forme: People oftenused AC here.](b) Suppose f : X → Y is a continuous map, and F is a flasque sheaf on X. Show that f∗F is aflasque sheaf on Y .(If 0→ F ′ → F → F ′′ → 0 is exact, and F ′ is flasque, then 0→ F ′(U)→ F(U)→ F ′′(U)→ 0 isexact, i.e. the global section functor is exact here, even on the right. Similarly, for any continuousmap f : X → Y , 0→ f∗F ′ → f∗F → f∗F ′′ → 0 is exact. I haven’t thought about how hard thisis yet, so I haven’t made this part of the exercise. But it is good to know, and gives a reason to

like flasque sheaves.) [forme: This parenthetical bit isn’t too bad; use Zorn.] 52 AC

flasque sheaf[forme: In the etale cohomology group, Jack showed how to show that the cat-egory of sheaves on X with values in an abelian category is an abelian category: itis additive (because it is a full category of the category of presheaves, and is closedunder direct sum). So we hove to show that kernels and cokernels exist, which wedo as usual, and we need to show that image = coimage. We have a map coimage toimage, that is an isomorphism on stalks, hence it is an isomorphism as desired.]

]

4.6 The inverse image sheaf

50tensorsheaf1

51H.II.1.14

52H.E.II.1.16, (c) is H(b), (a) is H(b), (b) is H(c)

86

53inverse image sheaf

[forme: Use “inverse image” not “pullback”. Be very careful about this.] Wenext describe a notion that is rather fundamental, but is still a bit intricate. Weinverse image sheaf —

requires sheafification won’t need it (at least for a long while), so this may be best left for a secondreading. 54 Suppose we have a continuous map f : X → Y . If F is a sheaf on X ,we have defined the pushforward or direct image sheaf f∗F , which is a sheaf on Y .There is also a notion of inverse image sheaf. (We won’t call it the pullback sheaf,reserving that name for a later construction, involving quasicoherent sheaves.) Thisis a covariant functor f−1 from sheaves on Y to sheaves on X . If the sheaves on Yhave some additional structure (e.g. group or ring), then this structure is respectedby f−1.

4.6.1. Definition by adjoint: elegant but abstract. Here is a categorical definitionof the inverse image: f−1 is left-adjoint to f∗.

55

This isn’t really a definition; we need a construction to show that the adjointexists. (Also, for pedants, this won’t determine f−1F ; it will only determine itup to unique isomorphism.) Note that we then get canonical maps f−1f∗F → F(associated to the identity in MorY (f∗F , f∗F)) and G → f∗f

−1G (associated to theidentity in MorX(f−1G, f−1G)).

4.6.2. Construction: concrete but ugly. Define the temporary notation f−1Gpre(U) =lim−→V ⊃f(U)

G(V ). (Recall the explicit description of direct limit: sections are sections

on open sets containing f(U), with an equivalence relation.)

4.6.A. Exercise. Show that this defines a presheaf on X . [forme: It is not a

sheaf: consider 2 points mapping to 1 point. Espace etale construction works.]ex.done

Now define the inverse image of G by f−1G := (f−1Gpre)sh.inverse image

You will show that this construction satisfies the universal property in Exer-cise 4.6.F. For the exercises before that, feel free to use either the adjoint descriptionor the construction.

4.6.B. Exercise. Show that the stalks of f−1G are the same as the stalks ofG. More precisely, if f(x) = y, describe a natural isomorphism Gy

∼= (f−1G)x.(Possible hint: use the concrete description of the stalk, as a direct limit. Recallthat stalks are preserved by sheafification, Exercise 4.4.M.) 56 [forme: Jarod usedadjoints! His argument: let i : x → X. The functors (AbGps) ↔ sheaves of abeliangroups on X given by i∗ and stalkifying are adjoint, i.e. HomAb(Fx, A) ∼= HomX(F , i∗A).We also know that f−1 and f∗ are adjoint; in fact this is a generalization. If j : y → Y ,then we have

HomAb((f−1G)x, A) ∼= HomX(f−1G, i∗A) ∼= HomY (G, f∗i∗A) ∼= HomY (G, j∗A) ∼= HomAb(G, A).

Since these are natural bijections for all A, by appplying Yoneda’s lemma to Abop,

we get a natural isomorphism (f−1G)x ∼= Gx. ]

53s:inverseimagesheaf

54pullback

55pushforwardadjoint

56pullbackstalkssame

87

4.6.C. Exercise (easy but useful). If U is an open subset of Y , i : U → Yis the inclusion, and G is a sheaf on Y , show that i−1G is naturally isomorphic toG|U . 57 [primordial: (i) the direct limit is easy, and (ii) you don’t need to sheafify.] ]

4.6.D. Exercise (easy but useful). If y ∈ Y , i : y → Y is the inclusion,and G is a sheaf on Y , show that i−1(G) is naturally isomorphic to the stalk Gy.

4.6.E. Exercise. Show that f−1 is an exact functor from sheaves of abeliangroups on Y to sheaves of abelian groups on X . (Hint: exactness can be checkedon stalks, and by Exercise 4.6.B, the stalks are the same.) The identical argumentwill show that f−1 is an exact functor from OY -modules (on Y ) to f−1OY -modules(on X), but don’t bother writing that down. (Remark for experts: f−1 is a left-adjoint, hence right-exact by abstract nonsense, §3.7.6. The left-exactness is truefor “less categorical” reasons.)

4.6.F. Important Exercise: the construction satisfies the universalproperty. If f : X → Y is a continuous map, and F is a sheaf on X and G is asheaf on Y , describe a bijection58

MorX(f−1G,F)↔ MorY (G, f∗F).

Observe that your bijection is “natural” in the sense of the definition of adjoints.[forme: Method: The one on the right is a compatible collection sequence of maps,for each open V ⊂ Y , G(V ) → F(f−1(V )). The one on the right is a “bigger” com-patible collection of maps. So we quickly get a map from the left to right; wejust then construct the map from right to left, and verify that everything com-mutes.] [Chad’s argument: We identify Mor(f−1G,F), the set of sheaf morphisms,with Mor(f−1Gpre,F), the set of presheaf morphisms. This is possible by the univer-sal definition of sheafification. Let γ : f−1Gpre → F . We define φ(γ) ∈ Mor(G, f∗F) asfollows: for each V ∈ O(Y ), there is a natural map σV : G(V ) → f−1Gpre(f−1[V ]). Byidentifying F(f−1 [V ]) with f∗F(V ), we can define (φ(γ))V = γf−1[V ] σV . This com-

mutes with restriction; this is straightforward but requires lots of notation, so we’llskip it.

ψ will be the map in the other direction. Let δ : G → f∗F . Let U ∈ O(X).f−1Gpre(U) is a direct limit over open V ⊃ f [U ]. For each such V , there is a map δV :G(V )→ F(f−1[V ]); compose with restriction to get a map resf−1[V ],U δV to F(U). Since

δ “commutes” with restriction, we get a map ψ(δ)U : f−1Gpre(U) = lim−→V⊃F [U]G(V ) →

F(U). The ψ(δ)U also behave properly with respect to restriction, but this is alsoannoying to prove.

We must now show that φ ψ and ψ φ are the respective identities. Let δ ∈Hom(G, f∗F). Then φ(ψ(δ))V = ψ(δ)f−1 [V ] σV , which is actually δV since ψ(δ)f−1 [V ] was

defined as a direct limit of maps, one of which is δV . Thus φ ψ is the identity. Nowlet γ ∈ Hom(f−1Gpre,F). We want to show that ψ(φ(γ))U = γU for all U . It suffices toshow that ψ(φ(γ))U σV = γU σV for all open V ⊃ f [U ], by the direct-limit definitionof f−1Gpre. But

ψ(φ(γ))U σV = ψ(γ)V = γf−1[V ] σV = γU σV ,the last because γ commutes with restrictions. Thus ψ φ is also the identity. Kate

Gruher also gave a good argument.]]

4.6.G. Exercise. [forme: Do I need this? I could move the definition of support

elsewhere.] (a) Suppose Z ⊂ Y is a closed subset, and i : Z → Y is the inclusion.If F is a sheaf on Z, then show that the stalk (i∗F)y is a one element-set if y /∈ Z,

57easypullback

58inverseimage2defs

88

and Fy if y ∈ Z.(b) Important definition: Define the support of a sheaf F of sets, denoted SuppF ,as the locus where the stalks are not the one-element set:

SuppF := x ∈ X | |Fx| 6= 1.

(More generally, if the sheaf has value in some category, the support consists ofpoints where the stalk is not the final object. For sheaves of abelian groups, thesupport consists of points with non-zero stalks.) 59 Suppose SuppF ⊂ Z wheresupport of a sheaf

Z is closed. Show that the natural map F → i∗i−1F is an isomorphism. Thus a

sheaf supported on a closed subset can be considered a sheaf on that closed subset.[forme: Answer: show isomorphism on stalks.]

4.7 Recovering sheaves from a “sheaf on a base”

[forme: This must set the reader up for the theorem on sheaves on a distinguished

affine base, Theorem 17.1.1. I also hope that I’ve done it correctly (that the exercises

work out — gluability gave me a bit of a surprise), and that it gives me what I want.]60

Sheaves are natural things to want to think about, but hard to get one’s handson. We like the identity and gluability axioms, but they make proving things trickierthan for presheaves. We have discussed how we can understand sheaves using stalks.We now introduce a second way of getting a hold of sheaves, by introducing thenotion of a sheaf on a base.sheaf on a base

First, let me define the notion of a base of a topology. Suppose we havebase of a topology

a topological space X , i.e. we know which subsets of X are open Ui. Then abase of a topology is a subcollection of the open sets Bj ⊂ Ui, such that eachUi is a union of the Bj . There is one example that you have seen early in yourmathematical life. Suppose X = Rn. Then the way the usual topology is oftenfirst defined is by defining open balls Br(x) = y ∈ Rn | |y − x| < r, and declaringthat any union of open balls is open. So the balls form a base of the usual topology.Equivalently, we often say that they generate the usual topology. As an applicationof how we use them, to check continuity of some map f : X → Rn, you need onlythink about the pullback of balls on Rn.

[forme: There is a slightly nicer notion I want to use. A base is particularly

pleasant if the intersection of any two elements is also an element of the base. I will

call this a nice base. [Does anyone know if this has a name?] For example if X = Rn,

then a base would be convex open sets. Certainly the intersection of two convex open

sets is another convex open set. Also, this certainly forms a base, because it includes

the balls. ]nice base of a topology

Now suppose we have a sheaf F on X , and a base Bi on X . Then con-sider the information (F(Bi), resBi,Bj : F(Bi) → F(Bj)), which is a subsetof the information contained in the sheaf — we are only paying attention to theinformation involving elements of the base, not all open sets.

59d:support

60s:sheafonbase

89

We can recover the entire sheaf from this information. Reason: we can de-termine the stalks from this information, and we can determine when germs arecompatible.

4.7.A. Exercise. Make this precise. [primordial: Proof:

F(U) = (fi ∈ F(Bi))Bi⊂U | resBi,Bj(fi) = fj.

The map from the left side to the right side is clear. We get a map from the right side to the left

side as follows. By gluability, each element gives at least one element of the left side. By identity,

it gives a unique element. Conclusion: we can recover a sheaf from less information. ]This suggests a notion, that of a sheaf on a base. A sheaf of sets (rings etc.)

on a base Bi is the following. For each Bi in the base, we have a set F(Bi).If Bi ⊂ Bj , we have maps resji : F(Bj) → F(Bi). (Things called B are alwaysassumed to be in the base.) If Bi ⊂ Bj ⊂ Bk, then resBk ,Bi = resBj ,Bi resBk,Bj .So far we have defined a presheaf on a base Bi.

We also require base identity: If B = ∪Bi, then if f, g ∈ F(B) such thatresB,Bi f = resB,Bi g for all i, then f = g. sheaf on a base

We require base gluability too: If B = ∪Bi, and we have fi ∈ F(Bi) such thatfi agrees with fj on any basic open set contained in Bi ∩ Bj (i.e. resBi,Bk

fi =resBj ,Bk

fj for all Bk ⊂ Bi ∩ Bj) then there exist f ∈ F(B) such that resB,Bi = fi

for all i.

4.7.1. Theorem. — Suppose Bi is a base on X, and F is a sheaf of setson this base. Then there is a unique sheaf F extending F (with isomorphismsF(Bi) ∼= F (Bi) agreeing with the restriction maps).61 sheaf on base determines

sheafProof. We will define F as the sheaf of compatible germs of F .

Define the stalk of F at x ∈ X by

Fx = lim−→F (Bi)

where the colimit is over all Bi (in the base) containing x.We’ll say a family of germs in an open set U is compatible near x if there is a

section s of F over some Bi containing x such that the germs over Bi are preciselythe germs of s. More formally, define

F(U) := (fx ∈ Fx)x∈U : ∀x ∈ U, ∃B with x ⊂ B ⊂ U, s ∈ F (B) | sy = fy∀y ∈ Bwhere each B is in our base.

This is a sheaf (for the same reasons as the sheaf of compatible germs wasearlier, cf. [forme: XXX] ).

I next claim that if U is in our base, the natural map F (B) → F(B) is anisomorphism.

4.7.B. Tricky exercise. Describe the inverse map F(B) → F (B), and verifythat it is indeed inverse. [forme: a bit subtle. Possible hint: elements of F(U) are

determined by stalks, as are elements of F (U).]

Thus sheaves on X can be recovered from their “restriction to a base”. This isa statement about objects in a category, so we should hope for a similar statementabout morphisms.

61t:sheafonbase

90

4.7.C. Important exercise: morphisms of sheaves correspond to mor-phisms of sheaf on a base. Suppose Bi is a base for the topology of X .(a) Verify that a morphism of sheaves is determined by the induced morphism ofsheaves on the base.(b) Show that a morphism of sheaves on the base (i.e. such that the diagram

F (Bi) //

G(Bi)

F (Bj) // G(Bj)

commutes for all Bj → Bi) gives a morphism of the induced sheaves. 62

[forme: Possible thing to add: OX-modules vs. OX-modules on a base.]

4.7.D. Important Exercise. Suppose X = ∪Ui is an open cover of X , andwe have sheaves Fi on Ui along with isomorphisms φij : Fi|Ui∩Uj → Fj |Ui∩Uj thatagree on triple overlaps (i.e. φij φjk = φij on Ui ∩ Uj ∩ Uk). Show that thesesheaves can be glued together into a unique sheaf F on X , such that Fi = F|Ui ,and the isomorphisms over Ui ∩ Uj are the obvious ones. (Thus we can “gluesheaves together”, using limited patching information.) (You can use the ideas ofthis section to solve this problem, but you don’t necessarily need to. Hint: As thebase, take those open sets contained in some Ui.) [forme: [EH] do this] 63

4.7.E. Unimportant exercise. Suppose a morphism of sheaves F → G on abase Bi is surjective for all Bi (i.e. F(Bi) → G(Bi) is surjective for all i). Showthat the morphism of sheaves (not on the base) is surjective. The converse is nottrue, unlike the case for injectivity. This gives a useful criterion for surjectivity(“surjectivity on small enough open sets”).64

4.7.2. Remark for experts. This almost says that the “set” of sheaves forms asheaf itself, but not quite. Making this precise leads one to the notion of a stack.stack

[forme: Perhaps say more?]

[primordial:

4.7.3. Remark. Suppose you have a presheaf you want to sheafify, and when restricted to a base

it is already a sheaf. Then the sheafification is obtained by taking this process. 65

Example: Let X = C, and consider the sequence

1 // Z ×2πi // OXexp // O∗X // 1.

Let’s check that it is exact, using our new knowledge. We instead work on the base of convexopen sets. then on these open sets, this is indeed exact. The key fact here is that on any convexopen set B, every element of O∗X(B) has a logarithm, so we have surjectivity here.

[primordial: Notes to keep in FOAG.: Another variant where this turns up: stacks, wherethe open sets correspond to schemes, but the stack isn’t a scheme.

[There is a cut section “Aside: Sheafification exists” here.] ] ]

62e:morphismonbase

63e:gluesheaves

64surjectivityonbase

65r:sheafifyifniceonbase

Part II

Schemes and morphisms ofschemes

CHAPTER 5

Toward affine schemes: the underlying set, and

the underlying topological space

5.1 Toward schemes

[forme: At some point in this chapter, refer forward to an example of an affine

open set that is not a distinguished open: line in plane minus conic ??.]

[forme: Complex varieties! We will keep returning to these. Pictures.]

We are now ready to consider the notion of a scheme, which is the type ofgeometric space considered by algebraic geometry. We should first think throughwhat we mean by “geometric space”. You have likely seen the notion of a manifold,and we wish to abstract this notion so that it can be generalized to other settings,notably so that we can deal with non-smooth and arithmetic objects.

The key insight behind this generalization is the following: we can understanda geometric space (such as a manifold) well by understanding the functions onthis space. More precisely, we will understand it through the sheaf of functionson the space. If we are interested in differentiable manifolds, we will considerdifferentiable functions; if we are interested in smooth manifolds, we will considersmooth functions and so on.

Thus we will define a scheme to be the following data

• The set: the points of the scheme• The topology: the open sets of the scheme• The structure sheaf: the sheaf of “algebraic functions” (a sheaf of rings)

on the scheme.

Recall that a topological space with a sheaf of rings is called a ringed space (§4.2.12).ringed space

structure sheafWe will try to draw pictures throughout, so our geometric intuition can guidethe algebra development (and, eventually, vice versa). Pictures can help developgeometric intuition. [forme: We learn to draw them; the algebra tells how to think

about them geometrically. So these comments are saying: “this is a good way to

think”. Eventually the picture tells you some algebra. [Whenever someone talks

about primes etc., you should often immediately try to make a picture.]] Somereaders will find the pictures very helpful, while others will find the opposite.

We will try to make all three notions as intuitive as possible. For the set, inthe key example of complex (affine) varieties (roughly, things cut out in Cn bypolynomials), we’ll see that the points are the “old-fashioned points” (n-tuples ofcomplex numbers), plus some extra points that will be handy to have around. Forthe topology, we’ll require that functions vanish on closed sets, and require nothing

93

94

else. For the sheaf of algebraic functions (the structure sheaf), we’ll expect that inthe complex plane, (3x2 + y2)/(2x + 4xy + 1) should be an algebraic function onthe open set consisting of points where the denominator doesn’t vanish, and thiswill largely motivate our definition.

5.1.1. Example: Differentiable manifolds. As motivation, we return to ourexample of differentiable manifolds, reinterpreting them in this light. 1 We willbe quite informal in this section. Suppose X is a manifold. It is a topologicalspace, and has a sheaf of differentiable functions OX (see §4.1). This gives X thestructure of a ringed space. [forme: Make sure to define OX,p as (OX)p somewhere.

I’m putting this in the index and notation to be sure I don’t forget.] We haveOX,pobserved that evaluation at p gives a surjective map from the stalk to R

OX,p // // R,

so the kernel, the (germs of) functions vanishing at p, is a maximal ideal mX (see§4.1.1).

We could define a differentiable real manifold as a topological space X witha sheaf of rings such that there is a cover of X by open sets such that on eachopen set the ringed space is isomorphic to a ball around the origin in Rn with thesheaf of differentiable functions on that ball. With this definition, the ball is thebasic patch, and a general manifold is obtained by gluing these patches together.(Admittedly, a great deal of geometry comes from how one chooses to patch theballs together!) In the algebraic setting, the basic patch is the notion of an affinescheme, which we will discuss soon.

Functions are determined by their values at points. This is an obvious state-ment, but won’t be true for schemes in general. We will see an example in Ex-ercise 5.3.A(a), and discuss this behavior further in §XXXX. [primordial: Thetangent space in terms of the stalk. (This notion is subtle, and you shouldn’t understand thisfor quite some time.) We can interpret the tangent space at p in terms of the stalk OX,p. It is

naturally isomorphic to (mx/m2x)∨. Let’s make this more explicit. Every function vanishing at p

canonically gives a functional on the tangent space to X at p. If X = R2, the function sinx−y+y2

gives the functional x− y. The function sin2 x+ y2 gives the functional 0. If you can, you shouldtry to convince yourself that this is reasonable. This will involve figuring out what “naturallyisomorphic” means — there is always more to this statement than meets the eye. One of thereasons this insight is hard to grasp is because we believe we have a good intuitive understandingof what “tangent space” means. However, making this precise always requires some new intuition.

It is an interesting fact that the cotangent space m/m2 is algebraically more natural than the

tangent space (m/m2)∨ (for example, it requires fewer symbols to describe), despite the fact that

tangent spaces are more geometrically intuitive. We will later see this when we define the tangent

bundle and cotangent bundle: if a scheme is singular, the notion of “cotangent bundle” (really,

cotangent sheaf) is much more natural than “tangent bundle”. ]Morphisms of manifolds. How can we describe differentiable maps of manifolds

X → Y ? They are certainly continuous maps — but which ones? We can pull backfunctions along continuous maps. Differentiable functions pull back to differentiablefunctions. More formally, we have a map f−1OY → OX . (The inverse image sheaff−1 was defined in §4.6) Inverse image is left-adjoint to pushforward, so we get amap f# : OY → f∗OX .

1diffmotivation

95

Certainly given a differentiable map of manifolds, differentiable functions pull-back to differentiable functions. It is less obvious that this is a sufficient conditionfor a continuous function to be differentiable.

5.1.A. Important exercise for those with a little experience with man-ifolds. Prove that a continuous function of differentiable manifolds f : X → Yis differentiable if differentiable functions pull back to differentiable functions, i.e.if pullback by f gives a map OY → f∗OX . (Hint: check this on small patches.Once you figure out what you are trying to show, you’ll realize that the result isimmediate.)

5.1.B. Exercise. Show that a morphism of differentiable manifolds f : X →Y with f(p) = q induces a morphism of stalks f# : OY,q → OX,p. Show thatf#(mY,q) ⊂ mX,p. In other words, if you pull back a function that vanishes at q,you get a function that vanishes at p — not a huge surprise.

Here is a little more for experts: Notice that this induces a map on tangentspaces

(mX,p/m2X,p)

∨ → (mY,q/m2Y,q)

∨.

This is the tangent map you would geometrically expect. Again, it is interestingthat the cotangent map mY,q/m2

Y,q → mX,p/m2X,p is algebraically more natural than

the tangent map.Experts are now free to try to interpret other differential-geometric information

using only the map of topological spaces and map of sheaves. For example: howcan one check if f is a submersion? How can one check if f is an immersion? (Wewill see that the algebro-geometric version of these notions are smooth morphismsand locally closed immersion, [forme: ADD REF!] .)

[primordial: Then we have a normal exact sequence. Vector bundle can be rewritten in normal exact sequence

terms of sheaves; explain how. ]

5.1.2. Side Remark. Manifolds are covered by disks that are all isomorphic.Schemes (or even complex algebraic varieties) will not have isomorphic open sets.We’ll see an example in [forme: GET REF] . Informally, this is because in thetopology on schemes, all non-empty open sets are “huge” and have more “struc-ture”.

[primordial: This bit is complicated to say correctly!

5.1.3. Variation: complex analytic varieties. We could repeat all of the above discussionfor holomorphic manifolds, and holomorphic maps. This leads us so close to the definition ofcomplex analaytic varieties that it is a shame not state it here. Complex analytic varieties are ageneralization of complex manifolds; certain singularities are allowed. Classically (and informally),they were defined by defining the atomic pieces as subsets of an open subest U ⊂ Cn cut out byholomorphic equations, and then gluing them together along “holomorphic maps”. We may definethem as follows.

We begin with the Zariski topology on ... Define (for this paragraph only) a patch of acomplex analytic variety as a ringed space as follows. We begin with the data of an open subsetU ⊂ Cn, and Z ⊂ U cut out by holomorphic equations (in the n variables of Cn). This gives usthe set. The topology is that induced from U (or Cn). The structure sheaf is also that inducedfrom U (or Cn): if V ⊂ Z is open, meaning that it can be written as V ′ ∩ Z where V ′ is open

in Cn, then the holomorphic functions on V are defined to be the restrictions of the holomorphicfunctions on some V ′. (There is a great deal to be checked. For example, this indeed forms a ring.Also, if V is a complex manifold in the old-fashioned sense, the sheaf of holomorphic functionsdefined in this way should agree with the old definition.) Then a complex analytic manifold

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I’ll continue to talk vaguely. Affine complex varieties, classically: things cut out in Cn.Their Zariski topology: closed sets are those cut out by algebraic equations. Immediately impliespolynomial maps are continuous. (Conversely, requiring this leads us to the definition of algebraicsets.) Sheaf of functions: things that look like quotients of polynomials.

We could even define complex algebraic varieties in general. They are topological spaces,along with a sheaf of complex-valued functions, that are covered by open sets that are affinevarieties.

We will keep complex varieties in mind when defining schemes in general, and use them asa running example. We’re doing this for several reasons: these are among the most importantexamples. And they give good geometric intuition even for other cases. Several complications willappear, that just take getting used to: we have new kinds of points in addition to the old-fashionedones, even for complex varieties; we will later have more general schemes, we will have functionsthat are not “determined by their values at points”!

Why not real varieties? Non-algebraically closed fields are more complicated, as we’ll soonsee. We’ll indeed have schemes over R, but they will behave a little less intuitively than schemesover C.

][primordial: DO I WANT TO INCLUDE THIS? LIKELY INCLUDE IT

AFTER THE SET

5.2 Motivating examples II: “Pictures of rings”

In this section, we use lots of words we haven’t defined yet, and draw pictures.C[x]. C. Picture. Smooth curve. Primes are all the points. Maximal ideals: (x− a). Value

at a point: modulo that maximal ideal. A function vanishes at the point if it is in the maximalideal. Picture: smooth curve. (x− 3)2/(x+ 1)2. Triple 0, double pole.

Z. Picture. Smooth curve. 17. Value at a point 2. 27/4. Value at a point? Where does 95vanish? 3/2 + 4/5.

Can you guess the stalk of this sheaf in both cases??Pictures associated to various operations. Quotient, and localization: get both subsets in

both cases.Closed subschemes. Value at a point. Instead: modulo x(x − 1). Restrict to two points

simultaneously. Closed subschemes.

Interesting variation: x2. Read off the first derivative! Spec C[x]/x2 will indeed be an inter-esting scheme. Note: we have a function that is non-zero, but whose square is zero. New feature!Functions are not quite determined by their values at points. (However, they are determined bygerms: sections of a sheaf are determined by their stalks, as we’ve seen!)

Open subschemes. Instead, into localization. Open subscheme (distinguished).Vanishing set. Zariski topology: declare just these to be closed. very coarse/blunt. Cofinite

topology.How to draw the extra points? C[x]. One extra point (0). It is in every open set but the

empty set.C[x, y]. Fact: primes are of 3 forms.Again, what happens with standard ring operations? Quotient: closed subset. Localization:

induced topology. Divide by one thing: get an open set. In general for localization: hard to say.For example, picture of local rings; that’s the second standard localization do.

Points: seem to have dimension. “One in the closure of the other” means containment. Ledto Krull dimension.

Morphisms: Line to parabola: x→ t, y → t2.Generic point: certain things true there are also “generically true”. Does x = 3 at the generic

point? No. Also not true generically.Often: you could just look at closed points. That’s true for classical things (in n-space

for example). For example, in C3, you’re only allowed to talk about sets you can describe withpolynomials.

Not true for local rings! for example, k[x](x). ]

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5.3 The underlying set of affine schemes

2 [forme: Complex varieties mixed in more! ADD SOON: The outline of this.

Definition is very short. We concentrate on the geometric interpretation of these

things. In particular, analogue of n-space. We have additional points, and we inves-

tigate that. We state that Z behaves like a smooth curve.]

[forme: [Alan Weinstein told me that this is related to actual spectrum: If A is a

matrix with distinct eigenvalues, then the algebra of the matrix is naturally functions

on the eigenvalues.]] 3For any ring A, we are going to define something calledSpec A, the spectrum of A. In this section, we will define it as a set, but we willsoon endow it with a topology, and later we will define a sheaf of rings on it (thestructure sheaf). Such an object is called an affine scheme. In the future, Spec A affine scheme

will denote the set along with the topology. (Indeed, it will often implicitly includethe data of the structure sheaf.) But for now, as there is no possibility of confusion,Spec A will just be the set. spectrum; SpecA; [p]

The set Spec A is the set of prime ideals of A. The point of Spec A correspondingto the prime ideal p will be denoted [p].

We now give some examples. Here are some temporary definitions to help usunderstand these examples. Elements a ∈ A will be called functions on Spec A,and their value at the point [p] will be a (mod p). This is weird: a function cantake values in different places in different points — the function 7 on Spec Z takesthe value 1 (mod 2) at [(2)] and 1 (mod 3) at [(3)]. “An element a of the ringlying in a prime ideal p” translates to “a function a that is 0 at the point [p]” or “afunction a vanishing at the point [p]”, and we will use these phrases interchangeably.Notice that if you add or multiply two functions, you add or multiply their valuesat all points; this is a translation of the fact that A → A/p is a homomorphismof rings. These translations are important — make sure you are very comfortablewith them!

Example 1: A1C := Spec C[x]. This is known as “the affine line” or “the affine

line over C”. Let’s find the prime ideals. As C[x] is an integral domain, 0 is prime. affine line, A1

Also, (x− a) is prime, where a ∈ C: it is even a maximal ideal, as the quotient bythis ideal is field:

0 // (x− a) // C[x]f 7→f(a) // C // 0

(This exact sequence should remind you of (??) in our motivating example ofmanifolds.)

We now show that there are no other prime ideals. We use the fact that C[x]has a division algorithm, and is a unique factorization domain. Suppose p is a primeideal. If p 6= 0, then suppose f(x) ∈ p is a non-zero element of smallest degree. Itis not constant, as prime ideals can’t contain 1. If f(x) is not linear, then factorf(x) = g(x)h(x), where g(x) and h(x) have positive degree. Then g(x) ∈ p orh(x) ∈ p, contradicting the minimality of the degree of f . Hence there is a linearelement x − a of p. Then I claim that p = (x − a). Suppose f(x) ∈ p. Thenthe division algorithm would give f(x) = g(x)(x − a) + m where m ∈ C. Thenm = f(x)− g(x)(x − a) ∈ p. If m 6= 0, then 1 ∈ p, giving a contradiction.

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Thus we have a picture of Spec C[x] (see Figure 1). There is one point foreach complex number, plus one extra point. The point [(x− a)] we will reasonablyassociate to a ∈ C. Where should we picture the point [(0)]? Where is it? Thebest way of thinking about it is somewhat zen. It is somewhere on the complexline, but nowhere in particular. Because (0) is contained in all of these primes,we will somehow associate it with this line passing through all the other points.[(0)] is called the “generic point” of the line; it is “generically on the line” but youcan’t pin it down any further than that. We’ll place it far to the right for lack ofanywhere better to put it. You will notice that we sketch A1

C as one-dimensional inthe real sense; this is to later remind ourselves that this will be a one-dimensionalspace, where dimensions are defined in an algebraic (or complex-geometric) sense.

(0)(x) (x− 1) (x− a)

Figure 1. A picture of A1C = Spec C[x]

To give you some feeling for this space, let me make some statements that arecurrently undefined, but suggestive. The functions on A1

C are the polynomials. Sof(x) = x2 − 3x + 1 is a function. What is its value at [(x − 1)], which we think ofas the point 1 ∈ C? Answer: f(1)! Or equivalently, we can evalute f(x) modulox− 1 — this is the same thing by the division algorithm. (What is its value at (0)?It is f(x) (mod 0), which is just f(x).)

Here is a more complicated example: g(x) = (x − 3)3/(x − 2) is a “rationalfunction”. It is defined everywhere but x = 2. (When we know what the structuresheaf is, we will be able to say that it is an element of the structure sheaf on theopen set A1

C − 2.) g(x) has a triple zero at 3, and a single pole at 2.Example 2: A1

k := Spec k[x] where k is an algebraically closed field. This iscalled the affine line over k. All of our discussion in the previous example carriesover without change. We will use the same picture, which is after all intended tojust be a metaphor.A1

k, affine line

Example 3: Spec Z. One amazing fact is that from our perspective, this willSpec Zlook a lot like the affine line. This is another unique factorization domain, witha division algorithm. The prime ideals are: (0), and (p) where p is prime. Thuseverything from Example 1 carries over without change, even the picture. Ourpicture of Spec Z is shown in Figure 2.

· · ·(2) (3) (5) (0)

Figure 2. A “picture” of Spec Z, which looks suspiciously like Figure 1

Let’s blithely carry over our discussion of functions on this space. 100 is afunction on Spec Z. It’s value at (3) is “1 (mod 3)”. It’s value at (2) is “0 (mod 2)”,

99

and in fact it has a double zero. 27/4 is a rational function on Spec Z, defined awayfrom (2). It has a double pole at (2), a triple zero at (3). Its value at (5) is

27× 4−1 ≡ 2× (−1) ≡ 3 (mod 5).

Example 4: stupid examples. Spec k where k is any field is boring: only onepoint. Spec 0, where 0 is the zero-ring, is the empty set, as 0 has no prime ideals.

5.3.A. A small exercise about small schemes. (a) Describe the set Spec k[ε]/(ε2).4

This is called the ring of dual numbers, and will turn out to be quite useful. Youshould think of ε as a very small number, so small that its square is 0 (althoughit itself is not 0). It is a non-zero function whose value at all points is zero, thus dual numbers

giving our first example of functions not being determined by their values at points.We will discuss this phenomenon further in §5.3.5.(b) Describe the set Spec k[x](x). (We will see this scheme again in Exercise 5.4.J.)

In Example 2, we restricted to the case of algebraically closed fields for a reason:things are more subtle if the field is not algebraically closed.

Example 5: R[x]. Using the fact that R[x] is a unique factorization domain, we A1R

see that the primes are (0), (x−a) where a ∈ R, and (x2 +ax+b) where x2 +ax+bis an irreducible quadratic. The latter two are maximal ideals, i.e. their quotientsare fields. For example: R[x]/(x− 3) ∼= R, R[x]/(x2 + 1) ∼= C.

5.3.B. Unimportant Exercise. Show that for the last type of prime, of theform (x2 + ax + b), the quotient is always isomorphic to C.

ex.doneSo we have the points that we would normally expect to see on the real line,

corresponding to real numbers; the generic point 0; and new points which we mayinterpret as conjugate pairs of complex numbers (the roots of the quadratic). Thislast type of point should be seen as more akin to the real numbers than to thegeneric point. You can picture A1

R as the complex plane, folded along the real axis.But the key point is that Galois-conjugate points are considered glued.

Let’s explore functions on this space; consider the function f(x) = x3 − 1. Itsvalue at the point [(x − 2)] is f(x) = 7, or perhaps better, 7 (mod x − 2). Howabout at (x2 + 1)? We get

x3 − 1 ≡ x− 1 (mod x2 + 1),

which may be profitably interpreted as i− 1.One moral of this example is that we can work over a non-algebraically closed

field if we wish. It is more complicated, but we can recover much of the informationwe wanted.

5.3.C. Exercise. Describe the set A1Q. (This is harder to picture in a way

analogous to A1R; but the rough cartoon of points on a line, as in Figure 1, remains

a reasonable sketch.) A1Q

ex.doneExample 6: Fp[x]. As in the previous examples, this has a division algorithm,so the prime ideals are of the form (0) or (f(x)) where f(x) ∈ Fp[x] is an irreduciblepolynomials, which can be of any degree. Irreducible polynomials correspond tosets of Galois conjugates in Fp.

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Note that Spec Fp[x] has p points corresponding to the elements of Fp, but also(infinitely) many more. This makes this space much richer than simply p points.For example, a polynomial f(x) is not determined by its values at the p elementsof Fp, but it is determined by its values at the points of Spec Fp. (As we havementioned before, this is not true for all schemes.)

You should think about this, even if you are a geometric person — this intuitionwill later turn up in geometric situations. Even if you think you are interestedonly in working over an algebraically closed field (such as C), you will have non-algebraically closed fields (such as C(x)) forced upon you.

Example 7: A2C = Spec C[x, y]. [forme: “Example 7” is referred to inaffine plane, A2

k

§5.5.3.] (As with Examples 1 and 2, discussion will apply with C replaced by anyalgebraically closed field.) Sadly, C[x, y] is not a Principal Ideal Domain: (x, y) isnot a principal ideal. We can quickly name some prime ideals. One is (0), whichhas the same flavor as the (0) ideals in the previous examples. (x−2, y−3) is prime,and indeed maximal, because C[x, y]/(x − 2, y − 3) ∼= C, where this isomorphismis via f(x, y) 7→ f(2, 3). More generally, (x − a, y − b) is prime for any (a, b) ∈ C2.Also, if f(x, y) is an irreducible polynomial (e.g. y − x2 or y2 − x3) then (f(x, y))is prime.

5.3.D. Exercise. (Feel free to skip this exercise, as we will see a different proofof this in §15.4.3.) Show that we have identified all the prime ideals of C[x, y].[forme: Can we do this?] 5

We can now attempt to draw a picture of this space. The maximal primescorrespond to the old-fashioned points in C2: [(x−a, y− b)] corresponds to (a, b) ∈C2. We now have to visualize the “bonus points”. [(0)] somehow lives behind all ofthe old-fashioned points; it is somewhere on the plane, but nowhere in particular.So for example, it does not lie on the parabola y = x2. The point [(y − x2)] lieson the parabola y = x2, but nowhere in particular on it. You can see from thispicture that we already want to think about “dimension”. The primes (x−a, y− b)are somehow of dimension 0, the primes (f(x, y)) are of dimension 1, and (0) issomehow of dimension 2. (All of our dimensions here are complex or algebraicdimensions. The complex plane C2 has real dimension 4, but complex dimension2. Complex dimensions are in general half of real dimensions.) We won’t definedimension precisely until Chapter 15, but you should feel free to keep it in mindbefore then.

Note too that maximal ideals correspond to “smallest” points. Smaller idealscorrespond to “bigger” points. “One prime ideal contains another” means that thepoints “have the opposite containment.” All of this will be made precise once wehave a topology. This order-reversal is a little confusing, and will remain so evenonce we have made the notions precise.

We now come to the obvious generalization of Example 7, affine n-space.Example 8: An

C := Spec C[x1, . . . , xn]. (More generally, AnA is defined to beAn; affine space.

Spec A[x1, . . . , xn], where A is an arbitrary ring.)For concreteness, let’s consider n = 3. We now have an interesting question in

algebra: What are the prime ideals of C[x, y, z]? Analogously to before, (x− a, y−b, z− c) is a prime ideal. This is a maximal ideal, with residue field C; we think of

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these as “0-dimensional points”. We will often write (a, b, c) for [(x−a, y−b, z−c)]because of our geometric interpretation of these ideals.

There are no more maximal ideals, by Hilbert’s Nullstellensatz. (This is some-times called the “weak version” of the Nullstellensatz.) You may have already seenthis result. We will prove it later (as well as a slightly stronger form, stated inExercise 7.3.I), in §15.4.A, so we will simply state it here.

5.3.1. Hilbert’s Nullstellensatz. — Suppose A = k[x1, . . . , xn], where k is an Hilbert’s Nullst., nullst.

algebraically closed field. Then the maximal ideals are precisely those of the form(x1 − a1, . . . , xn − an), where ai ∈ k. 6

[forme: Place these and call them statements of the Nullstellensatz. Have them

prove them as exercises. (Dummit and Foote) k is a field. Then closed subsets of

closed points correspond to radical ideals (I and V again). Yet another Nullstellensatz

statement. Suppose k is a field, B a finitely generated k-algebra. Then if B is a field,

then it is a finite algebraic extension of k. [AM, Cor. 5.24, p. 67] This comes up in

7.3.I, although I phrased it differently. ] [forme: Greg has some notes on the

Nullstellensatz, which I have in my 2007 210B directory.] There are other primeideals too. We have (0), which is corresponds to a “3-dimensional point”. We have(f(x, y, z)), where f is irreducible. To this we associate the hypersurface f = 0,so this is “2-dimensional” in nature. But we have not found them all! One clue:we have prime ideals of “dimension” 0, 2, and 3 — we are missing “dimension1”. Here is one such prime ideal: (x, y). We picture this as the locus wherex = y = 0, which is the z-axis. This is a prime ideal, as the corresponding quotientC[x, y, z]/(x, y) ∼= C[z] is an integral domain (and should be interpreted as thefunctions on the z-axis). There are lots of one-dimensional primes, and it is notpossible to classify them in a reasonable way. It will turn out that they correspondto things that we think of as irreducible curves: the natural answer to this algebraicquestion is geometric.

5.3.2. Important fact: Maps of rings induce maps of spectra (as sets). 7

We now make an observation that will later grow up to be morphisms of schemes.If φ : B → A is a map of rings, and p is a prime ideal of A, then φ−1(p) is a primeideal of B (check this!). Hence a map of rings φ : B → A induces a map of setsSpec A → Spec B “in the opposite direction”. This gives a contravariant functorfrom the category of rings to the category of sets: the composition of two maps ofrings induces the composition of the corresponding maps of spectra. [forme: Next

2 in sequence: 5.4.G, 8.3.2.]

[forme: I REALLY NEED TO DRAW SOME PICTURES HERE, and do some

examples.]

We now describe two important cases of this: maps of rings inducing inclusionsof sets. There are two particularly useful ways of producing new rings from a ringA. One is by taking the quotient by an ideal I . The other is by localizing at amultiplicative set. We’ll see how Spec behaves with respect to these operations. Inboth cases, the new ring has a Spec that is a subset of Spec of the old ring.

First important example (quotients): Spec B/I in terms of Spec B. Asa motivating example, consider Spec B/I where B = C[x, y], I = (xy). We have a

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picture of Spec B, which is the complex plane, with some mysterious extra “higher-dimensional points”. It is an important fact that the primes of B/I are in bijectionwith the primes of B containing I .8 (If you do not know why this is true, you shouldalgebra prereq

prove it yourself.) [forme: Make exercise in class; ditto for localization.] Thus wecan picture Spec B/I as a subset of Spec B. We have the “0-dimensional points”(a, 0) and (0, b). We also have two “1-dimensional points” (x) and (y).

We get a bit more: the inclusion structure on the primes of B/I correspondsto the inclusion structure on the primes containing I . More precisely, if J1 ⊂ J2 inB/I , and Ki is the ideal of B corresponding to Ji, then K1 ⊂ K2. (Again, provethis yourself if you have not seen it before.)

So the minimal primes of C[x, y]/(xy) are the “biggest” points we see, and thereare two of them: (x) and (y). Thus we have the intuition that will later be madeprecise (Exercise 5.5.O): the minimal primes of A correspond to the “components”of Spec A.

As an important motivational special case, you now have a picture of “com-plex affine varieties”. Suppose A is a finitely generated C-algebra, generated byx1, . . . , xn, with relations f1(x1, . . . , xn) = · · · = fr(x1, . . . , xn) = 0. Then thisdescription in terms of generators and relations naturally gives us an interpretationof Spec A as a subset of An

C, which we think of as “old-fashioned points” (n-tuples ofcomplex numbers) along with some “bonus” points. To see which subsets of the old-fashioned points are in Spec A, we simply solve the equations f1 = · · · = fr = 0. Forexample, Spec C[x, y, z]/(x2 + y2− z2) may be pictured as shown in Figure 3. (Ad-mittedly this is just a “sketch of the R-points”, but we will still find it helpful later.)This entire picture carries over (along with the Nullstellensatz) with C replaced byany algebraically closed field. Indeed, the picture of Figure 3 can be said to rep-resent k[x, y, z]/(x2 + y2 − z2) for most algebraically closed fields k (although it ismisleading in characteristic 2, because of the coincidence x2+y2−z2 = (x+y+z)2).

5.3.E. Exercise. Ring elements that have a power that is 0 are called nilpotents.If I is an ideal of nilpotents, show that Spec B/I → Spec B is a bijection. Thusnilpotents don’t affect the underlying set. (We will soon see in §5.4.5 that theywon’t affect the topology either — the difference will be in the structure sheaf.) 9

Second important example (localization): Spec S−1B in terms of Spec B,10

where S is a multiplicative subset of B. There are two particularly important flavorsof multiplicative subsets. The first is B \ p, where p is a prime ideal. This localiza-tion S−1B is denoted Bp. A motivating example is B = C[x, y], S = B−(x, y). Thesecond is 1, f, f2, . . . , where f ∈ B. This localization is denoted Bf . (Notationalwarning: If p is a prime ideal, then Bp means you’re allowed to divide by elementsnot in p. However, if f ∈ B, Bf means you’re allowed to divide by f . This canbe confusing. For example, if (f) is a prime ideal, then Bf 6= B(f).) A motivatingexample is B = C[x, y], f = x.

5.3.3. Essential algebra fact (to review and know). The map Spec S−1B → Spec Bgives an order-preserving bijection of the primes of S−1B with the primes of B thatdon’t meet the multiplicative set S.

8primesofAmodI

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Figure 3. A “picture” of Spec C[x, y, z]/(x2 + y2 − z2)

So if S = B − p where p is a prime ideal, the primes of S−1B are just theprimes of B contained in p. If S = 1, f, f 2, . . . , the primes of S−1B are just thoseprimes not containing f (the points where “f doesn’t vanish” — draw a picture ofSpec C[x]x2−x to see how this works).

5.3.4. Warning. sometimes localization is first introduced in the special casewhere B is an integral domain. In this example, B → Bf , but this isn’t true whenone inverts zero-divisors.11 (A zero-divisor of a ring B is an element a such thatthere is a non-zero element b with ab = 0. The other elements of B are called non-zero-divisors.) One definition of localization is as follows. The elements of S−1B 0-div, non-0-div; 0 is

a 0-divare of the form a/s where r ∈ B and s ∈ S, and (a1/s1) × (a2/s2) = (a1a2/s1s2),and (a1/s1)+(a2/s2) = (a1s2+s1a2)/(s1s2). We say that a1/s1 = a2/s2 if for somes ∈ S s(a1s2 − a2s1) = 0. So for example, B[1/0] ∼= 0. [forme: (Geometrically,

this is good: the locus of points where 0 doesn’t vanish is the empty set, so certainly

D(0) = SpecB0.)] [forexperts: 0 is a zero-divisor. Is that horrible?] [forme:

Make sure I follow this convention throughout.]

[primordial: I CERTAINLY WANT THIS! I WANT MY TWO EXAMPLES.Pictures of localization. In each of these two cases, a picture is worth a thousand words. In

these notes, I’m not making pictures unfortunately. But I’ll try to describe them in less than athousand words. The case of S = 1, f, f2, . . . is easier: we just throw out those points where fvanishes. (We will soon call this a distinguished open set, once we know what open sets are.) In distinguished open setour example of B = k[x, y], f = x, we throw out the y-axis.

In general, inverting zero-divisors can make things behave a bit oddly. Example: R =k[x, y]/(xy). f = x. What do you get? It’s actually a straightforward ring, and we’ll use somegeometric intuition to figure out what it is. Spec k[x, y]/(xy) “is” the union of the two axes inthe plane. Localizing means throwing out the locus where x vanishes. So we’re left with the

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x-axis, minus the origin, so we expect Spec k[x]x. So there should be some natural isomorphism(k[x, y]/(xy))x ∼= k[x]x.

5.3.F. F. igure out why these two rings are isomorphic. (You’ll see that y on the left goes to 0on the right.)

In the case of S = R − p, we keep only those primes contained in p. In our example

R = k[x, y], p = (x, y), we keep all those points corresponding to “things through the origin”,

i.e. the 0-dimensional point (x, y), the 2-dimensional point (0), and those 1-dimensional points

(f(x, y)) where f(0, 0) = 0, i.e. those “irreducible curves through the origin”. (There is a picture

of this in Mumford’s Red Book.) ]

5.3.5. Important comment: functions are not determined by their valuesat points. 12We are developing machinery that will let us bring our geometricintuition to algebra. There is one point where your intuition will be false, so youshould know now, and adjust your intuition appropriately. Suppose we have afunction (ring element) vanishing at all points. Then it is not necessarily the zerofunction! The translation of this question is: is the intersection of all prime idealsnecessarily just 0? The answer is no, as is shown by the example of the ring of dualnumbers k[ε]/(ε2): ε 6= 0, but ε2 = 0. (We saw this scheme in Exercise 5.3.A(a).)dual numbers

Any function whose power is zero certainly lies in the intersection of all primeideals. The converse is also true: the intersection of all the prime ideals consists offunctions for which some power is zero, otherwise known as the nilradical N. 13nilradical, N

(You should check that the nilpotents indeed form an ideal. For example, the sumof two nilpotents is always nilpotent.)nilpotents

5.3.6. Theorem. The nilradical N(A) is the intersection of all the primes of A.nilradical14

5.3.G. Exercise. If you don’t know this theorem, then look it up, or even better,prove it yourself. (Hint: one direction is easy. The other will require knowing thatany proper ideal of A is contained in a maximal ideal, which requires the axiom ofchoice.) [forme: For hard direction: suppose x /∈ N(A). We wish to show that thereAC

is a prime ideal not containing x. Show that Ax is not the 0-ring, by showing that

1 6= 0.]

5.3.7. In particular, although it is upsetting that functions are not determined bytheir values at points, we have precisely specified what the failure of this intuitionis: two functions have the same values at points if and only if they differ by anilpotent. And if there are no non-zero nilpotents — if N = 0 — then functionsare determined by their values at points. If a ring has no non-zero nilpotents, wesay that it is reduced. 15reduced ring

[primordial:

5.3.H. Easy fun unimportant exercise. Suppose we have a polynomial f(x) ∈ k[x]. Instead,

we work in k[x, ε]/ε2. What then is f(x + ε)? (Do a couple of examples, and you will see the

pattern. For example, if f(x) = 3x3 + 2x, we get f(x + ε) = (3x3 + 2x) + ε(9x2 + 2). Prove

the pattern!) Useful tip: the dual numbers are a good source of (counter)examples, being the

12functionsnotdeterminedbyvalues

13nilradicaldef

14nilradicalprime

15d:reducedring

105

“smallest ring with nilpotents”. They will also end up being important in defining differential

information (chapter 22). ]

5.4 The Zariski topology: The underlying topological spaceof an affine scheme

[forme: Quick explanation: functions vanish on closed sets.]

We next introduce the Zariski topology on the spectrum of a ring. At first itseems like an odd definition, but in retrospect it is reasonable. For example, con-sider A2

C = Spec C[x, y], the complex plane (with a few extra points). In algebraicgeometry, we will only be allowed to consider algebraic functions, i.e. polynomialsin x and y. The locus where a polynomial vanishes should reasonably be a closedset, and the Zariski topology is defined by saying that the only sets we should con-sider closed should be these sets, and other sets forced to be closed by these. Inother words, it is the coarsest topology where these sets are closed. Zariski topology

In particular, although topologies are often described using open subsets, it willmore convenient for us to define this topology in terms of closed subsets. If S is asubset of a ring A, define the V anishing set of S by vanishing set V (S)

V setV (S) := [p] ∈ Spec A | S ⊂ p.It is the set of points on which all elements of S are zero. (It should now be secondnature to equate “vanishing at a point” with “contained in a prime”.) We declarethat these (and no other) are the closed subsets.

For example, consider V (xy, xz) ⊂ A3 = Spec C[x, y, z]. Which points arecontained in this locus? We think of this as solving xy = yz = 0. Of the “old-fashioned” points (interpreted as ordered triples of complex numbers, thanks tothe Hilbert’s Nullstellensatz 5.3.1), we have the points where y = 0 or x = z = 0:the xz-plane and the y-axis respectively. Of the “new” points, we have the genericpoint of the xz-plane (also known as the point [(y)]), and the generic point of they-axis (also known as the point [(x, z)]). You might imagine that we also have anumber of “one-dimensional” points contained in the xz-plane.

5.4.A. Easier Exercise. Check that the x-axis is contained in this set.ex.done

Let’s return to the general situation. The following exercise lets us restrictattention to vanishing sets of ideals.

5.4.B. Easier Exercise. Show that if (S) is the ideal generated by S, thenV (S) = V ((S)). 16

ex.doneWe define the Zariski topology by declaring that V (S) is closed for all S. Let’s d:Zariski top

check that this is a topology. We have to check that the empty set and the totalspace are open; the union of an arbitrary collection of open sets are open; and theintersection of two open sets are open.

5.4.C. Exercise. (a) Show that ∅ and Spec A are both open.(b) Show that V (I1) ∪ V (I2) = V (I1I2). Hence show that the intersection of anyfinite number of open sets is open.

16jan2807

106

(c) (The union of any collection of open sets is open.) If Ii is a collection of ideals(as i runs over some index set), check that ∩iV (Ii) = V (

∑

i Ii).17

5.4.1. Properties of “vanishing set” function V (·). The function V (·) isobviously inclusion-reversing: If S1 ⊂ S2, then V (S2) ⊂ V (S1). Warning: Wecould have equality in the second inclusion without equality in the first, as the nextexercise shows.

5.4.D. Exercise/Definition. If I ⊂ R is an ideal, then define its radical byradical√

I := r ∈ R | rn ∈ I for some n ∈ Z≥0.For example, the nilradical N (§5.3.5) is

√

(0). Show that V (√

I) = V (I). We say√I

an ideal is radical if it equals its own radical.radical ideal, nilradi-

cal Here are two useful consequences. As (I ∩ J)2 ⊂ IJ ⊂ I ∩ J , we have thatV (IJ) = V (I ∩J) (= V (I)∪V (J) by Exercise 5.4.C(b)). Also, combining this with

Exercise 5.4.B, we see V (S) = V ((S)) = V (√

(S)).

5.4.E. Exercise (practice with the concept). If I1, . . . , In are ideals of aring A, show that

√

∩ni=1Ii = ∩n

i=1

√Ii. (We will use this property without referring

back to this exercise.)

5.4.F. Exercise for future use. Show that√

I is the intersection of all theprime ideals containing I . (Hint: Use Theorem 5.3.6 on an appropriate ring.)

5.4.2. Examples. Let’s see how this meshes with our examples from the previoussection.

Recall that A1C, as a set, was just the “old-fashioned” points (corresponding to

maximal ideals, in bijection with a ∈ C), and one “new” point (0). The Zariskitopology on A1

C is not that exciting: the open sets are the empty set, and A1C minus

a finite number of maximal ideals. (It “almost” has the cofinite topology. Noticethat the open sets are determined by their intersections with the “old-fashionedpoints”. The “new” point (0) comes along for the ride, which is a good sign that itis harmless. Ignoring the “new” point, observe that the topology on A1

C is a coarsertopology than the analytic topology.)

The case Spec Z is similar. The topology is “almost” the cofinite topology inthe same way. The open sets are the empty set, and Spec Z minus a finite numberof “ordinary” ((p) where p is prime) primes.

5.4.3. Closed subsets of A2C. 18The case A2

C is more interesting. You should thinkthrough where the “one-dimensional primes” fit into the picture. In Exercise 5.3.D,we identified all the primes of C[x, y] (i.e. the points of A2

C) as the maximal ideals(x−a, y−b) (a, b ∈ C), the “one-dimensional points” (f(x, y)) (f(x, y) irreducible),and the “two-dimensional point” (0).

Then the closed subsets are of the following form:

(a) the entire space, and(b) a finite number (possibly zero) of “curves” (each of which is the closure of

a “one-dimensional point”) and a finite number (possibly zero) of closedpoints.

17H.L.II.2.1ab

18closedA2C

107

5.4.4. Important fact: Maps of rings induce continuous maps of topo-logical spaces. We saw in §5.3.2 that a map of rings φ : B → A induces a mapof sets π : Spec A→ Spec B.

5.4.G. Important exercise. By showing that closed sets pull back to closedsets, show that π is a continuous map. 19 [forme: Other 2 in sequence: 5.3.2,

8.3.2.]

Not all continuous maps arise in this way. Consider for example the continuousmap on A1

C that is the identity except 0 and 1 (i.e. [(x)] and [(x−1)] are swapped);there is no polynomial that can manage this.

In §5.3.2, we saw that Spec B/I and Spec S−1B are naturally subsets of Spec B.It is natural to ask if the Zariski topology behaves well with respect to these inclu-sions, and indeed it does.

5.4.H. Important exercise. Suppose that I, S ⊂ B are an ideal and multi-plicative subset respectively. Show that Spec B/I is naturally a closed subset ofSpec B. Show that the Zariski topology on Spec B/I (resp. Spec S−1B) is the sub-space topology induced by inclusion in Spec B. (Hint: compare closed subsets.)20

5.4.5. In particular, if I ⊂ N is an ideal of nilpotents, the bijection Spec B/I →Spec B (Exercise 5.3.E) is a homeomorphism. Thus nilpotents don’t affect thetopological space. (The difference will be in the structure sheaf.)21

5.4.I. Useful exercise for later. Suppose I ⊂ B is an ideal. Show that fvanishes on V (I) if and only if fn ∈ I for some n. 22 (If you are stuck, you will geta hint when you see Exercise 5.7.E.) [forme: Why is this here?]

5.4.J. Exercise. (Exercise 5.3.A is related.) Describe the topological spaceSpec k[x](x).

23

5.5 Topological definitions

A topological space is said to be irreducible if it is not the union of two properclosed subsets. In other words, X is irreducible if whenever X = Y ∪Z with Y andZ closed, we have Y = X or Z = X . 24 irr

5.5.A. Easy Exercise. Show that on an irreducible topological space, anynonempty open set is dense. (The moral: unlike in the classical topology, in theZariski topology, non-empty open sets are all “huge”.)

19morphismtop

20Ztoplocalization

21nilpotenttop

22fninI

23smallscheme2

24topdefs

108

5.5.B. Exercise. Show that Spec A is irreducible if and only if A has only oneminimal prime. (Minimality is with respect to inclusion.) In particular, if A is anintegral domain, then Spec A is irreducible. 25 26minimal prime

ex.doneA point of a topological space x ∈ X is said to be closed if x is a closed

subset. In the old-fashioned topology on Cn, all points are closed.cl point

5.5.C. Exercise. Show that the closed points of Spec A correspond to themaximal ideals.

ex.done

Thus Hilbert’s Nullstellensatz lets us associate the closed points of AnC with n-

tuples of complex numbers. Hence from now on we will say “closed point” instead of“old-fashioned point” and “non-closed point” instead of “bonus” or “new-fangled”point when discussing subsets of An

C. [primordial: Exercise; Show that the closed

points of AnZ can be identified with the disjoint union of the closed points of AnFpfor all prime p.

]

5.5.1. Quasicompactness. A topological space X is quasicompact if givenquasicompact topologi-

cal space any cover X = ∪i∈IUi by open sets, there is a finite subset S of the index set Isuch that X = ∪i∈SUi. Informally: every cover has a finite subcover. Dependingon your definition of “compactness”, this is the definition of compactness, minuspossibly a Hausdorff condition. We will like this condition, because we are afraidof infinity.

5.5.D. Exercise. (a) Show that Spec A is quasicompact. (Hint: Exercise 5.7.C.)27

(b) Show that if A is Noetherian, every open subset of A is quasicompact. [pri-Noetherian topological

space mordial: Maybe this is iff. We probably don’t care though. ](c) Show that Spec A can have nonquasicompact open sets. (Possible hint: letA = k[x1, x2, x3, . . . ] and m = (x1, x2, . . . ) ⊂ A, and consider the complement ofV (m).) [forme: Check out H.E.II.2.13(b).] 28 [forme: This example will be

useful to construct other enlightening examples later, e.g. Exercise 12.1.F.]

5.5.E. Exercise. (a) If X is a finite union of quasicompact spaces, show that Xis quasicompact.29

(b) Show that every closed subset of a quasicompact topological space is quasicom-pact. 30

5.5.2. Specialization and generization. Given two points x, y of a topologicalspace X , we say that x is a specialization of y, and y is a generization of x,if x ∈ y. This now makes precise our hand-waving about “one point containedanother”. It is of course nonsense for a point to contain another. But it is notnonsense to say that the closure of a point contains another. For example, inA2

C = Spec C[x, y], [(y − x2)] is a generization of (2, 4) = [(x − 2, y − 4)], and (2, 4)is a specialization of [(y − x2)]. [forme: Add figure.]specialization, gener-

alization

25irreducibleaffine

26specdomainirreducible

27SpecAqc

28nonqcqa

29unionofquasicompact

30unionofaffines

109

5.5.F. Exercise. If X = Spec A, show that [p] is a specialization of [q] if andonly if q ⊂ p. Verify to your satisfaction that we have made our intuition of“containment of points” precise: it means that the one point is contained in theclosure of another.

We say that a point x ∈ X is a generic point for a closed subset K if x = K. gen pt

5.5.G. Exercise. Verify that [(y − x2)] ∈ A2 is a generic point for V (y − x2).[forme: A closed set is sometimes called an algebraic set in the geometric literature.

We will avoid this terminology. ] algebraic set

ex.doneWe will soon see (Exercise 5.6.E) that there is a natural bijection between points

of Spec A and irreducible closed subsets of Spec A. You know enough to show thisnow, although we’ll wait until we have developed some convenient terminology.

5.5.H. Less important exercise. (a) Suppose I = (wz−xy, wy−x2, xz−y2) ⊂k[w, x, y, z]/I . Show that Spec k[w, x, y, z] is irreducible, by showing that I is prime.(One possible approach: Show that the quotient ring is a domain, by showing thatit is isomorphic to the subring of k[a, b] including only monomials of degree divisibleby 3. There are other approaches as well, some of which we will see later. This isan example of a hard question: how do you tell if an ideal is prime?) [primordial:Another approach, taking advantage of dimension theory: show that D(w) is irreducible, simlarly

for the others, show that they are the same irreducible component. All that is missing is a point.

Use Krull. Advantage of the first: it generalizes in an important special case. Advantage of

the second: it works much more generally. One possible answer to hard question: identify the

quotient. Another, less obvious but more general: look at it locally. ... Refer also to affine cone

] We will later see this as the cone over the twisted cubic curve (twisted cubic isdefined in Exercise 9.1.T. twisted cubic curve

(b) Note that the ideal of part (a) may be rewritten as

rank

(

w x yx y z

)

= 1,

i.e. that all determinants of 2× 2 submatrices vanish. Generalize this to the idealof rank 1 2×n matrices. This notion will correspond to the cone (§10.2.3) over thedegree n rational normal curve (Exercise 10.5.G). rational normal curve,

real def’n later5.5.3. Noetherian conditions.

In the examples we have considered, the spaces have naturally broken up intosome obvious pieces. Let’s make that a bit more precise.

A topological space X is called Noetherian if it satisfies the descendingchain condition for closed subsets: any sequence Z1 ⊇ Z2 ⊇ · · · ⊇ Zn ⊇ · · · ofclosed subsets eventually stabilizes: there is an r such that Zr = Zr+1 = · · · . N topological space

desc chain cond31 The following exercise may be enlightening.

5.5.I. Exercise. Show that any decreasing sequence of closed subsets of A2C =

Spec C[x, y] must eventually stabilize. Note that it can take arbitrarily long tostabilize. (The closed subsets of A2

C were described in §5.4.3.)

5.5.4. It turns out that all of the spectra we have considered have this property,but that isn’t true of the spectra of all rings. The key characteristic all of our

31e7ref

110

examples have had in common is that the rings were Noetherian. Recall that aring is Noetherian if every ascending sequence I1 ⊂ I2 ⊂ · · · of ideals eventuallystabilizes: there is an r such that Ir = Ir+1 = · · · . (This is called the ascendingNoetherian ring, ascend-

ing chain condition chain condition on ideals.)32

N ring

N ring, asc chain con-

dition

Here are some quick facts about Noetherian rings. You should be able to provethem all.

Noetherian rings, im-

portant facts about

• Fields are Noetherian. Z is Noetherian.• If A is Noetherian, and I is any ideal, then A/I is Noetherian.• If A is Noetherian, and S is any multiplicative set, then S−1A is Noether-

ian.• Any submodule of a finitely generated module over a Noetherian ring is

finitely generated. (Hint: prove it for An, and use the next exercise.)

5.5.J. Exercise. Show that a ring A is Noetherian if and only if every ideal ofA is finitely generated. [forme: Make sure I understand the reverse implication!]

5.5.K. Exercise. Suppose 0 → M ′ → M → M ′′ → 0, and M ′ and M ′′ satisfythe ascending chain condition for modules. Show that M does too. (The conversealso holds; we won’t use this, but you can show it if you wish.)33

ex.done[forme: I have more facts about modules over Noetherian rings near 17.7.2.

Perhaps collect all these in one place. For example, if I don’t need the previous

exercise here, I can bring it up later.] The next fact is non-trivial.

5.5.5. The Hilbert basis theorem. — Show that if A is Noetherian, then sois A[x]. 34Hilbert basis theorem

Using these results, then any polynomial ring over any field, or over the integers,is Noetherian — and also any quotient or localization thereof. Hence for exampleany finitely-generated algebra over k or Z, or any localization thereof is Noetherian.Most “nice” rings are Noetherian, but not all rings are Noetherian, e.g. k[x1, x2, . . . ]because m = (x1, x2, . . . , ) is not finitely generated.Proof of the Hilbert Basis Theorem 5.5.5. We show that any ideal I ⊂ A[x] isfinitely-generated. We inductively produce a set of generators f1, . . . as follows. Forn > 0, if I 6= (f1, . . . , fn−1), let fn be any non-zero element of I − (f1, . . . , fn−1)of lowest degree. Thus f1 is any element of I of lowest degree, assuming I 6= (0).If this procedure terminates, we are done. Otherwise, let an ∈ A be the initialcoefficient of fn for n > 0. Then as A is Noetherian, (a1, a2, . . . ) = (a1, . . . , aN ) for

some N . Say aN+1 =∑N

i=1 biai. Then

fN+1 −N∑

i=1

bifixdeg fN+1−deg fi

is an element of I that is nonzero (as fN+1 /∈ (f1, . . . , fN )) of lower degree thanfn+1, yielding a contradiction. [forme: [E] p. 27, very short proof. Greg Brumfiel

has a short proof, along with R[[x]], which I have in my 210B file (2007).]

32d:Nring

33acces

34Hilbertbasistheorem

111

5.5.L. Unimportant exercise. Show that if A is Noetherian, then so is A[[x]],the ring of power series in x.

[primordial:

5.5.6. The Game of Chomp (unimportant but fun example). (The game of finite chomp is a

first-person winner. Give ref to Conway.) The fact that the game of infinite chomp is guaranteed

to end is an analog of the Hilbert basis theorem. In fact, this is a consequence of the Hilbert basis

theorem — the fact that infinite chomp is guaranteed to end corresponds to the fact that any

ascending chain of monomial ideals in k[x, y] must eventually stabilize. I learned of this cute fact

from Rahul Pandharipande. If you prove the Chomp problem, you’ll understand how to prove the

Hilbert basis theorem. ]

5.5.M. Exercise. If A is Noetherian, show that Spec A is a Noetherian topolog-ical space. 35

5.5.N. Less important exercise. Show that the converse is not true: if Spec Ais a Noetherian topological space, A need not be Noetherian. [forme: Possible

answer: k[xε](xε).] Describe a ring A such that Spec A is not a Noetheriantopological space. [forme: I just told them one!]

ex.doneIf X is a topological space, and Z is a maximal irreducible subset (an irreducible

closed subset not contained in any larger irreducible closed subset), Z is said to bean irreducible component of X . [forme: Hartshorne has a different definition irreducible component

of irreducible component.] We think of these as the “pieces of X” (see Figure 4).

Figure 4. This closed subset of A2 has six irreducible components

5.5.O. Exercise. If A is any ring, show that the irreducible components ofSpec A are in bijection with the minimal primes of A.36 minimal prime

ex.done

35NimpliesN

36minprimecomponent

112

For example, the only minimal prime of k[x, y] is (0). What are the minimalprimes of k[x, y]/(xy)?

5.5.7. Proposition. — Suppose X is a Noetherian topological space. Thenevery non-empty closed subset Z can be expressed uniquely as a finite union Z =Z1 ∪ · · · ∪ Zn of irreducible closed subsets, none contained in any other. 37

Translation: any non-empty closed subset Z has of a finite number of pieces.As a corollary, this implies that a Noetherian ring A has only finitely many

minimal primes.

Proof. The following technique is often called Noetherian induction, for reasonsthat will become clear. 38 [forme: Justin prefers Noetherian descent. I’ll avoid itNoetherian induction

for various reasons.]

Consider the collection of closed subsets of X that cannot be expressed as afinite union of irreducible closed subsets. We will show that it is empty. Otherwise,let Y1 be one such. If it properly contains another such, then choose one, and call itY2. If this one contains another such, then choose one, and call it Y3, and so on. Bythe descending chain condition, this must eventually stop, and we must have someYr that cannot be written as a finite union of irreducible closed subsets, but everyclosed subset contained in it can be so written. But then Yr is not itself irreducible,so we can write Yr = Y ′∪Y ′′ where Y ′ and Y ′′ are both proper closed subsets. Bothof these by hypothesis can be written as the union of a finite number of irreduciblesubsets, and hence so can Yr, yielding a contradiction. Thus each closed subset canbe written as a finite union of irreducible closed subsets. We can assume that noneof these irreducible closed subsets contain any others, by discarding some of them.

We now show uniqueness. Suppose

Z = Z1 ∪ Z2 ∪ · · · ∪ Zr = Z ′1 ∪ Z ′

2 ∪ · · · ∪ Z ′s

are two such representations. Then Z ′1 ⊂ Z1 ∪ Z2 ∪ · · · ∪ Zr, so Z ′

1 = (Z1 ∩ Z ′1) ∪

· · · ∪ (Zr ∩ Z ′1). Now Z ′

1 is irreducible, so one of these is Z ′1 itself, say (without

loss of generality) Z1 ∩ Z ′1. Thus Z ′

1 ⊂ Z1. Similarly, Z1 ⊂ Z ′a for some a; but

because Z ′1 ⊂ Z1 ⊂ Z ′

a, and Z ′1 is contained in no other Z ′

i, we must have a = 1,and Z ′

1 = Z1. Thus each element of the list of Z’s is in the list of Z ′’s, and viceversa, so they must be the same list.

5.5.8. Definition. We recall two definitions from topology. A topological spaceX is connected if it cannot be written as the disjoint union of two non-empty opensets. A subset Y of X is a connected component if it is a maximal connectedconnected

subset.

5.5.P. Exercise. Show that an irreducible topological space is connected.

5.5.Q. Exercise. Give (with proof!) an example of a scheme that is connectedbut reducible. (Possible hint: a picture may help. The symbol “×” has two “pieces”yet is connected.)

37H.P.I.1.5

38Ninduction

113

5.5.R. Exercise. If A is a Noetherian ring, show that the connected componentsare unions of the irreducible components. Show that the connected components arethe subsets that are simultaneously open and closed. connected component

5.5.S. Exercise. If A =∏

A1×A2×· · ·×An, describe an isomorphism Spec A =Spec A1

∐

Spec A2

∐ · · ·∐ Spec An. Show that each Spec Ai is a distinguished opensubset D(fi) of Spec A. (Hint: let fi = (0, · · · , 0, 1, 0, · · ·0) where the 1 is in theith component.) In other words,

∐ni=1 Spec Ai = Spec

∏ni=1 Ai.

5.5.9. Fun but irrelevant remark. The previous exercise shows that∐n

i=1 Spec Ai∼=

Spec∏n

i=1 Ai, but this can’t hold if “n is infinite” as Spec of any ring is quasicom-pact (Exercise 5.5.D(a)). This leads to an interesting phenomenon. We show thatSpec

∏∞i=1 Ai is “strictly bigger” than

∐∞i=1 Spec Ai where each Ai is isomorphic

to the field k. First, we have an inclusion of sets∐∞

i=1 Spec Ai → Spec∏∞

i=1 Ai,as there is a maximal ideal of Spec

∏

Ai corresponding to each i (precisely thoseelements 0 in the ith component.) But there are other maximal ideals of

∏

Ai.Hint: describe a proper ideal not contained in any of these maximal ideal. (Oneidea: consider elements

∏

ai that are “eventually zero”, i.e. ai = 0 for i 0.)This leads to the notion of ultrafilters, which are very useful, but irrelevant to ourcurrent discussion. [forme: Brian Conrad points out: Interesting fact: we have ultrafilter

infinitely man y connected components, as factor rings are always closed. In the case

k = F2, all elements are idempotents, which implies all local rings are fields.]

5.6 The function I(·), taking subsets of Spec A to ideals of A

39We now introduce a notion that is in some sense “inverse” to the vanishingset function V (·). Given a subset S ⊂ Spec A, I(S) is the set of functions vanishingon S. I(S)

We make three quick observations:

• I(S) is clearly an ideal.• I(S) = I(S).• I(·) is inclusion-reversing: if S1 ⊂ S2, then I(S2) ⊂ I(S1).

5.6.A. Exercise/Example. Let A = k[x, y]. If S = [(x)], [(x − 1, y)] (seeFigure 5), then I(S) consists of those polynomials vanishing on the y axis, and atthe point (1, 0). Give generators for this ideal. [forme: Answer: (y(x − 1), y2)] 40

5.6.B. Tricky exercise. Suppose X ⊂ A3 is the union of the three axes. (Thex-axis is defined by y = z = 0, and the y-axis and z-axis are deined analogously.)Give generators for the ideal I(X). Be sure to prove it! We will see in Exercise ??that this ideal is not generated by less than three elements. [forme: Answer:

xy, yz, zx.] 41

ex.done

39s:ifunction

40Iexample

41threeaxes

114

[(x− 1, y)]

[(x)]

Figure 5. The set S of Exercise/example 5.6.A, pictured as asubset of A2

5.6.C. Exercise. Show that V (I(S)) = S. Hence V (I(S)) = S for a closed setS. 42 (Compare this to Exercise 5.6.D below.)

Note that I(S) is always a radical ideal — if f ∈√

I(S), then fn vanishes onS for some n > 0, so then f vanishes on S, so f ∈ I(S).

5.6.D. Exercise. Prove that if I ⊂ A is an ideal, then I(V (I)) =√

I . 43

ex.doneThis exercise and Exercise 5.6.C suggest that V and I are “almost” inverse.

More precisely:

5.6.1. Theorem. — V (·) and I(·) give a bijection between closed subsets ofSpec A and radical ideals of A (where a closed subset gives a radical ideal by I(·),and a radical ideal gives a closed subset by V (·)). 44

5.6.E. Important exercise. Show that V (·) and I(·) give a bijection betweenirreducible closed subsets of Spec A and prime ideals of A. From this concludethat in Spec A there is a bijection between points of Spec A and irreducible closedsubsets of Spec A (where a point determines an irreducible closed subset by takingthe closure). Hence each irreducible closed subset of Spec A has precisely one generic

point — any irreducible closed subset Z can be written uniquely as z. [forme:

Argument: clearly prime gives irreducible. Suppose S irreducible closed. The key

step is to show that I(S) is prime.] 45 [forme: Useful fact to mention: M.II.2.P.246]

5.7 Distinguished open sets

42VI, used to get bijection

43IV

44nilradicalfact

45bijectionirrpts

46M.II.2.P.2

115

47If f ∈ A, define the distinguished open set D(f) = [p] ∈ Spec A | f /∈ p. distinguished open set

It is the locus where f doesn’t vanish. (I often privately write this as D(f 6= 0) toremind myself of this. I also privately call this a “Doesn’t-vanish set” in analogywith V (f) being the Vanishing set.) We have already seen this set when discussingSpec Af as a subset of Spec A. For example, we have observed that the Zariski-topology on the distinguished open set D(f) ⊂ Spec A coincides with the Zariskitopology on Spec Af (Exercise 5.4.H).

Here are some important but not difficult exercises to give you a feel for theseimportant open sets.

5.7.A. Exercise. 48 Show that the distinguished open sets form a base for theZariski topology. (Hint: Given an ideal I , show that the complement of V (I) is∪f∈ID(f).)

5.7.B. Exercise. Suppose fi ∈ A as i runs over some index set J . Show that∪i∈JD(fi) = Spec A if and only if (fi) = A. (One of the directions will use the factthat any proper ideal of A is contained in some maximal ideal.) [primordial:One direction: Suppose [p] /∈ ∪D(fi). You can unwind this to get an algebraic statement. I think

of it as follows. All of the fi vanish at [p], i.e. all fi ∈ p, so then (f1, . . . ) ⊂ p and hence this ideal

can’t be all of R. Conversely, consider the ideal (f1, . . . ). If it isn’t R, then it is contained in a

maximal ideal. But then there is some prime ideal containing all the fi. Translation: [m] /∈ D(fi)

for any i. ] 49

5.7.C. Exercise. Show that if Spec A is an infinite union ∪i∈JD(fi), then in factit is a union of a finite number of these. (Hint: use the previous exercise 5.7.B.)Show that Spec A is quasicompact. [forme: We’ll use this very soon in the proof of

base identity of the structure sheaf, Theorem 6.1.1. This is also the key to proving

quasicompactness.] 50

5.7.D. Exercise. Show that D(f) ∩ D(g) = D(fg). [forme: Hence the

distinguished base is a nice base.] 51

5.7.E. Exercise. Show that if D(f) ⊂ D(g), if and only if fn ∈ (g) for somen if and only if g is a unit in Af .52 (Hint for the first equivalence: f ∈ I(V ((g))).Compare this to Exercise 5.4.I.) We will use this shortly. [primordial: This

is a special case of earlier exercise 5.4.I, and vice versa. Here is how I think of it geometrically.

Draw a picture of SpecR. Draw the closed subset V (g) = SpecR/(g). That’s where g vanishes,

and the complement is D(g), where g doesn’t vanish. Then f is a function on this closed subset,

and it vanishes at all points of the closed subset. (Translation: Consider f as an element of the

ring R/(g).) Now any function vanishing at every point of Spec a ring must have some power

which is 0. Translation: there is some n such that fn = 0 in R/(g), i.e. fn ≡ 0 (mod g) in R, i.e.

fn ∈ (g).] ]

5.7.F. Exercise. Show that D(f) = ∅ if and only if f ∈ N. [forme: Reason:

Theorem 5.6.1]

47s:dos

48distinguishedbase

49e:prepreqc

50e:preqc

51DfintersectDg

52DfDg

116

ex.done

CHAPTER 6

The structure sheaf of an affine scheme, and the

definition of schemes in general

6.1 The structure sheaf

1The final ingredient in the definition of an affine scheme is the structure sheafOSpecA, which we think of as the “sheaf of algebraic functions”. As motivation, in structure sheaf on

SpecA, OSpecAA2, we expect that on the open set D(xy) (away from the two axes), (3x4+y+4)/x7

should be an algebraic function.These functions will have values at points, but won’t be determined by their

values at points. But like all sheaves, they will indeed be determined by their germs.This is discussed in Section 6.1.4.

It suffices to describe the structure sheaf as a sheaf (of rings) on the base ofdistinguished open sets (Theorem 4.7.1 and Exercise 5.7.A). Our strategy is asfollows. We will define the sections on the base by 2

(21) OSpecA(D(f)) = Af

We need to make sure that this is well-defined, i.e. that we have a natural isomor-phism Af → Ag if D(f) = D(g). We will define the restriction maps resD(g),D(f)

as follows. If D(f) ⊂ D(g), then we have shown that D(fg) = D(f). There is anatural map Ag → Afg given by r/gm 7→ (rfm)/(fg)m, and we will define

resD(g),D(fg)=D(f) : OSpecA(D(g))→ OSpecA(D(fg))

to be this map. But it will be cleaner to state things a little differently.If D(f) ⊂ D(g), then by Exercise 5.7.E, g is a unit in Af Thus by the universal

property of localization, there is a natural map Ag → Af which we temporarilydenote resg,f , but which we secretly think of as resD(g),D(f). If D(f) ⊂ D(g) ⊂D(h), then these restriction maps commute:3

(22) Ah

resh,g //

resh,f AAA

AAAA

AAg

resg,f~~

Af

commutes. (The map Ah → Af is defined by universal property, and the composi-tion resg,f resh,g satisfies this universal property.)

1s:ss

2Odf

3dfdgdh

117

118

In particular, if D(f) = D(g), then resg,f resf,g is the identity on Af , (takeh = f in the above diagram (22)), and similarly resf,g resg,f = idAg . Thus we candefine OSpecA(D(f)) = Af , and this is well-defined (independent of the choice off).

By (22), we have defined a presheaf on the distinguished base.[forme: Joe suggested an alternate definition: OSpecA(D(f)) should consist of the

localization of A at the multiplicative set of all functions that vanish only in V (f),

the complement of D(f). Then it is immediate from the definition that this depends

only on D(f) and not just on f , but you’ll still want to verify that this is canonically

Af .]

[primordial: If D(f) ⊂ D(g), then we have shown that fn ∈ (g), i.e. we can writefn = ag by Exercise 5.7.E. There is a natural map Ag → Af given by r/gm 7→ (ram)/(fmn),and we define

resD(g),D(f) : OSpecA(D(g)) →OSpecA(D(f))

to be this map. Exercise: (a) Verify that (21) is well-defined, i.e. if D(f) = D(f ′) then Af iscanonically isomorphic to Af ′ .(b) Show that resD(g),D(f) is well-defined, i.e. that it is independent of the choice of a and n, and

if D(f) = D(f ′) and D(g) = D(g′), then

Ag

∼

resD(g),D(f) // Af

∼

Ag′

resD(g′),D(f′) // Af ′

commutes. 4 (Possible hint: if D(f) ⊂ D(g), use Ag → Af → Afg .) ]We now come to a key theorem.

6.1.1. Theorem. — The data just described gives a sheaf on the distinguishedbase, and hence determines a sheaf on the topological space Spec A. 5

This sheaf is called the structure sheaf, and will be denoted OSpecA, orstructure sheaf on

SpecA sometimes O if the scheme in question is clear from the context. Such a topologicalspace, with sheaf, will be called an affine scheme. The notation Spec A willaffine scheme

hereafter denote the data of a topological space with a structure sheaf.[primordial: EXAMPLES AT THIS POINT JUST CONFUSED THEM. Before we

begin the proof, you may want to try a couple of examples such as Spec C[x] and Spec Z, to

see that nothing fancy is going on. [forme: [This isn’t appropriate yet...] Later, when

teaching quasicoherent sheaves, I realized that there are at least two other ways to go

here. Both rely on showing that 0→M → QMfi

→QMfifj

is exact. One shows this

either by checking stalk-locally [word not yet defined] or by checking affine-locally. I

like the stalk-local argument best (right now). It requires knowing that if M is 0 at

all primes, then it is 0. This boils down to showing that if m ∈ M is 0 at all primes,

then it is 0. This turns into a partition of unity argument, but a simpler one. (I

should then mention that one need only check at the maximal primes. I should also

emphasize the geometry.) The affine-local argument is similar, and the partition-of-

unity argument requires “powers”. You take this sequence, and localize at f1. Then

you show that the resulting thing can be obtained from a double complex, each row

of which is exact by the inductive hypothesis.] ]

4ssrm

5OSpecA

119

Proof. We first check identity on the base. We deal with the case of a cover ofthe entire space A, and let you verify that essentially the same argument holds fora cover of some Af . Suppose that Spec A = ∪i∈ID(fi) where i runs over someindex set I . Then there is some finite subset of I , which we name 1, . . . , n,such that Spec A = ∪n

i=1D(fi), i.e. (f1, . . . , fn) = A (quasicompactness of Spec A,Exercise 5.7.C). Suppose we are given s ∈ A such that resSpecA,D(fi) s = 0 inAfi for all i. (We wish to show that s = 0.) Hence there is some m such thatfor each i ∈ 1, . . . , n, fm

i s = 0. [forme: When teaching: A → Af . What goes

to 0? Precisely things killed by some power of f .] Now (fm1 , . . . , fm

n ) = A(Spec A = ∪D(fi) = ∪D(fm

i )), so there are ri ∈ A with∑n

i=1 rifmi = 1 in A, from

which

s =(

∑

rifmi

)

s =∑

ri(fmi s) = 0.

Thus we have checked the “base identity” axiom for Spec A. (Serre has describedthis as a “partition of unity” argument, and if you look at it in the right way, hisinsight is very enlightening.)

6.1.A. Exercise. Make the tiny changes to the above argument to show baseidentity for any distinguished open D(f). (Possible strategy: show that the argu-ment is the same as the previous argument for Spec Af .) 6

ex.doneWe next show base gluability. As with base identity, we deal with the case

where we wish to glue sections to produce a section over Spec A. As before, weleave the general case where we wish to glue sections to produce a section overD(f) to the reader (Exercise 6.1.B).

Suppose ∪i∈ID(fi) = Spec A, where I is a index set (possibly horribly un-countably infinite). Suppose we are given elements in each Afi that agree on theoverlaps Afifj . (Note that intersections of distinguished open sets are also distin-guished open sets.)

Aside: experts might realize that we are trying to show exactness of

0→ A→∏

i

Afi →∏

i6=j

Afifj .

(What is the right-hand map?) Base identity corresponds to injectivity at A. Thecomposition of the right two morphisms is trivially zero, and gluability is verifyingexactness at

∏

i Afi .Choose a finite subset 1, . . . , n ⊂ I with (f1, . . . , fn) = A (i.e. use quasi-

compactness of Spec A to choose a finite subcover by D(fi)). We have elements

ai/f lii ∈ Afi agreeing on overlaps Afifj . Letting gi = f li

i , using D(fi) = D(gi), wecan simplify notation by considering our elements as of the form ai/gi ∈ Agi .

The fact that ai/gi and aj/gj “agree on the overlap” (i.e. in Agigj ) means thatfor some mij ,

(gigj)mij (gjai − giaj) = 0

in A. By taking m = maxmij (here we use the finiteness of I), we can simplifynotation:

(gigj)m(gjai − giaj) = 0

6baseidentitydistinguished

120

for all i, j. Let bi = aigmi for all i, and hi = gm+1

i (so D(hi) = D(gi)). Then wecan simplify notation even more: on each D(hi), we have a function bi/hi, and theoverlap condition is hjbi − hibj = 0

Now ∪iD(hi) = A, implying that 1 =∑n

i=1 rihi for some ri ∈ A. Definer =

∑

ribi. This will be the element of A that restricts to each bj/hj . Indeed,

rhj − bj =∑

i

ribihj −∑

i

bjrihi =∑

i

ri(bihj − bjhi) = 0.

We are not quite done! We are supposed to have something that restricts toai/f li

i for all i ∈ I , not just i = 1, . . . , n. But a short trick takes care of this. Wenow show that for any α ∈ I−1, . . . , n, r restricts to the desired element aα Afα .Repeat the entire process above with 1, . . . , n, α in place of 1, . . . , n, to obtainr′ ∈ A which restricts to αα for i ∈ 1, . . . , n, α. Then by base identity, r′ = r.(Note that we use base identity to prove base gluability. This is an example of howbase identity is “prior” to base gluability.) Hence r restricts to aα/f lα

α as desired.

6.1.B. Exercise. Alter this argument appropriately to show base gluability forany distinguished open D(f). 7 [forme: Same trick as Exercise 6.1.A.]

ex.doneWe have now completed the proof of Theorem 6.1.1.

The proof of Theorem 6.1.1 immediately generalizes, as the following exerciseshows. This exercise will be essential for the definition of a quasicoherent sheaflater on [say where].

6.1.C. Important exercise/definition. Suppose M is an A-module. Show

that the following construction describes a sheaf M on the distinguished base. ToD(f) we associate Mf = M⊗AAf ; the restriction map is the “obvious” one. 8 ThisM

is an OSpecA-module! This sheaf M will be very important soon; it is an exampleof a quasicoherent sheaf. [forme: We get a natural bijection: rings, modules ↔qcoh sheaf

Affine schemes, M.]ex.done

Here is a useful fact for later: As a consequence, note that if (f1, . . . , fr) = A,we have identified M with a specific submodule of Mf1 × · · · ×Mfr . Even thoughM →Mfi may not be an inclusion for any fi, M →Mf1×· · ·×Mfr is an inclusion.We don’t care yet, but we’ll care about this later, and I’ll invoke this fact. (Reason:we’ll want to show that if M has some nice property, then Mf does too, which willbe easy. We’ll also want to show that if (f1, . . . , fn) = R, then if Mfi have thisproperty, then M does too.)

6.1.2. Definition. We can now define scheme in general. First, define anscheme

isomorphism of ringed spaces (X,OX) and (Y,OY ) as (i) a homeomorphismf : X → Y , and (ii) an isomorphism of sheaves OX and OY , considered to be on thesame space via f . (Condition (ii), more precisely: an isomorphism OX → f−1OY

of sheaves on X , or f∗OX → OY of sheaves on Y .) In other words, we have acorrespondence of sets, topologies, and structure sheaves. An affine scheme is aringed space that is isomorphic to (Spec A,OSpecA). A scheme (X,OX) is a ringed

7affinebaseglue

8Mtilde

121

space such that any point x ∈ X has a neighborhood U such that (U,OX |U ) is anaffine scheme. The scheme can be denoted (X,OX), although it is often denotedX , with the structure sheaf implicit.

An isomorphism of two schemes (X,OX) and (Y,OY ) is an isomorphismas ringed spaces. 9 isomorphism of schemes

6.1.3. Remark. From this definition of the structure sheaf on an affine scheme,several things are clear. First of all, if we are told that (X,OX) is an affine scheme,we may recover its ring (i.e. find the ring A such that Spec A = X) by taking thering of global sections, as X = D(1), so:

Γ(X,OX) = Γ(D(1),OSpecA) as D(1) = Spec A

= A1 (i.e. allow 1’s in the denominator) by definition

= A.

(You can verify that we get more, and can “recognize X as the scheme Spec A”:we get a natural isomorphism f : (Spec Γ(X,OX),OSpecΓ(X,OX)) → (X,OX). Forexample, if m is a maximal ideal of Γ(X,OX), f([m]) = V (m).) More generally,given f ∈ A, Γ(D(f),OSpecA) ∼= Af . Thus under the natural inclusion of setsSpec Af → Spec A, the Zariski topology on Spec A restricts to give the Zariskitopology on Spec Af (Exercise 5.4.H), and the structure sheaf of Spec A restrictsto the structure sheaf of Spec Af , as the next exercise shows.

6.1.D. Important but easy exercise. Suppose f ∈ A. Show that underthe identification of D(f) in Spec A with Spec Af , there is a natural isomorphismof sheaves (D(f),OSpecA|D(f)) ∼= (Spec Af ,OSpecAf

). [forme: This is because

distinguished open sets of SpecRf are already distinguished open sets in SpecR. If

r ∈ R, then D(r/fn) in SpecRf is D(rf) in SpecR.] 10

6.1.E. Exercise. Show that if X is a scheme, then the affine open sets forma base for the Zariski topology. [forme: Warning: they don’t form a nice base,

as we’ll see in Exercise 6.1.J below. However, in “most nice situations” this will be

true, as we will later see, when we define the analogue of “Hausdorffness”, called

separatedness.] separatedness

6.1.F. Exercise. If X is a scheme, and U is any open subset, prove that(U,OX |U ) is also a scheme. [forme: Answer: [draw picture] given any point x ∈ U ,

there is an affine open set SpecA ⊂ X containing x, by the definition of a scheme.

Because the distinguished open sets of SpecA form a base for the topology, there is a

distinguished open SpecAf containing x and contained in U . In other words, we have

verified that U satisfies the definition of a scheme.]ex.done

(U,OX |U ) is called an open subscheme of U . If U is also an affine scheme, open subscheme

we often say U is an affine open subset, or an affine open subscheme, orsometimes informally just an affine open. For an example, D(f) is an affine opensubscheme of Spec A.

6.1.4. Stalks of the structure sheaf: germs, and values at a point. Likeevery sheaf, the structure sheaf has stalks, and we shouldn’t be surprised if they

9isoscheme

10DfAffine

122

are interesting from an algebraic point of view. In fact, we have seen them before.11value of function at a

point, germ near a point6.1.G. Important exercise. Show that the stalk of OSpecA at the point [p] isthe ring Ap. [forme: (Hint: use distinguished open sets in the direct limit you use to

define the stalk. In the course of doing this, you’ll discover a useful principle. In the

concrete definition of stalk, the elements were sections of the sheaf over some open

set containing our point, and two sections over different open sets were considered

the same if they agreed on some smaller open set. In fact, you can just consider

elements of your base when doing this. This is called a cofinal system in the directed

set.) This is yet another reason to like the notion of a sheaf on a base. [I shouldcofinal system

probably talk about cofinal systems clearly at some point, possibly here. Justin says

that this is defined for both direct and inverse systems, and gives as refs Dugundji’s

topology, and probably Rotman’s Advanced Modern Algebra.]]ex.done

Essentially the same argument will show that the stalk of the sheaf M , definedin Exercise 6.1.C at [p] is Mp. Here is an interesting consequence, or if you prefer,a geometric interpretation of an algebraic fact. A section is determined by it stalks(Exercise 4.4.A), meaning that M → ∏

p Mp is an inclusion. So for example anA-module is zero if and only if all its localizations at primes are zero.

The residue field of a scheme at a point is the local ring modulo itsmaximal ideal.residue field of a scheme

at a point So now we can make some of our vague discussion in the examples of §5.3precise. Suppose [p] is a point in some open set U of Spec A. For example, sayA = k[x, y], p = [(x)] [draw picture], and U = A2 − (0, 0).

Then a function on U , i.e. a section of OSpecA over U , has a germ near [p],which is an element of Ap. This stalk Ap is a local ring, with maximal ideal pAp.In our example, consider the function (3x4 + x2 + xy + y2)/(3x2 + xy + y2 + 1),which is defined on the open set D(3x2 +xy +y2 +1). Because there are no factorsof x in the denominator, it is indeed in Ap.

A germ has a value at [p], which is an element of Ap/pAp. This is isomorphicto FF(A/p), the fraction field of the quotient domain. It is useful to note thatlocalization at p and taking quotient by p “commute”, i.e. the following diagramcommutes.

Ap

''PPPPPPPPPPPPP

A

>>

AAA

AAAA

A Ap/pAp = FF (A/p)

A/p

FF (·)

77nnnnnnnnnnnnn

So the value of a function at a point always takes values in a field. In our example,to see the value of our germ at x = 0, we simply set x = 0. So we get thevalue y2/(y2 + 1), which is certainly in FF(k[y]). (If you think you care only aboutcomplex schemes, and hence only about algebraically closed fields, let this be a firstwarning: Ap/pAp won’t be algebraically closed in general, even if A is a finitelygenerated C-algebra!)

11valueatpoint

123

We say that the germ vanishes at p if the value is zero. In our example, thegerm doesn’t vanish at p.

If anything makes you nervous, you should make up an example to assuageyour nervousness. (Example: 27/4 is a regular function on Spec Z − [(2)], [(7)].What is its value at [(5)]? Answer: 2/(−1) ≡ −2 (mod 5). What is its value atthe generic point [(0)]? Answer: 27/4. Where does it vanish? At [(3)].)

We now give three extended examples. Our short term goal is to see that wecan really work with this sheaf, and can compute the ring of sections of interestingopen sets that aren’t just distinguished open sets of affine schemes. Our long-termgoal is to see interesting examples that will come up repeatedly in the future. Allthree examples are non-affine schemes, so these examples are genuinely new to us.

6.1.5. Example: The plane minus the origin. I now want to work throughan example with you, to show that this distinguished base is indeed something thatyou can work with. Let A = k[x, y], so Spec A = A2

k. If you want, you can let k beC, but that won’t be relevant. Let’s work out the space of functions on the openset U = A2 − (0, 0).12

It is a non-obvious fact that you can’t cut out this set with a single equation,so this isn’t a distinguished open set. We’ll see why fairly soon [forme: where?]

. But in any case, even if we’re not sure that this is a distinguished open set, wecan describe it as the union of two things which are distinguished open sets. IfI throw out the x axis, i.e. the set y = 0, I get the distinguished open set D(y).If I throw out the y axis, i.e. x = 0, I get the distinguished open set D(x). SoU = D(x)∪D(y). (Remark: U = A2−V (x, y) and U = D(x)∪D(y). Coincidence?I think not!) We will find the functions on U by gluing together functions on D(x)and D(y).

What are the functions on D(x)? They are, by definition, Ax = k[x, y, 1/x].In other words, they are things like this: 3x2 + xy + 3y/x + 14/x4. What arethe functions on D(y)? They are, by definition, Ay = k[x, y, 1/y]. Note thatA → Ax, Ay. This is because x and y are not zero-divisors. (A is an integral domain— it has no zero-divisors, besides 0 — so localization is always an inclusion.) Sowe are looking for functions on D(x) and D(y) that agree on D(x)∩D(y) = D(xy),i.e. they are just the same Laurent polynomial. Which things of this first form arealso of the second form? Just old-fashioned polynomials — 13

(23) Γ(U,OA2) ≡ k[x, y].

In other words, we get no extra functions by throwing out this point. Notice howeasy that was to calculate!

6.1.6. (Aside: Notice that any function on A2 − (0, 0) extends over all of A2.This is an analog of Hartogs’ Lemma in complex geometry: you can extend aholomorphic function defined on the complement of a set of codimension at leasttwo on a complex manifold over the missing set. This will work more generally inthe algebraic setting: you can extend over points in codimension at least 2 not onlyif they are smooth, but also if they are mildly singular — what we will call normal.We will make this precise in §15.6.6. This fact will be very useful for us.) 14 normal

12planeminusorigin

13oldfashioned

14firstHartogs

124

We can now verify an interesting fact: (U,OA2 |U ) is a scheme, but it is not anaffine scheme. Here’s why: otherwise, if (U,OA2 |U ) = (Spec A,OSpecA), then wecan recover A by taking global sections:

A = Γ(U,OA2 |U ),

which we have already identified in (23) as k[x, y]. So if U is affine, then U = A2k.

But we get more: we can recover the points of Spec A by taking the primes of A.In particular, the prime ideal (x, y) of A should cut out a point of Spec A. Buton U , V (x) ∩ V (y) = ∅. Conclusion: U is not an affine scheme. (If you are everlooking for a counterexample to something, and you are expecting one involving anon-affine scheme, keep this example in mind!)

You’ve seen two examples of non-affine schemes: an infinite disjoint union ofnon-empty schemes (Exercise 6.1.N), and now A2−(0, 0). I want to give you twomore important examples. They are important because they are the first examplesof fundamental behavior, the first pathological, and the second central.

First, I need to tell you how to glue two schemes together. And before that,you should review how to glue topological spaces together along isomorphic opensets. Given two topological spaces X and Y , and open subsets U ⊂ X and V ⊂ Yalong with a homeomorphism U ∼= V , we can create a new topological space W ,that we think of as gluing X and Y together along U ∼= V . It is the quotient ofthe disjoint union X

∐

Y by the equivalence relation U ∼= V , where the quotientis given the quotient topology. Then X and Y are naturally (identified with) opensubsets of W , and indeed cover W . Can you restate this with an arbitrary (notnecessarily finite) number of topological spaces?

Now that we have discussed gluing topological spaces, let’s glue schemes to-gether. Suppose you have two schemes (X,OX) and (Y,OY ), and open sub-

sets U ⊂ X and V ⊂ Y , along with a homeomorphism f : U∼ // V , and an

isomorphism of structure sheaves OX∼= f∗OY (i.e. an isomorphism of schemes

(U,OX |U ) ∼= (V,OY |V )). Then we can glue these together to get a single scheme.Reason: let W be X and Y glued together using the isomorphism U ∼= V . ThenExercise 4.7.D shows that the structure sheaves can be glued together to get a sheafof rings. Note that this is indeed a scheme: any point has a neighborhood that isan affine scheme. (Do you see why?)

6.1.H. Exercise. For later reference, show that you can glue together an arbi-trary number of schemes together. Suppose we are given:

• schemes Xi (as i runs over some index set I , not necessarily finite),• open subschemes Xij ⊂ Xi,• isomorphisms fij : Xij → Xji with fii the identity• (the cocycle condition) such that the isomorphisms “agree along triple

intersections”, i.e. fik|Xij∩Xik= fjk|Uji∩Ujk

fij |Xij∩Xik.

(The cocycle condition ensures that fij and fji are inverses.) Show that there is aunique scheme X (up to unique isomorphism) along with open subset isomorphicto Xi respecting this gluing data in the obvious sense. 15

ex.done

15e:gluingschemes

125

I’ll now give you two non-affine schemes. In both cases, I will glue togethertwo copies of the affine line A1

k. Again, if it makes you feel better, let k = C, butit really doesn’t matter.

Let X = Spec k[t], and Y = Spec k[u]. Let U = D(t) = Spec k[t, 1/t] ⊂ X andV = D(u) = Spec k[u, 1/u] ⊂ Y . We will get both examples by gluing X and Ytogether along U and V . The difference will be in how we glue.

6.1.7. Extended example: the affine line with the doubled origin. Con-sider the isomorphism U ∼= V via the isomorphism k[t, 1/t] ∼= k[u, 1/u] given byt ↔ u. Let the resulting scheme be X . This is called the affine line with doubledorigin. Figure 1 is a picture of it. 16 affine line with doubled

origin

Figure 1. The affine line with doubled origin

As the picture suggests, intuitively this is an analogue of a failure of Hausdorff-ness. A1 itself is not Hausdorff, so we can’t say that it is a failure of Hausdorffness.We see this as weird and bad, so we’ll want to make up some definition that willprevent this from happening. This will be the notion of separatedness. This will separated

answer other of our prayers as well. For example, on a separated scheme, the “affinebase of the Zariski topology” is nice — the intersection of two affine open sets willbe affine.

6.1.I. Exercise. Show that X is not affine. Hint: calculate the ring of globalsections, and look back at the argument for A2 − (0, 0).

6.1.J. Exercise. Do the same construction with A1 replaced by A2. You’ll havedefined the affine plane with doubled origin. Use this example to show that theaffine base of the Zariski topology isn’t a nice base, by describing two affine opensets whose intersection is not affine. 17

6.1.8. Example 2: the projective line. Consider the isomorphism U ∼= V viathe isomorphism k[t, 1/t] ∼= k[u, 1/u] given by t ↔ 1/u. Figure 2 is a suggestivepicture of this gluing. Call the resulting scheme the projective line over thefield k, P1

k. projective line P1

Notice how the points glue. Let me assume that k is algebraically closed forconvenience. (You can think about how this changes otherwise.) On the first affineline, we have the closed (= “old-fashioned”) points [(t − a)], which we think of as“a on the t-line”, and we have the generic point. On the second affine line, we haveclosed points that are “b on the u-line”, and the generic point. Then a on the t-lineis glued to 1/a on the u-line (if a 6= 0 of course), and the generic point is gluedto the generic point (the ideal (0) of k[t] becomes the ideal (0) of k[t, 1/t] uponlocalization, and the ideal (0) of k[u] becomes the ideal (0) of k[u, 1/u]. And (0) ink[t, 1/t] is (0) in k[u, 1/u] under the isomorphism t↔ 1/u).

16linewithdoubledorigin

17afftopnotnice

126

Figure 2. Gluing two affine lines together to get P1

We can interpret the closed (“old-fashioned”) points of P1 in the following way,which may make this sound closer to the way you have seen projective space definedearlier. The points are of the form [a; b], where a and b are not both zero, and [a; b]is identified with [ac; bc] where c ∈ k∗. Then if b 6= 0, this is identified with a/b onthe t-line, and if a 6= 0, this is identified with b/a on the u-line.

6.1.9. Proposition. — P1k is not affine.

ex.done

Proof. We do this by calculating the ring of global sections.The global sections correspond to sections over X and sections over Y that

agree on the overlap. A section on X is a polynomial f(t). A section on Y is apolynomial g(u). If I restrict f(t) to the overlap, I get something I can still callf(t); and ditto for g(u). Now we want them to be equal: f(t) = g(1/t). How manypolynomials in t are at the same time polynomials in 1/t? Not very many! Answer:only the constants k. Thus Γ(P1,OP1) = k. If P1 were affine, then it would beSpec Γ(P1,OP1) = Spec k, i.e. one point. But it isn’t — it has lots of points.

Note that we have proved an analog of a theorem: the only holomorphic func-tions on CP1 are the constants!

6.1.10. Important example: Projective space. We now make a preliminarydefinition of projective n-space Pn

k , by gluing together n + 1 open sets each iso-morphic to An

k . Judicious choice of notation for these open sets will make our lifeeasier. Our motivation is as follows. In the construction of P1 above, we thoughtof points of projective space as [x0; x1], where (x0, x1) are only determined up toscalars, i.e. (x0, x1) is considered the same as (λx0, λx1). Then the first patch canbe interpreted by taking the locus where x0 6= 0, and then we consider the points[1; t], and we think of t as x1/x0; even though x0 and x1 are not well-defined, x1/x0

is. The second corresponds to where x1 6= 0, and we consider the points [u; 1], andwe think of u as x0/x1. It will be useful to instead use the notation x1/0 for t and

x0/1 for u. 18

For Pn, we glue together n + 1 open sets, one for each of i = 0, . . . , n + 1. Theith open set Ui will have co-ordinates x0/i, . . . , x(i−1)/i, x(i+1)/i, . . . , xn/i. It will

18projpatches

127

be convenient to write this as

Spec k[x0/i, x1/i, . . . , xn/i]/(xi/i − 1)

(so we have introduced a “dummy variable” xi/i which we set to 1). We glue thedistinguished open set D(xj/i) of Ui to the distinguished open set D(xi/j) of Uj ,by identifying these two schemes by describing the identification of rings

Spec k[x0/i, x1/i, . . . , xn/i, 1/xj/i]/(xi/i − 1) ∼=Spec k[x0/j , x1/j , . . . , xn/j , 1/xi/j ]/(xj/j − 1)

via xk/i = xk/j/xi/j and xk/j = xk/i/xj/i (which implies xi/jxj/i = 1). We need

to check that this gluing information agrees over triple overlaps. 19

6.1.K. Exercise. Check this, as painlessly as possible. (Possible hint: the tripleintersection is affine; describe the corresponding ring.)

ex.doneNote that our definition doesn’t use the fact that k is a field. Hence we may as

well define PnA for any ring A. This will be useful later. 20 PnA

6.1.L. Exercise. Show that the only global sections of the structure sheaf areconstants, and hence that Pn

k is not affine if n > 0. (Hint: you might fear that youwill need some delicate interplay among all of your affine open sets, but you willonly need two of your open sets to see this. There is even some geometric intuitionbehind this: the complement of the union of two open sets has codimension 2. But“Algebraic Hartogs’ Lemma” (discussed informally in 6.1.6, to be stated rigorouslyin 15.6.6) says that any function defined on this union extends to be a function onall of projective space. Because we’re expecting to see only constants as functionson all of projective space, we should already see this for this union of our two affineopen sets.) 21 Pnk , proj. space

Algebraic Hartogs’

Lemma6.1.M. Exercise. Show that if k is algebraically closed, the closed points ofPn

k may be interpreted in the same way as we interpreted the points of P1k. (The

points are of the form [a0; . . . ; an], where the ai are not all zero, and [a0; . . . ; an] isidentified with [λa0; . . . ; λan] where λ ∈ k∗.) [forme: [Define projective coordinates

on projective space? We’ll make this more precisely later.]] 22 projective coordinates

ex.doneWe will later give another definition of projective space [forme: Add ref] .

Our definition (from §6.1.10) will often be handy for computing things. But thereis something unnatural about it — projective space is highly symmetric, and thatisn’t clear from your point of view.

[forme: At some point, remark that in each of the above examples, life was easy,

as restriction maps are inclusions.]

6.1.11. Fun aside: The Chinese Remainder Theorem is a geometric fact.I want to show you that the Chinese Remainder theorem is embedded in whatwe’ve done, which shouldn’t be obvious to you. I’ll show this by example. TheChinese Remainder Theorem says that knowing an integer modulo 60 is the sameas knowing an integer modulo 3, 4, and 5. Here’s how to see this in the language

19d1:projectivespace

20d:PnA

21Pnglobalsections

22Pnkbar

128

of schemes. What is Spec Z/(60)? What are the primes of this ring? Answer:those prime ideals containing (60), i.e. those primes dividing 60, i.e. (2), (3), and(5). Figure 3 is a sketch of Spec Z/(60). They are all closed points, as these areall maximal ideals, so the topology is the discrete topology. What are the stalks?You can check that they are Z/4, Z/3, and Z/5. My The nilpotents “at (2)” areindicated by the “fuzz” on that point. So what are global sections on this scheme?They are sections on this open set (2), this other open set (3), and this third openset (5). In other words, we have a natural isomorphism of ringsChinese Remainder

Theorem Z/60→ Z/4× Z/3× Z/5.

[forme: Clean this up]

[(5)][(2)] [(3)]

Figure 3. A picture of the scheme Spec Z/(60)

On a related note:

6.1.N. Exercise. (a) Show that the disjoint union of a finite number of affineschemes is also an affine scheme.23 (Hint: say what the ring is.)(b) Show that an infinite disjoint union of (non-empty) affine schemes is not anaffine scheme. 24

6.1.12. ? Example. Here is an example of a function on an open subset of ascheme that is a bit surprising. On X = Spec k[w, x, y, z]/(wx − yz), consider theopen subset D(y)∪D(w). Show that the function x/y on D(y) agrees with z/w onD(w) on their overlap D(y)∩D(w). Hence they glue together to give a section. Youmay have seen this before when thinking about analytic continuation in complexgeometry — we have a “holomorphic” function which has the description x/y on anopen set, and this description breaks down elsewhere, but you can still “analyticallycontinue” it by giving the function a different definition on different parts of thespace.

Follow-up for curious experts:25 This function has no “single description” as acone over smooth

quadric surface well-defined expression in terms of w, x, y, z! There is lots of interesting geometryhere. This example will be a constant source of interesting examples for us. Wewill later recognize it as the cone over the quadric surface. Here is a glimpse,in terms of words we have not yet defined. Spec k[w, x, y, z] is A4, and is, notsurprisingly, 4-dimensional. We are looking at the set X , which is a hypersurface,and is 3-dimensional. It is a cone over a smooth quadric surface in P3 [show themhyperboloid of one sheet, and point out the two rulings, I have a fig at 1]. D(y)is X minus some hypersurface, so we are throwing away a codimension 1 locus.D(z) involves throwing another codimension 1 locus. You might think that theirintersection is then codimension 2, and that maybe failure of extending this weird

23finiteunionofaffines, hint is referred to

24infinitedisjointunionnotaffine

25coneoverquadric

129

function to a global polynomial comes because of a failure of our Hartogs’ Lemma-type theorem, which will be a failure of normality. But that’s not true — V (y)∩V (z)is in fact codimension 1 — so no Hartogs-type theorem holds. Here is what isactually going on. V (y) involves throwing away the (cone over the) union of twolines l and m1, one in each “ruling” of the surface, and V (z) also involves throwingaway the (cone over the) union of two lines l and m2. The intersection is the(cone over the) line l, which is a codimension 1 set. Neat fact: despite being “purecodimension 1”, it is not cut out even set-theoretically by a single equation. (It ishard to get an example of this behavior. This example is the simplest example Iknow.) This means that any expression f(w, x, y, z)/g(w, x, y, z) for our functioncannot correctly describe our function on D(y) ∪D(z) — at some point of D(y) ∪D(z) it must be 0/0. Here’s why. Our function can’t be defined on V (y) ∩ V (z),so g must vanish here. But g can’t vanish just on the cone over l — it must vanishelsewhere too. (For the experts among the experts: here is why the cone over lis not cut out set-theoretically by a single equation. If l = V (f), then D(f) isaffine. Let l′ be another line in the same ruling as l, and let C(l) (resp. l′) be thecone over l (resp. l′). Then C(l′) can be given the structure of a closed subschemeof Spec k[w, x, y, z], and can be given the structure of A2. Then C(l′) ∩ V (f) is aclosed subscheme of D(f). Any closed subscheme of an affine scheme is affine. Butl ∩ l′ = ∅, so the cone over l intersects the cone over l′ in a point, so C(l′) ∩ V (f)is A2 minus a point, which we’ve seen is not affine, so we have a contradiction.)

CHAPTER 7

Some properties of schemes

1

7.1 Topological properties

We will now going to define some useful properties of schemes. The definitionsof irreducible, irreducible component, closed point, specialization, generalization,generic point, connected, connected component, and quasicompact were given in§5.5–5.7. You should have pictures in your mind of each of these notions. irreducible, closed

point, specialization,

generalization, generic

point, connected, qua-

sicompact irreducible

components

Exercise 5.5.B shows that An is irreducible (it was easy). This argument “be-haves well under gluing”, yielding:

7.1.A. Exercise. Show that Pnk is irreducible.

7.1.B. Exercise. Exercise 5.6.E showed that there is a bijection between irre-ducible closed subsets and points. Show that this is true of schemes as well.

7.1.C. Exercise. Prove that if X is a scheme that has a finite cover X =∪n

i=1 Spec Ai where Ai is Noetherian, then X is a Noetherian topological space.(We will soon call such a scheme a Noetherian scheme, §7.3.5.)2 [forexperts: Is Noetherian scheme

there a counterexample to converse?]ex.done

Thus Pnk and Pn

Z are Noetherian topological spaces: we built them by gluingtogether a finite number of Spec’s of Noetherian rings.

7.1.D. Easy exercise. Show that a scheme X is quasicompact if and only ifit can be written as a finite union of affine schemes (Hence Pn

k is quasicompact.)[forme: Answer: Ex. 5.5.D]

7.1.E. Exercise: Quasicompact schemes have closed points. Show that ifX is a nonempty quasicompact scheme, then it has a closed point. (Warning: thereexist non-empty schemes with no closed points, so your argument had better usethe quasicompactness hypothesis! [forme: [Refer to Karl Schwede’s article. Is the

example locally Noetherian? Is there a locally Noetherian scheme where the closed

points aren’t dense?]] 3 We will see that in good situations, the closed points aredense, Exercise 7.3.I.) [forme: There are four references to closed points: 7.4.D,

7.3.I, 7.1.E, 16.4.A. I should smooth this discussion out.]

1propertiesofaffines

2Nschemetopspace

3qcclosedpoints

131

132

7.1.1. Quasiseparatedness.4Quasiseparatedness is a weird notion that comes in handy for certain kinds of

people. Most people, however, can ignore this notion. A scheme is quasiseparatedif the intersection of any two quasicompact sets is quasicompact, or equivalently,qsep

if the intersection of any two affine open subsets is a finite union of affine opensubsets.

7.1.F. Short exercise. Prove this equivalence.ex.done

We will see later that this will be a useful hypothesis in theorems (in conjunc-tion with quasicompactness), and that various interesting kinds of schemes (affine,locally Noetherian, separated, see Exercise 7.1.G, Exercise 7.3.C, and [forme: ???]

resp.) are quasiseparated, and this will allow us to state theorems more succinctly(e.g. “if X is quasicompact and quasiseparated” rather than “if X is quasicompact,and either this or that or the other thing hold”). [forme: Give refs for all this]

[forme: locally Noetherian schemes; affine schemes; separated schemes. really this

will be a hypothesis on morphisms]

7.1.G. Exercise. Show that affine schemes are quasiseparated. 5

“Quasicompact and quasiseparated” means something rather down to earth:

7.1.H. Exercise. Show that a scheme X is quasicompact and quasiseparated ifand only if X can be covered by a finite number of affine open subsets, any two ofwhich have intersection also covered by a finite number of affine open subsets.6

[primordial: One very important topological notion is dimension. (It is likely not clear

to you that this is a topological idea.) But this is more difficult, so we will put it off until later. ]

7.2 Reducedness and integrality

Recall that one of the alarming things about schemes is that functions are notdetermined by their values at points, and that was because of the presence ofnilpotents (§5.3.5).nilpotents

7.2.1. Definition. Recall that a ring is reduced if it has no nonzero nilpotents(§5.3.7). A scheme X is reduced if OX(U) has no nonzero nilpotents for any openset U of X . 7reduced

An example of a nonreduced affine scheme is Spec k[x, y]/(y2, xy). A usefulrepresentation of this scheme is given in Figure 1, although we will only explain in§7.5 why this is a good picture. The fuzz indicates that there is some nonreduced-ness going on at the origin. Here are two different functions: x and x + y. Theirvalues agree at all points (all closed points [(x − a, y)] = (a, 0) and at the genericpoint [(y)]). They are actually the same function on the open set D(x), which isnot surprising, as D(x) is reduced, as the next exercise shows. (This explains whythe fuzz is only at the origin, where y = 0.)

4s:qsepscheme

5affqsep

6qcqsdowntoearth

7d:reduced

133

Figure 1. A picture of the scheme Spec k[x, y]/(xy, y2)

7.2.A. Exercise. Show that(

k[x, y]/(y2, xy))

xhas no nilpotents. (Possible hint:

show that it is isomorphic to another ring, by considering the geometric picture.)[forme: Refer back to when we looked at D(y) on the union of axes — that was an

earlier exercise. Where did it go?] 8

7.2.B. Exercise (reducedness is stalk-local). Show that a scheme is re-duced if and only if none of the stalks have nilpotents. Hence show that if f and gare two functions on a reduced scheme that agree at all points, then f = g. (Twohints: OX(U) →∏

x∈U OX,x from Exercise 4.4.A, and the nilradical is intersection

of all prime ideals from Theorem 5.3.6.) 9

ex.done

reduced (is stalk-local)Warning: if a scheme X is reduced, then it is immediate from the definitionthat its ring of global sections is reduced. However, the converse is not true; wewill meet an example in [forme: ADD LATER] .

7.2.C. Exercise. Suppose X is quasicompact, and f is a function (a globalsection of OX) that vanishes at all points of x. Show that there is some n suchthat fn = 0. Show that this may fail if X is not quasicompact. (This exerciseis less important, but shows why we like quasicompactness, and gives a standardpathology when quasicompactness doesn’t hold.)10 Hint: take an infinite disjointunion of Spec An with An := k[ε]/εn.

ex.done[primordial:[forme: CUT: Perhaps include? Perhaps hard! If R is a reduced scheme, and U

is an open subset of X, and f and g are two sections of the structure sheaf over U

that agree on a dense open subset of U , then f = g. We’ll generalize this property:

“separated”. We can define a “rational section” of the structure sheaf. It has a

maximal domain of definition” if X is reduced.] ]Definition. A scheme X is integral if OX(U) is an integral domain for every

open set U of X . reduced, integral

7.2.D. Important Exercise. Show that a scheme X is integral if and only if itis irreducible and reduced. 11 [forme: Int → irr, red immediate. Other direction:

Suppose fg = 0 in OX(X). Then V (f) = X or V (g) = X, say the former. Caution:

can’t say that fn = 0, see Ex. 7.2.C. But f is 0 in each stalk. [This is harder than it

looks!!]]

7.2.E. Exercise. Show that an affine scheme Spec A is integral if and only if Ais an integral domain.

8e:fuzfig

9reducednessstalklocal

10redqccaution

11H.P.II.3.1

134

7.2.F. Exercise. Suppose X is an integral scheme. Then X (being irreducible)has a generic point η. Suppose Spec A is any non-empty affine open subset of X .Show that the stalk at η, OX,η, is naturally FF(A), the fraction field of A. Thisis called the function field FF(X) of X . It can be computed on any non-emptyopen set of X , as any such open set contains the generic point. 12 The symbolfunction field

FF (X), function field FFis deliberately ambiguous — it may stand for fraction field or function field.

7.2.G. Exercise. Suppose X is an integral scheme. Show that the restrictionmaps resU,V : OX(U) → OX(V ) are inclusions so long as V 6= ∅. Suppose Spec Ais any non-empty affine open subset of X (so A is an integral domain). Show thatthe natural map OX (U)→ OX,η = FF(A) (where U is any non-empty open set) isan inclusion. Thus irreducible varieties (an important example of integral schemesdefined later) have the convenient that sections over different open sets can beconsidered subsets of the same thing. This makes restriction maps and gluing easyto consider; this is one reason why varieties are usually introduced before schemes.13

[forme: Referred to in “associated point” section.]ex.done

An almost-local criterion for integrality is given in 7.3.G.[forme: Another characterizations of integrality in Exercises ??]

7.3 Properties of schemes that can be checked“affine-locally”

[forme: Should I define locally of finite presentation?] This section is intendedto address something tricky and annoying in the definition of schemes. We’vedefined a scheme as a topological space with a sheaf of rings, that can be coveredby affine schemes. Hence we have all of the affine open sets in the cover, butwe don’t know how to communicate between any two of them. Somewhat moreexplicitly, if I have an affine cover, and you have an affine cover, and we want tocompare them, and I calculate something on my cover, there should be some way ofus getting together, and figuring out how to translate my calculation over onto yourcover. The Affine Communication Lemma 7.3.3 will provide a convenient machinefor doing this.

Thanks to this lemma, we can define a host of important properties of schemes.All of these are “affine-local” in that they can be checked on any affine cover, i.e.a covering by open affine sets. We like such properties because we can check themusing any affine cover we like. If the scheme in question is quasicompact, then weneed only check a finite number of affine open sets.

7.3.1. Warning. In our limited examples so far, any time we’ve had an affine opensubset of an affine scheme Spec B ⊂ Spec A, in fact Spec B = D(f) for some f .But this is not always true, and we will eventually have an example, using ellipticcurves (Example 25.15.1). 14

12d:functionfield

13e:inFF

14affineinaffine

135

7.3.2. Proposition. — Suppose Spec A and Spec B are affine open subschemes ofa scheme X. Then Spec A∩Spec B is the union of open sets that are simultaneouslydistinguished open subschemes of Spec A and Spec B. 15

ex.done

[p]

SpecASpecBSpecAf

SpecBg

Figure 2. Trick to show that the intersection of two affine opensets may be covered by open sets that are simultaneously open inboth affines

Proof. (See Figure 2 for a sketch.) Given any point [p] ∈ Spec A ∩ Spec B, weproduce an open neighborhood of [p] in Spec A ∩ Spec B that is simultaneouslydistinguished in both Spec A and Spec B. Let Spec Af be a distinguished opensubset of Spec A contained in Spec A ∩ Spec B. Let Spec Bg be a distinguishedopen subset of Spec B contained in Spec Af . Then g ∈ Γ(Spec B,OX) restricts toan element g′ ∈ Γ(Spec Af ,OX) = Af . The points of Spec Af where g vanishes areprecisely the points of Spec Af where g′ vanishes, so

Spec Bg = Spec Af \ [p] | g′ ∈ p= Spec(Af )g′ .

If g′ = g′′/fn (g′′ ∈ A) then Spec(Af )g′ = Spec Afg′′ , and we are done.

The following easy result will be crucial for us.

7.3.3. Affine Communication Lemma. — Let P be some property enjoyed bysome affine open sets of a scheme X, such that

(i) if an affine open set Spec A → X has P then for any f ∈ A, Spec Af → Xdoes too.

(ii) if (f1, . . . , fn) = A, and Spec Afi → X has P for all i, then so doesSpec A → X.

Suppose that X = ∪i∈I Spec Ai where Spec Ai is an affine, and Ai has property P .Then every other open affine subscheme of X has property P too. 16 Affine communication

lemma

ex.doneWe say such a property is affine-local. Note that any property that is stalk-

affine-locallocal (a scheme has property P if and only if all its stalks have property Q) isnecessarily affine-local (a scheme has property P if and only if all of its affines haveproperty R, where an affine scheme has property R if and only if and only if all itsstalks have property Q), but it is sometimes not so obvious what the right definitionof Q is; see for example the discussion of normality in the next section. stalk-local

15necessarylemma

16t:ACT

136

Proof. Let Spec A be an affine subscheme of X . Cover Spec A with a finite numberof distinguished open sets Spec Agj , each of which is distinguished in some Spec Ai.This is possible by Proposition 7.3.2 and the quasicompactness of Spec A. By (i),each Spec Agj has P . By (ii), Spec A has P .

7.3.A. Warm-up Exercise. Show that the notion of quasiseparatedness of ascheme is affine-local. [forme: Easy]

By choosing property P appropriately, we define some important properties ofschemes.

7.3.4. Proposition. — Suppose A is a ring, and (f1, . . . , fn) = A. 17

(a) If A is a Noetherian ring, then so is Afi . If each Afi is Noetherian, thenso is A.

(b) If A is reduced, then Afi is also reduced. If each Afi is reduced, then sois A.

(c) Suppose B is a ring, and A is an B-algebra. (Hence Ag is a B-algebrafor all B.) If A is a finitely generated B-algebra, then so is Afi . If eachAfi is a finitely-generated B-algebra, then so is A.

[primordial: Suppose R is an integral domain. If R is integrally closed, then so is Rfi.

If each Rfiis integrally closed, then so is R. I have a proof in class 9 notes from the first FOAG

class. ] We’ll prove these shortly. But let’s first motivate you to read the proof bygiving some interesting definitions assuming Proposition 7.3.4 is true.

7.3.5. Important Definitions. Suppose X is a scheme.18 If X can be coveredby affine open sets Spec A where A is Noetherian, we say that X is a locallyNoetherian scheme. If in addition X is quasicompact, or equivalently can becovered by finitely many such affine open sets, we say that X is a Noetherianscheme. By Exercise 7.1.C, the underlying topological space of a Noetherian(l) Noetherian scheme

scheme is Noetherian. (We will see a number of definitions of the form “if X hasthis property, we say that it is locally Q; if further X is compact, we say that it isQ.”) [forme: Caution: [M] seems to define Noetherian schemes in a different way:

if the underlying topological space is Noetherian. 19]

7.3.B. Exercise. Show that all open subsets of a Noetherian topological space(hence a Noetherian scheme) are quasicompact. 20

7.3.C. Exercise. Show that locally Noetherian schemes are quasiseparated. 21

7.3.D. Exercise. Show that a Noetherian scheme has a finite number of ir-reducible components. Show that a Noetherian scheme has a finite number ofconnected components, each a finite union of irreducible components.

17actapp

18importantdefinitions

19M.III.2.D.1

20Noetherianopenqc

21lNqsep

137

7.3.E. Exercise. If X is a Noetherian scheme, show that every point p has aclosed point in its closure. (In particular, every non-empty Noetherian scheme hasclosed points; this is not true for every scheme, as remarked in Exercise 7.1.E.) 22

7.3.F. Exercise. If X is an affine scheme or Noetherian scheme, show that itsuffices to check reducedness at closed points. (Hint: For the Noetherian case, recallExercise 7.3.E.) [forme: Have some comment about things that are closed-point-

local.]

Integrality is not stalk-local, but it almost is, as is shown in the followingbelievable exercise.

7.3.G. Unimportant exercise. Show that a locally Noetherian scheme X isintegral if and only if X is connected and all stalks OX,p are integral domains(informally: “the scheme is locally integral”). Thus in “good situations” (whenthe scheme is Noetherian), integrality is the union of local (stalks are domains)and global (connected) conditions. [forexperts: Is there a counterexampleif the Noetherian hypothesis is removed?] 23 24 [forexperts: “LocallyNoetherian” is not a stalk-local condition, and Joe Rabinoff has givenme a short neat example of why.] [forme: Joe’s counterexample: Let k be analgebraically closed field, let b1, b2, b3, ... ∈ k be a sequence of distinct elements, andlet

A = k[s, a1, a2, ...]/((s− bi)ai+1 − ai, a2i )i=1,2,...

I claim that A is not noetherian, but that Ap is noetherian for every prime ideal. Itsuffices to check for maximal ideals, as Noetherianness is preserved by localization..The nilradical N of A is (a1, a2, ...) (as the ai clearly lie in the nilradical, and A/(a1, . . . )is a domain so we’ve found it all), and A/N = k[s], so the maximal ideals of A are theideals of the form m = (s− b, a1, a2, ...) for b ∈ k Let m be such an ideal.

• Suppose that b = bn for some n. For i 6= n, we have ai+1 = ai/(s− bi) in Am.

Hence Am is the localization of a ring generated by the two variables s andan, so it’s Noetherian.

• If b is distinct from all the bi, then Am is the localization of a ring generatedby s and a1, as above.

Hence all stalks are Noetherian, but clearly the nilradical of A is not finitely gener-ated.

False argument: “Locally Noetherian” is in fact a stalk-local condition. Here is

a hint of why. If SpecA is Noetherian, then the proof of Proposition 7.3.4(a) shows

that any localization is Noetherian, hence the stalks are Noetherian. If SpecA is not

Noetherian, then let I1 ⊂ I2 ⊂ · · · be a strictly ascending infinite chain of ideals in

A. Then ∪Ii 6= A, so this ideal is contained in some maximal ideal m. Then this

corresponds to a strictly ascending chain of ideals in Am.]

7.3.H. Exercise. Show that X is reduced if and only if X can be covered byaffine open sets Spec A where A is reduced (nilpotent-free). reduced

ex.doneOur earlier definition required us to check that the ring of functions over any

open set is nilpotent free. Our new definition lets us check a single affine cover.Hence for example An

k and Pnk are reduced.

22noetherianclosedpoints

23integralconnectedstalk. This requires affine-locality of integrality: X is integral if and only if for every

affine open U ⊂ X, OX (U) is an integral domain. Another statement, from Jarod: If SpecA is connected, thenit is integral iff all stalks are integral domains — with no Noetherian hypothesis.

24integralitybonus1

138

7.3.6. A-schemes. 25Suppose X is a scheme, and A is a ring (e.g. A is a fieldk), and Γ(U,OX) has an A-algebra for all A, and the restriction maps respectthe A-algebra structure. Then we say that X is an A-scheme, or a schemeover A. Suppose X is an A-scheme. If X can be covered by affine open setsA-scheme, scheme over

A Spec Bi where each Bi is a finitely generated A-algebra, we say that X is locallyof finite type over A, or that it is a locally of finite type A-scheme. (This isadmittedly cumbersome terminology; it will make more sense later, once we knowabout morphisms in §9.3.1.) If furthermore X is quasicompact, X is finite typeover A, or a finite type A-scheme. Note that a scheme locally of finite type overlft, ft A-scheme

k or Z (or indeed any Noetherian ring) is locally Noetherian, and similarly a schemeof finite type over any Noetherian ring is Noetherian. As our key “geometric”example: if I ⊂ C[x1, . . . , xn] is an ideal, then Spec C[x1, . . . , xn]/I is a finite-typeC-scheme.

7.3.7. We now make a connection to the language of varieties. An affine schemethat is reduced and finite type k-scheme is said to be an affine variety (over k),or an affine k-variety. We are not yet ready to define varieties in general; we willneed the notion of separatedness first, to exclude abominations of nature like theline with the doubled origin (Example 6.1.7). We will define projective k-varieties(§10.2.2) before defining varieties in general (as separated finite type k-schemes,§12.1.7). (Warning: in the literature, it is sometimes also required that theaffine variety

scheme be irreducible, or that k be algebraically closed.) 26

7.3.I. Exercise. Show that a point of a locally finite type k-scheme is a closedpoint if and only if the residue field of the stalk of the structure sheaf at that pointis a finite extension of k. (Recall the following form of Hilbert’s Nullstellensatz,richer than the version stated in §5.3.1: the maximal ideals of k[x1, . . . , xn] areprecisely those with residue of the form a finite extension of k. We will provethis in Exercise 15.4.A.) Show that the closed points are dense on such a scheme.(For another exercise on closed points, see 7.1.E.) 27 [forme: Four refs to closedNullstellensatz

points:7.4.D,7.3.I,7.1.E,16.4.A. I should smooth this discussion out.]

7.3.8. Proof of Proposition 7.3.4. (a) (i) If I1 ( I2 ( I3 ( · · · is a strictlyincreasing chain of ideals of Af , then we can verify that J1 ( J2 ( J3 ( · · · is astrictly increasing chain of ideals of A, where

Jj = r ∈ A | r ∈ Ijwhere r ∈ Ij means “the image in Af lies in Ij”. (We think of this as Ij ∩A, exceptin general A needn’t inject into Afi .) Clearly Jj is an ideal of A. If x/fn ∈ Ij+1 \Ij

where x ∈ A, then x ∈ Jj+1, and x /∈ Jj (or else x(1/f)n ∈ Jj as well). (ii) SupposeI1 ( I2 ( I3 ⊂ · · · is a strictly increasing chain of ideals of A. Then for each1 ≤ i ≤ n,

Ii,1 ⊂ Ii,2 ⊂ Ii,3 ⊂ · · ·is an increasing chain of ideals in Afi , where Ii,j = Ij ⊗A Afi . It remains to showthat for each j, Ii,j ( Ii,j+1 for some i; the result will then follow.

25Ascheme

26d:affinevariety

27lftclosedpoints

139

7.3.J. Exercise. Finish this argument. [forme: Answer [include?]: Suppose

I ( J. We show that for some i, Ii ( Ji (where as before Ii = I ⊗A Afiand similarly

for Ji). Assume first that I = (0). Choose any r ∈ J \ I. If r = 0 in Afifor all i, then

there would be some m such that fmi r = 0 for all i. As (fm1 , . . . , fmn ) = A, we can choose

ai ∈ A withPaif

m1 = 1; multiplying this by r, we obtain r = 0, giving a contradiction.

Hence for some i, r 6= 0 in Afi. The general case (for arbitrary I) follows by applying

the preceding argument to A/I instead of A.]

7.3.K. Exercise. Prove (b). [forme: Answer: (i) If Rf is not radical, then thereis some x/fn ∈ Rf (with x ∈ R) with x/fn 6= 0 (i.e. fax 6= 0 for all a) but (x/fn)m = 0

for some m (i.e. fbxm = 0 for b 0). Then (fax)m′= 0 for m′ 0, so R is not radical

either.

[Better for (ii): follow (c).] (ii) follows in the same way as in (a): if (0) (p

(0),

then for some fi, (0) ⊗R Rfi(p

(0) ⊗R Rfi, i.e. (0) (

p(0) in Rfi

.]

(c) (i) is clear: if A is generated over B by r1, . . . , rn, then Af is generatedover B by r1, . . . , rn, 1/f .

(ii) Here is the idea. We have generators of Ai: rij/f ji , where rij ∈ A. I claim

that rijij ∪ fii generate A as a B-algebra. Here’s why. Suppose you have anyr ∈ A. Then in Afi , we can write r as some polynomial in the rij ’s and fi, dividedby some huge power of fi. So “in each Afi , we have described r in the desired way”,except for this annoying denominator. Now use a partition of unity type argumentto combine all of these into a single expression, killing the denominator. Show thatthe resulting expression you build still agrees with r in each of the Afi . Thus it isindeed r.

7.3.L. Exercise. Make this argument precise.ex.done

This concludes the proof of Proposition 7.3.4

7.4 Normality and factoriality

7.4.1. Normality.[forme: Four properties: exactness, reducedness, normality, factoriality. I should

have mentioned surjectivity. Later examples: torsion-freeness.] 28 We can now de-fine a property of schemes that says that they are “not too far from smooth”, callednormality, which will come in very handy. We will see later that “locally Noether-ian normal schemes satisfy Hartogs’ Lemma” (Algebraic Hartogs’ Lemma 15.6.6for Noetherian normal schemes): functions defined away form a set of codimension2 extend over that set, (2) Rational functions that have no poles are defined every-where. We need definitions of dimension and/or poles to make this precise. (Wesaw a first glimpse of this in §6.1.6.) Hartogs

A scheme X is normal if all of its stalks OX,x are normal (i.e. are domains, andintegrally closed in their fraction fields). As reducedness is a stalk-local property normal

(Exercise 7.2.B), normal schemes are reduced.

28s:stalklocal

140

7.4.A. Exercise. Show that integrally closed domains behave well under local-ization: if A is an integrally closed domain, and S is a multiplicative subset, showthat S−1A is an integrally closed domain. (The domain portion is easy. Hint forintegral closure: assume that xn + an−1x

n−1 + · · · + a0 = 0 where ai ∈ S−1A hasa root in the fraction field. Turn this into another equation in A[x] that also has aroot in the fraction field.)29

ex.doneIt is no fun checking normality at every single point of a scheme. Thanks to this

exercise, we know that if A is an integrally closed domain, then Spec A is normal.Also, for Noetherian schemes, normality can be checked at closed points, thanks tothis exercise, and the fact that for such schemes, any point is a generization of aclosed point (see Exercise 7.3.E)

It is not true that normal schemes are integral. For example, the disjointunion of two normal schemes is normal. Thus Spec k

∐

Spec k ∼= Spec(k × k) ∼=Spec k[x]/(x(x − 1)) is normal, but its ring of global sections is not a domain.

7.4.B. Unimportant exercise. Show that a Noetherian scheme is normal ifand only if it is the finite disjoint union of integral Noetherian normal schemes.[forme: Answer: divide into irreducible components. No two can meet, basically by

that Exercise on connected schemes being irreducible iff locally irreducible.]Noetherian normal

schemes are finite

disjoint unions of ir-

reducible Noetherian

normal schemes

ex.done

We are close to proving a useful result in commutative algebra, so we may aswell go all the way.

7.4.2. Proposition. — If A is an integral domain, then the following areequivalent. 30

(1) A integrally closed.(2) Ap is integrally closed for all prime ideals p ⊂ A.(3) Am is integrally closed for all maximal ideals m ⊂ A.

Proof. Clearly (2) implies (3). Exercise 7.4.A shows that integral closure is pre-served by localization, so (1) implies (2).

It remains to show that (3) implies (1). This argument involves a very niceconstruction that we will use again. Suppose A is not integrally closed. We showthat there is some m such that Am is also not integrally closed. Suppose

(24) xn + an−1xn−1 + · · ·+ a0 = 0

(with ai ∈ A) has a solution s in FF(A). Let I be the ideal of denominators ofideal of denominators

s:

I := r ∈ A | rs ∈ A.(Note that I is clearly an ideal of A.) Now I 6= A, as 1 /∈ I . Thus there is somemaximal ideal m containing I . Then s /∈ Am, so equation (24) in Am[x] shows thatAm is not integrally closed as well, as desired.

7.4.C. Unimportant Exercise. If A is an integral domain, show that A = ∩Am,where the intersection runs over all maximal ideals of A. (We won’t use this exercise,but it gives good practice with the ideal of denominators.)

29iclocalization

30normalitystalklocal

141

7.4.D. Unimportant Exercise relating to the ideal of denominators.One might naively hope from experience with unique factorization domains that theideal of denominators is principal. This is not true. As a counterexample, considerour new friend A = k[a, b, c, d]/(ad− bc) (which we last saw in Example 6.1.12, andwhich we will later recognize as the cone over the quadric surface), and a/c = b/d ∈FF(A). Show that I = (c, d), which is not principal. We’ll soon see that it is notprincipal (Exercise 16.1.C). 31 [forme: The cone over the quadric surface..] cone over the quadric

surface, quadric surface[primordial: Remark on closed points Earlier algebraic geometry dealt with varietiesover a field, usually algebraically closed. The set was just the closed points. To compare: locallyfinite type k-schemes: open sets are determined by their intersections with the subset of closedpoints. (Hint: reduce to affine case. Show first that closed points are dense. Use Nullstellensatz.)In particular, closed points are dense. (The reader can also check that this is also true forquasicompact schemes, although we won’t use this fact. Warning: it is not true for schemes ingeneral — there are in fact (non-empty) schemes with no closed points! Hence there are variousalgebraic facts that will be of the form: a ring has property p if and only if its localization atmaximal ideals does if and only if its localization at all prime ideals do. Usually, it is well-behavedunder localization. Then all we have to do is show that if if Rm has the property for all m, thenso does R. I’d kind of want the finite type k-algebra statements before this. 32 [forme: Thereare four references to closed points: 7.4.D, 7.3.I, 7.1.E, 16.4.A. I should smooth thisdiscussion out.]

]

7.4.3. Factoriality.We define a notion which implies normality.

7.4.4. Definition. If all the stalks of a scheme X are unique factorizationdomains, we say that X is factorial. [forme: I am following Mumford. Hartshorne factorial

calls this locally factorial. Danger of this is that it sounds as though there are small

enough neighborhoods of each open sets that are UFD’s. But an elliptic curve is an

example: 25.5.2.]

7.4.E. Exercise. Show that any localization of a Unique Factorization Domainis a Unique Factorization Domain.

ex.doneThus if A is a unique factorization domain, then Spec A is factorial. (The

converse need not hold. Hence this property is not affine-local, as we will verifyin Exercise 18.1.E. Here is a counter-example without proof: Z[

√17].) [forme:

Perhaps add elliptic curve example, 25.5.2?] Hence it suffices to check factorialityby finding an appropriate affine cover. 33

7.4.5. ?? How to check if a ring is a Unique Factorization Domain. We note herethat there are very few means of checking that a Noetherian domain is a uniquefactorization domain. Some useful ones are: (1) the previous exercise, that thelocalization of a unique factorization domain is also a unique factorization domain.(2) height 1 primes are principal (Exercise 15.6.8). (3) Nagata’s Lemma 18.1.I. (4)normal and Cl = 0 (Exercise 18.2). 34

One of the reasons we like factoriality is that it implies normality.

31rnotp

32remarkonclosedpoints

33localizationufd

34ufd4

142

7.4.F. Important exercise. Show that unique factorization domains are in-tegrally closed. Hence factorial schemes are are normal, and if A is a unique fac-torization domain, then Spec A is normal. (However, rings can be integrally closedwithout being unique factorization domains, as we’ll see in Exercise 7.4.I. An ex-ample without proof: Z[

√17] again.) 35

7.4.G. Easy Exercise. Show that the following schemes are normal: Ank , Pn

k ,Spec Z.

7.4.H. Exercise (which will give us a number of enlightening exampleslater). Suppose A is a Unique Factorization Domain with 2 invertible, f ∈A has no repeated prime factors, and z2 − f is irreducible in A[z]. Show thatSpec A[z]/(z2−f) is normal. Show that if f is not square-free, then Spec A[z]/(z2−f) is not normal. (Hint: B := A[z]/(z2 − f) is a domain, as (z2 − f) is prime inA[z]. Suppose we have monic F (T ) = 0 with F (T ) ∈ B[T ] which has a solutionα in FF(B). Then by replacing F (T ) by F (T )F (T ), we can assume F (T ) ∈ A[T ].Also, α = g + hz where g, h ∈ FF(A). Now α is the solution of monic Q(T ) =T 2 − 2gT + (g2 − h2f)T ∈ FF(A)[T ], so we can factor F (T ) = P (T )Q(T ) in K[T ].By Gauss’ lemma, 2g, g2−h2f ∈ A. Say g = r/2, h = s/t (s and t have no commonfactors, r, s, t ∈ A). Then g2−h2f = (r2t2− rs2f)/4t2. Then t = 1, and r is even.)

7.4.I. Exercise. Show that the following schemes are normal:

(a) Spec Z[x]/(x2−n) where n is a square-free integer congruent to 3 (mod 4);(b) Spec k[x1, . . . , xn]/x2

1 + x22 + · · ·+ x2

m where chark 6= 2, m ≥ 3;(c) Spec k[w, x, y, z]/(wz−xy) where char k 6= 2 and k is algebraically closed.

This is our cone over a quadric surface example from Exercises 6.1.12and 7.4.D. (Hint: the side remark below may help.)

36 37 [forme: Make link to show that normality can be checked after finite changesmooth quadric surface

of base field, §??. There is even an exercise there about the cone over the quadric

surface. Is it true that if R is normal then so is R[x]? Should be, using R1 and S2 —

add ref to this later.]

7.4.6. Side remark: diagonalizing quadrics. Suppose k is an algebraically closedfield of characteristic not 2. Then any quadratic form in n variables can be “di-agonalized” by changing coordinates to be a sum of squares (e.g. uw − v2 =((u + v)/2)2 + (i(u − v)/2)2 + (iv)2), and the number of such squares is invari-ant of the change of coordinates. (Reason: write the quadratic form on x1, . . . , xn

as

(

x1 · · · xn

)

M

x1

...xn

where M is a symmetric matrix — here you are using characteristic 6= 2. Thendiagonalize M — here you are using algebraic closure.) 38diagonalizing quadrics

35ufdic

36quadricconenormal

37H.E.II.6.4 and 6.5a

38diagonalizequadric

143

7.4.J. Exercise. Suppose A is a k-algebra where chark = 0, and l/k is a finitefield extension. Show that A is normal if and only if A ⊗k l is normal. Show thatSpec k[w, x, y, z]/(wz−xy) is normal if k is characteristic 0. (In fact the hypothesison the characteristic is unnecessary.) Possible hint: reduce to the case where l/k isGalois. 39

[primordial: We conclude with a discussion of a necessary and sufficient criterion ofhow to check normality on an affine cover. If you can cover a scheme with affines such that eachaffine corresponds to an integrally closed domain, then the scheme is affine.

7.4.7. Note that not every cover of a normal scheme has this property: if R and S are integrallyclosed, then SpecR

‘Spec S = SpecR× S [ref?] is normal, but R× S is not an integral domain.

40

Suppose R is a Noetherian ring. Here is a characterization for when SpecR is normal. (i)R is reduced. By Noetherianness, R has a finite number of minimial primes. I should verify thatR→ S :=

QpminimalRp is an inclusion [now verified in the next section]. The thing on the

the right is the right generalization of fraction fields for reduced rings. (ii) We require that anysolution of a monic algebraic f(x) ∈ R[x] in S is actually in the image of R.

Verify that the irreducible components of SpecR correspond to the connected componentsof SpecR. Thus R is a product of a finite number of rings, each of which is an integrally closeddomain. Thus the example of 7.4.7 is really the only bad behaviour that can happen. Moreexplicitly:

7.4.K. Exercise (unimportant). Show that SpecA is normal if and only if A can be writtenas A1 × · · · × An where A1, . . . , An are integrally closed domains.

Perhaps it is true that a scheme is normal if it can be covered by affines that are Spec’s ofintegrally closed domains. We have just verified that this is the case if X is locally Noetherian.

Older notes: Normality satisfies the hypotheses of the Affine CommunicationLemma, fairly tautologically, because it is a stalk-local property. We can say moreexplicitly and ring-theoretically what it means for SpecA to be normal, at least whenA is Noetherian. Show that SpecA is normal if and only if A is reduced, and it isintegrally closed in its ring of fractions. (Constructions that make sense for domains ring of fractionsand involve function fields often generalize to Noetherian rings, where the role of the“function field” should be played by the “ring of fractions”.)

Perhaps this should wait until after the next section?

Warning to myself: Matsumura 2nd ed p. 117: There exists a normal ring A

such that A[[X]] is not normal. Credited to Seidenberg. ]

7.5 Associated points of (locally Noetherian) schemes, anddrawing fuzzy pictures

41 [forme: Ref: [AM] p. 50-.] Recall from just after Definition 7.2.1 (ofreduced) our “fuzzy” pictures of the non-reduced scheme Spec k[x, y]/(y2, xy) (seeFigure ??). When this picture was introduced, we mentioned that the “fuzz” atthe origin indicated that the non-reduced behavior was concentrated there; this wasverified in Exercise 7.2.A, and indeed the origin is the only point where the stalkof the structure sheaf is non-reduced.

You might imagine that in a bigger scheme, we might have different closedsubsets with different amount of “non-reducedness”. This intuition will be madeprecise in this section. We will define associated points of a scheme, which will

39e:normaldescent

40normalnondomain

41s:associated

144

be the most important points of a scheme, encapsulating much of the interestingbehavior of the structure sheaf. These will be defined for any locally Noetherianscheme. The primes corresponding to the associated points of an affine schemeSpec A will be called associated primes of A. (In fact this is backwards; we willdefine associated primes first, and then define associated points.)

The four properties about associated points that it will be most important toremember are as follows. Frankly, it is much more important to remember thesefour facts than it is to remember their proofs.

(1) The generic points of the irreducible components are associated points. Theother associated points are called embedded points. 42embedded points

(2) If X is reduced, then X has no embedded points. (This jibes with theintuition of the picture of associated points described earlier.)

(3) Recall that one nice property of integral schemes X (such as irreducibleaffine varieties) not shared by all schemes is that for any open U ⊂ X , the naturalmap Γ(U,OX) → FF(X) is an inclusion (Exercise 7.2.G). Thus all sections overany open set (except ∅) and stalks can be thought of as lying in a single fieldFF(X), which is the talk at the generic point.

More generally, if X is a locally Noetherian scheme, then for any U ⊂ X , thenatural map 43

(25) Γ(U,OX)→∏

associated p in U

OX,p

is an injection.We define a rational function44 on a locally Noetherian scheme to be anrational function

element of the image of Γ(U,OU ) in (25) for some U containing all the associ-ated points. (This definition makes sense more generally, for example for integralschemes.) [forme: We need this comment because in Exercise 13.1.A, we want to

interpret rational functions on int. sep. schemes as rational maps to A1Z.] The ra-

tional functions form a ring, called the total fraction ring of X , denoted FF(X).total fraction ring

If X = Spec A is affine, then this ring is called the total fraction ring of A,FF(A). Note that if X is integral, this is the function field FF(X), so this extendsour earlier definition 7.2.F of FF (·). [forme: E calls this the total quotient ring.function field

Hartshorne might call it the ring of total fractions. Maybe in the affine case they

apply this to the full right side? These two notions are the same if A is reduced.

] It can be more conveniently interpreted as follows, using the injectivity of (25).A rational function is a function defined on an open set containing all associatedpoints, i.e. and ordered pair (U, f), where U is an open set containing all associatedpoints, and f ∈ Γ(U,OX). Two such data (U, f) and (U ′, f ′) define the same openrational function if and only if the restrictions of f and f ′ to U ∩ U ′ are the same.If X is reduced, this is the same as requiring that they are defined on an open setof each of the irreducible components. A rational function has a maximal domainof definition, because any two actual functions on an open set (i.e. sections of thestructure sheaf over that open set) that agree as “rational functions” (i.e. on smallenough open sets containing associated points) must be the same function, by theinjectivity of (25). We say that a rational function f is regular at a point p if p

42d:embeddedpoint

43associatedmap

44d:rationalfunction

145

is contained in this maximal domain of definition (or equivalently, if there is someopen set containing p where f is defined). regular function

The previous facts are intimately related to the following one.(4) A function on X is a zero divisor if and only if it vanishes at an associated

point of X.Motivated by the above four properties, when sketching (locally Noetherian)

schemes, we will draw the irreducible components (the closed subsets correspondingto maximal associated points), and then draw “additional fuzz” precisely at theclosed subsets corresponding to embedded points. All of our earlier sketches wereof this form. (See Figure 3.)

Figure 3. This scheme has 6 associated points, of which 3 areembedded points. A function is a zero-divisor if it vanishes atone of these six points. It is nilpotent if it vanishes at all six ofthese points. (In fact, it suffices to vanish at the non-embeddedassociated points.)

Let’s now get down to business of defining associated points, and showing thatthey the desired properties (1) through (4).

We say an ideal I ⊂ A in a ring is primary if I 6= A and if xy ∈ I implieseither x ∈ I or yn ∈ I for some n > 0. primary ideal

It is useful to interpret maximal ideals as “the quotient is a field”, and primeideals as “the quotient is an integral domain”. We can interpret primary idealssimilarly as “the quotient is not 0, and every zero-divisor is nilpotent”. Thus thenotion of “primary” should be seen as a condition on A/I , not on I .

7.5.A. Exercise. Show that if q is primary, then√

q is prime. [forme: [AM]

Prop. 4.1] If p =√

q, we say that q is p-primary. (Caution:√

q can be primewithout q being primary — consider our example (y2, xy) in k[x, y].)

7.5.B. Exercise. Show that if q and q′ are p-primary, then so is q ∩ q′.45

7.5.C. Exercise (reality check). Find all the primary ideals in Z. (Answer:(0) and (pn).)

7.5.1. ? Unimportant warning for experts (all others should skip this). Aprime power need not be primary. [forme: [AM, p. 51]] For example, let

45pprimary

146

A = k[x, y, z]/(xy − z2), and p = (x, z). Then p is prime but p2 is not primary.(Verify this — the algebra is easy! Why is (x2, xz, z2, xy − z2) not primary ink[x, y, z]?) We will soon be able to interpret Spec A as a “cone”, and V (x, z) as the“ruling” of the cone, see Figure 4, and the corresponding picture gives a geometrichint that there is something going on. [forme: We’ll come back to this LATER

THIS SECTION.]

Figure 4. V (x, z) ⊂ Spec k[x, y, z]/(xy−z2) is a ruling on a cone;(x, z)2 is not (x, z)-primary.

7.5.2. Primary decompositions.A primary decomposition of an ideal I ⊂ A is an expression of the ideal as

a finite intersection of primary ideals.primary decomposition

I = ∩ni=1qi

If there are “no redundant elements” (the√

qi are all distinct, and for no i isqi ⊃ ∩j 6=iqj), we say that the decomposition is minimal. Clearly any idealwith a primary decomposition has a minimal primary decomposition (using Ex-ercise 7.5.B).minimal primary decom-

position7.5.D. Important Exercise (existence of primary decomposition forNoetherian rings). Suppose A is a Noetherian ring. Show that every properideal I ⊂ A has a primary decomposition. (Hint: mimic the Noetherian inductionargument of Proposition 5.5.7.)

7.5.E. Exercise. (a) Find a minimal primary decomposition of (y2, xy).46 (b)Find another one. [forme: Possible answers include (y) ∩ (y2, xy, xn) for any n, and

46xsquaredxy

147

(y) ∩ (y2, x).] (Possible hint: see Figure 1. You might be able to draw sketches ofyour different primary decompositions.)

ex.doneIn order to study these objects, we’ll need a useful fact and a definition.

7.5.F. Essential Exercise. (a) If p, p1, . . . , pn are prime ideals, and p = ∩pi,show that p = pi for some i. (Hint: assume otherwise, choose fi ∈ pi − p, andconsider

∏

fi.)(b) If p ⊃ ∩pi, then p ⊃ pi for some i. 47 [forme: (a) and (b) are essentially the

same.]

(c) Suppose I ⊆ ∪ni=1pi. (The right side is not an ideal!) Show that I ⊂ pi for

some i. (Hint: by induction on n. Don’t look in the literature — you might find amuch longer argument! See Exercise 15.6.C for a related problem.)48 [forme: By prime avoidance (temp.

notation), [E, p. 90]induction on i. The base case i = 1 is trivial. Suppose the result fails for n, but holds

for n− 1. Then I ( ∪n−1j=1 pj , and I ( pn. Let f ∈ I− ⊂n−1

j=1 pj (so f ∈ pn) and g ∈ I − pn

(so g ∈ ∪n−1j=1 pj). Then f + g must be in I −∪n−1

j=1 pj − pn, contradiction. Which of these

is called prime avoidance?]

Parts (a) and (b) are “geometric facts”; try to draw pictures of what they mean.If I ⊂ A is an ideal, and x ∈ A, then define the colon ideal (I : x) := a ∈ A :

ax ∈ I. (We will use this terminology only for this section.) For example, x is a (I : x)

zero-divisor if (0 : x) 6= 0. zero-divisor, (I : x)

colon ideal7.5.3. Theorem (“uniqueness” of primary decomposition). — SupposeI ⊂ A has a minimal primary decomposition

I = ∩ni=1qi.

(For example, this is always true if A is Noetherian.) Then the√

qi are preciselythe prime ideals that are of the form

√

(I : x)

for some x ∈ A. Hence this list of primes is independent of the decomposition.49

[forme: Iitaka p. 109 calls this Lasker’s theorem, presumably after the world

chess champion. More precisely, any proper ideal of a NOetherian ring can be irre-

dundantly as the intersection of primary ideals with distinct primes.]

These primes are called the associated primes of the ideal I . The asso-ciated primes of A are the associated primes of 0. [forme: Hartshorne defines

associated primes differently, p. 257.]

Proof. [forme: Following [AM] Theorem 4.5] We make a very useful observation:for any x ∈ A,

(I : x) = (∩qi : x) = ∩(qi : x),

from which50

(26)√

(I : x) = ∩√

(qi : x) = ∩x/∈qjpj .

Now we prove the result.

47usefulexercise

48primeavoidance

49UPD

50unbelievable

148

Suppose first that√

(I : x) is prime, say p. Then p = ∩x/∈qjpj by (26), and by

Exercise 7.5.F(a), p = pj for some j.

Conversely, given qi, we find an x such that√

(I : x) =√

qi (= pi). Takex ∈ ∩j 6=iqj − qi (which is possible by minimality of the primary decomposition).Then by (26), we’re done.

7.5.G. Exercise (associated primes behave well with respect to local-ization). Show that if A is a Noetherian ring, and S is a multiplicative subset (sothere is an inclusion-preserving correspondence between the primes of S−1A andthose primes of A not meeting S, 5.3.E), then the associated primes of S−1A arejust the associated primes of A not meeting S. 51

ex.done

We then define the associated points of a locally Noetherian scheme X tobe those points p ∈ X such that, on any affine open set Spec A containing p, pcorresponds to an associated prime of A. [forme: [AM] Prop. 4.8.] Note that thisassociated point

notion is well-defined: If p has two affine open neighborhoods Spec A and Spec B(say corresponding to primes p ⊂ A and q ⊂ B respectively), then p correspondsto an associated prime of A if and only if it corresponds to an associated prime ofAp = OX,p = Bq if and only if it corresponds to an associated prime of B.

If furthermore X is quasicompact (i.e. X is a Noetherian scheme), then thereare a finite number of associated points.

7.5.H. Exercise. (a) Show that the minimal primes of A are associated primes.We have now proved important fact (1). (Hint: suppose p ⊃ ∩n

i=1qi. Then p =√p ⊃

√

∩ni=1qi = ∩n

i=1

√qi = ∩n

i=1pi, so by Exercise 7.5.F(b), p ⊃ pi for some i. Ifp is minimal, then as p ⊃ pi ⊂ (0), we must have p = pi.)(b) Show that there can be other associated primes that are not minimal. (Hint:Exercise 7.5.E.) Your argument will show more generally that the minimal primesof I are associated primes of I .

7.5.I. Unimportant exercise (interesting algebraic fact we won’t use).Show that the primary ideal corresponding to a minimal prime is independent ofprimary decomposition. (Thus this part of the primary decomposition is unique,and if there are no embedded primes then the entire primary decomposition isunique. You might have observed this in a special case in Exercise 7.5.E. You mayfind this fact believable from your sketches.) [forme: Iitaka p. 110]

7.5.J. Exercise. Show that if A is reduced, then the only associated primes arethe minimal primes. (This establishes (2).)

ex.doneThe qi corresponding to minimal primes are unique, but the qi corresponding

to other associated primes are not unique. We will not need this fact, and hencewon’t prove it.

7.5.4. Proposition. — The set of zero-divisors is the union of the associatedprimes. [forme: Part of [AM] Proposition 4.7] 52

ex.done

51assloc, used

52earlyeffectiveCartier

149

This establishes (4): a function is a zero-divisor if and only if it vanishes at anassociated point. Thus (for a Noetherian scheme) a function is a zero divisor if andonly if its zero locus contains one of a finite set of points.

You may wish to try this out on the example of Exercise 7.5.E.

Proof. If pi is an associated prime, then pi =√

(0 : x) from the proof of Theo-rem 7.5.3, so ∪pi is certainly contained in the set Z of zero-divisors.

For the converse:

7.5.K. Exercise. Show that

Z = ∪x6=0(0 : x) ⊆ ∪x6=0

√

(0 : x) ⊆ Z.

HenceZ = ∪x6=0

√

(0 : x) = ∪x

(

∩x/∈qjpj

)

⊆ ∪pj

using (26).

7.5.L. Unimportant exercise. Here is an interesting variation on (4): showthat a ∈ A is nilpotent if and only if it vanishes at the associated points of Spec A.Algebraically: we know that the nilpotents are the intersection of all prime ideals(Theorem 5.3.6); now show that in the Noetherian case, the nilpotents are in factthe intersection of the (finite number of) associated prime ideals.

7.5.5. Proposition. — The natural map A→ ∏

associated p Ap is an inclusion.53

ex.done

Proof. Suppose r 6= 0 maps to 0 under this map. Then there are si ∈ A − p withsir = 0. Then I := (s1, . . . , sn) is an ideal consisting only of zero-divisors. HenceI ⊆ ∪pi. Then I ⊂ pi for some i by Exercise 7.5.F(c), contradicting si /∈ pi.

7.5.M. Easier and less important exercise. Prove fact (3). (The previousProposition establishes it for affine open sets.)

[forme: Examples to work out privately with students: intersect two planes in A4

with a hyperplane. We expect that intersection to have an embedded point, hence

to be nonreduced. The cone example.]

[primordial: Note for experts from Kirsten and Joe: Let X be a locally Noetherianscheme, x ∈ X. Then x is an associated point of X if and only if every nonunit of OX,x is azero-divisor. Proof: We must show that a prime ideal p of a Noetherian ring A is associated ifand only if every nonunit of Ap is a zero-divisor, i.e., if and only if pAp is an associated prime inAp. But this is obvious since primary decompositions respect localization.

Associated primes of modules!!! [E, §3.1]What facts do I need? torsion part of a subsheaf; the cokernel is torsion-free.I is a special case, by consider A/I as the module.Here is a summary. P is an associated prime if it assassinates (precisely kills) an element.Theorem. A Noetherian, M finitely-generated A-module.

(z) They exist(a) [E, Thm. 3.1] AssM is finite. It contains all primes minimal among those containing annM .Hence nonempty if M 6= 0.(b) union of associated primes are the zero-divisors.(c) commutes with localization.

53ass3

150

(d) [E, Cor 3.2(d)] Every ideal of zero-divisors actually kills some element.(e) [E, Cor 3.5(a)] M →Q

Mp over associated primes is an inclusion.(f) 0→M ′ →M →M ′′ → 0 Then AssM ′ ⊂ AssM ⊂ AssM ∪ AssM ′′. [E, Lemma 3.6(b)].(g) Can filter so each piece is a R/p.

See [M] around p. 136. 54 ]

54M.II.4.C

Part III

Morphisms of schemes

CHAPTER 8

Morphisms of schemes

8.1 Introduction

[forme: We have some choices in defining “morphisms”. (1a) gluing affines

together. (1b) temporarily including local-ringed spaces (in either [H] or [EH] man-

ners). (2a) OY → f∗OX (2b) f−1OX →OY . I am going with (1b), (2b).] 1Wheneveryou learn about a new type of object in mathematics, you should naturally becurious about maps between them, which means understanding how they form acategory. [forme: Teaching: take a vote] In order to satisfy this curiosity, we’llintroduce the notion of morphism of schemes now, and at the same time we may aswell define some easy-to-state properties of morphisms. However, we’ll leave moresubtle properties of morphisms for [forme: later] .

Recall that a scheme is (i) a set, (ii) with a topology, (iii) and a (structure)sheaf of rings, and that it is sometimes helpful to think of the definition as havingthree steps. In the same way, the notion of morphism of schemes X → Y maybe defined (i) as a map of sets, (ii) that is continuous, and (iii) with some furtherinformation involving the sheaves of functions. In the case of affine schemes, wehave already seen the map as sets (§5.3.2) and later saw that this map is continuous(§5.4.G).

Here are two motivations for how morphisms should behave. The first is alge-braic, and the second is geometric.

(a) We’ll want morphisms of affine schemes Spec B → Spec A to be preciselythe ring maps A → B. We have already seen that ring maps A → B inducemaps of topological spaces in the opposite direction (Exercise 5.4.G); the main newingredient will be to see how to add the structure sheaf of functions into the mix.Then a morphism of schemes should be something that “on the level of affines,looks like this”.

(b) We are also motivated by the theory of differentiable manifolds (Section 5.1.1).Notice that if π : X → Y is a map of differentiable manifolds, then a differentiablefunction on Y pulls back to a differentiable function on X . More precisely, givenan open subset U ⊂ Y , there is a natural map Γ(U,OY ) → Γ(π−1(U),OX ). Thisbehaves well with respect to restriction (restricting a function to a smaller openset and pulling back yields the same result as pulling back and then restricting),so in fact we have a map of sheaves on Y : OY → π∗OX . Similarly a morphism ofschemes X → Y should induce a map OY → π∗OX . But in fact in the categoryof differentiable manifolds a continuous map X → Y is a map of differentiable

1s:morphismsintro

153

154

manifolds precisely when differentiable functions on Y pull back to differentiablefunctions on X (i.e. the pullback map from differentiable functions on Y to func-tions on X in fact lies in the subset of differentiable functions, i.e. the continuousmap X → Y induces a pullback of differential functions OY → OX), so this mapof sheaves characterizes morphisms in the differentiable category. So we could usethis as the definition of morphism in the differentiable category.

But how do we apply this to the category of schemes? In the category ofdifferentiable manifolds, a continuous map X → Y induces a pullback of (thesheaf of) functions, and we can ask when this induces a pullback of differentiablefunctions. However, functions are odder on schemes, and we can’t recover thepullback map just from the map of topological spaces. A reasonable patch is tohardwire this into the definition of morphism, i.e. to have a continuous map f :X → Y , along with a pullback map f# : OY → f∗OX . This leads to the definitionof the category of ringed spaces.

One might hope to define morphisms of schemes as morphisms of ringed spaces.This isn’t quite right, as then motivation (a) isn’t satisfied: as desired, to eachmorphism A → B there is a morphism Spec B → Spec A, but there can be addi-tional morphisms of ringed spaces Spec B → Spec A not arising in this way (Exer-cise 8.3.C). A revised definition as morphisms of ringed spaces that locally looksof this form will work, but this is awkward to work with, and we take a differenttack. However, we will check that our eventual definition actually is equivalent tothis (Exercise XXXX).

We’ll begin by discussing morphisms of ringed spaces.Before we do, we take this opportunity to use motivation (a) to motivate the

definition of equivalence of categories. We wish to say that the category of ringsand the category of affine schemes are opposite categories, i.e. that the “oppositecategory of affine schemes” (where all the arrows are reversed) is “essentially thesame” as the category of rings. We indeed have a functor from rings to affineschemes (sending A to Spec A), and a functor from affine schemes to rings (sendingX to Γ(X,OX)). But if you think about it, you’ll realize their composition isn’texactly the identity. (It all boils down to the meaning of “is” or “same”, and thiscan get confusing.) Rather than trying to set things up so the composition is theidentity, we just don’t let this bother us, and make precise the notion that thecomposition is “essentially” the identity.

Suppose F and G are two functors from A to B. A natural transformationof functors F → G is the data of a morphism ma : F (a) → G(a) for each a ∈ Asuch that for each f : a→ a′ in A, the diagram

F (a)F (f) //

ma

F (a′)

ma′

G(a)

G(f)// G(a′)

A natural isomorphism of functors is a natural transformation such that eachma is an isomorphism. The data of functors F : A → B and F ′ : B → A suchthat F F ′ is naturally isomorphic to the identity IB on B and F ′ F is naturallyisomorphic to IA is said to be an equivalence of categories. This is the “right”notion of isomorphism of categories.

155

Two examples might make this strange concept more comprehensible. Thedouble dual of a finite-dimensional vector space V is not V , but we learn early tosay that it is canonically isomorphic to V . We make can that precise as follows.Let f.d.Veck be the category of finite-dimensional vector spaces over k. Note thatthis category contains oodles of vector spaces of each dimension.

8.1.A. Exercise. Let ∨∨ : f.d.Veck → f.d.Veck be the double dual functor fromthe category of vector spaces over k to itself. Show that ∨∨ is naturally isomorphicto the identity. (Without the finite-dimensional hypothesis, we only get a naturaltransformation of functors from id to ∨∨.)

ex.doneLet V be the category whose objects are kn for each n (there is one vector space

for each n), and whose morphisms are linear transformations. This latter space canbe thought of as vector spaces with bases, and the morphisms are honest matrices.There is an obvious functor V → f.d.Veck, as each kn is a finite-dimensional vectorspace.

8.1.B. Exercise. Show that V → f.d.Veck gives an equivalence of categories,by describing an “inverse” functor. (You’ll need the axiom of choice, as you’llsimultaneously choose bases for each vector space in f.d.Veck!)

ex.doneOnce you have come to terms with the notion of equivalence of categories, you

will quickly see that rings and affine schemes are basically the same thing, with thearrows reversed:

8.1.C. Exercise. Assuming that morphisms of schemes are defined so thatMotivation (a) holds, show that the category of rings and the opposite category ofaffine schemes are equivalent.

[forme: Andrew Blumberg explains that this definition is the same as an alterna-

tive one about essential surjectivity and full faithfulness: “so indeed (in the presence

of the axiom of choice), the various criteria for a functor F : C → D to induce an

equivalence of categories are equivalent. basically, the point is that given a fully

faithful and essentially surjective F, you can explicitly construct an adjoint — for

a particular object x of D, by essential surjectivity you can choose an isomorphism

x to Fc, and so define Gx to be c. the isomorphism x → FGx is in fact universal

(i.e. initial in the appropriate over-category), since F is fully faithful. therefore (and

here’s where the axiom of choice comes in) you can assemble this assignment on

objects into an actual functor in a unique way, and it specifies an adjunction.” ]

8.2 Examples

[forme: In this section, I should assume motivation (a), and see how maps of

rings corresponds to maps of schemes in a concrete way.]

8.3 Morphisms of ringed spaces

156

8.3.1. Definition. A morphism π : X → Y of ringed spaces is a contin-uous map of topological spaces (which we unfortunately also call π) along with a“pullback map” OY → π∗OX . By adjointness (§4.6.1), this is the same as a mapπ−1OY → OX . There is an obvious notion of composition of morphisms; hencethere is a category of ringed spaces. Hence we have notion of automorphisms andisomorphisms. You can easily verify that an isomorphism f : (X,OX ) → (Y,OY )morphism of ringed

space is a homeomorphism f : X → Y along with an isomorphism OY → f∗OX (orequivalently f−1OY → OX).

If U ⊂ Y is an open subset, then there is a natural morphism of ringed spaces(U,OY |U )→ (Y,OY ). (Check this! The f−1 interpretation is cleaner to use here.)This is our model for an open immersion. More precisely, if U → Y is an iso-morphism of U with an open subset V of Y , and we are given an isomorphism(U,OU ) ∼= (V,OV ) (via the isomorphism U ∼= V ), then the resulting map of ringedspaces is called an open immersion of ringed spaces.

8.3.A. Exercise (morphisms of ringed spaces glue). Suppose (X,OX) and(Y,OY ) are ringed spaces, X = ∪iUi is an open cover of X , and we have morphismsof ringed spaces fi : Ui → Y that “agree on the overlaps”, i.e. fi|Ui∩Uj = fj |Ui∩Uj .Show that there is a unique morphism of ringed spaces f : X → Y such thatf |Ui = fi. (Exercise 4.2.D essentially showed this for topological spaces.) 2

8.3.B. Easy important exercise. Given a morphism of ringed spaces f : X →Y with f(p) = q, show that there is a map of stalks (OY )q → (OX)p.

3 [forme:

fancy answer: f# : OY → f∗OX . Take adjoints, and take stalks.]

8.3.2. Key Exercise. 4Suppose f# : B → A is a morphism of rings. Define amorphism of ringed spaces f : Spec A→ Spec B as follows. The map of topologicalspaces was given in Exercise 5.4.G. To describe a morphism of sheaves OB → f∗OA

on Spec B, it suffices to describe a morphism of sheaves on the distinguished baseof Spec B. On D(g) ⊂ Spec B, we define

OB(D(g))→ OA(f−1D(g)) = OA(D(f#g))

by Bg → Af#g . Verify that this makes sense (e.g. is independent of g), and thatthis describes a morphism of sheaves on the distinguished base. (This is the thirdin a series of exercises. We showed that a morphism of rings induces a map of setsin Exercise 5.3.2, a map of topological spaces in Exercise 5.4.G, and now a map ofringed spaces here.)

This will soon be an example of morphism of schemes! In fact we could makethat definition right now.

8.3.3. Definition we won’t start with. A morphism of schemes f : (X,OX)→(Y,OY ) is a morphism of ringed spaces that “locally looks like” the maps of affineschemes described in Key Exercise 8.3.2. Precisely, for each choice of affine opensets Spec A ⊂ X , Spec B ⊂ Y , such that f(Spec A) ⊂ Spec B, the induced map ofringed spaces should be of the form shown in Key Exercise 8.3.2. 5

2e:morphismsrsglue

3rsstalk

4ie

5baddef

157

We would like this definition to be checkable on an affine cover, and we mighthope to use the affine communication lemma to develop the theory in this way.This works, but it will be more convenient to use a clever trick: in the next section,we will use the notion of locally ringed spaces, and then once we have used it, wewill discard it like yesterday’s garbage.

The map of ringed spaces of Key Exercise 8.3.2 is really not complicated. Hereis an example. Consider the ring map C[x]→ C[y] given by x 7→ y2 (see Figure 1).We are mapping the affine line with co-ordinate y to the affine line with co-ordinatex. The map is (on closed points) a 7→ a2. For example, where does [(y− 3)] go to?Answer: [(x − 9)], i.e. 3 7→ 9. What is the preimage of [(x − 4)]? Answer: thoseprime ideals in C[y] containing [(y2−4)], i.e. [(y−2)] and [(y +2)], so the preimageof 4 is indeed ±2. This is just about the map of sets, which is old news, so let’s nowthink about functions pulling back. What is the pullback of the function 3/(x− 4)on D([(x− 4)]) = A1 − 4? Of course it is 3/(y2 − 4) on A1 − −2, 2.

double cover map

Figure 1. Spec C[y]→ Spec C[x] given by x 7→ y2

We conclude with an example showing that not every morphism of ringed spacesbetween affine schemes is of the form of Key Exercise 8.3.2.

8.3.C. Unimportant Exercise. Recall6 (Exercise 5.4.J) that Spec k[x](x) hastwo points, corresponding to (0) and (x), where the second point is closed, and thefirst is not. Consider the map of ringed spaces Spec k(x) → Spec k[x](x) sending

the point of Spec k(x) to [(x)], and the pullback map f#OSpeck(x) → OSpeck[x](x)

is induced by k → k(x). Show that this map of ringed spaces is not of the formdescribed in Key Exercise 8.3.2.

8.4 From local-ringed spaces to morphisms of schemes

[forexperts: I am toying with unusual terminology: local-ringedspace rather than locally ringed space. See below for an explanation.How offended are you?]7In order to prove that morphisms behave in a way wehope, we will introduce the notion of a local-ringed space. It will not be used later,although it is useful elsewhere in geometry. The notion of local-ringed spaces isinspired by what we know about manifolds. If π : X → Y is a morphism of mani-folds, with π(p) = q, and f is a function on Y vanishing at q, then the pulled backfunction π#f on X should vanish on p. Put differently: germs of functions (atq ∈ Y ) vanishing at q should pull back to germs of functions (at p ∈ X) vanishingat p.

A local-ringed space is a ringed space (X,OX) such that the stalks OX,x are local-ringed space

all local rings. (The motivation for the terminology comes from thinking of sheavesin terms of stalks. A ringed space is a sheaf whose stalks are rings. A local-ringedspace is a sheaf whose stalks are local rings. The traditional terminology is locally

6e:notmorphism

7s:lrs

158

ringed space, but in light of the other uses of the adjective “locally” in algebraicgeometry, this suggests an unintended definition, that the space is locally a ringedspace.) A morphism of local-ringed spaces f : X → Y is a morphism of ringedspaces such that the induced map of stalks OY,q → OX,p (Exercise 8.3.B) sends themaximal ideal of the former into the maximal ideal of the latter (a “local morphismof local rings”). This means something rather concrete and intuitive: “if p 7→ q,and g is a function vanishing at q, then it will pull back to a function vanishing atp.” Note that local-ringed spaces form a category.category of local-ringed

spaces8.4.A. Exercise. Show that morphisms of local-ringed spaces glue8 (cf. Exer-cise 8.3.A). (Hint: Basically, the proof of Exercise 8.3.A works.)

8.4.B. Easy important exercise. (a) Show that Spec A is a local-ringed space.(b) The morphism of ringed spaces f : Spec A→ Spec B defined by a ring morphismf# : B → A (Exercise 5.4.G) is a morphism of locally ringed spaces. 9 [forme: [H,

Prop. 2.3(a–b)]]

8.4.1. Key Proposition. — If f : Spec A → Spec B is a morphism of local-ringed spaces then it is the morphism of locally ringed spaces induced by the mapf# : B = Γ(Spec B,OSpecB) → Γ(Spec A,OSpecA) = A as in Exercise 8.4.B(b).[forme: [H, Prop. 2.3(c)]]

ex.done

Proof. Suppose f : Spec A→ Spec B is a morphism of locally ringed spaces. Thenwe wish to show that f# : OSpecB → f∗OSpecA is the morphism of sheaves given byExercise 8.3.2 (cf. Exercise 8.4.B(b)). It suffices to checked this on the distinguishedbase (Exercise 4.7.C(a)). We now want to check that for any map of local-ringedspaces including map of sheaves OSpecB → f∗OSpecA, the map of sections on anydistinguished open set D(b) ⊂ Spec B is determined by the map of global sectionsB → A.

Note that if b ∈ B, f−1(D(b)) = D(f#b); this is where we use the hypothesisthat f is a morphism of local-ringed spaces.

The commutative diagram

Γ(Spec B,OSpecB)f#Spec B //

⊗BBb

Γ(Spec A,OSpecA)

⊗AAf#b

=⊗BBb

Γ(D(b),OSpecB)

f#D(b) // Γ(D(f#b),OSpecA)

may be written as

Bf#Spec B //

⊗BBb

A

⊗AAf#b

=⊗BBb

Bb

f#D(b) // Af#b.

But then the lower map f#D(b) is indeed determined by the rest of the diagram.

8e:morphismslrsglue

9e:mlocringedsp

159

We are ready for our definition.

8.4.2. Definition. If X and Y are schemes, then a morphism of local-ringedspaces is called a morphism of schemes. We have thus defined a category ofschemes. (We then have notions of isomorphism — just the same as before,§6.1.2 — and automorphism.)

The definition in terms of local-ringed spaces easily implies tentative defini-tion 8.3.3:

8.4.C. Important Exercise. Show that a morphism of schemes f : X →Y is a morphism of ringed spaces that looks locally like morphisms of affines.Precisely, if Spec A is an affine open subset of X and Spec B is an affine opensubset of Y , and f(Spec A) ⊂ Spec B, then the induced morphism of ringed spaces(Exercise 9.1.2) is a morphism of affine schemes. Show that it suffices to check ona set (Spec Ai, Spec Bi) where the Spec Ai form an open cover X . 10 morphism of schemes

ex.doneIn practice, we will use the fact the affine cover interpretation, and forget

completely about local-ringed spaces.It is also clear (from the corresponding facts about local-ringed spaces) that

morphisms glue (Exercise 8.4.A), and the composition of two morphisms is a mor-phism. Isomorphisms in this category are precise what we defined them to be earlier(§6.1.2).

8.4.3. The category of schemes (or k-schemes, or A-schemes, or Z-schemes). 11 It is often convenient to consider subcategories. For example,the category of k-schemes (where k is a field) is defined as follows. The objects aremorphisms of the form

X

structure morphism

Spec k

(This is the same definition as in §7.3.6, but in a more satisfactory form.) The structure morphism

morphisms in the category of k-schemes are commutative diagrams

X

// Y

Spec k

= // Spec k

which is more conveniently written as a commutative diagram

X //

##GGG

GGGG

GGY

xxxx

xxxx

x

Spec k.

For example, complex geometers may consider the category of C-schemes.When there is no confusion, simply the top row of the diagram is given. More

generally, if A is a ring, the category of A-schemes is defined in the same way, with

10morphismdef

11categoryofschemes

160

A replacing k. And if Z is a scheme, the category of Z-schemes is defined in thesame way, with Z replacing Spec k.

8.4.4. Examples.

8.4.D. Important exercise. (This exercise will give you some practice withunderstanding morphisms of schemes by cutting up into affine open sets.) Make

sense of the following sentence: “An+1 \ ~0 → Pn given by

(x0, x1, . . . , xn+1) 7→ [x0; x1; . . . ; xn]

is a morphism of schemes.” Caution: you can’t just say where points go; youhave to say where functions go. So you’ll have to divide these up into affines, anddescribe the maps, and check that they glue.

8.4.E. Important exercise. Show that morphisms X → Spec A are in naturalbijection with ring morphisms A → Γ(X,OX). 12 (Hint: Show that this is truewhen X is affine. Use the fact that morphisms glue.)

ex.doneIn particular, there is a canonical morphism from a scheme to Spec of its space

of global sections. (Warning: Even if X is a finite-type k-scheme, the ring ofglobal sections might be nasty! In particular, it might not be finitely generated,see 25.15.1.)

ex.done

8.4.5. ? Side fact for experts: Γ and Spec are adjoints. We have a functor Specfrom rings to local-ringed spaces, and a functor Γ from local-ringed spaces to rings.In fact (Γ, Spec) is an adjoint pair! Thus we could have defined Spec by requiringit to be adjoint to Γ. [forme: This is mostly implied by Exercise 8.4.E, but not

completely. This might be important for the definition of Spec.]

8.4.F. Easy Exercise. Show that Spec Z is the final object in the categoryof schemes. In other words, if X is any scheme, there exists a unique morphismto Spec Z. (Hence the category of schemes is isomorphic to the category of Z-schemes.)13

8.4.G. Unimportant easy exercise. Show that morphisms X → A1Z =

Spec Z[t] correspond to global sections of the structure sheaf. [forme: Possible

picture here: C[t].] 14

ex.done

8.4.6. ? Representable functors. [forme: This isn’t so great. Also, checking

that something is a ring-scheme is best done in terms of the functor. Also, the main

point here is that we get a group object. Later discussion: this should be the next

(second?) optional starred bit involving Yoneda. I should define a group object in

a category. Then they will see that they can struggle with annoyances showing that

A1 is a group scheme, or they can do it Yoneda-style. The next example might be

projective space; or else it might be fibered products.] This is one of our firstexplicit examples of an important idea, that of representable functors. This is avery useful idea, whose utility isn’t immediately apparent. We have a contravariant

12morphismtoaffine

13Zfinal

14A1Z

161

functor from schemes to sets, taking a scheme to its set of global sections. We haveanother contravariant functor from schemes to sets, taking X to Hom(X, Spec Z[t]).This is describing a (natural) isomorphism of the functors. More precisely, we aredescribing an isomorphism Γ(X,OX) ∼= Hom(X, Spec Z[t]) that behaves well withrespect to morphisms of schemes: given any morphism f : X → Y , the diagram

Γ(Y,OY )∼ //

f∗

Hom(Y, Spec Z[t])

f

Γ(X,OX)

∼// Hom(X, Spec Z[t])

commutes. Given a contravariant functor from schemes to sets, by the usual uni-versal property argument, there is only one possible scheme Y , up to isomorphism,such that there is a natural isomorphism between this functor and Hom(·, Y ). Ifthere is such a Y , we say that the functor is representable.

Here are a couple of examples of something you’ve seen to put it in context. (i)The contravariant functor hY = Hom(·, Y ) (i.e. X 7→ Hom(X, Y ), Example 3.2.19)is representable by Y . (ii) Suppose we have morphisms X, Y → Z. The contravari-ant functor Hom(·, X)×Hom(·,Z) Hom(·, Y ) is representable if and only if the fiberedproduct X ×Z Y exists (and indeed then the contravariant functor is representedby Hom(·, X ×Z Y )). This is purely a translation of the definition of the fiberedproduct — you should verify this yourself.

Remark for experts: The global sections form something better than a set —they form a scheme. You can define the notion of ring scheme, and show that ifa contravariant functor from schemes to rings is representable (as a contravariantfunctor from schemes to sets) by a scheme Y , then Y is guaranteed to be a ringscheme. The same is true for group schemes. [forme: Define group scheme for

them? This is now done later, in 25.14.6. (For experts: you know enough to define

ring scheme, or indeed a ring object in any category, by “categorifying” the definition

of a ring. Whoops, we need to do this over a base.)]

8.4.H. Related easy exercise. Show that invertible global sections of OX

correspond naturally to maps to Spec Z[t, t−1]. [forme: Method 1: Z[t, t−1]. Method

2: by hand.] (Spec Z[t, t−1] is a group scheme.) [forme: Group schemes are group scheme

discussed/defined in 25.14.6. There I should ask them: what is a group manifold =

Lie group?]

[primordial: Omit:

8.4.I. Exercise. (Morphisms from Spec of a local ring to X) Suppose X is a scheme, and (R,m)is a local ring. Suppose we have a scheme morphism π : SpecR→ X sending [m] to x. Show thatany open set containing x contains the image of f . (That’s basically the previous exercise!) Showthat there is a bijection between Hom(SpecR,X) and x ∈ X, local homomorphismsOx,X → R.[This currently isn’t used, but I think it might be.] 15

8.4.7. Radiciel morphisms ??. Joe Rabinoff writes: According to EGA I, section 3.5, amorphism f : X → Y of preschemes is radiciel if any of the following equivalent conditions hold:

(1) f is a universally injective morphism of schemes(2) f : X(K)→ Y (K) is injective for any field K(3) f : X(Ω)→ Y (Ω) is injective for any algebraically closed field Ω (d) f is injective and

for every x ∈ X, the extension of residue fields k(f(x))→ k(x) is radiciel

15M.II.6.P.3

162

I didn’t actually know what a radiciel extension of fields was, but one can derive it using the fact

that the above conditions are equivalent. Indeed, the equivalence of the conditions implies that an

extension K ⊂ L of fields is radiciel if and only if the morphism SpecL→ SpecK is radiciel, which

is true if and only if, for every algebraically closed field Ω, two homomorphisms L → Ω agree if

and only if they agree on K. I think this is the same thing as L/K being a purely inseparable

field extension, no? ]radiciel

CHAPTER 9

Some properties of morphisms of schemes

We now define a bunch of types of morphisms.

9.1 Quasicompact; open immersion; affine, finite, closedimmersion; locally closed immersion

In this section, we’ll give some analogues of open subsets, closed subsets, andlocally closed subsets. This will also give us an excuse to define quasicompact, affineand finite morphisms (closed immersions are a special case). It will also give us anexcuse to define some important special closed immersions, in the next section. Insection after that, we’ll define some more types of morphisms.

9.1.1. Quasicompact and quasiseparated morphisms.[forme: We need this before locally closed immersions, because we use this as

a hypothesis in the next section.] A morphism f : X → Y is quasicompact quasicompact morphism

if for every open affine subset U of Y , f−1(U) is quasicompact. Equivalently, thepreimage of any quasicompact open subset is quasicompact. We will like this notionbecause (i) we know how to take the maximum of a finite set of numbers, and (ii)most reasonable schemes will be quasicompact. [forme: Really, we will take max

over a finite set. Also, this sounds like the notion of proper morphism in ordinary

topology, but don’t say that.]

9.1.A. Easy exercise. Show that the composition of two quasicompact mor-phisms is quasicompact.

9.1.B. Exercise. Show that any morphism from a Noetherian scheme is quasi-compact. 1

9.1.C. Exercise (quasicompactness is affine-local on the target). Showthat a morphism f : X → Y is quasicompact if there is cover of Y by open affinesets Ui such that f−1(Ui) is quasicompact. [forme: Ex. H.II.3.2] (Hint: easyapplication of the affine communication lemma!) 2

ex.doneAlong with quasicompactness comes the weird notion of quasiseparatedness.

A morphism f : X → Y is quasiseparated if for every open affine subset U of quasiseparated mor-

phismY , f−1(U) is a quasiseparated scheme (§7.1.1). This will be a useful hypothesisin theorems (in conjunction with quasicompactness), and that various interestingkinds of morphisms (locally Noetherian source, affine, separated, see Exercise 9.1.D,

1Nqc

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164

Exercise 9.1.J, and [forme: ???] resp.) are quasiseparated, and this will allow usto state theorems more succinctly.

9.1.D. Exercise. Show that any morphism from a locally Noetherian schemeis quasiseparated. (Hint: Exercise 7.3.C.) Thus those readers working only withNoetherian schemes may take this as a standing hypothesis. 3

9.1.E. Easy exercise. Show that the composition of two quasiseparated mor-phisms is quasiseparated.

9.1.F. Exercise (quasiseparatedness is affine-local on the target). Showthat a morphism f : X → Y is quasiseparated if there is cover of Y by open affinesets Ui such that f−1(Ui) is quasiseparated. (Hint: easy application of the affinecommunication lemma!)

ex.doneFollowing Grothendieck’s philosophy of thinking that the important notions are

properties of morphisms, not of objects, we can restate the definition of quasicom-pact (resp. quasiseparated) scheme as a scheme that is quasicompact (resp. qua-siseparated) over the final object Spec Z in the category of schemes (Exercise 8.4.F).

9.1.2. Open immersions.An open immersion of schemes is defined to be an open immersion as ringed

spaces. In other words, a morphism f : (X,OX) → (Y,OY ) of schemes is an openimmersion if f factors as

(X,OX)g

∼// (U,OY |U )

h // (Y,OY )

where g is an isomorphism, and U → Y is an inclusion of an open set. It isimmediate that isomorphisms are open immersions. We say that (U,OY |U ) is anopen subscheme of (Y,OY ), and often sloppily say that (X,OX) is an opensubscheme of (Y,OY ). 4 [primordial: We make an obvious observation: if W ⊂ Xopen subscheme

open immersion (of

ringed spaces and

schemes)

and Y ⊂ Z are both open immersions of ringed spaces, show that any morphism of ringed spaces

X → Y induces a morphism of ringed spaces W → Z, by composition W → X → Y → Z.5

[forme: This doesn’t deserve to be said!] ]

open imm9.1.G. Important but easy exercise. Suppose i : U → Z is an open im-mersion, and f : Y → Z is any morphism. Show that U ×Z Y exists. (Hint: I’lleven tell you what it is: (f−1(U),OY |f−1(U)).) [forme: This is important for the

existence of fibered product.] 6

9.1.H. Easy exercise. If U → Z and V → Z are open immersions, describe anisomorphism U ∩ V ∼= U ×Z V .

9.1.I. Easy exercise. Suppose f : X → Y is an open immersion. Show thatif Y is locally Noetherian, then X is too. Show that if Y is Noetherian, then Xis too. However, show that if Y is quasicompact, X need not be. (Hint: let Y beaffine but not Noetherian, see [forme: XXX] .)

ex.done

3lNmqsep

4d:openimm

5ringedspacerestriction

6fiberedopen

165

“Open immersions” are scheme-theoretic analogues of open subsets. “Closedimmersions” are scheme-theoretic analogues of closed subsets, but they are of aquite different flavor, as we’ll see soon.

9.1.3. Affine morphisms.A morphism f : X → Y is affine if for every affine open set U of Y , f−1(U) is

an affine scheme. We have immediately that affine morphisms are quasicompact.7 affine morphism

9.1.J. Fast Exercise. Show that affine morphisms are quasiseparated. (Hint:Exercise 7.1.G.) 8

9.1.4. Proposition (the property of “affineness” is affine-local on thetarget). — A morphism f : X → Y is affine if there is a cover of Y by affineopen sets U such that f−1(U) is affine. 9 [forme: Ex. H.II.17(b) — which section

of II?]ex.done

For part of the proof, it will be handy to have a lemma.

9.1.5. Lemma. — If X is a quasicompact quasiseparated scheme and s ∈Γ(X,OX), then the natural map Γ(X,OX)s → Γ(Xs,OX) is an isomorphism.10

[forme: Jarod gave this, and he says: “I think the lemma is interesting on its own

and has the generalization (Ln≥0 Γ(X, F⊗Ln))(s) = Γ(Xs,F) for quasi-coherent sheaves

F , line bundles L and sections s ∈ Γ(X,Ln) for n ≥ 0.”]ex.done

A brief reassuring comment on the “quasicompact quasiseparated” hypothesis:This just means that X can be covered by a finite number of affine open subsets,any two of which have intersection also covered by a finite number of affine opensubsets (Exercise 7.1.H). The hypothesis applies in lots of interesting situations,such as if X is affine (Exercise 7.1.G) or Noetherian (Exercise 7.3.C).

Proof. Cover X with finitely many affine open sets Ui = Spec Ai. Let Uij = Ui∩Uj .Then

Γ(X,OX)→∏

i

Ai ⇒∏

i,j

Γ(Uij ,OX)

is exact. Localizing at s gives

Γ(X,OX)s →(

∏

i

Ai

)

s

⇒

∏

i,j

Γ(Uij ,OX)

s

As localization commutes with finite products,

Γ(X,OX)s →∏

i

(Ai)si ⇒∏

i,j

Γ(Uij ,OX)s

is exact, where the global function s induces functions si ∈ Ai. If Γ(Uij ,OX)s∼=

Γ((Uij)s,OX), then it is clear that Γ(X,Ox)s are the sections over Xs. Note that Uij

are quasicompact, by the quasiseparatedness hypothesis, and also quasiseparated,as open subsets of quasiseparated schemes are quasiseparated. Therefore we canreduce to the case where X ⊆ Spec A is a (quasicompact quasiseparated) open

7M.II.7.D.3b

8affmqsep

9affineaffinelocal

10affineaffinelocallemma

166

subset of an affine scheme. Then Uij = Spec Afifj is affine and Γ(Uij ,Ox)s =Γ((Uij)s,OX) so the same exact sequence implies the result.

Proof of Proposition 9.1.4. As usual, we use the Affine Communication Lemma 7.3.3.We check our two criteria. First, suppose f : X → Y is affine over Spec B, i.e.f−1(Spec B) = Spec A. Then f−1(Spec Bs) = Spec Af#s.

Second, suppose we are given f : X → Spec B and (f1, . . . , fn) = B with Xfi

affine (Spec Ai, say). We wish to show that X is affine too. X is quasi-compact(as it is covered by n affine open sets). Let ti ∈ Γ(X,OX) be the pullback of thesections si ∈ B. The morphism f factors as h g where g : X → Spec Γ(X,OX)and h : Spec Γ(X,OX)→ Spec B are the natural maps. Then Lemma 9.1.5 impliesthat g|f−1(SpecBsi

) : Xti → Spec Γ(X,OX)ti are isomorphisms. Therefore, g is an

isomorphism and X is affine.

9.1.6. Finite morphisms.11 [forme: Perhaps Give the example of the Frobenius morphism. Ideally give

the correct name to the various Frobenii.] An affine morphism f : X → Y isfinite if for every affine open set Spec B of Y , f−1(Spec B) is the spectrum of anB-algebra that is a finitely-generated B-module. Warning about terminology (finitefinite morphism (is

affine) vs. finitely-generated): Recall that if we have a ring homomorphism A → B suchthat B is a finitely-generated A-module then we say that B is a finite A-algebra.This is stronger than being a finitely-generated A-algebra.finite module

By definition, finite morphisms are affine.12 [forme: Soon: how to picture it.]

9.1.K. Exercise (the property of finiteness is affine-local on the tar-get). Show that a morphism f : X → Y is finite if there is a cover of Y by affineopen sets Spec A such that f−1(Spec A) is the spectrum of a finite A-algebra.

9.1.L. Easy Exercise (cf. Exercise 15.2.C). Show that the composition oftwo finite morphisms is also finite.13 [forme: extremely easy]finite extension of rings

We now give four examples of finite morphisms, to give you some feeling forhow finite morphisms behave. In each example, you’ll notice two things. In eachcase, the maps are always finite-to-one. We’ll verify this in Exercise 9.3.E. You’llalso notice that the morphisms are closed, i.e. the image of closed sets are closed.We’ll show that finite morphisms are always closed in Corollary 14.1.6. Intuitively,you should think of finite as being closed plus finite fibers, although this isn’t quitetrue. We’ll make this precise later. [forme: Where?] [forme: Maybe I should

include [M] I.7 Prop. 3.3, finite f is surjective iff f∗ is injective; maybe that means

on all affines of the target.14]

Example 1: Branched covers. If p(t) ∈ k[t] is a non-zero polynomial, thenSpec k[t]→ Spec[u] given by u 7→ p(t) is a finite morphism, see Figure 1.

Example 2: Closed immersions (to be defined soon, in §9.1.8). The morphismSpec k → Spec k[t] given by t 7→ 0 is a finite morphism, see Figure 2.

11d:finite

12M.II.7.D.3a

13finitefinite

14M.I.7.P.3.3

167

Figure 1. The “branched cover” of A1 given by u 7→ p(t) is finite

0

Figure 2. The “closed immersion” Spec k → Spec k[t] is finite

Example 3: Normalization (to be defined later). [forme: Where?] Themorphism Spec k[t]→ Spec k[x, y]/(y2 − x2 − x3) given by (x, y) 7→ (t2 − 1, t3 − t)(check that this is a well-defined ring map!) is a finite morphism, see Figure 3.

9.1.M. Important Exercise (Example 4, finite morphisms to Spec k).Show that if X → Spec k is a finite morphism, then X is a discrete finite union ofpoints, each with residue field a finite extension of k, see Figure 4. (An example isSpec F8 × F4[x, y]/(x2, y4)× F4[t]/t9 × F2 → Spec F2.)

15

9.1.7. Example. The natural map A2 − (0, 0) → A2 is an open immersion, andhas finite fibers, but is not affine (as A2−(0, 0) isn’t affine, §6.1.5) and hence notfinite. 16

We will later (§15.2) define the related notion of integral morphisms. integral morphism

9.1.8. Closed immersions and closed subschemes.17Just as open immersions (the scheme-theoretic version of open set) are locally

modeled on open sets U ⊂ Y , the analogue of closed subsets also has a local model.This was foreshadowed by our understanding of closed subsets of Spec B as roughlycorresponding to ideals. If I ⊂ B is an ideal, then Spec B/I → Spec B is amorphism of schemes, and this is our prototypical example of a closed immersion.18

15finiteoverfield

16qfnotf

17cidef

18d:climm

168

Figure 3. The “normalization” Spec k[t]→ Spec k[x, y]/(y2−x2−x3) given by (x, y) 7→ (t2 − 1, t3 − t) is finite

A morphism f : X → Y is a closed immersion if it is an affine morphism,and for each open subset Spec B ⊂ Y , with f−1(Spec B) ∼= Spec A, B → A is asurjective map (i.e. of the form B → B/I , our desired local model). We often saythat X is a closed subscheme of Y .cl.imm, cl subscheme

9.1.N. Easy exercise. 19 Show that closed immersions are finite.

9.1.O. Easy exercise. Show that the composition of two closed immersions isa closed immersion. 20

9.1.P. Exercise. Show that the property of being a closed immersion is affine-local on the target. [forme: Ex. H.II.3.11(b) implied by this] 21closed immersion affine-

local19

H.E.II.5.5b20

climmcomp21

climmaffinelocal

169

Figure 4. A picture of a finite morphism to Spec k. Notice thatbigger fields are written as bigger dots. [I’d like to add some fuzzto some of these points at some point.]

A closed immersion f : X → Y determines an ideal sheaf on Y , as the kernelIX/Y of the map of OY -modules

OY → f∗OX

(An ideal sheaf on Y is what it sounds like: it is a sheaf of ideals. It is a sub-OY -module I → OY . On each open subset, it gives an ideal I(U) → OY (U).) We ideal sheaf

thus have an exact sequence 0→ IX/Y → OY → f∗OX → 0.

9.1.Q. Important exercise: a useful criterion for when ideals in affineopen sets define a closed subscheme. It will be convenient (for example in§9.2) to define certain closed subschemes of Y by defining on any affine open subsetSpec B of Y an ideal IB ⊂ B. Show that these Spec B/IB → Spec B glue togetherto form a closed subscheme precisely if for each affine open subset Spec B → Y andeach f ∈ B, I(Bf ) = (IB)f . 22 [forme: Diogo’s comments show me that this is a

hard question! Explain more about what it means.]

Warning: you might hope that closed subschemes correspond to ideal sheavesof OY . Sadly not every ideal sheaf arises in this way. Here is an example.

9.1.R. Unimportant exercise. Let X = Spec k[x](x), the germ of the affineline at the origin, which has two points, the closed point and the generic point η.Define I(X) = 0 ⊂ OX(X) = k[x](x), and I(η) = k(x) = OX(η). Show that this

sheaf of ideals does not correspond to a closed subscheme (see Exercise 17.11). 23

We will see later [forme: REF] that closed subschemes correspond to qua-sicoherent sheaves of ideals; the mathematical content of this statement will turnout to be precisely Exercise 17.11.

9.1.S. Important Exercise. (a) In analogy with closed subsets, define thenotion of a finite union of closed subschemes of X , and an arbitrary intersection ofclosed subschemes. [forme: Interesting fact: this construction works for arbitrary

unions of closed subschemes.]

22qcsideals

23nonqsci, [EH, p. 24]

170

(b) Describe the scheme-theoretic intersection of (y−x2) and y in A2. See Figure 5for a picture. (For example, explain informally how this corresponds to two curvesmeeting at a single point with multiplicity 2 — notice how the 2 is visible in youranswer. Alternatively, what is the non-reducedness telling you — both its “size”and its “direction”?) Describe the scheme-theoretic union.(c) Describe the scheme-theoretic intersection of (y2 − x2) and y in A2. Draw apicture. (Are you surprised? Did you expect the intersection to be multiplicity oneor multiplicity two?) Hence show that if X , Y , and Z are closed subschemes of W ,then (X ∩ Z) ∪ (Y ∩ Z) 6= (X ∪ Y ) ∩ Z in general.(d) Show that the underlying set of a finite union of closed subschemes is the finiteunion of the underlying sets, and similarly for arbitrary intersections. [forme:

Perhaps enumerate the desired properties. For experts: infinite scheme-theoretic

union still makes sense: intersection of any number of ideals is still an ideal.]

=intersect

Figure 5. The scheme-theoretic intersection of the parabola y =x2 and the x-axis is a non-reduced scheme (with fuzz in the x-direction)

9.1.9. Example: Closed immersions of projective space PnA. 24Consider the defi-

nition of projective space PnA given in §6.1.K (along with the terminology defined

there). Any homogeneous polynomial f in x0, . . . , xn defines a closed subscheme.(Thus even though x0, . . . , xn don’t make sense as functions, their vanishing locusstill makes sense.) On open set Ui, the closed subscheme is f(x0/i, . . . , xn/i), which

we think of as f(x0, . . . , xn)/xdeg fi . On the overlap

Ui ∩ Uj = Spec A[x0/i, . . . , xn/i, x−1j/i]/(xi/i − 1),

these functions on Ui and Uj don’t exactly agree, but they agree up to a non-vanishing scalar, and hence cut out the same subscheme of Ui ∩ Uj :

f(x0/i, . . . , fn/i) = xdeg fj/i f(x0/j , . . . , xn/j).

Thus by intersecting such closed subschemes, we see that any collection of homo-geneous polynomials in A[x0, . . . , xn] cut out a closed subscheme of Pn

A. We couldtake this as a provisional definition of a projective A-scheme (or a projective schemeover A). (We’ll give a better definition in the next Chapter.)

Notice: piggybacking on the annoying calculation that Pn consists of n + 1pieces glued together nicely is the fact that any closed subscheme of Pn cut out bya bunch of homogeneous polynomials consists of n + 1 pieces glued together nicely.

24climmPn

171

Notice also that this subscheme is not in general cut out by a single globalfunction on Pn

A. For example, if A = k, there are no nonconstant global functions(Exercise 6.1.L). We take this opportunity to introduce some related terminology.A closed subscheme is locally principal if on each open set in a small enoughopen cover it is cut out by a single equation. Thus each homogeneous polynomialin n + 1 variables defines a locally principal closed subscheme. (Warning: onecan check this on a fine enough affine open cover, but this is not an affine-localcondition, see Exercise 10.3.C!) A case that will be important later is when theideal sheaf is not just locally generated by a function, but is generated by a functionthat is not a zero-divisor. For reasons that will become a little clearer later, we callsuch a closed subscheme an effective Cartier divisor.25 Warning! We will usethis terminology before we explain where it came from! [forme: The corresponding

ideal is called an invertible ideal sheaf.] effective Cartier divisor,

invertible ideal sheafA closed subscheme cut out by a single (homogeneous) equation is called ahypersurface in Pn

k . The degree of a hypersurface is the degree of the poly-nomial. (Implicit in this is that this notion can be determined from the subschemeitself; we haven’t yet checked this.) A hypersurface of degree 1 (resp. degree 2, 3,. . . ) is called a hyperplane (resp. quadric, cubic, quartic, quintic, sextic,septic, octic, . . . hypersurface). If n = 2, a degree 1 hypersurface is called aline, and a degree 2 hypersurface is called a conic curve, or a conic for short.If n = 3, a hypersurface is called a surface.) (In Chapter 15, we will justify theterms curve and surface.) 26 hypersurface, hyper-

plane, quadric, cubic,

quartic, quintic, sextic,

septic, octic, conic

9.1.T. Exercise. (a) Show that wz = xy, x2 = wy, y2 = xz describes anirreducible curve in P3

k. This curve is called the twisted cubic. The twisted cubicis a good non-trivial example of many things, so it you should make friends withit as soon as possible. 27 [forme: One start: Show that I is the kernel of the twisted cubic

morphism k[w,x, y, z] → k[s, t], w 7→ s, x 7→ st, y 7→ st2, z 7→ st3.] (b) Show that thetwisted cubic is isomorphic to P1

k. (If you are stuck, you will eventually get a hintin Exercise 10.5.F.)

9.1.U. Unimportant Exercise. The usual definition of a closed immersion isa morphism f : X → Y such that f induces a homeomorphism of the underlyingtopological space of Y onto a closed subset of the topological space of X , and theinduced map f# : OX → f∗OY of sheaves on X is surjective. Show that thisdefinition agrees with the one given above. (To show that our definition involvingsurjectivity on the level of affine open sets implies this definition, you can usethe fact that surjectivity of a morphism of sheaves can be checked on a base,Exercise 4.7.E.) [forme: [H, p. 85]] [forme: Diogo’s comments show that this is a

hard exercise. Perhaps give more of a hint. The fact that this is hard is one reason

for me to prefer the definition I give.]

9.1.10. ? A fun example. The affine-locality of affine morphisms (Proposi-tion 9.1.4) has some non-obvious consequences, as shown in the next exercise.

[forme: effective Cartier divisor defined in §9.1.9.]

25d:effectivecartierdivisor

26firstirrhyp

27d:twistedcubic

172

9.1.V. Exercise. Suppose X is an affine scheme, and Y is a locally principalclosed subscheme. Show that X −Y is affine. (This is clear if Y is globally cut outby one equation f ; then if X = Spec A then Y = Spec Af . However, Y is not alwaysof this form, see Exercise 10.3.C.) [forme: [Example is given in Example 25.15.1.]

] 28

[forme: Now this is in the wrong order.]

9.1.11. Example. Here is an explicit consequence. We showed earlier that onthe cone over the smooth quadric surface Spec k[w, x, y, z]/(wz−xy), the cone overa ruling w = x = 0 is not cut out scheme-theoretically by a single equation, byconsidering Zariski-tangent spaces. We now show that it isn’t even cut out set-theoretically by a single equation. For if it were, its complement would be affine.But then the closed subscheme of the complement cut out by y = z = 0 wouldbe affine. But this is the scheme y = z = 0 (also known as the wx-plane) minusthe point w = x = 0, which we’ve seen is non-affine (§6.1.5). (For comparison, onthe cone Spec k[x, y, z]/(xy − z2), see Figure 4, the ruling x = z = 0 is not cutout scheme-theoretically by a single equation, but it is cut out set-theoretically byx = 0.) Verify all this! (Hint: Use Exercise 9.1.4.) 29 [forme: Other hint: f∗OX is

finite type. I can’t give this because we don’t know this yet.]

We have now defined the analog of open subsets and closed subsets in the land ofschemes. Their definition is slightly less “symmetric” than in the usual topologicalsetting: the “complement” of a closed subscheme is a unique open subscheme, butthere are many “complementary” closed subschemes to a given open subscheme ingeneral. (We’ll soon define one that is “best”, that has a reduced structure, §9.2.6.)

9.1.12. Locally closed immersions and locally closed subschemes.[forme: [From Joe R: EGA I.4 has 8 pages on the notion of “immersion”, which

is a closed immersion followed by an open immersion. He proves: an immersion is a

map that’s a homeomorphism of the topological space onto its image, such that the

map of local rings are surjective (this is the hard one); a composition of immersions

is an immersion; being an immersion is local on the base, and is preserved under

base change; conditions for a morphism to factor through an immersion; if X → X ′

is any morphism then Y ×X Y → Y ×X′ Y ′ is an immersion (when such a thing makes

sense).]] [forme: Add: Interesting example of something that is not a locally

closed immersion: normalization of nodal curve in plane. Figure ??.] Now that wehave defined analogs of open and closed subsets, it is natural to define the analogof locally closed subsets. Recall that locally closed subsets are intersections ofopen subsets and closed subsets. Hence they are closed subsets of open subsets, orequivalently open subsets of closed subsets. That equivalence will be a little subtlein the land of schemes.

We say a morphism X → Y is a locally closed immersion if it can factored

into Xf // Z

g // Y where f is a closed immersion and g is an open immersion.(Warning: The term immersion is often used instead of locally closed immersion,but this is unwise terminology, the differential geometric notion of immersion iscloser to the what algebraic geometers call unramified, which we’ll define in XXXX.

28affineminusCartier

29M.II.7.P.5f, he says he’ll prove it III.1

173

The algebro-geometric notion of locally closed immersion is closest to the differentialgeometric notion of embedding.)

It is often said that X is a locally closed subscheme of Y .For example, Spec k[t, t−1] → Spec k[x, y] where (x, y) 7→ (t, 0) is a locally

closed immersion (see Figure 6). locally closed immersion

Figure 6. The locally closed immersion Spec k[t, t−1] → k[x, y](t 7→ (t, 0) = (x, y), i.e. (x, y)→ (t, 0))

We can make sense of finite intersections of locally closed immersions, and theyare also locally closed immersions.

Clearly a open subscheme U of a closed subscheme V of X can be interpretedas a closed subscheme of an open subscheme: as the topology on V is induced fromthe topology on X , the underlying set of U is the intersection of some open subsetU ′ on X with V . We can take V ′ = V ∩U , and then V ′ → U ′ is a closed immersion,and U ′ → X is an open immersion.

It is less clear that a closed subscheme V ′ of an open subscheme U ′ can beexpressed as an open subscheme U of a closed subscheme V . In the category oftopological spaces, we would take V as the closure of V ′, so we are now motivated todefine the analogous construction, which will give us an excuse to introduce severalrelated ideas, in the next section. We will then resolve this issue in good cases (e.g.if X is Noetherian) in Exercise 9.2.D.

9.2 Constructions related to “smallest closed subschemes”:scheme-theoretic image, scheme-theoretic closure, induced

reduced subscheme, and the reduction of a scheme

30We now define a series of notions that are all of the form “the smallest closedsubscheme such that something or other is true”. One example will be the notion ofscheme-theoretic closure of a locally closed immersion, which will allow us to inter-pret locally closed immersions in three equivalent ways (open subscheme intersect

30s:stimage

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closed subscheme; open subscheme of closed subscheme; and closed subscheme ofopen subscheme).

9.2.1. Scheme-theoretic image.[forme: [Brian O writes: “Interesting cultural note: EGA seems to have missed

the fact that scheme-theoretic image always exists, although the proof isn’t too hard.”

See EGA I, section 9.5, and there it is called “closed image”.] We start with thenotion of scheme-theoretic image. If f : X → Y is a morphism of schemes, thenotion of the image of f as sets is clear: we just take the points in Y that are theimage of points in X . But if we would like the image as a scheme, then the notionbecomes more problematic. (For example, what is the image of A2 → A2 given by(x, y) 7→ (x, xy)?) We will come back to the notion of image later, but for now wewill define the “scheme-theoretic image”. This will incorporate the notion that theimage of something with non-reduced structure (“fuzz”) can also have non-reducedstructure.

Definition. Suppose i : Z → Y is a closed subscheme, giving an exact sequence0 → IZ/Y → OY → i∗OZ → 0. We say that the image of f : X → Y lies in Zif the composition IZ/Y → O)Y → f∗OX is zero. Informally, locally functionsvanishing on Z pull back to the zero function on X . If the image of f lies intwo subschemes Z1 and Z2, it clearly lies in their intersection Z1 ∩ Z2. We thendefine the scheme-theoretic image of f of f , a closed subscheme on Y , as the“smallest closed subscheme containing the image”, i.e. the intersection of all closedsubschemes containing the image.

Example 1. Consider Spec k[ε]/ε2 → Spec k[x] = A1k given by x 7→ ε. Then the

scheme-theoretic image is given by k[x]/x2 (the polynomials pulling back to 0 areprecisely multiples of x2). Thus the image of the fuzzy point still has some fuzz.

[forme: Figure here]

Example 2. Consider f : Spec k[ε]/ε2 → Spec k[x] = A1k given by x 7→ 0. Then

the scheme-theoretic image is given by k[x]/x: the image is reduced. In this picture,the fuzz is “collapsed” by f .

Example 3. Consider f : Spec k[t, t−1] = A1 − 0 → A1 = Spec k[u] given byu 7→ t. Any function g(u) which pulls back to 0 as a function of t must be thezero-function. Thus the scheme-theoretic image is everything. The set-theoreticimage, on the other hand, is the distinguished open set A1 − 0. Thus in not-too-pathological cases, the underlying set of the scheme-theoretic image is not theset-theoretic image. But the situation isn’t terrible: the underlying set of thescheme-theoretic image must be closed, and indeed it is the closure of the set-theoretic image. We might imagine that in reasonable cases this will be true, andin even nicer cases, the underlying set of the scheme-theoretic image will be set-theoretic image. We will later [forme: when?] see that this is indeed thecase.

But we feel obliged to show that pathologies can happen.Example 4. Let X =

∐

k[εn]/(εnn) and Y = Spec k[x], and define X → Y by

x → εn on the nth component of X . Then if a function g(x) on Y pulls back to 0on X , then its Taylor expansion is 0 to order n (by examining the pullback to thenth component of X , so g(x) must be 0. Thus the scheme-theoretic image is Y ,while the set-theoretic image is easily seen to be just the origin.

This example clearly is weird though, and we can show that in “reasonablecircumstances” such pathology doesn’t occur. It would be great to compute the

175

scheme-theoretic image affine-locally. On affine open set Spec B ⊂ Y , define theideal IB ⊂ B of functions which pullback to 0 on X . (Formally, IB := ker(B →Γ(f∗(OX), Spec B).) Then if for each such B, and each g ∈ B, IB ⊗B Bg →IBg is an isomorphism, then we will have defined the pushforward subscheme (seeExercise 17.11). Clearly each function on Spec B that vanishes when pulled backto f−1(Spec B) also vanishes when restricted to D(g) and then pulled back tof−1(D(g)). So the question is: given a function r/gn on D(g) that pulls back tof−1D(g), is it true that for some m, rgm = 0 when pulled back to f−1(Spec B)?(i) The answer is clearly yes if f−1(Spec B) is reduced: we simply take rg. (ii) Theanswer is also yes if f−1(Spec B) is affine, say Spec A: if r′ = f#r and g′ = f#gin A, then if r′ = 0 on D(g′), then there is an m such that r′(g′)m = 0: r′ = 0in D(g′), which means precisely this fact. (iii) Furthermore, the answer is yes iff−1(Spec B) is quasicompact: cover f−1(Spec B) with finitely many affine opensets. For each one there will be some mi so that rgmi = 0 when pulled back tothis open set. Then let m = max(mi). (We now see why quasicompactness is ourfriend!)

In conclusion, we have proved the following theorem.

9.2.2. Theorem. — Suppose f : X → Y is a morphism of schemes. If X isreduced or f is quasicompact (e.g. if X is Noetherian, Exercise 9.1.B), then thescheme-theoretic image of f may be computed affine-locally. 31

9.2.3. Corollary. — Under the hypotheses of the previous theorem, the closureof the set-theoretic image of f is the underlying set of the scheme-theoretic image.

Example 4 above shows that we cannot excise these hypotheses.

Proof. The set-theoretic image is clearly in the underlying set of the scheme-theoretic image. The underlying set of the scheme-theoretic image is closed, sothe closure of the set-theoretic image is contained in underlying set of the scheme-theoretic image. On the other hand, if U is the complement of the closure of theset-theoretic image, f−1(U) = ∅. As under these hypotheses, the scheme theoreticimage can be computed locally, the scheme-theoretic image is the empty set onU .

We conclude with a few stray remarks.

9.2.A. Easy Exercise. If X is reduced, show that the scheme-theoretic imageof f : X → Y is also reduced.

More generally, you might expect there to be no unnecessary non-reduced struc-ture on the image not forced by non-reduced structure on the source. We make thisprecise in the locally Noetherian case, when we can talk about associated points.

9.2.B. ? Unimportant exercise. If f : X → Y is a morphism of locallyNoetherian schemes, show that the associated points of the image subscheme are asubset of the image of the associated points of X . 32

9.2.4. Aside: set-theoretic images can be nice too. I want to say a littlemore on what the set-theoretic image of a morphism can look like, although we’ll

31niceschemetheoreticimage

32imageassociated, used shortly

176

hold off before proving these statements. We know that the set-theoretic imagecan be open (open immersion), and closed (closed immersions), and locally closed(locally closed immersions). But it can be weirder still: consider the exampleA2 → A2 given by (x, y) 7→ (x, xy) mentioned earlier. The image is the plane,minus the y-axis, plus the origin. The image can be stranger still, and indeed if Sis any subset of a scheme Y , it can be the image of a morphism: let X be the disjointunion of spectra of the residue fields of all the points of S, and let f : X → Y be thenatural map. This is quite pathological, and in fact that if we are in any reasonablesituation, the image is essentially no worse than arose in the previous example.

We define a constructable subset of a Noetherian scheme to be a subsetwhich belongs to the smallest family of subsets such that (i) every open set is inthe family, (ii) a finite intersection of family members is in the family, and (iii) thecomplement of a family member is also in the family. So for example the image of(x, y) 7→ (x, xy) is constructable.constructable subset

Note that if X → Y is a morphism of Noetherian schemes, then the preimageof a constructable set is a constructable set.

9.2.C. Exercise. Suppose X is a Noetherian scheme. Show that a subset of Xis constructable if and only if it is the finite disjoint union of locally closed subsets.[forme: Luis Diogo says: this is true on an arbitrary topological space. His argument

is in the file diogoconstr.pdf in my FOAG/”other references” directory.]ex.done

Then if f : X → Y is a finite type morphism of Noetherian schemes, theimage of any constructable set is constructable. This is Chevalley’s Theorem (The-orem 15.5.1), and we will prove it later. We will also have reasonable criteria forwhen the image is closed. 33Chev’s Thm

(For hardened experts only: [EGA, 0III.9.1] gives a definition of constructablein more generality. A constructable subset of a topological space X is a memberof the Boolean algebra generated by open subsets U of X such that the inclusionU → X is quasicompact. If X is an affine scheme, or more generally quasicompactand quasiseparated, this is equivalent to U being quasicompact. A subset Z ⊂ Xis locally constructable if X admits an open covering Vi such that Z ∩ Vi ⊂ Vi

is constructable for each i. If X is quasicompact and quasiseparated, this is thesame as Z ⊂ X being constructable, so if X is a scheme, then it is equivalent tosay that Z ∩ V is constructable for every affine open set V . The general form ofChevalley’s constructibility theorem [EGA, IV1.1.8.4] is that the image of a locallyconstructable set under a finitely presented map, is also locally constructibility.)

9.2.5. Scheme-theoretic closure of a locally closed subscheme.We define the scheme-theoretic closure of a locally closed immersion f :

X → Y as the scheme-theoretic image of X .

9.2.D. Exercise. If X → Y is quasicompact (e.g. if X is Noetherian, Exer-cise 9.1.B) or if X is reduced, show that the following three notions are the same.(Hint: Theorem niceschemetheoreticimage.)

(a) V is an open subscheme of X intersect a closed subscheme of X(b) V is an open subscheme of a closed subscheme of X(c) V is a closed subscheme of an open subscheme of X .

33Chevalley1

177

(Hint: it will be helpful to note that the scheme-theoretic image may be computedon each open subset of the base.)

9.2.E. Unimportant exercise useful for intuition. If f : X → Y is a locallyclosed immersion into a locally Noetherian scheme (so X is also locally Noetherian),then the associated points of the scheme-theoretic image are (naturally in bijectionwith) the associated points of X . (Hint: Exercise 9.2.B.) Informally, we get nonon-reduced structure on the scheme-theoretic closure not “forced by” that on X .

9.2.6. The induced reduced subscheme structure on a closed subset.[forexperts: This is called the reduced induced subscheme structure

elsewhere, but I think this order makes more sense.]34Suppose Xset is aclosed subset of a scheme Y . Then we can define a canonical scheme structure Xon Xset, that is reduced. We could describe it as being cut out by those functionswhose values are zero at all the points of X set. On affine open set Spec B of Y , ifthe set Xset corresponds to the radical ideal I = I(X set) (recall the I(·) functionof §5.6!), the scheme X corresponds to Spec B/I . We could also consider thisconstruction as an example of a scheme-theoretic image in the following crazy way:let W be the scheme that is a disjoint union of all the points of X set, where thepoint corresponding to p in X set is Spec of the residue field of OY,p. Let f : W → Ybe the “canonical” map sending “p to p”, and giving an isomorphism on residuefields. Then the scheme structure on X is the scheme-theoretic image of f . A thirddefinition: it is the smallest closed subscheme whose underlying set contains X set.

This construction is called the induced reduced subscheme structure onthe closed subset Xset. (Vague exercise: Make a definition of the induced reducedsubscheme structure precise and rigorous to your satisfaction.) [forme: I think

this will be used in “separated morphisms” bit. Also in proof of Chevalley 15.5.1.]

[forme: The induced reduced subscheme structure is handy basically because it induced reduced sub-

scheme structuregives a way of giving a subscheme structure to a closed subset. I’m not sure we care

about the fact that it is reduced, except in the case of the reduction.] 35 induced reduced sub-

scheme structure9.2.F. Exercise. Show that the underlying set of the induced reduced subschemeX → Y is indeed the closed subset Xset. Show that X is reduced.

9.2.7. Reduced version of a scheme.In the special case where X set all of Y , we obtain a reduced closed subscheme

Y red → Y , called the reduction of Y . On affine open subset Spec B → Y , Y red →Y corresponds to the nilradical N(B) of B. The reduction of a scheme is the“reduced version” of the scheme, and informally corresponds to “shearing off thefuzz”. 36 reduction

An alternative equivalent definition: on the affine open subset Spec B → Y ,the reduction of Y corresponds to the idealN (B) ⊂ Y . As for any f ∈ B, N (B)f =N (Bf ), by Exercise 17.11 this defines a closed subscheme.

9.2.G. Unimportant exercise (but useful for visualization). Show thatif Y is a locally Noetherian scheme, the “reduced locus” of Y (where Y red → Y isan isomorphism) is an open subset of Y . (In fact the non-reduced locus is a closure

34d:riss

35reducedinduced

36reddef

178

of certain associated points.) [forme: I’m not convinced we need a proof of this

right now.]

[primordial: Exercise. By Exercise 17.9.A, we have that the reduced locus of a locally

Noetherian scheme is open. More precisely: Let I be the ideal sheaf of Xred, so on X we have anexact sequence

0→ I → OX →OXred → 0

of quasicoherent sheaves on X. Then I is coherent as X is locally Noetherian. Hence the support

of I is closed. The complement of the support of I is the reduced locus. Geometrically, this saysreduced locus (closed in

lN case) that “the fuzz is on a closed subset”. [forme: [Do we really need this locally Noetherian

hypothesis?!]] [Picture!!] [forme: In fact we have that it is the closure of certain

associated points] ][primordial:

9.2.H. Unimportant exercise.

(a) Give an example to show that it is not true that Γ(Xred,OXred ) = Γ(X,OX)/p

Γ(X,OX).(Hint:

‘n>0 Spec k[t]/(tn) with global section (t, t, t, . . . ).) [The definition in Hartshorne

involves doing this on all open subsets and sheafifying. Reader may verify that this isequivalent. Hence we have 3 definitions; the right one (universal property), the affineone, and Hartshorne’s.]

(b) Show that the equality in (b) does hold if X is Noetherian. (Reason: Because X isNoetherian, every open subset is quasicompact 5.5.D.) [forme: Exercise 7.3.B]I’m not sure if this is true! I can show that there is an inclusion from theright side to the left, but it isn’t clear that any section of the right lifts toa section of the left!

][primordial: Non-example: Give them the example of something that is not a locally

closed immersion: take the normalization of the node, and erase one of the node branches. Use a

picture for this. See Figure... ]

9.3 More finiteness conditions on morphisms: (locally) offinite type, quasifinite, (locally) of finite presentation

9.3.1. Morphisms (locally of) finite type.A morphism f : X → Y is locally of finite type if for every affine open set

Spec B of Y , f−1(Spec B) can be covered with open sets Spec Ai so that the inducedmorphism B → Ai expresses Ai as a finitely generated B-algebra. By the affine-locality of finite-typeness of B-schemes (see Proposition 7.3.4), this is equivalentto: for every affine open set Spec Ai in X , Ai is a finitely generated B-algebra. 37locally of finite type

A morphism is of finite type if it is locally of finite type and quasicompact.finite type

Translation: for every affine open set Spec B of Y , f−1(Spec B) can be coveredwith a finite number of open sets Spec Ai so that the induced morphism B → Ai

expresses Ai as a finitely generated B-algebra. [forme: Teaching: X= locally X

plus quasicompact.]

9.3.A. Exercise (the notions “locally of finite type” and “finite type”are affine-local on the target). Show that a morphism f : X → Y is

37finitetype

179

locally of finite type if there is a cover of Y by affine open sets Spec Bi such thatf−1(Spec Bi) is locally of finite type over Bi.

9.3.B. Exercise. Show that a morphism f : X → Y is locally of finite type iffor every affine open subsets Spec A ⊂ X , Spec B ⊂ Y , with f(Spec A) ⊂ Spec B,A is a finitely generated B-algebra. (Hint: use the affine communication lemma onf−1(Spec B).)

ex.doneExample: the “structure morphism” Pn

A → Spec A is of finite type, as PnA is

covered by n+1 open sets of the form Spec A[x1, . . . , xn]. More generally, ProjS∗ →Spec A (where S0 = A) is of finite type.

More generally still: our earlier definition of schemes of “finite type over k” (or“finite type k-schemes”) from ?????? is now a special case of this more generalnotion: a scheme X is of finite type over k means that we are given a morphismX → Spec k (the “structure morphism”) that is of finite type.

Here are some properties enjoyed by morphisms of finite type.

9.3.C. Easy exercise. Show that finite morphisms are of finite type. Henceclosed immersions are of finite type.

9.3.D. Exercises (not hard, but important).

(a) Show that an open immersion is locally of finite type. Show that anopen immersion into a locally Noetherian scheme is of finite type. Moregenerally, show that a quasicompact open immersion is of finite type.

(b) Show that the composition of two morphisms of locally finite type is lo-cally of finite type. (Hence as quasicompact morphisms also compose, thecomposition of two morphisms of finite type is also of finite type.)

(c) Suppose we have morphisms Xf // Y

g // Z , with f quasicompact,

and g f of finite type. Show that f is finite type. 38

(d) Suppose f : X → Y is locally of finite type, and Y is locally Noetherian.Show that X is also locally Noetherian. If X → Y is a morphism of finitetype, and Y is Noetherian, show that X is Noetherian. 39

A morphism f is quasifinite if it is of finite type, and for all y ∈ Y , f−1(y) isa finite set. [forme: [H] omits requirement of finite type.] The main point of this quasifinite

definition is the “finite fiber” part; the “finite type” part is there so this notion is“preserved by fibered product”, Exercise 11.4.C.

9.3.E. Exercise. Show that finite morphisms are quasifinite. (This is a useful finite implies quasifinite

exercise, because you will have to figure out how to figure out how to get at pointsin a fiber of a morphism: given f : X → Y , and y ∈ Y , what are the pointsof f−1(y)? Here is a hint: if X = Spec A and Y = Spec B are both affine, andy = [p], then we can throw out everything in A outside y by modding out by p;you can show that the preimage is A/p. Then we have reduced to the case whereY is the Spec of an integral domain, and [p] = [0] is the generic point. We canthrow out the rest of the points by localizing at 0. You can show that the preimageis (Ap)/pAp. [forme: I have some earlier picture of this] Then, once you have

38H.E.II.3.13; rewrite!

39finitetypeoverN, used in discrete valuative criterion of separatedness, Theorem 16.8.1

180

shown that finiteness behaves well with respect to the operations you made done,you have reduced the problem to Exercise 9.1.M.) 40

There are quasifinite morphisms which are not finite, for example A2−(0, 0 →A2 (Example 9.1.7). The key example of a morphism with finite fibers that is notquasifinite is Spec Q→ Spec Q.

How to picture quasifinite morphisms. If X → Y is a finite morphism, thenquasi-compact open subset U ⊂ X is quasi-finite over Y . In fact every reasonablequasifinite morphism arises in this way. [forme: (Correct hypotheses: Y separated,

X/Y qcqs? separated? Uses Zariski’s Main Theorem. See 29.14.4.] Thus the rightway to visualize quasifiniteness is as a finite map with some (closed locus of) pointsremoved.

9.3.2. ? Morphisms (locally) of finite presentation. There is a variant oftenof use to non-Noetherian people. A morphism f : X → Y is locally of finitepresentation (or locally finitely presented) if for each affine open set Spec B ofY , f−1(Spec B) == ∪i Spec Ai with B → Ai finitely presented (finitely generatedwith a finite number of relations). A morphism is of finite presentation (orfinitely presented) if it is locally of finite presentation and quasicompact.(locally of) finite presen-

tation If X is locally Noetherian, then locally of finite presentation is the same aslocally of finite type, and finite presentation is the same as finite type. So if youare a Noetherian person, you needn’t worry about this notion.

9.3.F. Exercise. Show that the notion of “locally finite presentation” is affine-local. [forme: I haven’t done this!]

9.3.G. ?? Exercise: locally of finite presentation is a purely cate-gorical notion. Show that “locally of finite presentation” is equivalent to thefollowing. If F : (Sch/Y ) → (Sets), S 7→ HomY (S, X), we require F to commutewith direct limits, i.e. if Ai is a direct system, then F (lim−→Ai) = lim−→F (Ai).

[primordial:

9.4 Left over

Degree of a finite morphism at a point. Suppose f : X → Y is a finite morphism.

f∗OX is a finite type (quasicoherent) sheaf on Y , and the rank of this sheaf at a point p is calledthe degree of the finite morphism at p. This is a upper semicontinuous function (we’ve shown thatthe rank of a finite type sheaf is uppersemicontinuous in an exercise when we discussed rank).[forme: Give ref. Give also a picture of two lines mapping to the union of axes sothey can see the uppersemicontinuity.]degree of a finite mor-

phism9.4.A. Exercise. Show that the rank at p is non-zero if and only if f−1(p) is non-empty.

[forme: [EXERCISE SOMEWHERE: For schemes of finite type, surjectivity isthe same as surjectivity on closed points. This could result from Chevalley.]]

[forme: Thus if f : X → Y is a finite morphism, then f−1(y) is a finite set for ally. More is true in good situations. Suppose Y (and hence X) is affine, say X = SpecA,Y = SpecB, and f induced by φ : B → A. Suppose B (and hence A) is Noetherian.Suppose that φ is an inclusion. Then we have the useful Going up theorem. [That isnow moved from here, back to 15.2.2.]

40M.I.7.P.3.2

181

9.4.B. Less important exercise. Suppose that f : SpecA→ SpecB is a finite morphism,corresponding to φ : B → A. Show that A is surjective if and only if φ is injective. 41

(One direction is the going-up theorem.)][forme: Omit in class:

9.4.C. Exercise. “Why varieties are easier”: maps from a reduced scheme aredetermined by points. (Prove it.) Note that a special case was done earlier, when weshowed that two functions from a reduced scheme to A1 were the same if their valuesagreed. Example of more interesting behavior when you have nilpotents: Zariskitangent space example. (Residue field extension?) Perhaps give example in the caseof prevarieties; then we need only closed points. [Refer to this in varieties above.]]

]

41M.I.7.P.3.3

CHAPTER 10

Projective schemes

10.1 Introduction

[forme: (A reminder of why we like projective schemes. (i) it is an easy way

of getting interesting non-affine schemes. (ii) we get lots of schemes of classical

interest. (iii) we have a hard time thinking of anything that isn’t projective or an

open subset of a projective. (iv) a k-scheme is a first approximation of what we

mean by compact.)] [forme: After this chapter, the reader should be comfortable

with doing explicit calculations with projective A-schemes.] At this point, weknow that we can construct schemes by gluing affine schemes together. If a largenumber of affine schemes are involved, this can obviously be a laborious and tediousprocess. Example 9.1.9 of closed subschemes of projective space showed that wecould piggyback on the construction of projective space to produce complicatedand interesting schemes. In this chapter, we formalize this notion of projectiveschemes. Projective schemes over the complex numbers give good examples (inthe classical topology) of compact complex varieties. In fact they are such goodexamples that it is quite hard to come up with an example of a compact complexvariety that is provably not projective. (We will see examples in §29.3.4 and §27.5.A,although we won’t concern ourselves with the relationship to the classical topology.)Similarly, it is quite hard to come up with an example of a complex variety that isprovably not an open subset of a projective variety. In particular, most examplesof complex varieties that come up in nature are of this form. More generally,projective schemes will be the key example of the algebro-geometric analogue ofcompactness (properness, see Chapter 14). Thus one advantage of the notion ofprojective scheme is that it encapsulates much of the algebraic geometry arising innature.

In fact Example 9.1.9 already gives the notion of projective A-schemes in fullgenerality. Recall that any collection of homogeneous elements of A[x0, . . . , xn]describes a closed subscheme of Pn

A. Any closed subscheme of PnA cut out by a set

of homogeneous polynomials will be called a projective A-scheme. (You may beinitially most interested in the “classical” case where A is an algebraically closedfield.) If I is the ideal in A[x0, . . . , xn] generated by these homogeneous polynomials,the scheme we have constructed will be called ProjA[x0, . . . , xn]/I . Then x0, . . . ,xn are informally said to be projective coordinates on the scheme. [forme: Be sure projective coordinates

that this notation is defined well in these notes, perhaps move this back.] Warning:they are not functions on the scheme. (We will later interpret them as sections of aline bundle.) This chapter [forme: lecture] will reinterpret this example in a moreuseful language. For example, just as there is a rough dictionary between rings

183

184

and affine schemes, we will have an analogous dictionary between graded rings andprojective schemes. Just as one can work with affine schemes by instead workingwith rings, one can work with projective schemes by instead working with gradedrings.

10.1.1. A motivating picture from classical geometry.1We motivate a useful way of picturing projective schemes by recalling how

one thinks of projective space “classically” (in the classical topology, over the realnumbers). Pn can be interpreted as the lines through the origin in Rn+1. Thussubsets of Pn correspond to unions of lines through the origin of Rn+1, and closedsubsets correspond to such unions which are closed. (The same is not true with“closed” replaced by “open”!)

One often pictures Pn as being the “points at infinite distance” in Rn+1, wherethe points infinitely far in one direction are associated with the points infinitely farin the opposite direction. We can make this more precise using the decomposition2

(27) Pn+1 = Rn+1∐

Pn

by which we mean that there is an open subset in Pn+1 identified with Rn+1 (thepoints with last projective co-ordinate non-zero), and the complementary closedsubset identified with Pn (the points with last projective co-ordinate zero).

Then for example any equation cutting out some set V of points in Pn will alsocut out some set of points in Rn that will be a closed union of lines. We call thisthe affine cone of V . These equations will cut out some union of P1’s in Pn+1, andwe call this the projective cone of V . The projective cone is the disjoint union ofthe affine cone and V . For example, the affine cone over x2 + y2 = z2 in P2 is justthe “classical” picture of a cone in R2, see Figure 1.

We will make this analogy precise in our algebraic setting in §10.2.3.

10.2 The Proj construction

Let’s abstract these notions, just as we abstracted the notion of the Spec of aring with given generators and relations over k to the Spec of a ring in general.

In the examples we’ve seen, we have a graded ring A[x0, . . . , xn]/I where I is ahomogeneous ideal (i.e. I is generated by homogeneous elements of A[x0, . . . , xn]).Here we are taking the usual grading on A[x0, . . . , xn], where each xi has weight1. Then A[x0, . . . , xn]/I is also a graded ring S•, and we’ll call its graded piecesS0, S1, etc. (The subscript • in S• is intended to remind us of the indexing. In agraded ring, multiplication sends Sm×Sn to Sm+n. Note that S0 is a subring, andS is a S0-algebra.) In our examples that S0 = A, and S• is generated over S0 byhomogeneous ideal

S1.

10.2.1. Standing assumptions about graded rings. We make some standingassumptions on graded rings. Fix a ring A (the base ring). Our motivatingexample is S• = A[x0, x1, x2], with the usual grading. Assume that S• is gradedby Z≥0, with S0 = A. Hence each Sn is an A-module. The subset S+ :=generated in deg 1;

assumptions on graded

rings 1classicalaffinecone

2nicecone

185

projective cone in $\projˆ3$

$xˆ2+yˆ2=zˆ2$ in $\projˆ2$

affine cone: $xˆ2+yˆ2=zˆ2$ in $\Rˆ3$

Figure 1. The affine and projective cone of x2 + y2 = z2 in clas-sical geometry

⊕i>0Si ⊂ S• is an ideal, called the irrelevant ideal. The reason for the name irrelevant ideal

“irrelevant” will be clearer soon. Assume that the irrelevant ideal S+ is afinitely-generated ideal.

10.2.A. Exercise. Show that S• is a finitely-generated graded ring if and onlyif S• is a finitely-generated graded A-algebra, i.e. generated over A = S0 by afinite number of homogeneous elements of positive degree. (Hint for the forwardimplication: show that the generators of S+ as an ideal are also generators of S•

as an algebra.)ex.done

If these assumptions hold, we say that S• is a finitely generated gradedring. [forexperts: This seems an imperfect phrase. Anything better?]

We now define a scheme ProjS•. You won’t be surprised that we will define itas a set, with a topology, and a structure sheaf. proj A-scheme

The set. The points of ProjS• are defined to be those homogeneous primeideals not containing the irrelevant ideal S+. The homogeneous primes containingthe irrelevant ideal are irrelevant.

For example, if S• = k[x, y, z] with the usual grading, then (z2 − x2 − y2) is ahomogeneous prime ideal. We picture this as a subset of Spec S•; it is a cone (seeFigure 1). We picture P2

k as the “plane at infinity”. Thus we picture this equationas cutting out a conic “at infinity”. We will make this intuition somewhat moreprecise in §10.2.3.

The topology. As with affine schemes, we define the Zariski topology bydescribing the closed subsets. They are of the form V (I), where I is a homogeneous

186

ideal. (Here V (I) has essentially the same definition as before: those homogeneousprime ideals containing I .) Particularly important open sets will the distinguishedopen sets D(f) = ProjS• \ V (f), where f ∈ S+ is homogeneous.

10.2.B. Easy exercise. Verify that the distinguished open sets form a base ofthe topology. (The argument is essentially identical to the affine case.)

ex.doneAs with the affine case (Exercise 5.7.E), if D(f) ⊂ D(g), then fn ∈ (g) for

some n, and vice versa. Clearly D(f) ∩ D(g) = D(fg), by the same immediateargument as in the affine case (Exercise 5.7.D).

The structure sheaf. We define OProjS•(D(f)) = ((S•)f )0 where ((S•)f )0

means the 0-graded piece of the graded ring (S•)f . (The notation ((S•)f )0 isadmittedly unfortunate — the first and third subscripts refer to the grading, and thesecond refers to localization.) As in the affine case (see §6.1), we define restrictionmaps, and verify that this is well-defined (i.e. if D(f) = D(f ′), then we are definingthe same ring, and that the restriction maps are well-defined).

For example, if S• = k[x0, x1, x2] and f = x0, we get (k[x0, x1, x2]x0)0 :=k[x1/0, x2/0] (using the language of Example 6.1.10).

We now check that this is a sheaf. We could show that this is a sheaf onthe base, and the argument would be as in the affine case (which was not easy).Here instead is a sneakier argument. We first note that the topological spaceD(f) and Spec((S•)f )0 are canonically homeomorphic: they have matching dis-tinguished bases. (To the distinguished open D(g) ∩ D(f) of D(f), we associateD(gdeg f/fdeg g) in Spec(Sf )0. To D(h) in Spec(Sf )0, we associate D(fnh) ⊂ D(f),where n is chosen large enough so that fnh ∈ S•.) Second, we note that the sheaf ofrings on the distinguished base of D(f) can be associated (via this homeomorphismjust described) with the sheaf of rings on the distinguished base of Spec((S•)f )0: thesections match (the ring of sections ((S•)fg)0 over D(g) ∩D(f) ⊂ D(f), those ho-mogeneous degree 0 quotients of S• with f ’s and g’s in the denominator, is naturallyidentified with the ring of sections over the corresponding open set of Spec((S•)f )0)and the restriction maps clearly match (think this through yourself!). Thus we havedescribed an isomorphism of schemes

(D(f),OProj S•)∼= Spec((S•)f )0.

10.2.C. Easy Exercise. Describe a natural “structure morphism” ProjS• →Spec A. 3

10.2.2. Projective and quasiprojective schemes.We call a scheme of the form ProjS• (where S0 = A) a projective scheme

over A, or a projective A-scheme. A quasiprojective A-scheme is an opensubscheme of a projective A-scheme. The “A” is omitted if it is clear from thequasiprojective scheme

context; often A is some field.We now make a connection to classical terminology. A projective variety

(over k), or an projective k-variety is a reduced projective k-scheme. (Warning:in the literature, it is sometimes also required that the scheme be irreducible, or thatk be algebraically closed.) A quasiprojective k-variety is an open subscheme ofa projective k-variety. We defined affine varieties in §7.3.7, and you can check that

3e:ps

187

affine open subsets of projective k-varieties are affine k-varieties. We will definevarieties in general in [forme: XXXX] . 4 projective variety

The notion of quasiprojective k-scheme is a good one, covering most interestingcases which come to mind. We will see before long that the affine line with thedoubled origin (Example 6.1.7) is not quasiprojective for somewhat silly reasons(“non-Hausdorffness”, [forme: REF] ), but we’ll call that kind of bad behavior“non-separated”. Here is a surprisingly subtle question: Are there quasicompactk-schemes that are not quasiprojective? Translation: if we’re gluing together afinite number of schemes each sitting in some An

k , can we ever get something notquasiprojective? We will finally answer this question in the negative in §29.3.4.

10.2.D. Easy Exercise. Show that all projective A-schemes are quasicompact.(Translation: show that any projective A-scheme is covered by a finite number ofaffine open sets.) Show that ProjS• is finite type over A = S0. If S0 is a Noetherianring, show that ProjS• is a Noetherian scheme, and hence that ProjS• has a finitenumber of irreducible components. Show that any quasiprojective scheme is locallyof finite type over A. If A is Noetherian, show that any quasiprojective A-schemeis quasicompact, and hence of finite type over A. Show this need not be true if Ais not Noetherian. Better: give an example of a quasiprojective A-scheme that isnot quasicompact (necessarily for some non-Noetherian A). (Hint: Flip ahead tosilly example 10.3.2.) 5

[forme: Show it is quasiseparated]

10.2.3. Affine and projective cones.6If S• is a finitely-generated graded ring, then the affine cone of ProjS•

is Spec S•. Note that this construction depends on S•, not just of ProjS•. Asmotivation, consider the graded ring S• = C[x, y, z]/(z2 − x2 − y2). Figure 2 is asketch of Spec S•. (Here we draw the “real picture” of z2 = x2 + y2 in R3.) It is acone in the most traditional sense; the origin (0, 0, 0) is the “cone point”. 7 affine cone

This gives a useful way of picturing Proj (even over arbitrary rings than C).Intuitively, you could imagine that if you discarded the origin, you would get some-thing that would project onto ProjS•. The following exercise makes that precise.

10.2.E. Exercise. If S• is a projective scheme over a field k, Describe a naturalmorphism Spec S• \ 0 → ProjS•. [forme: The argument is not a trivial variation

on things that have come before.]ex.done

This has the following generalization to A-schemes, which you might find geo-metrically reasonable. This again motivates the terminology “irrelevant”.

10.2.F. Exercise. If S• is a projective A-scheme, describe a natural morphismSpec S• \ V (S+)→ ProjS•.

[forme: IRRELEVANT IDEAL IS FUZZ NEAR THE ORIGIN, NOT NEAR

INFINITY. PICTURE (x2, xy).]

In fact, it can be made precise that ProjS• is the affine cone, minus the origin,modded out by multiplication by scalars. [forme: ref?] affine cone

4d:projectivevariety

5e:Pnft

6affinecone

7d:affinecone

188

Figure 2. A sketch of the cone Spec k[x, y, z]/(z2 − x2 − y2).

The projective cone of ProjS• is ProjS•[T ], where T is a new variable ofdegree 1. For example, the cone corresponding to the conic Proj k[x, y, z]/(z2 −x2 − y2) is Projk[x, y, z, T ]/(z2 − x2 − y2).

10.2.G. Exercise (cf. (27)). Show that the projective cone of ProjS•[T ] has aclosed subscheme isomorphic to ProjS• (corresponding to T = 0), whose comple-ment (the distinguished open set D(T )) is isomorphic to the affine cone Spec S•.affine and projective

cone

ex.doneYou can also check that ProjS• is a locally principal closed subscheme, and is

also locally not a zero-divisor (an effective Cartier divisor, §9.1.9).This construction can be usefully pictured as the affine cone union some points

“at infinity”, and the points at infinity form the Proj. The reader may which toponder Figure 2, and try to visualize the conic curve “at infinity”.

We have thus completely discribed the algebraic analog of the classical pictureof 10.1.1.

10.3 Examples

10.3.1. Example. We (re)define projective space by PnA := ProjA[x0, . . . , xn].

This definition involves no messy gluing, or choice of special patches.

10.3.A. Exercise. Check that this agrees with our earlier version of projectivespace (Definition 6.1.10).

189

10.3.2. Silly example. Note that P0A = ProjA[T ] ∼= Spec A. Thus “SpecA is a

projective A-scheme”.8 Here is a useful generalization of this example:

10.3.B. Exercise: finite morphisms to Spec A are projective. If B is afinitely generated A-algebra, define S• by S0 = A, and Sn = B for n > 0 (with theobvious graded ring structure). Describe an isomorphism

ProjS•oo ∼ //

%%KKKKKKKKK

Spec B

yyttttttttt

Spec A

10.3.C. Exercise. Show that X = P2k \ x2 + y2 = z2 is an affine scheme.

[forme: This comes for free with my approach; it is not so clear by Hartshorne.]

Show that x = 0 cuts out a locally principal closed subscheme that is not principal.9

[forme: Notice: Our tentative definition of Proj in the previous section a priori

relied on thinking of the ring as a quotient of a polynomial ring, i.e. choosing gener-

ators of S1. Our new definition depends only on the graded ring, with no choices of

generators.]

10.3.3. Example: PV . We can make this definition of projective space evenmore choice-free as follows. Let V be an (n + 1)-dimensional vector space overk. (Here k can be replaced by any ring A as usual.) Let Sym• V ∨ = k ⊕ V ∨ ⊕Sym2 V ∨ ⊕ · · · . (The reason for the dual is explained by the next exercise.) If forexample V is the dual of the vector space with basis associated to x0, . . . , xn, wewould have Sym• V ∨ = k[x0, . . . , xn]. Then we can define PV := Proj Sym• V ∨.In this language, we have an interpretation for x0, . . . , xn: they are the linearfunctionals on the underlying vector space V .

10.3.D. Unimportant exercise. Suppose k is algebraically closed. Describea natural bijection between one-dimensional subspaces of V and the points ofPV . Thus this construction canonically (in a basis-free manner) describes the one-dimensional subspaces of the vector space Spec V .

On a related note: you can also describe a natural bijection between points ofV and the points of Spec Sym• V ∨. This construction respects the affine/projectivecone picture of §10.2.3.

10.4 Maps of graded rings and maps of projective schemes

[At some point in this chapter, show that maps from schemes to projectivespace can be given by a bunch of functions nowhere vanishing. This should bean exercise. However, not all maps from schemes come in this way. We’ll see theright generalization (characterizing all maps from schemes to projective schemes) inXXX. This result will involve sections of line bundles, and our desire to understand

8P0A

9lineonplaneminusconic H.Ex.6.5.1

190

maps to projective schemes in a clean way will be one important motivation forunderstanding line bundles.

As maps of rings correspond to maps of affine schemes in the opposite direction,maps of graded rings sometimes give maps of projective schemes in the oppositedirection. Before we make this precise, let’s see an example to see what can gowrong. There isn’t quite a map P2

k → P1k given by [x; y; z] → [x; y], because this

alleged map isn’t defined only at the point [0; 0; 1]. What has gone wrong? Themap A3 = Spec k[x, y, z] → A2 = Spec k[x, y] makes perfect sense. However, thez-axis in A3 maps to the origin in A2, so the point of P2 corresponding to the z-axismaps to the “cone point” of the affine cone, and hence not to the projective scheme.The image of this point of A2 contains the irrelevant ideal. If this problem doesn’toccur, a map of rings gives a map of projective schemes in the opposite direction.

10.4.A. Important exercise. (a) Suppose that f : S•// R• is a morphism

of finitely-generated graded rings (i.e. a map of rings preserving the grading) overA. Suppose further that 10

(28) f(S+) ⊃ R+.

Show that this induces a morphism of schemes ProjR• → ProjS•. (Warning: notevery morphism arises in this way. [forme: EXAMPLE? Baby case... saturation.

But also others...] )

(b) Suppose further that S•// // R• is a surjection of finitely-generated graded

rings (so (28) is automatic). Show that the induced morphism ProjR• → ProjS•

is a closed immersion. (Warning: not every closed immersion arises in this way![forme: EXAMPLE?] ) 11

ex.done

Hypothesis (28) can be replaced with the weaker hypothesis√

f(S+) ⊃ R+,but in practice this hypothesis (28) suffices.

10.4.B. Exercise. Show that an injective linear map of k-vector spaces V →Winduces a closed immersion PV → PW . 12

ex.done

This closed subscheme is called a linear space. Once we know about dimen-linear space

sion, we will call this a linear space of dimension dim V − 1 = dim PV . A linearspace of dimension 1 (resp. 2, n, dim PW−1) is called a line (resp. plane, n-plane,hyperplane). (If the linear map in the previous exercise is not injective, then theline, plane, n-plane, hy-

perplane hypothesis (28) of Exercise 10.4.A fails.)

10.4.1. A particularly nice case: when S• is generated in degree 1. If S•

is generated by S1 as an A-algebra, we say that S• is generated in degree 1.gen. in deg. 1

10.4.C. Exercise. Suppose S• is a finitely generated graded ring generated indegree 1. Show that S1 is a finitely-generated module, and the irrelevant ideal S+

is generated in degree 1.

10.4.D. Exercise. 13 Show that if S• is generated by S1 (as an A-algebra) byn + 1 elements x0, . . . , xn, then ProjS• may be described as a closed subscheme of

10mopshypothesis

11morphismofprojectiveschemes

12linearmap

13gendeg1

191

PnA as follows. Consider An+1 as a free module with generators t0, . . . , tn associated

to x0, . . . , xn. The surjection of

Sym• An+1 = A[t0, t1, . . . , tn] // // S•

implies S• = A[t0, t1, . . . tn]/I , where I is a homogeneous ideal.ex.done

This is completely analogous to the fact that if R is a finitely-generated A-algebra, then choosing n generators of R as an algebra is the same as describingSpec R as a closed subscheme of An

A. In the affine case this is “choosing coordi-nates”; in the projective case this is “choosing projective coordinates”.

For example, Projk[x, y, z]/(z2 − x2 − y2) is a closed subscheme of P2k. (A

picture is shown in Figure 2.)[forme: Another example: P1 → P3 given by twisted cubic. Then edit the later

geometric examples.

Also before that, in the maps of schemes as sets, do some explicit examples. ]

Recall that we can interpret the closed points of Pn as the lines through theorigin in An+1. [forme: REF] The following exercise states this more generally.

10.4.E. Exercise. Suppose S• is a finitely-generated graded ring over an alge-braically closed field k, generated in degree 1 by x0, . . . , xn, inducing closed im-mersions ProjS• → Pn and Spec S• → An. Describe a natural bijection betweenthe closed points of ProjS• and the “lines through the origin” in Spec S• ⊂ An.

ex.done

10.5 Important exercises

There are many fundamental properties that are best learned by working throughproblems.

10.5.1. Analogues of results on affine schemes.

10.5.A. Exercise. 14

(a) Suppose I is any homogeneous ideal, and f is a homogeneous element.Show that f vanishes on V (I) if and only if fn ∈ I for some n. (Hint:Mimic the affine case; see Exercise 5.4.I.)

(b) If Z ⊂ ProjS•, define I(·). Show that it is a homogeneous ideal. For anytwo subsets, show that I(Z1 ∪ Z2) = I(Z1) ∩ I(Z2).

(c) For any subset Z ⊂ ProjS•, show that V (I(Z)) = Z.

[primordial: I earlier had: (d) For any homogeneous ideal I with V (I) 6= ∅, show that

I(V (I)) =√I. But that’s wrong. Dave Savitt: Probably you are aware of this, but I don’t think

this is quite right as stated (e.g. when S0 isn’t a field). A counterexample for (c) is the ideal (pt)

in Z[t], where I(V (I)) = (p), but p isn’t in the radical of (pt). I think one fix is as follows: in (a),

require that deg(f) > 0; in (b), define I(.) to be the intersection of S+ with one’s first guess for

I(.); in (c), that I(V (I)) = rad(I) ∩ S+. The point is that I and I ∩ S+ define the same subset

of Proj(S), which might be a worthwhile exercise on its own. ]

14H.E.I.2.1 is a

192

10.5.B. Exercise. Show that the following are equivalent. (This is more moti-vation for the S+ being “irrelevant”: any ideal whose radical contains it is “geo-metrically irrelevant”.)

(a) V (I) = ∅

(b) for any fi (i in some index set) generating I , ∪D(fi) = ProjS•

(c)√

I ⊃ S+. [primordial: Anssi pointed out this is false: (d) I ⊃ Sd for some

d > 0. ]15

[forme:

10.5.2. Closed subschemes and saturated ideals.

FIX!!! For affine schemes, there was a natural bijection between closedsubsets and radical ideals. In the previous series of problems, we saw a naturalbijection between closed subsets and radical ideals not containing the irrelevant ideal.[forme: Is this right now?] For affine schemes, there is a natural bijection betweenclosed subschemes and ideals [forme: [refs for this paragraph!]] . Similarly,for projective schemes, there is a natural bijection between closed subschemes andsaturated homogeneous ideals.

[forme: WRITE THIS! I want later the same thing for modules, so I may as well

do it here. Perhaps postpone this to 19.3.3. ] ]

10.5.3. Scaling the grading, and the Veronese embedding.Here is a useful construction. Define Sn• = ⊕∞

j=0Snj . (We could rescale ourdegree, so “old degree” n is “new degree” 1.)

10.5.C. Exercise. Show that ProjSn• is isomorphic to ProjS•.16

10.5.D. Exercise. Suppose S• is generated over S0 by f1, . . . , fn. Find a dsuch that Sd• is generated in “new” degree 1 (= “old” degree d). [forme: Older:Suppose

d = lcm(deg f1, . . . , deg fn).

Show that Sd• is generated in “new” degree 1 (= “old” degree d). (Hint: I like to

show this by induction on the number of distinct elements of deg f1, . . . ,deg fn.)][forme: Dave Savitt’s sol’n Nov. 13 ’07: Claim: Sn lcm(fi)•

is generated in new degree

1. If k lcm(fi) = c1degf1 + · · ·+ cndegfn with k ≥ n and each ci ≥ 0, then there is some

choice of nonnegative di’s with di ≤ ci and d1degf1 + ...dndegfn = n lcm(fi). (This is

trivially true for k = n, then induct on k using the fact that some cidegfi ≥ d.)] Thisis handy, as it means that, using the previous Exercise 10.5.C, we can assume thatany finitely-generated graded ring is generated in degree 1. In particular, we canplace every Proj in some projective space via the construction of Exercise 10.4.D.17

ex.done

Example: Suppose S• = k[x, y], so ProjS• = P1k. Then S2• = k[x2, xy, y2] ⊂

k[x, y]. We identify this subring as follows.

10.5.E. Exercise. Let u = x2, v = xy, w = y2. Show that S2• = k[u, v, w]/(uw−v2).

ex.done

15e:projanalog

16Veronesetrick

17generateddeg1

193

We have a graded ring generated by three elements in degree 1. Thus we thinkof it as sitting “in” P2, via the construction of §10.4.D. This can be interpreted as“P1 as a conic in P2”.

10.5.4. 18Thus if k is algebraically closed of characteristic not 2, using the factthat we can diagonalize quadrics (Remark 7.4.6), the conics in P2, up to change ofco-ordinates, come in only a few flavors: sums of 3 squares (e.g. our conic of theprevious exercise), sums of 2 squares (e.g. y2−x2 = 0, the union of 2 lines), a singlesquare (e.g. x2 = 0, which looks set-theoretically like a line, and is non-reduced),and 0 (not really a conic at all). Thus we have proved: any plane conic (over analgebraically closed field of characteristic not 2) that can be written as the sum ofthree squares is isomorphic to P1.

We now soup up this example.

10.5.F. Exercise. Show that ProjS3• is the twisted cubic “in” P3. (The twisted twisted cubic

cubic was defined in Exercise 9.1.T, and in fact this exercise will give a solution toExercise 9.1.T(b).) 19

10.5.G. Exercise. Show that ProjSd• is given by the equations that(

y0 y1 · · · yd−1

y1 y2 · · · yd

)

is rank 1 (i.e. that all the 2×2 minors vanish). This is called the degree d rationalnormal curve “in” Pd. 20 rational normal curve

ex.done

10.5.5. The Veronese embedding. 21More generally, if S• = k[x0, . . . , xn], thenProjSd• ⊂ PN−1 (where N is the number of degree d polynomials in x0, . . . , xn) iscalled the d-uple embedding or d-uple Veronese embedding. It is enlighteningto interpret this closed immersion as a map of graded rings. 22 V embedding, d-uple

embedding10.5.H. Combinatorial exercise. Show that N =

(

n+dd

)

.

10.5.I. Unimportant exercise. Find five linearly independent quadric equa-tions vanishing on the Veronese surface ProjS2• where S• = k[x0, x1, x2], whichsits naturally in P5. (You needn’t show that these equations generate all the equa- Veronese surface

tions cutting out the Veronese surface, although this is in fact true.)

10.5.6. Entertaining geometric exercises.

10.5.J. Useful geometric exercise. Describe all the lines on the quadricsurface wz − xy = 0 in P3

k. (Hint: they come in two “families”, called the rulingsof the quadric surface.) This construction arises all over the place in nature. quadric surface, rulings

Hence by Remark 7.4.6, if we are working over an algebraically closed field ofcharacteristic not 2, we have shown that all rank 4 quadric surfaces have two rulingsof lines.

18smoothconicrational

19e2:twistedcubic

20d:rationalnormalcurve

21s:Veronese

22d:Veronese

194

Figure 3. The two rulings on the quadric surface V (wz − xy) ⊂P3. One ruling contains the line V (w, x) and the other containsthe line V (w, y).

10.5.K. Exercise. Show that Pnk is normal. More generally, show that Pn

A isnormal if A is a Unique Factorization Domain. [forme: We don’t know that R[X]

is integrally closed if R is yet. This could have gone in the first section, but I didn’t

want to distract from the central point of that section.]

[primordial:

10.5.L. Exercise. If S• is an integrally closed Noetherian domain, show that ProjS• is normal.

(Four hints: Assume S• is generated in degree 1. Look at the proof of the previous exercise

[Whoops, that was Exercise 15.12.B.] Recall Proposition 7.4.2. Show that if A[x] is integrally

closed, then so is A.) [forme: Later define projective normality, when we have closed

subschemes, and refer back to here. Show them that the converse is not always true;

state that here...] 23 ]

10.5.7. Example. If we put a non-standard weighting on the variables ofk[x1, . . . , xn] — say we give xi degree di — then Proj k[x1, . . . , xn] is called weightedprojective space P(d1, d2, . . . , dn).24 [primordial: Weighted projective space is nor-weighted Pn

mal, by Exercise 10.5.L. ]

10.5.M. Exercise. Show that P(m, n) is isomorphic to P1. Show that P(1, 1, 2) ∼=Projk[u, v, w, z]/(uw − v2). Hint: do this by looking at the even-graded parts ofk[x0, x1, x2], cf. Exercise 10.5.C. [forme: This is a projective cone over a conic

curve.]

23e:projnorm

24weightedprojectivespace

CHAPTER 11

Fibered products of schemes

[forme: Notes from Sept. 12, 2005. Neron models 2.2 and 2.3. Brian Conrad’s

RR question — what was that? Also do: analytification. ]

11.0.N. Exercise. Given 2 functions X → An, agreeing on a dense open set, showthat they are the same morphism. (Hint 1: directly. Hint 2: An = A1 × · · · × A1.Use earlier exercise.) [False unless X is reduced.]

[forme:

11.0.O. Very unimportant exercise. [Perhaps move it a bit later, after C⊗R C. In that

example —State: The isomorphism L⊗K L→ LG is just a restatement of the classical

form of linear independence of characters. (Pointed out by Anatoly Pregel.)] This is

an example that if we have a category with fibered product, and a subcategory, the

subcategory may also have a fibered product, that may differ from that of the larger

category! Motivation: k(u1/p) ⊗k(u) k(u1/p) is nonreduced. Our big category is the

category of k-schemes, where k is a field of characteristic p. We have a subcategory of

reduced k-schemes, and again we get a fibered product, but it disagrees with fibered

products of schemes in the case of positive characteristic (David Sheppard). The

reduced fibered product is the reduction of the scheme-theoritic fibered product.

Algebraic fact that this can’t happen in characteristic 0.]

Useful fact: product of 2 domains over k is a domain. (Prove this? I needflatness, and I would like to excise it.) But notice C⊗RC example below! Mumford’sred book p. 34 has a link to where irreducible times irreducible = irreducible. Fact:product of irreducible varieties over an algebraic closed field are still irreducible, inaffine case. From where? See Joe R’s summary of my argument.

Have an exercise on fiber cubes early on.

11.1 Fibered products of schemes exist

We will now construct the fibered product in the category of schemes. In otherwords, given X, Y → Z, we will show that X×Z Y exists. (Recall that the absoluteproduct in a category is the fibered product over the final object, Easy Exercise 3.3.Jso X × Y = X ×Z Y in the category of schemes, and X × Y = X ×S Y if we areimplicitly working in the category of S-schemes, for example if S is the spectrum ofa field.) Notational warning: lazy people wanting to save chalk and ink will write fibered product of

schemes, absolute

product, ×k, ×Z

×k for ×Speck, and similarly for ×Z. It already happened in the paragraph above!Before we get started, we’ll make a few random remarks.

195

196

Remark 1. We’ve made a big deal about schemes being sets, endowed with atopology, upon which we have a structure sheaf. So you might think that we’ll con-struct the product in this order. However, here is a sign that something interestinghappens at the level of sets that will mess up this strategy. you should believe thatif we take the product of two affine lines (over your favorite algebraically closed fieldk, say), you should get the affine plane: A1

k×k A1k should be A2

k. And we’ll see thatthis is indeed true. But the underlying set of the latter is not the underlying set ofthe former —- we get additional points! Thus products of schemes do something alittle subtle on the level of sets.

11.1.A. Exercise. If k is algebraically closed, describe a natural map of setsA1

k × A1k → A2

k. Show that this map is not surjective. On the other hand, showthat it is a bijection on closed points. 1

Remark 2. Recall that the diagram of a fibered square

W //

Y

X

f // Z

goes by a number of names, including fibered diagram, Cartesian diagram, fiberedsquare, and Cartesian square. [forme: See §3.3.7] Because of its geometric inter-pretation, in algebraic geometry it is often called a base change diagram or apullback diagram, and W → X is called the pullback of Y → Z by f , and Wis called the pullback of Y by f .

The reason for the phrase “base change” or “pullback” is the following. If Xis a point of Z (i.e. f is the natural map of Spec of the residue field of a point of Zinto Z), then W is interpreted as the fiber of the family.

11.1.B. Exercise. Show that in the category of topological spaces, this is true,i.e., if Y → Z is a continuous map, and X is a point p of Z, then the fiber of Yover p is naturally identified with X ×Z Y .

More generally, for general X → Z, the fiber of W → X over a point p of X isnaturally identified with the fiber of Y → Z over f(p).fiber digram, pullback

diagram, Cartesian dia-

gram, base change dia-

gram

Let’s now show that fibered products always exist in the category of schemes.

11.1.1. Big Theorem (fibered products always exist). — Suppose f : X →Z and g : Y → Z are morphisms of schemes. Then the fibered product

X ×Z Yf ′

//

g′

Y

g

X

f // Z

exists in the category of schemes.2

As always when showing that certain objects defined by universal propertiesexist, we have two ways of looking at the objects in practice: by using the universalproperty, or by using the details of the construction.

1A1A1A2a

2fiberedproductsexist

197

The key idea, roughly, is this: we cut everything up into affine open sets,do fibered products in that category (where it turns out we have seen the conceptbefore in a different guise), and show that everything glues nicely. The conceptuallydifficult part of the proof comes from the gluing, and realizing that we have to checkalmost nothing.

The proof will be a little long, but you will notice that we repeat a kind ofargument several times. A much shorter proof is possible by interpreting this inthe language of representable functors, and we give this proof afterward for experts.

Proof. We have an extended proof by universal property. We divide the proof upinto a number of bite-sized pieces. Between bites, we will often take a break forsome side comments.

Step 1: everything affine. First, if X, Y, Z are affine schemes, say X = Spec A,Y = Spec B, Z = Spec C, the fibered product exists, and is Spec A ⊗C B. Here’swhy. Suppose W is any scheme, along with morphisms f ′′ : W → X and g′′ : W →Y such that f f ′′ = g g′′ as morphisms W → Z. We hope that there exists aunique h : W → Spec A⊗C B such that f ′′ = g′ h and g′′ = f ′ h.

W∃!?

&&LLLLLLLLLLg′′

++VVVVVVVVVVVVVVVVVVVVVVV

f ′′

:::

::::

::::

::::

::

Spec A⊗C B

g′

f ′// Spec B

g

Spec A

f // Spec C

But maps to affine schemes correspond precisely to maps of global sections inthe other direction (Exercise 8.4.E):

Γ(W,OW )

A⊗C B

∃!?ffMMMMMMMMMM

Bf ′oo

g′′jjUUUUUUUUUUUUUUUUUUU

A

g′

OOf ′′

^^<<<<<<<<<<<<<<<<<C

foo

g

OO

But this is precisely the universal property for tensor product! (The tensorproduct is the cofibered product in the category of rings.)

11.1.2. Side remark (cf. Exercise 11.1.A). Thus indeed A1 ×A1 ∼= A2, and moregenerally (A1)n ∼= An. 3

Step 2: fibered products with open immersions. Second, we note that the fiberedproduct with open immersions always exists: if Y → Z an open immersion, thenfor any f : X → Z, X ×Z Y is the open subset f−1(Y ). (More precisely, this opensubset satisfies the universal property.) This was Exercise 9.1.G (which wasn’t

3A1A1A2

198

hard).

f−1(Y ) _

// Y _

X

f // ZStep 3: fibered products of affine with almost-affine over affine. We can combine

steps 1 and 2 as follows. Suppose X and Z are affine, and Y → Z factors as

Y i // Y ′

g // Z where i is an open immersion and Y ′ is affine. Then X ×Z Yexists. This is because if the two smaller squares of

W //

Y

W ′ //

Y ′

X // Z

are fibered diagrams, then the “outside rectangle” is also a fibered diagram. (Thiswas Exercise 3.3.K, although you should be able to see this on the spot.)

Key Step 4: fibered product of affine with arbitrary over affine exists. We nowcome to the key part of the argument: if X and Z are affine, and Y is arbitrary.This is confusing when you first see it, so we’ll first deal with a special case, whenY is the union of two affine open sets Y1 ∪ Y2. Let Y12 = Y1 ∩ Y2.

Now for i = 1, 2, X ×Z Yi exists by Step 1; call this Wi. Also, X ×Z Y12 existsby Step 3 (call it W12), and comes with natural open immersions into W1 and W2.Thus we can glue W1 to W2 along W12; call this resulting scheme W .

We’ll check that this is the fibered product by verifying that it satisfies theuniversal property. Suppose we have maps f ′′ : V → X , g′′ : V → Y that compose(with f and g respectively) to the same map V → Z. We need to construct aunique map h : V →W , so that f ′ h = g′′ and g′ h = f ′′.

V∃!?

AAA

AAAA

Ag′′

((PPPPPPPPPPPPPPP

f ′′

00000000000000

W

g′

f ′// Y

g

X

f // Z

For i = 1, 2, define Vi := (g′′)−1(Yi). Define V12 := (g′′)−1(Y12) = V1 ∩ V2. Thenthere is a unique map Vi →Wi such that the composed maps Vi → X and Vi → Yi

are desired (by the universal product of the fibered product X ×Z Yi = Wi), hencea unique map hi : Vi → W . Similarly, there is a unique map h12 : V12 → W suchthat the composed maps V12 → X and V12 → Y are as desired. But the restrictionof hi to V12 is one such map, so it must be h12. Thus the maps h1 and h2 agreeon V12, and glue together to a unique map h : V → W . We have shown existenceand uniqueness of the desired h. (We are using the fact that “morphisms glue”, see§8.4, which corresponds to the fact that maps to a scheme form a sheaf. This leadsto a shorter explanation of the proof, which we give at the end of this long proof.)

199

We have thus shown that if Y is the union of two affine open sets, and X andZ are affine, then X ×Z Y exists.

We now tackle the general case. (The reader may prefer to first think throughthe case where “two” is replaced by “three”.) We now cover Y with open sets Yi,as i runs over some index set (not necessarily finite!). As before, we define Wi andWij . We can glue these together to produce a scheme W along with open sets weidentify with Wi (Exercise 6.1.H).

As in the two-affine case, we show that W is the fibered product by showingthat it satisfies the universal property. Suppose we have maps f ′′ : V → X ,g′′ : V → Y that compose to the same map V → Z. We construct a unique maph : V → W , so that f ′ h = g′′ and g′ h = f ′′. Define Vi = (g′′)−1(Yi) andVij := (g′′)−1(Yij) = Vi ∩ Vj . Then there is a unique map Vi → Wi such that thecomposed maps Vi → X and Vi → Yi are desired, hence a unique map hi : Vi →W .Similarly, there is a unique map hij : Vij → W such that the composed mapsVij → X and Vij → Y are as desired. But the restriction of hi to Vij is one suchmap, so it must be hij . Thus the maps hi and hj agree on Vij . Thus the hi gluetogether to a unique map h : V → W . We have shown existence and uniqueness ofthe desired h, completing this step.

Side remark. One special case of it is called extending the base field: if X isa k-scheme, and k′ is a field extension (often k′ is the algebraic closure of k), thenX ×Speck Spec k′ (sometimes informally written X ×k k′ or Xk′) is a k′-scheme.Often properties of X can be checked by verifying them instead on Xk′ . This is thesubject of descent — certain properties “descend” from Xk′ to X . We have alreadyseen that the property of being normal descends in this way (Exercise 7.4.J). descent, extending the

base fieldStep 5: Z affine, X and Y arbitrary. We next show that if Z is affine, and Xand Y are arbitrary schemes, then X ×Z Y exists. We just follow Step 4, with theroles of X and Y reversed, using the fact that by the previous step, we can assumethat the fibered product with an affine scheme with an arbitrary scheme over anaffine scheme exists.

Step 6: Z admits an open immersion into an affine scheme Z ′, X and Yarbitrary. This is akin to Step 3: X×Z Y satisfies the universal property of X×Z′ Y .

Step 7: the general case. We again employ the trick from Step 4. Say f :X → Z, g : Y → Z are two morphisms of schemes. Cover Z with affine opensubsets Zi. Let Xi = f−1Xi and Yi = g−1Yi. Define Zij = Zi ∩ Zj , and Xij

and Yij analogously. Then Wi := Xi ×Zi Yi exists for all i, and has as open setsWij := Xij ×Zij Yij along with gluing information satisfying the cocycle condition(arising from the gluing information for Z from the Zi and Zij). Once again, weshow that this satisfies the universal property. Suppose V is any scheme, alongwith maps to X and Y that agree when they are composed to Z. We need to showthat there is a unique morphism V → W completing the diagram

V∃!?

AAA

AAAA

Ag′′

((PPPPPPPPPPPPPPP

f ′′

00000000000000

W

g′

f ′// Y

g

X

f // Z.

200

Now break V up into open sets Vi = g′′ f−1

(Zi). Then by the universal propertyfor Wi, there is a unique map Vi →Wi (which we can interpret as Vi →W ). Thuswe have already shown uniqueness of V →W . These must agree on Vi∩Vj , becausethere is only one map Vi∩Vj to W making the diagram commute. Thus all of thesemorphisms Vi →W glue together, so we are done.

11.1.3. ?? Describing the existence of fibered products using high-falutin’language.

The previous proof can be described more cleanly in the language of repre-sentable functors. You’ll find this enlightening only after you have absorbed theargument above and meditated on it for a long time. For experts, we include themore abstract picture here. You might find that this is most useful to shed lighton representable functors, rather than on the existence of the fibered product.

Recall that to each scheme X we have a contravariant functor hX from thecategory of schemes Sch to the category of Sets, taking a scheme Y to Mor(Y, X)(§3.2.19). It may be more convenient to think of it as a covariant functor hX :Schopp → Sets.

But this functor hX is better than a functor. We know that if Ui is an opencover of Y , a morphism Y → X is determined by its restrictions Ui → X , andgiven morphisms Ui → X that agree on the overlap Ui ∩ Uj → X , we can gluethem together to get a morphism Y → X . (This is roughly our statement that“morphisms glue”, §??.) In the language of equalizer exact sequences (§4.2.7),

· // Hom(Y, X) // ∏Hom(Ui, X) // //∏

Hom(Ui ∩ Uj , X) .

Thus morphisms to X (i.e. the functor hX) form a sheaf on every scheme X . Ifthis holds, we say that the functor is a sheaf. (If you want to impress your friendsand frighten your enemies, you can tell them that this is a sheaf on the big Zariskisite.)

We can repeat this discussion for the category SchS of schemes over a givenbase scheme S.

Notice that the definition of fibered product also gives a contravariant functor

hX×ZY : Sch→ Sets :

to the scheme W we associate the set of commutative diagrams

W

Y

X // Z

(What is the image of W → W ′ under this functor?) The existence of fiberedproduct is precise the statement that there is a natural isomorphism of functorshX×ZY

∼= hW for some scheme W . In that case, we say that hX×ZY is a rep-resentable functor, and that it is representable by W . The usual universalproperty argument shows that this determines W up to unique isomorphism.

We can now interpret Key Step 4 of the proof of Theorem 11.1.1 as follows.Suppose X and Z are affine, and Yi is an affine open cover of Y . Suppose thecovariant functor FY : (SchY )opp → Sets is a sheaf on the category of Y -schemesSchY , and FYi is the “restriction of the sheaf to Yi” (where we include only those

201

Y -schemes that are in fact Yi-schemes, i.e. those T → Y whose structure morphismsfactor through Yi, T → Yi → Y ).

11.1.C. Exercise. Show that if FYi is representable, then so is FY . (Hint: thisis basically just the proofs of Steps 3 and 4.)

ex.doneWe then apply this in the special case where FY is given by

( Tf // Y ) 7→

Tf //

Y

X // Z

.

11.1.D. Exercise. Check that this FY is a sheaf. (This is not hard once yourealize what this is asking.)4

ex.doneThen Steps 5 through 7 are one-liners; you should think these through. (For

Step 5, you’ll replace Y by X . For Steps 6/7, you’ll replace Y by Z.)We can make this argument slicker still (and not have to repeat three similar

arguments) as follows. (This is frighteningly abstract.) One of Grothendieck’sinsights is that we should hope to treat contravariant functors Sch → Sets as“geometric spaces”, even if we don’t know if they are representable. For this reason,I’ll call such a functor (for this section only!) a functor-space, to emphasize thatwe are thinking of it as some sort of spaces. Many notions carry over to thismore general setting without change, and some notions are easier. For example,a morphism of functor-spaces h → h′ is just a natural transformation of functors.The following exercise shows that this extends the notion of morphisms of schemes.

11.1.E. Exercise. Show that if X and Y are schemes, then there is a naturalbijection between morphisms of schemes X → Y and morphisms of functor spaceshX → hY . (Hint: this has nothing to do with schemes; your argument will work inany category.)

Also, fibered products of functor-spaces always exist: h×h′′ h′ may be definedby

h×h′′ h′(W ) = h(W )×h′′(W ) h′(W )

(where the fibered product on the right is a fibered product of sets, and those alwaysexist). Notice that this didn’t use any properties of schemes; this works with Schreplaced by any category.

We can make some other definitions that extend notions from schemes tofunctor-spaces. We say that h → h′ express h as an open subfunctor of h′

if for all representative morphisms hX and maps hX → h′, the fibered producthX ×h′ h is representable, by u say, and hU → hX is an open immersion. the

4fibersheaf

202

following fibered square may help.

hY //

open

h

hX // h′

Notice that a morphism of representable functor spaces hW → hZ is an openimmersion if and only if W → Z is an open immersion, so this indeed extends thenotion of open immersion to these functors.

A collection hi of open subfunctors of h′ is said to cover h′ if for each maphX → h′ from a representable subfunctor, the corresponding open subsets Ui → Xcover X .

11.1.F. Key exercise. If a functor-space h is a sheaf that has an open cover byrepresentable functor-spaces (“is covered by schemes”), then h is representable.

ex.doneGiven this formalism, we can now give a quick description of the proof of the

existence of fibered products. Exercise 11.1.D showed that hX×ZY is a sheaf.

11.1.G. Exercise. Suppose (Zi)i is an affine cover of Z, (Xij)j is an affine coverof the preimage of Zi in X , and (Yik)k is an affine cover of the preimage of Zi inY . Show that (hXij×Zi

Yik)ijk is an open cover of the functor hX×ZY . (Hint: use

the definition of open covers!)ex.done

But (hXij×ZiYik

)ijk is representable (fibered products of affines over and affine

exist, Step 1 of the proof of Theorem 11.1.1), so we are done.

11.2 Computing fibered products in practice

5Before giving a bunch of examples, we should first see how to actually computefibered products in practice.

There are four types of morphisms that it is particularly easy to take fiberedproducts with, and all morphisms can be built from these four atomic components.(1) Base change by open immersions.

We’ve already done this (Exercise 9.1.G), and we used it in the proof thatfibered products of schemes exist.

f−1(Y ) _

// Y _

X

f // Z

I’ll describe the remaining three on the level of affine open sets, because weobtain general fibered products by gluing.(2) Adding an extra variable.

11.2.A. Easy algebra exercise.. Show that B ⊗A A[t] ∼= B[t].ex.done

5fiberedinpractice

203

Hence the following is a fibered diagram.

Spec B[t]

// Spec A[t]

Spec B // Spec A

(3) Base change by closed immersions[primordial: Here is a useful algebraic fact.

11.2.B. Exercise. Suppose C → A,B are two ring morphisms, so in particular A and B areC-modules. Let I be an ideal of A. Let Ie be the extension of I to A ⊗C B. (These are theelements

Pj ij ⊗ bj where ij ∈ I, bj ∈ B.) Show that there is a natural isomorphism

(A/I) ⊗C B ∼= (A⊗C B)/Ie.

(Hint: consider I → A→ A/I → 0, and use the right-exactness of ⊗CB, Exercise 3.3.B.) 6

ex.doneThus the natural morphism A⊗C B → A/I ⊗C B is a surjection, and each rectangle in

Spec(A⊗C B)/Ie _

// SpecA/I _

SpecA⊗C B

// SpecA

SpecB // SpecC

is a fibered diagram. ]

11.2.C. Exercise. Suppose φ : A → B is a ring homomorphism, and I ⊂ A isan ideal. Let Ie := 〈φ(i)〉i∈I ⊂ B be the extension of I to B. Describe a natural extension of an ideal

isomorphism B/Ie ∼= B⊗A (A/I). (Hint: consider I → A→ A/I → 0, and use theright-exactness of ⊗AB, Exercise 3.3.B.) 7

ex.doneAs an immediate consequence: the fibered product with a subscheme is the sub-

scheme of the fibered product in the obvious way. We say that “closed immersionsare preserved by base change”.

As an application, we can compute tensor products of finitely generated kalgebras over k. For example, we have a canonical isomorphism

k[x1, x2]/(x21 − x2)⊗k k[y1, y2]/(y3

1 + y32)∼= k[x1, x2, y1, y2]/(x2

1 − x2, y31 + y3

2).

11.2.1. Example. We can also use now compute C⊗R C:

C⊗R C ∼= C⊗R (R[x]/(x2 + 1))

∼= (C⊗R R[x])/(x2 + 1) by (3)

∼= C[x]/(x2 + 1) by (2)∼= C[x]/(x− i)(x + i)∼= C× C

Thus Spec C ×R Spec C ∼= Spec C∐

Spec C. This example is the first example ofmany different behaviors. Notice for example that two points somehow correspond

6tensorclosed

7tensorclosed

204

to the Galois group of C over R; for one of them, x (the “i” in one of the copies ofC) equals i (the “i” in the other copy of C), and in the other, x = −i. 8 [forme:

Hint at what: Galois group; complex conjugation.]

(4) Base change of affine schemes by localization.[primordial:

11.2.D. Exercise. Suppose C → B,A are two morphisms of rings. Suppose S is a multiplicativeset of A. Then (S ⊗ 1) is a multiplicative set of A⊗C B. Show that there is a natural morphism(S−1A)⊗C B ∼= (S ⊗ 1)−1(A⊗C B).

ex.doneHence we have a fiber diagram:

Spec(S ⊗ 1)−1(A⊗C B)

// Spec S−1A

SpecA⊗C B

// SpecA

SpecB // SpecC

(where each rectangle is a fiber diagram). ]

11.2.E. Exercise. Suppose φ : A→ B is a ring homomorphism, and S ⊂ A is amultiplicative subset of A, which implies that φ(S) is a multiplicative subset of B.Describe a natural isomorphism φ(S)−1B ∼= B ⊗A (S−1A).

ex.doneTranslation: the fibered product with a localization is the localization of the

fibered product in the obvious way. We say that “localizations are preserved by basechange”. This is handy if the localization is of the form A → Af (correspondingto taking distinguished open sets) or A → FF (A) (from A to the fraction field ofA, corresponding to taking generic points), and various things in between.

These four facts let you calculate lots of things in practice, as we will seethroughout the rest of this chapter.

11.2.F. Exercise: the three important types of monomorphisms of schemes.Show that the following are monomorphisms: open immersions, closed immersions,and localization of affine schemes. As monomorphisms are closed under compo-sition, Exercise 3.3.N, compositions of the above are also monomorphisms (e.g.locally closed immersions, or maps from Spec of stalks at points of X to X).9

11.2.G. Exercise. If X, Y → Z are two locally closed immersions, show thatX ×Z Y is naturally isomorphic to X ∩ Y .10

11.3 Pulling back families and fibers of morphisms

11.3.1. Pulling back families.

8ccr

9e:nicemonomorphisms

10lcimmintersectionfiberedproduct

205

We can informally interpret fibered product in the following geometric way.Suppose Y → Z is a morphism. We interpret this as a “family of schemes parametrizedby a base scheme (or just plain base) Z.” Then if we have another morphismX → Z, we interpret the induced map X ×Z Y → X as the “pulled back family”.

X ×Z Y //

pulled back family

Y

family

X // Z

We sometimes say that X×Z Y is the scheme-theoretic pullback of Y , scheme-theoretic inverse image, or inverse image scheme of Y . [forme: I drew a

picture of this on the blackboard. I discussed the example: the family y2z = x3 + txz2

of cubics in P2 parametrized by the affine line, and what happens if you pull back to

the affine plane via t = uv, to get the family y2z = x3 + uvxz2.] For this reason, base scheme, base, base

change, change of basefibered product is often called base change or change of base or pullback.inverse image ideal

sheaf, scheme-theoretic

pullback, inverse im-

age scheme, inverse

image scheme, scheme-

theoretic inverse image

11.3.2. Fibers of morphisms.Suppose p→ Z is the inclusion of a point (not necessarily closed). (If K is the

residue field of a point, we mean the canonical map Spec K → Z.) Then if g : Y →Z is any morphism, the base change with p→ Z is called the fiber of g above por the preimage of p, and is denoted g−1(p). If Z is irreducible, the fiber abovethe generic point is called the generic fiber. In an affine open subscheme Spec A generic fiber, fiber above

a pointcontaining p, p corresponds to some prime ideal p, and the morphism correspondsto the ring map A → Ap/pAp. this is the composition if localization and closedimmersion, and thus can be computed by the tricks above.

(Quick remark: p→ Z is a monomorphism, by Exercise 11.2.F.)

11.3.3. Example. The following example has many enlightening aspects. Considerthe projection of the parabola y2 = x to the x axis over Q, corresponding to themap of rings Q[x] → Q[y], with x 7→ y2. (If Q alarms you, replace it with yourfavorite field and see what happens.)11

Then the preimage of 1 is two points:

Spec Q[x, y]/(y2 − x)⊗Q Spec Q[x]/(x− 1) ∼= Spec Q[x, y]/(y2 − x, x− 1)

∼= Spec Q[y]/(y2 − 1)

∼= Spec Q[y]/(y − 1)∐

Spec Q[y]/(y + 1).

The preimage of 0 is one nonreduced point:

Spec Q[x, y]/(y2 − x, x) ∼= Spec Q[y]/(y2).

The preimage of −1 is one reduced point, but of “size 2 over the base field”.

Spec Q[x, y]/(y2 − x, x + 1) ∼= Spec Q[y]/(y2 + 1) ∼= Spec Q[i].

The preimage of the generic point is again one reduced point, but of “size 2over the residue field”, as we verify now.

Spec Q[x, y]/(y2 − x)⊗Q(x) ∼= Spec Q[y]⊗Q(y2)

11Qdoublecover

206

i.e. you take elements polynomials in y, and you are allowed to invert polynomialsin y2. A little thought shows you that you are then allowed to invert polynomialsin y, as if f(y) is any polynomial in y, then

1

f(y)=

f(−y)

f(y)f(−y),

and the latter denominator is a polynomial in y2. Thus

Spec Q[x, y]/(y2 − x)⊗Q(x) ∼= Q(y)

which is a degree 2 field extension of Q(x).Notice the following interesting fact: in each case, the number of preimages

can be interpreted as 2, where you count to two in several ways: you can countpoints (as in the case of the preimage of 1); you can get non-reduced behavior (asin the case of the preimage of 0); or you can have a field extension of degree 2 (asin the case of the preimage of −1 or the generic point). In each case, the fiber is anaffine scheme whose dimension as a vector space over the residue field of the pointis 2. Number theoretic readers may have seen this behavior before. This is goingto be symptomatic of a very special and important kind of morphism (a finite flatmorphism).

Try to draw a picture of this morphism if you can, so you can develop a pictoralshorthand for what is going on.

Here are some other examples.

11.3.A. Exercise. Prove that AnA∼= An

Z ×SpecZ Spec A. Prove that PnA∼=

PnZ ×SpecZ Spec A.

11.3.B. Exercise. Show that the underlying topological space of the (scheme-theoretic) fiber X → Y above a point p is naturally identified with the topologicalfiber of X → Y above p. [forme: localization and quotient]

11.3.C. Exercise. Show that for finite-type schemes over C, the closed points(=complex-valued points by the Nullstellensatz, Exercise 7.3.I) of the fibered prod-uct correspond to the fibered product of the complex-valued points. (You will justuse the fact that C is algebraically closed.) [forme: Luis Diogo found this confusing.

hence earlier, show that for lft schemes, we can identify closed points with points

with residue field f.d. over C. This requires some thought!]ex.done

11.3.4. 12Here is a definition in common use. The terminology is a bit unfortunate,because it is a second (different) meaning of “points of a scheme”. (Sadly, we’ll evensee a third different meaning soon, §11.4.2.) If T is a scheme, the T -valued pointsof a scheme X are defined to be the morphism T → X . They are sometimesdenoted X(T ). If A is a ring (most commonly in this context a field), the A-valued points of a scheme X are defined to be the morphism Spec A → X .They are sometimes denoted X(A). For example, if k is an algebraically closedfield, then the k-valued points of a finite type scheme are just the closed points;but in general, things can be weirder. (When we say “points of a scheme”, andT -valued points of a

scheme, X(T ), ditto for

A

not A-valued points, we will always mean the usual meaning, not this meaning.)

12d:Tvaluedpoint

207

11.3.D. Exercise. Describe a natural bijection (X×ZY )(T ) ∼= X(T )×Z(T )Y (T ).(The right side is a fibered product of sets.) In other words, fibered productsbehaves well with respect to T -valued points. This is one of the motivations forthis notion.

11.3.E. Exercise. Consider the morphism of schemes X = Spec k[t] → Y =Spec k[u] corresponding to k[u] → k[t], t = u2, where char k 6= 2. Show thatX×Y X has 2 irreducible components. [forme: Ask them to see what is happening

over the generic point]ex.done

(What happens if char k = 2? See Exercise 11.4.F for a clue.)

11.3.F. Exercise generalizing C ⊗R C. Suppose L/K is a finite Galois fieldextension. What is L⊗K L?

11.3.G. Hard but fascinating exercise for those familiar with the Ga-lois group of Q over Q. Show that the points of Spec Q⊗Q Q are in natural

bijection with Gal(Q/Q), and the Zariski topology on the former agrees with theprofinite topology on the latter.13

11.3.H. Weird Exercise. Show that Spec Q(t) ⊗Q C has closed points in nat-ural correspondence with the transcendental complex numbers. (If the descriptionSpec C[t]⊗Q[t] Q(t) is more striking, you can use that instead.) This scheme doesn’tcome up in nature, but it is certainly neat! [forme: integral dimension one scheme;

unimportant]

11.4 Properties preserved by base change

We now discuss a number of properties that behave well under base change.We’ve already shown that the notion of “open immersion” is preserved by base

change (Exercise 9.1.G). We did this by explicitly describing what the fiberedproduct of an open immersion is: if Y → Z is an open immersion, and f : X → Zis any morphism, then we checked that the open subscheme f−1(Y ) of X satisfiesthe universal property of fibered products.

We have also shown that the notion of “closed immersion” is preserved by basechange (§11.2 (3)). In other words, given a fiber diagram

W //

X

Y

cl. imm.// Z

where Y → Z is a closed immersion, W → X is as well.

11.4.A. Easy Exercise. Show that locally principal closed subschemes pull backto locally principal closed subschemes.

ex.doneSimilarly, other important properties are preserved by base change.

13e:Qbar

208

11.4.B. Exercise. Show that the following properties of morphisms are preservedby base change. [forme: later order these]

(a) quasicompact(b) quasiseparated(c) affine morphism(d) finite(e) locally of finite type(f) finite type(g) locally of finite presentation(h) finite presentation

11.4.C. Exercise. Show that the notion of “quasifinite morphism” (finite type+ finite fibers) is preserved by base change. (Warning: the notion of “finite fibers”is not preserved by base change. Spec Q→ Spec Q has finite fibers, but Spec Q⊗Q

Q→ Spec Q has one point for each element of Gal(Q/Q), see Exercise 11.3.G.) 14

[forme: For this reason, Hartshorne’s definition isn’t great; EGA is right. How hard

is this exercise.]

11.4.D. Exercise. Show that surjectivity is preserved by base change. (Sur-jectivity has its usual meaning: surjective as a map of sets.) (You may end upsurjective

using the fact that for any fields k1 and k2 containing k3, k1 ⊗k3 k2 is non-zero,and also the axiom of choice.) [forme: Surjective morphisms are not necessarily

epimorphisms in the category of schemes. Example from Joe R: Spec k → Spec k[ε].

Consider maps of the former to Spec k[ε].]

11.4.E. Exercise. If P is a property of morphisms preserved by base change[forme: and composition] , and X → Y and X ′ × Y ′ are two morphisms of S-schemes with property P , show that X ×S X ′ → Y ×S Y ′ has property P as well.[forme: H.E.II.4.8(d)] 15

11.4.1. ? Properties not preserved by base change, and how to fix them.There are some notions that you should reasonably expect to be preserved

by pullback based on your geometric intuition. Given a family in the topologicalcategory, fibers pull back in reasonable ways. So for example, any pullback of afamily in which all the fibers are irreducible will also have this property; ditto forconnected. Unfortunately, both of these fail in algebraic geometry, as the Exam-ple 11.2.1 shows:

Spec C∐

Spec C //

Spec C

Spec C // Spec R

The family on the right (the vertical map) has irreducible and connected fibers, andthe one on the left doesn’t. The same example shows that the notion of “integralfibers” also doesn’t behave well under pullback.

14badqf

15productinP

209

11.4.F. Exercise. Suppose k is a field of characteristic p, so k(up)/k(u) is aninseparable extension. By considering k(up) ⊗k(u) k(up), show that the notion of“reduced fibers” does not necessarily behave well under pullback. (The fact thatI’m giving you this example should show that this happens only in characteristicp, in the presence of something as strange as inseparability.) 16

ex.doneWe rectify this problem as follows.

11.4.2. 17A geometric point of a scheme X is defined to be a morphism Spec k →X where k is an algebraically closed field. Awkwardly, this is now the third kindof “point” of a scheme! There are just plain points, which are elements of theunderlying set; there are T -valued points, which are maps T → X , §11.3.4; andthere are geometric points. Geometric points are clearly a flavor of a T -valuedpoint, but they are also an enriched version of a (plain) point: they are the data ofa point with an inclusion of the residue field of the point in an algebraically closedfield.

A geometric fiber of a morphism X → Y is defined to be the fiber overa geometric point of Y . A morphism has connected (resp. irreducible, inte-gral, reduced) geometric fibers if all its geometric fibers are connected (resp.irreducible, integral, reduced).

11.4.G. Exercise. Show that the notion of “connected (resp. irreducible, inte-gral, reduced)” geometric fibers behaves well under base change.

11.4.H. Exercise for the arithmetically-minded. Show that for the mor-phism Spec C → Spec R, all geometric fibers consist of two reduced points. (Cf.Example 11.2.1.)

11.4.I. Exercise. Recall Example 11.3.3, the projection of the parabola y2 = x tothe x axis, corresponding to the map of rings Q[x]→ Q[y], with x 7→ y2. Show thatthe geometric fibers of this map are always two points, except for those geometricfibers over 0 = [(x)].

Checking whether a k-scheme is geometrically connected etc. seems annoying:you need to check every single algebraically closed field containing k. However, ineach of these four cases, the failure of nice behavior of geometric fibers can alreadybe detected after a finite field extension. For example, Spec Q(i) → Spec Q is notgeometrically connected, and in fact you only need to base change by Spec Q(i) tosee this. We make this precise as follows.

11.4.J. Exercise. Suppose X is a k-scheme.

(a) Show that X is geometrically irreducible if and only if X×kks is irreducibleif and only if X ×k K is irreducible for all field extensions K/k. (Here ks

is the separable closure of k.)(b) Show that X is geometrically connected if and only if X×kks is connected

if and only if X ×k K is connected for all field extensions K/k.(c) Show that X is geometrically reduced if and only if X ×k kp is reduced if

and only if X×k K is reduced for all field extensions K/k. (Here kp is the

16e:reducedfibers

17d:geometricpoint

210

perfect closure of k.) Thus if chark = 0, then X is geometrically reducedif and only if it is reduced.

(d) Combining (a) and (c), show that X is geometrically integral if and onlyif X ×k K is geometrically integral for all field extensions K/k.

geo irreducible, con-

nected, reduced,

integral

geometric point, geo-

metric fiber

[forme: H.E.II.3.15]

11.5 Products of projective schemes: The Segre embedding

18I will next describe products of projective A-schemes over A. The case ofgreatest initial interest is if A = k. In order to do this, I need only describePm

A ×A PnA, because any projective scheme has a closed immersion in some Pm

A , andclosed immersions behave well under base change, so if X → Pm

A and Y → PnA are

closed immersions, then X ×A Y → PmA ×A Pn

A is also a closed immersion, cut outby the equations of X and Y .

We’ll describe PmA ×A Pn

A, and see that it too is a projective A-scheme.Before we do this, we’ll get some motivation from classical projective spaces

(non-zero vectors modulo non-zero scalars) in a special case. Our map will send[x0; x1; x2]× [y0; y1] to a point in P5, whose co-ordinates we think of as being entriesin the “multiplication table”

[ x0y0; x1y0; x2y0;x0y1; x1y1; x2y1 ]

This is indeed a well-defined map of sets. Notice that the resulting matrix is rankone, and from the matrix, we can read off [x0; x1; x2] and [y0; y1] up to scalars. Forexample, to read off the point [x0; x1; x2] ∈ P2, we just take the first row, unless itis all zero, in which case we take the second row. (They can’t both be all zero.) Inconclusion: in classical projective geometry, given a point of Pm and Pn, we haveproduced a point in Pmn+m+n, and from this point in Pmn+m+n, we can recoverthe points of Pm and Pn.

Suitably motivated, we return to algebraic geometry. We define a map

PmA ×A Pn

A → Pmn+m+nA

by([x0; . . . ; xm], [y0; . . . ; yn]) 7→ [z00; z01; · · · ; zij ; · · · ; zmn]

= [x0y0; x0y1; · · · ; xiyj ; · · ·xmyn].

More explicitly, we consider the map from the affine open set Ui × Vj (where Ui =D(xi) and Vj = D(yj) to the affine open set Wij = D(zij) by

(x0/i, . . . , xm/i, y0/j , . . . , yn/j) 7→ (x0/iy0/j ; . . . ; xi/iyj/j ; . . . ; xm/iyn/j)

or, in terms of algebras, zab/ij 7→ xa/iyb/j .

11.5.A. Exercise. Check that these maps glue to give a well-defined morphismPm

A ×A PnA → Pmn+m+n

A .[forme: (Aside: we now well know that a map to projective space corresponds

to an invertible sheaf with a bunch of sections. The invertible sheaf on this case is

π∗1OPmA

(1)⊗π∗2OPnA

(1), where πi are the projections of the product onto the two factors.

18s:segre

211

The notion is often used for this notion, when you pull back sheaves from each

factor of a product, and tensor. For example, this invertible sheaf could be written

O(1) O(1). People often write O(a) O(b) for O(a, b).) ] , O(a, b)

I claim this morphism is a closed immersion. [forme: cf Exercise 20.4.A] Wecan check this on an open cover of the target (the notion of being a closed immersionis affine-local, Exercise 9.1.P). Let’s check this on the open set where zij 6= 0. Thepreimage of this open set in Pm

A × PnA is the locus where xi 6= 0 and yj 6= 0, i.e.

Ui × Vj . As described above, the map of rings is given by zab/ij 7→ xa/iyb/j ; this isclearly a surjection, as zaj/ij 7→ xa/i and zib/ij 7→ yb/j .

This map is called the Segre morphism or Segre embedding. If A is a Segre product, Segre

embeddingfield, the image is called the Segre variety. [forme: [H, Ex. II.4.9]]

Here are some useful comments.

11.5.B. Exercise. Show that the Segre scheme (the image of the Segre mor-phism) is cut out by the equations corresponding to

rank

a00 · · · a0n

.... . .

...am0 · · · amn

= 1,

i.e. that all 2× 2 minors vanish. (Hint: suppose you have a polynomial in the aij

that becomes zero upon the substitution aij = xiyj . Give a recipe for subtractingpolynomials of the form monomial times 2× 2 minor so that the end result is 0.)

11.5.1. Important Example. Let’s consider the first non-trivial example, whenm = n = 1. We get P1 × P1 → P3. We get a single equation

rank

(

a00 a01

a10 a11

)

= 1,

i.e. a00a11 − a01a10 = 0. We get our old friend, the quadric surface! Hence: thenonsingular quadric surface wz − xy = 0 is isomorphic to P1 × P1 (Figure 3). Onefamily of lines corresponds to the image of x × P1 as x varies, and the othercorresponds to the image P1 × y as y varies.

Since (by diagonalizability of quadratics) all nonsingular quadratics over an al-gebraically closed field are isomorphic, we have that all nonsingular quadric surfacesover an algebraically closed field are isomorphic to P1 × P1.

Note that this is not true over a field that is not algebraically closed. Forexample, over R, w2 +x2 + y2 + z2 = 0 is not isomorphic to P1

R×R P1R. Reason: the

former has no real points, while the latter has lots of real points.

11.5.C. Exercise: A co-ordinate-free description of the Segre embed-ding. Show that the Segre embedding can be interpreted as PV ×PW → P(V ⊗W )via the surjective map of graded rings

Sym•(V ∨ ⊗W∨) // //∑∞i=0

(

Symi V ∨)

⊗(

Symi W∨)

“in the opposite direction”.Can you define the Segre embedding for the product of three projective spaces?

CHAPTER 12

Separated morphisms, and varieties

1

[forme: Place: if X is reduced and Y is separated, then a function is defined on

a dense open subset. We now get this.]

12.1 Separated morphisms

The notion of a separated morphism is fundamentally important. It looks weirdthe first time you see it, but it is highly motivated.

12.1.1. Motivation. Separatedness is the analogue of the Hausdorff condition formanifolds (see Exercise 12.1.A), so let’s review why we like Hausdorffness. Recallthat a topological space is Hausdorff if for every two points x and y, there aredisjoint open neighborhoods of x and y. The real line is Hausdorff, but the Hausdorff

“real line with doubled origin” is not. Many proofs and results about manifoldsuse Hausdorffness in an essential way. For example, the classification of compactone-dimensional real manifolds is very simple, but if the Hausdorff condition wereremoved, we would have a very wild set.[forexperts: “Separation” insteadof “separatedness”? No. Reason: closedness makes sense rather thanclosure; properness is better than propriety.]

So armed with this definition, we can cheerfully exclude the line with doubledorigin from civilized discussion, and we can (finally) define the notion of a variety,in a way that corresponds to the classical definition. 2 variety, Hausdorff

With our motivation from manifolds, we shouldn’t be surprised that all of ouraffine and projective schemes are separated: certainly, in the land of real manifolds,the Hausdorff condition comes for free for “subsets” of manifolds. (More precisely,if Y is a manifold, and X is a subset that satisfies all the hypotheses of a manifoldexcept possibly Hausdorffness, then Hausdorffness comes for free.)

As an unexpected added bonus, a separated morphism to an affine schemehas the property that the intersection of a two affine open sets in the source isaffine (Proposition 12.1.8). This will make Cech cohomology work very easily on(quasicompact) schemes. [forme: ref ] You should see this as the analogue of thefact that in Rn, the intersection of two convex sets is also convex. In fact affineschemes will be trivial from the point of view of quasicoherent cohomology, just asconvex sets in Rn are, so this metaphor is quite apt.

1sseparatedness

2Hausdorff

213

214

A lesson arising from the construction is the importance of the diagonal mor-phism. More precisely given a morphism X → Y , nice consequences can be lever-aged from good behavior of the diagonal morphism δ : X → X ×Y X , usuallydiagonal morphism δ

through fun diagram chases. This is a lesson that applies across many fields ofmathematics. (Another nice gift the diagonal morphism: it will soon give us agood algebraic definition of differentials, in Chapter 22.)

Grothendieck taught us that one should try to define properties of morphisms,not of objects; then we can say that an object has that property if the morphism tothe final object has that property. We saw this earlier with the notion of quasicom-pact. In this spirit, separatedness will be a property of morphisms, not schemes.

Before we define separatedness, we make an observation about all diagonalmaps.

12.1.2. Proposition. — Let X → Y be a morphism of schemes. Then thediagonal morphism δ : X → X×Y X is a locally closed immersion.3 [forme: Proof

of [H, Cor. II.4.2]]

This locally closed subscheme of X ×Y X (the diagonal) will be denoted ∆.

Proof. We will describe a union of open subsets of X ×Y X covering the image ofX , such that the image of X is a closed immersion in this union.

12.1.3. 4Say Y is covered with affine open sets Vi and X is covered with affineopen sets Uij , with π : Uij → Vi. Then the diagonal is covered by Uij×Vi Uij . (Anypoint p ∈ X lies in some Uij ; then δ(p) ∈ Uij ×Vi Uij . Figure 1 may be helpful.)As a reality check: Uij ×Vi Uij is indeed an affine open subscheme of X ×Y X , byconsidering the factorization

Uij ×Vi Uij → Uij ×Y Uij → Uij ×Y X → X ×Y X

where the first arrow is an isomorphism as Vi → Y is a monomorphism (as itis an open immersion, Exercise 11.2.F). The second and third arrows are openimmersions as open immersions are preserved by base change (Exercise 9.1.G).[forme: Note that δ−1(Uij ×Vi

Uij) = Uij : Uij ×ViUij ∼= Uij ×Y Uij because Vi → Y is a

monomorphism. [forme: Exercise?] Then because open immersions behave wellwith respect to base change, we have the fiber diagram

Uij //

X

Uij ×Y X // X ×Y X

from which δ−1(Uij ×Y X) = Uij . As δ−1(Uij ×Y Uij) contains Uij , we must have

δ−1(Uij ×Y Uij) = Uij . ]

Finally, we’ll check that Uij → Uij ×Vi Uij is a closed immersion. Say Vi =Spec S and Uij = Spec R. Then this corresponds to the natural ring map R×S R→R, which is obviously surjective.

The open subsets we described may not cover X ×Y X , so we have not shownthat δ is a closed immersion.

12.1.4. Definition. A morphism X → Y is separated if the diagonal morphismseparated

3Dlocallyclosed

4carefulhere

215

X ×Y X

X

X

Figure 1. A neighborhood of the diagonal is covered by Uij ×Vj Uij

δ : X → X ×Y X is a closed immersion. An A-scheme X is said to be separatedover A if the structure morphism X → Spec A is separated. When people say thata scheme (rather than a morphism) X is separated, they mean implicitly that somemorphism is separated. For example, if they are talking about A-schemes, theymean that X is separated over A. 5 sep over A

Thanks to Proposition 12.1.2, a morphism is separated if and only if the diag-onal is closed. This is reminiscent of a definition of Hausdorff, as the next exerciseshows. 6

12.1.A. Exercise (for those seeking topological motivation). Showthat a topological space X is Hausdorff if the diagonal is a closed subset of X ×X .(The reason separatedness of schemes doesn’t give Hausdorffness — i.e. that forany two open points x and y there aren’t necessarily disjoint open neighborhoods— is that in the category of schemes, the topological space X ×X is not in generalthe product of the topological space X with itself. For example, Exercise 11.1.Ashowed that A2

k does not have the product topology on A1k ×k A1

k.) separatedHausdorff

Hausdorff12.1.B. Important easy exercise. Show that open immersions and closedimmersions are separated. 7 (Hint: Just do this by hand. Alternatively, show thatmonomorphisms are separated. Open and closed immersions are monomorphisms,by Exercise 11.2.F.)

5d:sep

6H.C.II.4.2

7openclosedsep, H.C.II.4.6a

216

12.1.C. Important easy exercise. Show that every morphism of affine schemesis separated. 8 (Hint: this was essentially done in Proposition 12.1.2.)

ex.doneI’ll now give you an example of something separated that is not affine. The

following single calculation will imply that all quasiprojective A-schemes are sepa-rated (once we know that the composition of separated morphisms are separated,Proposition 12.1.13).

12.1.5. Proposition. — PnA → Spec A is separated. 9

ex.doneWe give two proofs. The first is by direct calculation. The second requires

no calculation, and just requires that you remember some classical constructionsdescribed earlier.Proof 1: direct calculation. We cover Pn

A×A PnA with open sets of the form Ui×Uj ,

where U0, . . . , Un form the “usual” affine open cover. The case i = j was takencare of before, in the proof of Proposition 12.1.2. If i 6= j then

Ui ×A Uj∼= Spec A[x0/i, . . . , xn/i, y0/j , . . . , yn/j ]/(xi/i − 1, yj/j − 1).

Now the restriction of the diagonal ∆ is contained in Ui (as the diagonal map com-posed with projection to the first factor is the identity), and similarly is containedin Uj . Thus the diagonal map over Ui×A Uj is Ui∩Uj → Ui×A Uj . This is a closedimmersion, as the corresponding map of rings

Spec A[x0/i, . . . , xn/i, y0/j , . . . , yn/j ]→ Spec A[x0/i, . . . , xn/i, x−1j/i]/(xi/i − 1)

(given by xk/i 7→ xk/i, yk/j 7→ xk/i/xj/i) is clearly a surjection (as each generator

of the ring on the right is clearly in the image — note that x−1j/i is the image of

yi/j).

Proof 2: classical geometry. Note that the diagonal map δ : PnA → Pn

A ×A PnA

followed by the Segre embedding S : PnA×A Pn

A → Pn2+n (§??, a closed immersion)

can also be factored as the second Veronese map ν2 : PnA → P(n+2

2 )−1 (§??) followed

by a linear map L : P(n+22 )−1 → Pn2+n (Exercise 10.4.B), both of which are closed

immersions. You should verify this. This forces δ to send closed sets to closed sets(or else S δ won’t, but L ν2 does).

PnA ×A Pn

A

S

%%LLLLLLLLLL

PnA

δ

::uuuuuuuuuu

ν2

$$IIIIIIIII Pn2+n

P(n+22 )−1

L

99ssssssssss

We note for future reference a minor result proved in the course of Proof 1.Figure 2 may help show why this is natural. You could also interpret this statementas

X ×X×AX (U ×A V ) ∼= U ×X V

8affinetoaffinesep, H.P.II.4.1

9p:Pnsep

217

which follows from the magic diagram, Exercise 3.3.L. [forme: teaching: point this

out during the proof]

U ∩ V ∼= (U × V ) ∩∆

U ×X

X × V

∆ U × V

Figure 2. Small Proposition 12.1.6

12.1.6. Small Proposition. — If U and V are open subsets of an A-scheme X,then ∆ ∩ (U ×A V ) ∼= U ∩ V . 10

12.1.D. Exercise. Show that the line with doubled origin X (Example 6.1.7) isnot separated, by verifying that the image of the diagonal morphism is not closed.11

(Another argument is given below, in Exercise 12.1.K.)ex.done

We finally define then notion of variety!

12.1.7. Definition. A variety over a field k, or k-variety, is a reduced, separatedscheme of finite type over k. For example, a reduced finite type affine k-scheme variety

is a variety. In other words, to check if Spec k[x1, . . . , xn]/(f1, . . . , fr) is a variety,you need only check reducedness. [forme: In FOAG: refer back to affine variety.

We don’t yet know that projective varieties are varieties.] 12

Notational caution: In some sources, the additional condition of irreducibilityis imposed. We will not do this. Also, it is often assumed that k is algebraicallyclosed. We will not do this either. [forme: Mumford red book?]

Here is a very handy consequence of separatedness.

12.1.8. Proposition. — Suppose X → Spec A is a separated morphism to anaffine scheme, and U and V are affine open sets of X. Then U ∩ V is an affineopen subset of X. 13

ex.done

10stupiddiagonalfact

11qwerty1

12d:variety

13H.E.II.4.3, affineintaffineisaffine

218

Before proving this, we state a consequence that is otherwise nonobvious. IfX = Spec A, then the intersection of any two affine open sets is open (just takeA = Z in the above proposition). This is certainly not an obvious fact! Weknow that the intersection of any two distinguished affine open sets is affine (fromD(f)∩D(g) = D(fg)), but we have very little handle on affine open sets in general.

Warning: this property does not characterize separatedness. For example, ifA = Spec k and X is the line with doubled origin over k, then X also has thisproperty. [forme: This will be generalized slightly in Exercise 12.1.M. But that’s

now commented out. ]

Proof. By Proposition 12.1.6, (U ×A V )∩∆ = U ∩V , where ∆ is the diagonal. ButU ×A V is affine (the fibered product is two affine schemes over an affine scheme isaffine, Step 1 of our construction of fibered products, Theorem 11.1.1), and ∆ is aclosed subscheme of an affine scheme, and hence affine.

12.1.9. Quasiseparated morphisms.[forme: Now this should be changed.] We now define a handy relative of

separatedness, that is also given in terms of a property of the diagonal morphism,and has similar properties. The reason it is less famous is because it automaticallyholds for the sorts of schemes that people usually deal with. We say a morphismf : X → Y is quasiseparated if the diagonal morphism δ : X → X ×Y X isquasiseparated

quasicompact.14 I’ll give a more insightful translation shortly, in Exercise 12.1.E.Most algebraic geometers will only see quasiseparated morphisms, so this may

be considered a very weak assumption. Here are two large classes of morphismsthat are quasiseparated. (a) As closed immersions are quasicompact (easy, seeExercise ??), separated implies quasiseparated. (b) If X is a Noetherian scheme,then any morphism to another scheme is quasicompact (easy, see Exercise 9.1.B), soany X → Y is quasiseparated. Hence those working in the category of Noetherianschemes need never worry about this issue. [forme: put that exercise back when

defining quasicompact morphisms]

The following characterization makes quasiseparatedness a useful hypothesis inproving theorems.

12.1.E. Exercise. Show that f : X → Y is quasiseparated if and only if forany affine open Spec A of Y , and two affine open subsets U and V of X mappingto Spec A, U ∩ V is a finite union of affine open sets. (Hint: compare this toProposition 12.1.8.) 15 [forme: Another hint: use the magic diagram Ex. 3.3.L.]

In particular, a morphism f : X → Y is quasicompact and quasiseparated ifand only if the preimage of any affine open subset of Y is a finite union of affineopen sets in X , whose pairwise intersections are all also finite unions of affine opensets. The condition of quasiseparatedness is often paired with quasicompactness inhypotheses of theorems.

12.1.F. Exercise (a nonquasiseparated scheme). Let X = Spec k[x1, x2, . . . ],and let U be X− [m] where m is the maximal ideal (x1, x2, . . . ). Take two copies of

14defqs

15niceqs

219

X , glued along U . Show that the result is not quasiseparated. 16 This open immer-sion U ⊂ X came up earlier in Exercise 5.5.D as an example of a nonquasicompactopen subset of an affine scheme. [forme: Older: (This answers Exercise 7.3.B(b).)

There is no (b) now.]

12.1.10. Theorem. — Both separatedness and quasiseparatedness are preservedby base change. 17

Proof. Suppose

W

// X

Y // Z

is a fiber square. We will show that if Y → Z is separated or quasiseparated, thenso is W → X . The reader should verify that

WδW //

W ×X W

Y

δY // Y ×Z Y

is a fiber diagram. (This is a categorical fact, and holds true in any category withfibered products.) As the property of being a closed immersion is preserved by basechange (§11.2 (3)), if δY is a closed immersion, so is δX .

Quasiseparatedness follows in the identical manner, as quasicompactness is alsopreserved by base change (Exercise 11.4.B).

12.1.11. Proposition. — The condition of being separated is local on the target.Precisely, a morphism f : X → Y is separated if and only if for any cover of Y byopen subsets Ui, f−1(Ui)→ Ui is separated for each i.18

ex.done

12.1.12. Hence affine morphisms are separated, by Exercise 12.1.C. In particular,finite morphisms are separated. 19 [forme: Part of H Ex. II.5.17(b)] aff, finite morphisms

separatedProof. If X → Y is separated, then for any Ui → Y , f−1(Ui) → Ui is separated,as separatedness is preserved by base change (Theorem 12.1.10). Conversely, tocheck if ∆ → X ×Y X is a closed subset, it suffices to check this on an opencover. If g : X ×Y X → Y is the natural morphism, our open cover Ui of Yinduces an open cover f−1(Ui) ×Ui f−1(Ui) of X ×Y X . Then f−1(Ui) → Ui

separated implies f−1(Ui)→ f−1(Ui)×Ui f(Ui) is a closed immersion by definitionof separatedness.

16nonqs, referred to in 7.3.B

17sepbasechange, H.C.II.4.6(c)

18seplocalontarget, H.C.II.4.6f

19affinesep

220

12.1.G. Exercise. Prove that the condition of being quasiseparated is local onthe target. (Hint: the condition of being quasicompact is local on the target byExercise 9.1.C; use a similar argument.)

12.1.13. Proposition. — (a) The condition of being separated is closed undercomposition. In other words, if f : X → Y is separated and g : Y → Z is separated,then g f : X → Z is separated.(b) The condition of being quasiseparated is closed under composition. 20

Proof. (a) [forme: teaching: start with magic diagram] We are given that δf :X → X ×Y X and δg : Y → Y ×Z Y are closed immersions, and we wish to showthat δh : X → X ×Z X is a closed immersion. Consider the diagram

Xδf // X ×Y X

c //

X ×Z X

Y

δg // Y ×Z Y.

The square is the magic fibered diagram of Exercise 3.3.L. As δg is a closed immer-sion, c is too (closed immersions are preserved by base change, §11.2 (3)). Thusc δf is a closed immersion (the composition of two closed immersions is also aclosed immersion, Exercise 9.1.O).

(b) The identical argument (with “closed immersion” replaced by “quasicom-pact”) shows that the condition of being quasiseparated is closed under composi-tion.

12.1.14. Proposition. — Any quasiprojective A-scheme is separated over A.[forme: part of H. Thm. II.4.9] 21quasiprojective is sepa-

rated As a corollary, any reduced quasiprojective k-scheme is a k-variety.

Proof. Suppose X → Spec A is a quasiprojective A-scheme. The structure mor-phism can be factored into an open immersion composed with a closed immersionfollowed by Pn

A → A. Open immersions and closed immersions are separated (Exer-cise 12.1.B), and Pn

A → A is separated (Proposition 12.1.5). Separated morphismsare separated (Proposition 12.1.13), so we are done.

[forme: [Another approach to do this is as follows. Suppose X is a scheme, such

that for all x, y ∈ X, there is an open affine containing both. The X is a scheme. Then

this can be applied to Pnk , or even PnZ . (Mild caution: a hyperplane may not work.)

[forme: M.I.6 Prop. 5 and corollary, M.I.6.P.5] 22]]

12.1.15. Proposition. — Suppose f : X → Y and f ′ : X ′ → Y ′ are separated(resp. quasiseparated) morphisms of S-schemes (where S is a scheme). Then theproduct morphism f × f ′ : X ×S X ′ → Y ×S Y ′ is separated (resp. quasiseparated).23

20sepcomposition, H.C.II.4.6b

21qprojsep

22M.I.6.P.5

23H.C.II.4.6d

221

Proof. Apply Exercise 11.4.E.

12.1.16. Applications.As a first application, we define the graph morphism.

12.1.17. Definition. Suppose f : X → Y is a morphism of Z-schemes. Themorphism Γf : X → X×Z Y given by Γf = (id, f) is called the graph morphism.Then f factors as pr2 Γf , where pr2 is the second projection (see Figure 3). 24 graph morphism

f

Γfpr1

pr2Y

X

X ×Z Y

Figure 3. The graph morphism

12.1.18. Proposition. — The graph morphism Γ is always a locally closedimmersion. If Y is a separated Z-scheme (i.e. the structure morphism Y → Z isseparated), then Γ is a closed immersion. 25

ex.doneThis will be generalized in Exercise 12.1.H. [forme: Similarly, if Y is quasisep-

arated, then the graph morphism is quasicompact.]

Proof by Cartesian diagram. [forme: special case of magic diagram, Ex. 3.3.L]

X //

X ×Z Y

Y

δ // Y ×Z Y

The notions of locally closed immersion and closed immersion are preserved by basechange, so if the bottom arrow δ has one of these properties, so does the top.

We now come to a very useful, but bizarre-looking, result. [forexperts: Is Cancellation Theorem

for morphismsthere a better name? Is this satisfactory?]

24d:graphmorphism

25p:graphmorphism

222

12.1.19. Cancellation Theorem for a Property P of Morphisms. — Let Pbe a class of morphisms that is preserved by base change and composition. Suppose

Xf //

h @@@

@@@@

Y

g~~

~~~~

~

Z

is a commuting diagram of schemes. 26 [forme: (a) implies H.C.II.4.6(e), (b)

implies H.Ex.II.4.8(e)]

(a) Suppose that the diagonal morphism δg : Y → Y ×Z Y is in P and h :X → Z is in P . The f : X → Y is in P .

(b) In particular, suppose that closed immersions are in P . Then if h is in Pand g is separated, then f is in P .

When you plug in different P , you get very different-looking (and non-obvious)consequences.

For example, locally closed immersions are separated, so by part (a), if youfactor a locally closed immersion X → Z into X → Y → Z, then X → Y must bea locally closed immersion.

Possibilities for P in case (b) include: finite morphisms, morphisms of finitetype, closed immersions, affine morphisms. [forme: Change that “needed exercise”

to a reference to the actual exercise.]

Proof of (a). By the fibered square

XΓf //

f

X ×Z Y

Y

δg // Y ×Z Y

we see that the graph morphism Γ : X → X ×Z Y is in P (Definition 12.1.17), asP is closed under base change. By the fibered square

X ×Z Yh′

//

Y

g

X

h // Z

the projection h′ : X ×Z Y → Y is in P as well. Thus f = h′ Γ is in P

Here now are some fun and useful exercises.

12.1.H. Exercise. Suppose π : Y → X is a morphism, and s : X → Y is asection of a morphism, i.e. π s is the identity on X . Show that s is a locally closedimmersion. Show that if π is separated, then s is a closed immersion. (This gen-eralizes Proposition 12.1.18.) 27 [forme: Proof: Consider X //

AAA

AAAA

Y

separated~~

X

26propertyP

27sectionlclimm

223

and use Cancellation Theorem 12.1.19(a) for a property P of morphisms, where P is

the class of locally closed immersions. Anssi instead shows that Xs //

s

Y

δ

Y

(sπ,id)// Y × Y

is a

pullback square.] Give an example to show that s needn’t be a closed immersionif π isn’t separated. [forme: Answer: Line with doubled origin projected to the

line.]

12.1.I. Exercise. Show that a A-scheme is separated (over A) if and only if itis separated over Z. (In particular, a complex scheme is separated over C if andonly if it is separated over Z, so complex geometers and arithmetic geometers cancommunicate about separated schemes without confusion.)

12.1.J. Useful exercise: The locus where two morphisms agree. Sup-pose f and g are two morphisms X → Y , over some scheme Z. We can now givemeaning to the phrase ’the locus where f and g agree’, and that in particular thereis a smallest locally closed subscheme where they agree. Suppose h : W → X issome morphism (perhaps a locally closed immersion). We say that f and g agreeon h if f h = g h. Show that there is a locally closed subscheme i : V → X suchthat any morphism h : W → X on which f and g agree factors uniquely throughi, i.e. there is a unique j : W → V such that h = i j. (You may recognize thisas a universal property statement.) Show further that if V → Z is separated, theni : V → X is a closed immersion. Hint: 28define V to be the following fiberedproduct:

V //

Y

δ

X

(f,g)// Y ×Z Y.

As δ is a locally closed immersion, V → X is too. Then if h : W → X is any schemesuch that g h = f h, then h factors through V . 29 [forme: Jack’s argument toan earlier wording: Suppose that f , g : X → Y are morphisms of Z-schemes, then wedefine the locus where f , g agree is defined as the fibered product:

V //

Y

δ

X

(f,g)// Y ×Z Y

.

Since δ is locally closed, then V → X is. We wish to prove that if h : W → X is anymap of Z-schemes such that g h = f h, then h factors through V . However, thisimplies that we have the following commutative diagram with the unique induced

28M.II.6.P.4

29where2morphismsagree, used in proof of valuative criterion of separatedness Theorem 16.8.1, also more

important, in rational map important theorem 13.1.1

224

W → V , as required.

W

h

BB

BB

$$

Y W T RP

MJ

h // X

g

V //

Y

δ

Id

##GGGGGGGGG

X

f **

(f,g)// Y ×Z Y

// Y

Y // Z

.

This diagram requires some explanation. Basically, the outside ring of maps creates a

unique map W → Y ×ZY and since the map Y → Y is the identity, then by uniqueness,

we must have W → Y → Y ×Z Y → Y and so we have the first induced map (the upper

dashed line). The induced map W → V now follows from the universal property. ]

Minor Remarks. 1) In the previous exercise, we are describing V → X by wayof a universal property. Taking this as the definition, it is not a priori clear that Vis a locally closed subscheme of X , or even that it exists.)

2) In the case of reduced finite type k-schemes, the locus where f and g agreecan be interpreted as follows. f and g agree at x if f(x) = g(x), and the twomaps of residue fields are the same. [forme: This is Mumford’s definition.30 Likely

exercise: two maps from variety where closed points go to same points, then they

are the same map.]

3) Notice that Z arises as part of the hypothesis, but is not present in theconclusion!

12.1.K. Exercise. Show that the line with doubled origin X (Example 6.1.7)is not separated, by finding two morphisms f1, f2 : W → X whose domain ofagreement is not a closed subscheme, cf. Proposition 12.1.2. 31 (Another argumentwas given above, in Exercise 12.1.D.)

12.1.L. Less important exercise. Suppose P is a class of morphisms such thatclosed immersions are in P , and P is closed under fibered product and composition.Show that if f : X → Y is in P then f red : Xred → Y red is in P . (Two examplesare the classes of separated morphisms and quasiseparated morphisms.) 32 Hint:

Xred //

%%LLLLLLLLLLL X ×Y Y red

// Y red

X // Y

[forme: Maybe!

12.1.M. Exercise. 33 Suppose π : X → Y is a morphism or a ring R, Y is a separated

R-scheme, U is an affine open subset of X, and V is an affine open subset of Y . Show

that U ∩ π−1V is an affine open subset of X. (Hint: this generalizes Proposition 1.9

30M.II.6.D.2

31qwerty2

32propertyPred, H.E.II.4.8(f)

33genaffineintaffineisaffine

225

of the Class 25 notes. Use Proposition 1.12 or 1.13.) This will be used in the proof

of the Leray spectral sequence, Section 23.9.]

CHAPTER 13

Rational maps

13.1 Rational maps

This is a historically ancient topic. It has appeared late for us because wehave just learned about separatedness. Informally: a rational map is a “morphismX → Y defined almost everywhere”. We will see that in good situations that wherea rational map is defined, it is uniquely defined.

For this chapter, unless otherwise stated, we will assume X and Y to be integraland separated, although the notions we will introduce can be useful in more generalcircumstances. [forme: Try to be careful and state these hypotheses in all theorems.]

The reader interested in more general notions should consider first the case wherethe schemes in question are reduced and separated, but not necessarily irreducible.Many notions can make sense in more generality (without reducedness hypothesesfor example), but I’m not sure if there is a widely accepted definition.

A key example will be irreducible varieties, and the language of rational mapsis most often used in this case.

A rational map from X to Y , denoted X 99K Y , is a morphism on a dense rational map

open set, with the equivalence relation: (f : U → Y ) ∼ (g : V → Y ) if there is adense open set Z ⊂ U ∩ V such that f |Z = g|Z . (In a moment, we will improvethis to: if f |U∩V = g|U∩V .) People often use the word “map” for “morphism”,which is quite reasonable. But then a rational map need not be a map. So to avoidconfusion, when one means “rational map”, one should never just say “map”.

An obvious example of a rational map is a morphism. Another example is thefollowing.

13.1.A. Easy exercise. Interpret rational functions on a separated integralscheme (§7.5) as rational maps to A1

Z. (This is analogous to functions correspondingto morphisms to A1

Z, Exercise 8.4.G.) 1

13.1.1. Important Theorem. — Two S-morphisms f1, f2 : U → Z from areduced scheme to a separated S-scheme agreeing on a dense open subset of U arethe same. 2

ex.done

[forme: Note that this generalizes the easy direction of the valuative criterion of

separatedness (which is the special case where U is Spec of a discrete valuation ring

— which consists of two points — and the dense open set is the generic point).]

1ratfun

2redsep

227

228

13.1.B. Exercise. Give examples to show how this breaks down when we give upreducedness of the base or separatedness of the target. Here are some possibilities.For the first, consider the two maps Spec k[x, y]/(y2, xy)→ Spec k[t], where we takef1 given by t 7→ x and f2 given by t 7→ x + y; f1 and f2 agree on the distinguishedopen set D(x). (See Figure 1.) For the second, consider the two maps from Spec k[t]to the line with the doubled origin, one of which maps to the “upper half”, and oneof which maps to the “lower half”. these to morphisms agree on the dense open setD(f). (See Figure 2.)

f1 f2

Figure 1. Two different maps from a nonreduced scheme agreeingon an open set

f2f1

Figure 2. Two different maps to a nonseparated scheme agreeingon an open set

Proof. Let V be the locus where f1 and f2 agree. It is a closed subscheme of U byExercise 12.1.J, which contains the generic point. But the only closed subschemeof a reduced scheme U containing the generic point is all of U .

Consequence 1. Hence (as X is reduced and Y is separated) if we have twomorphisms from open subsets of X to Y , say f : U → Y and g : V → Y , and theyagree on a dense open subset Z ⊂ U ∩ V , then they necessarily agree on U ∩ V .

Consequence 2. Also: a rational map has a largest domain of definition onwhich f : U 99K Y is a morphism, which is the union of all the domains of definition.

domain of definition

In particular, a rational function from a reduced scheme has a largest domainof definition.

229

13.1.2. The graph of a rational map.Define the graph of a rational map f : X 99K Y as follows. Let (U, f ′) be any graph

representative of this rational map (so f ′ : U → Y is a morphism). Let Γf be thescheme-theoretic closure of Γf ′ → U ×Y → X ×Y , where the first map is a closedimmersion, and the second is an open immersion.

13.1.C. Exercise. Show that the graph of a rational map is independent of thechoice of representative of the rational map.

[forme: Some notes from Rob’s solution. Definition 1. Let Ui ⊂ X be the (largest)

domain of definition of f , and let fi : Ui → Y represent f . Let γfi: Ui → U ×Z Y be

the graph morphism, and iUi: U ×Z Y → X ×Z Y be the open immersion induced

by the open immersion Ui → X. Let γfi= iUi

γfi| Ui → X ×Z Y . Then define the

graph of f to be the scheme-theoretic image of γ. This is clearly well-defined, but

possibly difficult to use in practice. [Maybe I should use this def!] Definition 2: Let

f1 : U1 → Y be any morphism representing f , with U ⊆ X a dense (=non-empty)open

subset of X. With the same notation as above, let Γf (f1) be the scheme-theoretic

image of γfi: U → X ×Z Y . Rob then fails to justify 2, and falls back on 1. ]

In analogy with graphs of morphisms (e.g. Figure 3), the following diagram ofa graph of a rational map can be handy. [forme: Do we use this?]

Γ // X × Y

wwww

wwww

w

##GGG

GGGG

GG

X

OO

Y.

13.2 Dominant and birational maps

A rational map f : X 99K Y is dominant if for some (and hence every) dominant

representative U → Y , the image is dense in Y . Equivalently, f is dominant ifit sends the generic point of X to the generic point of Y . [forme: sometimes

called dominating Iitaka p. 71 exercise: If φ : A → B is a ring morphism, show that dominating

the corresponding morphism of affine schemes SpecB → SpecA is dominant iff φ has

nilpotent kernel. ]

13.2.A. Exercise. Show that you can compose two rational maps f : X 99K Y ,g : Y 99K Z if f is dominant. dominant rational map

In particular, integral separated schemes and dominant rational maps betweenthem form a category which is geometrically interesting.

13.2.B. Easy exercise. Show that dominant rational maps give morphisms offunction fields in the opposite direction.

ex.doneIt is not true that morphisms of function fields give dominant rational maps,

or even rational maps. For example, Spec k[x] and Spec k(x) have the same func-tion field (k(x)), but there is no rational map Spec k[x] 99K Spec k(x). Reason:that would correspond to a morphism from an open subset U of Spec k[x], sayk[x, 1/f(x)], to k(x). But there is no map of rings k(x)→ k[x, 1/f(x)] for any onef(x).

230

However, maps of function fields indeed give dominant rational maps in thecase of varieties, see Proposition 13.2.1 below.

A rational map f : X → Y is said to be birational if it is dominant, and thereis another rational map (a “rational inverse”) that is also dominant, such that f gis (in the same equivalence class as) the identity on Y , and g f is (in the sameequivalence class as) the identity on X . This is the notion of isomorphism in thebirational (rational)

map

bir’l rat’l map

category of integral separated schemes and dominant rational maps.A morphism is birational if it is birational as a rational map. We say X

and Y are birational (to each other) if there exists a birational map X 99K Y .3 Birational maps induce isomorphisms of function fields. Proposition 13.2.1 willimply that a map between k-varieties that induces an isomorphism of function fieldsis birational.birational

We now prove a Proposition promised earlier.

13.2.1. Proposition. — Suppose X, Y are irreducible varieties, and we are givenf# : FF (Y ) → FF (Y ). Then there exists a dominant rational map f : X 99K Yinducing f#. 4

Proof. By replacing Y with an affine open set, we may assume Y is affine, sayY = Spec k[x1, . . . , xn]/(f1, . . . , fr). Then we have x1, . . . , xn ∈ K(X). Let U bean open subset of the domains of definition of these rational functions. Then weget a morphism U → An

k . But this morphism factors through Y ⊂ An, as x1, . . . ,xn satisfy the relations f1, . . . , fr.

13.2.C. Exercise. Let K be a finitely generated field extension of k. Showthere exists an irreducible k-variety with function field K. (Hint: let x1, . . . ,xn be generators for K over k. Consider the map k[t1, . . . , tn] → K given byti 7→ xi, and show that the kernel is a prime ideal p, and that k[t1, . . . , tn]/p

has fraction field K. This can be interpreted geometrically: consider the mapSpec K → Spec k[t1, . . . , tn] given by the ring map ti 7→ xi, and take the closure ofthe image.)

13.2.2. Proposition. — Suppose Y and Z are integral k-varieties. Then Y andZ are birational if and only if there is a dense (=non-empty) open subscheme U ofY and a dense open subscheme V of Z such that U ∼= V . [forme: The main part

of H.C.I.4.5; add a link.]ex.done

This gives you a good idea of how to think of birational maps.

Proof. I find this proof kind of surprising and unexpected.Clearly if Y and Z have isomorphic open sets U and V respectively, then they

are birational (with birational maps given by the isomorphisms U → V and V → Urespectively).

For the other direction, assume that f : Y 99K Z is a birational map, withinverse birational map g : Z 99K Y . Choose representatives for these rational mapsF : W → Y (where W is an open subscheme of Y ) and G : X → Z (where Z is anopen subscheme of Z). We will see that F−1(G−1(W ) ⊂ Y and G−1(F−1(X)) ⊂ Z

3d:birational2

4varietywithff

231

are isomorphic open subschemes.

F−1(G−1(W )) _

F

((QQQQQQQQQQQQG−1(F−1(X))

_

Gvvmmmmmmmmmmmm

F−1(X) _

F

((RRRRRRRRRRRRRRRG−1(W )

_

Gvvllllllllllllll

W _

F

))RRRRRRRRRRRRRRRRR X _

Guulllllllllllllllll

Y Z

The two morphisms G F and the identity from F−1(G−1(W )) → W representto the same rational map, so by Theorem 13.1.1 they are the same morphism.Thus G F gives the identity map from F−1(G−1(W )) to itself. Similarly F G gives the identity map on G−1(F−1(X)). All that remains is to show thatF maps F−1(G−1(W )) into G−1(F−1(X)), and that G maps G−1(F−1(X)) intoF−1(G−1(W )), and by symmetry it suffices to show the former. Suppose q ∈F−1(G−1(W )). Then F (G(F (q)) = F (q) ∈ X , from which F (q) ∈ G−1(F−1(X)).

13.3 Examples of rational maps

Here are some examples of rational maps. A recurring theme is that domainsof definition of rational maps to projective schemes extend over nonsingular codi-mension one points. We’ll make this precise when we discuss curves (Chapter 25.)

slope m

x

y

p

q

C

Figure 3. Finding primitive Pythagorean triples using geometry

232

The first example is how you find a formula for Pythagorean triples. Supposeyou are looking for rational points on the circle C given by x2 + y2 = 1 (Figure 3).One rational point is p = (1, 0). If q is another rational point, then pq is a lineof rational (non-infinite) slope. This gives a rational map from the conic C to A1.Conversely, given a line of slope m through p, where m is rational, we can recoverq as follows: y = m(x − 1), x2 + y2 = 1. We substitute the first equation into thesecond, to get a quadratic equation in x. We know that we will have a solutionx = 1 (because the line meets the circle at (x, y) = (1, 0)), so we expect to be ableto factor this out, and find the other factor. This indeed works:

x2 + (m(x − 1))2 = 1

⇒ (m2 + 1)x2 + (−2)x + (m2 − 1) = 0

⇒ (x− 1)((m2 + 1)x− (m2 − 1)) = 0

The other solution is x = (m2 − 1)/(m2 + 1), which gives y = 2m/(m2 + 1). Thuswe get a birational map between the conic C and A1 with coordinate m, givenby f : (x, y) 7→ y/(x − 1) (which is defined for x 6= 1), and with inverse rationalmap given by m 7→ ((m2 − 1)/(m2 + 1), 2m/(m2 + 1)) (which is defined away fromm2 + 1 = 0).

We can extend this to a rational map C 99K P1 via the inclusion A1 → P1.Then f is given by (x, y) 7→ [y; x− 1]. We then have an interesting question: whatis the domain of definition of f? It appears to be defined everywhere except forwhere y = x − 1 = 0, i.e. everywhere but p. But in fact it can be extended overp! Note that (x, y) 7→ [x + 1;−y] (where (x, y) 6= (−1, y)) agrees with f on theircommon domains of definition, as [x + 1;−y] = [y; x− 1]. Hence this rational mapcan be extended farther than we at first thought. This will be a special case of aresult we’ll see later [forme: REF!] .

(For the curious: we are working with schemes over Q. But this works for anyscheme over a field of characteristic not 2. What goes wrong in characteristic 2?)

13.3.A. Exercise. Use the above to find a “formula” yielding all Pythagoreantriples.

13.3.B. Exercise. Show that the conic x2 + y2 = z2 in P2k is isomorphic to P1

k

for any field k of characteristic not 2. (We’ve done this earlier in the case where kis algebraically closed, by diagonalizing quadrics, §10.5.4.) [forme: (Presumably

this is true for any ring in which 2 is invertible too.)]

In fact, any conic in P2k with a k-valued point (i.e. a point with residue field

k) is isomorphic to P1k. (This hypothesis is certainly necessary, as P1

k certainly hask-valued points. x2 + y2 + z2 = 0 over k = R gives an example of a conic that isnot isomorphic to P1

k.)

13.3.C. Exercise. Find all rational solutions to y2 = x3 + x2, by finding abirational map to A1, mimicking what worked with the conic. [forme: [Refer

forward: group law on them.]]ex.done

You will obtain a rational map to P1 that is not defined over the node x = y = 0,and can’t be extended over this codimension 1 set. This is an example of the limitsof our future result showing how to extend rational maps to projective space overcodimension 1 sets: the codimension 1 sets have to be nonsingular. More on thissoon!

233

13.3.D. Exercise. Use something similar to find a birational map from thequadric Q = x2 + y2 = w2 + z2 to P2. Use this to find all rational points onQ. (This illustrates a good way of solving Diophantine equations. You will find adense open subset of Q that is isomorphic to a dense open subset of P2, where youcan easily find all the rational points. There will be a closed subset of Q wherethe rational map is not defined, or not an isomorphism, but you can deal with thissubset in an ad hoc fashion.)

13.3.E. Important Concrete Exercise (a first view of a blow-up). Letk be an algebraically closed field. (We make this hypothesis in order to not needany fancy facts on nonsingularity.) Consider the rational map A2

k 99K P1k given

by (x, y) 7→ [x; y]. I think you have shown earlier that this rational map cannotbe extended over the origin. Consider the graph of the birational map, which wedenote Bl(0,0) A2

k. It is a subscheme of A2k×P1

k. Show that if the coordinates on A2

are x, y, and the coordinates on P1 are u, v, this subscheme is cut out in A2 × P1

by the single equation xv = yu. [forme: Show that Bl(0,0) A2k is nonsingular.]

Describe the fiber of the morphism Bl(0,0) A2k → P1

k over each closed point of P1k.

Describe the fiber of the morphism Bl(0,0) A2k → A2

k. Show that the fiber over (0, 0)is an effective Cartier divisor (a closed subscheme that is locally principal and nota zero-divisor, §9.1.9). It is called the exceptional divisor. [forme: Refer forward exceptional divisor

to blow-up section.] [forme: Make a picture!] blow-up

13.3.F. Exercise (the Cremona transformation, a useful classical con-struction). Consider the rational map P2 99K P2, given by [x; y; z]→ [1/x; 1/y; 1/z].What is the the domain of definition? (It is bigger than the locus where xyz 6= 0!)You will observe that you can extend it over codimension 1 sets. This will againforeshadow a result we will soon prove. 5 Cremona

[primordial:

13.3.G. Exercise. Show that P2 99K P4 given by [x, y, z] 99K [x2, xy, y2, xz, yz] is defined awayfrom [0; 0; 1]. In other words, show that this rational map cannot be extended to a morphism.[skip; this is boring]

13.3.H. Exercise. X any variety over an algebraically closed field, of dimension n; then X hasbirational morphism to a hypersurface of Pn+1. Don’t even need algebraically closed field.

13.3.I. Exercise. X normal projective n-dimensional variety. Then X is the normalization ofa hypersurface H ⊂ Pn+1. 6

13.3.1. Proposition. — If K is a function field of transcendence degree d over k, there is avariety with function field K. 7 [H, Lemma I.4.2], [H, Prop. I.4.9]

Proof. We will use the following algebraic fact, Proposition 13.3.1: If k is a perfect field (i.e. any

algebraic extension is separable), in particular if k = k, then any finitely generated field extensionK/k is separably generated.

That means that we can find a transcendence basis x1, . . . , xd, so that K/k(x1, . . . , xd) is afinite separable extension. By the theorem of the primitive element (I.4.6A), we can find a singleadditional element y such that K = k(x1, . . . , xr, y). y satisfies a polynomial equation in the x’s:f(x1, . . . , xd, y) = 0. This polynomial is irreducible. This is a hypersurface in Ad+1. Its functionfield is precisely K.

5cremona

6M.III.8.P.6

7fftovar

234

13.3.2. Trick question. Find all rational points on x2 + y2 = −1. (What goes wrong with theprevious example?)

Natural question: is there a rational map from any curve to any other curve? No, and we’llsee why later.

13.3.J. Exercise. In good circumstances a map of function fields gives rational maps. (“Uniquein good circumstances” = separable). Refer to “isomorphic stalks↔ birational” discussion earlier.8 [Used in proof of big curve theorem 25.3.1.]

13.3.K. Exercise. X normal projective n-dimensional variety. Then X is the normalization ofa hypersurface H ⊂ Pn+1. 9

X any variety; then X is birational to a hypersurface of Pn+1. ]

13.4 An important birational morphism: normalization

[primordial: Finiteness of integral closure: Greg’s argument (I have Kirsten’s notes,

see FiniteICGreg.pdf (Kirsten’s notes) in “other references” on Vaio in FOAG folder.), then the

full argument. Mention that this is false in general; that integral closures of Noetherian domains

need not be Noetherian! ]I now want to define other schemes using universal properties, in ways that are

vaguely analogous to fibered product.As a warm-up, I’d like to revisit an earlier topic: reduction of a scheme. Recall

that if X is a scheme, we defined a closed immersion X red → X . (See the commentjust before §1.4 in class 19.) I’d like to revisit this.

13.4.A. Potentially enlightening exercise. Show that X red → X satisfiesthe following universal property: any morphism from a reduced scheme Y to Xfactors uniquely through X red.

Y

@@@

@@@@

@∃! // Xred

||zzzz

zzzz

X

.

You can use this as a definition for X red → X . Let me walk you throughpart of this. First, prove this for X affine. (Here you use the fact that we knowthat maps to an affine scheme correspond to a maps of global sections in the otherdirection.) Then use the universal property to show the result for quasiaffine X .Then use the universal property to show it in general. Oops! I don’t think I’vedefined quasiaffine before. It is any scheme that can be expressed as anopen subset of an affine scheme. I should eventually put this definitionearlier in the course notes, but may not get a chance to. It may appearin the class 22 notes, which are yet to be written up. The concept isreintroduced yet again in Exercise 13.4.C below. [I actually did definequasiaffine before, in the fibered product section.]

SECTION STARTS HERE; EARLIER STUFF IS JUST ROUGH NOTES[forme: There is a normalization discussion around p. 277 of Mumford.] I now

want to tell you how to normalize a reduced Noetherian scheme. A normalization

8functionfieldtorationalmap

9M.III.8.P.6

235

of a scheme X is a morphism ν : X → X from a normal scheme, where ν induces abijection of components of X and X , and ν gives a birational morphism on each ofthe components; it will be nicer still, as it will satisfy a universal property. (I drewa picture of a normalization of a curve.) Oops! I didn’t define birational until[class 27]. Please just plow ahead! I may later patch this anachronism,but most likely I won’t get the chance.

[I have an interesting question for experts: there is a reasonable extension toschemes in general; does anything go wrong? I haven’t yet given this much thought,but it seems worth exploring.]

I’ll begin by dealing with the case where X is irreducible, and hence integral.(I’ll then deal with the more general case, and also discuss normalization in afunction field extension.)

In this case of X irreducible, the normalization satisfies dominant morphismfrom an irreducible normal scheme to X , then this morphism factors uniquelythrough ν:

Y

???

????

?∃! // X

ν~~

~~~~

~

X

.

Thus if it exists, then it is unique up to unique isomorphism. We now have toshow that it exists, and we do this in the usual way. We deal first with the casewhere X is affine, say X = Spec R, where R is an integral domain. Then let R bethe integral closure of R in its fraction field FF(R).

13.4.B. Exercise. Show that ν : Spec R → Spec R satisfies the universal prop-erty.

13.4.C. Exercise. Show that normalizations exist for any quasiaffine X (i.e. anyX that can be expressed as an open subset of an affine scheme). 10

13.4.D. Exercise. Show that normalizations exist in general.[I appear not to define the universal property in general!!][Transition needed here.]I described normalization last day in the case when X is irreducible, and hence

integral. In this case of X irreducible, the normalization satisfies the universalproperty, that if Y → X is any other dominant morphism from a normal schemeto X , then this morphism factors uniquely through ν:

Y

???

????

?∃! // X

ν~~

~~~~

~

X

.

Thus if it exists, then it is unique up to unique isomorphism. We then showedthat it exists, using an argument we saw for the third time. (The first time was inthe existence of the fibered product. The second was an argument for the existenceof the reduction morphism.) The ring-theoretic case got us started: if X = Spec R,

10quasiaffinedef

236

then and R is the integral closure of R in its fraction field FF(R), then I gave as

an exercise that ν : Spec R→ Spec R satisfies the universal property.[forme: Now might be a good place to do an example of normalization! Use

Spec k[t]→ Spec k[x, y]/(y2 − x2(x+ 1)). Show that it is the normalization.Do this again with a cuspidal curve. Refer to other appearances of node and

cusp.

Notice that it has “resolved the singularities = non-smooth points” (not that we

yet know what singularities are). In general, it will do so in dimension one (example

later). It won’t do so in higher dimension necessarily (example).]

13.4.E. Exercise. Show that the normalization morphism is surjective. (Hint:Going-up!)

We now mention some bells and whistles. The following fact is handy.

13.4.1. Theorem (finiteness of integral closure). — Suppose A is a domain,K = FF(A), L/K is a finite field extension, and B is the integral closure of A inL (“the integral closure of A in the field extension L/K”, i.e. those elements of Lintegral over A).(a) if A is integrally closed, then B is a finitely generated A-module.(b) if A is a finitely generated k-algebra, then B (the integral closure of A in itsfraction field) is a finitely generated A-module.

I hope to type up a proof of these facts at some point to show you that theyare not that bad. Much of part (a) was proved by Greg Brumfiel in 210B last year.

Warning: (b) does not hold for Noetherian A in general. I find this veryalarming. I don’t know an example offhand, but one is given in Eisenbud’s book.[Get ref]

13.4.F. Exercise. Show that dim X = dim X (hint: see our going-up discussion).

13.4.G. Exercise. Show that if X is an integral finite-type k-scheme, then itsnormalization ν : X → X is a finite morphism.

13.4.H. Exercise. Explain how to generalize the notion of normalization tothe case where X is a reduced Noetherian scheme (with possibly more than onecomponent). This basically requires defining a universal property. I’m not surewhat the “perfect” definition, but all reasonable universal properties should lead tothe same space.

13.4.I. Exercise. Show that if X is an integral finite type k-scheme, then itsnon-normal points form a closed subset. (This is a bit trickier. Hint: considerwhere ν∗OX has rank 1.) I haven’t thought through all the details recently, so Ihope I’ve stated this correctly. [forme: see [H, Ex. II.3.8]]

Here is an explicit example to think through some of these ideas.

13.4.J. Exercise. Suppose X = Spec Z[15i]. Describe the normalization X → X .(Hint: it isn’t hard to find an integral extension of Z[15i] that is integrally closed.By the above discussion, you’ve then found the normalization!) Over what pointsof X is the normalization not an isomorphism? [forme: [Can I ask the singular

structure?]] [The desired points p are those for which Z[15i]p 6= Z[i]p, i.e. thoseprimes containing 15. Is this rigorous?]

237

13.4.K. Exercise. (This is an important generalization!) Suppose X is an

integral scheme. Define the normalization of X, ν : X → X, in a given finite fieldextension of the function field of X . Show that X is normal. (This will be hard-wired into your definition.) Show that if either X is itself normal, or X is finite typeover a field k, then the normalization in a finite field extension is a finite morphism.11 [forme: This is used in the proof of the big curve theorem 25.3.1.] Again, thisis a finite morphism. (Again, for this we need finiteness of integral closure A.2.5.)

Let’s try this in a few cases.

13.4.L. Exercise. Suppose X = Spec Z (with function field Q). Find its integralclosure in the field extension Q(i). [Perhaps say, in response to a comment ofJustin’s: There is no “geometric” way to do this; it is purely an algebraic problem,although the answer should be understood geometrically.]

A finite extension K of Q is called a number field, and the integral closure ofZ in K the ring of integers of K, denoted OK . (This notation is a little awkwardgiven our other use of the symbol O.) By the previous exercises, SpecOK is a number field, ring of in-

tegers in a number fieldNoetherian normal domain of dimension 1 (hence regular). This is called a Dedekinddomain. We think of it as a smooth curve. [forme: Dan Bump says Lang defines Dedekind domain

a Dedekind domain to be a domain in which the monoid of fractional ideals forms a

group. Greg’s definition: Noetherian, integrally closed, all non-zero prime ideals are

maximal; this agrees with mine. It’s probably worth saying it in this way. Greg has

some notes, which I have in my 2007 210B directory.]

13.4.M. Exercise. (a) Suppose X = Spec k[x] (with function field k(x)). Findits integral closure in the field extension k(y), where y2 = x2 + x. (Again we get aDedekind domain.)(b) Suppose X = P1, with distinguished open Spec k[x]. Find its integral closurein the field extension k(y), where y2 = x2 + x. (Part (a) involves computingthe normalization over one affine open set; now figure out what happens over the“other” affine open set.) [This is hard; Jack consulted Hartshorne. Refer them tothat quadratic extension question.]

13.4.2. Gluing two schemes together along a closed subscheme (Less importantexercise). Glueing two affine schemes along a closed subscheme. Perhaps exampleof gluing an affine scheme to itself along a closed subscheme? Then an interestingarithmetic example of that? Sometimes you can’t glue a scheme to itself and get ascheme. Example later. Gluing 2 schemes along closed subscheme. [Do this.] Note gluing two schemes to-

gether along a closed

subscheme

that gluing along an open subscheme is no problem. We can even glue somethingto itself along an open subscheme. ]

[forme: At some point I should give an exercise on gluing two different schemes

together along a closed subscheme. I’ve put this in the index and in the notation to

make sure I don’t forget.] gluing two schemes

along closed subschemes[Coproduct is confusing. But gluing two schemes to each other along a closedsubscheme. (Warning: can’t glue a scheme to itself.) Idea on affines: product oftwo rings. want residues to be the same. It almost works for gluing something toitself. Exercise: come up with an interesting example of the following form. Drawpicture of quadratic extension. Glue it. You should get an order.

11Ninfieldextension

238

Michel Brion’s talk in Georgia: coproduct along two finite morphisms. Whathypotheses are necessary?]

CHAPTER 14

Proper morphisms

[forme: This is so short that it could well just go in the “separated” chapter.]1

14.1 Proper morphisms

I’ll now tell you about a new property of morphisms, the notion of properness.You can think about this in several ways.

Recall that a map of topological spaces (also known as a continuous map!)is said to be proper if the preimage of compact sets is compact. Properness ofmorphisms is an analogous property. For example, proper varieties over C willbe the same as compact in the “usual” topology. Alternatively, we will see thatprojective A-schemes are proper over A — this is the hardest thing we will prove— so you can see this as nice property satisfied by projective schemes, and quiteconvenient to work with. [forme: Indeed, in some sense, essentially all interesting

properties of projective morphisms that don’t explicitly involve O(1) turn out to be

properties of proper morphisms. The key tool in showing such results is Chow’s

Lemma, which I will state but not prove. Like separatedness, there is a valuative

criterion for properness.]

A (continuous) map of topological spaces f : X → Y is closed if for each closedsubset S ⊂ X , f(S) is also closed. [forme: teaching: This is the definition used

elsewhere in mathematics.] A morphism of schemes is closed if the underlyingcontinuous map is closed. We say that a morphism of schemes f : X → Y isuniversally closed if for every morphism g : Z → Y , the induced morphismZ ×Y X → Z is closed. In other words, a morphism is universally closed if itremains closed under any base change. (A note on terminology: if P is someproperty of schemes, then a morphism of schemes is said to be “universally P” ifit remains P under any base change.) (u) closed, proper

A morphism f : X → Y is proper if it is separated, finite type, and universallyclosed. A scheme X is often said to be proper if some implicit structure morphismis proper. For example, a k-scheme X is often described as proper if X → Spec kis proper. (A k-scheme is often said to be complete if it is proper. We will not usethis terminology.) complete

Let’s try this idea out in practice. We expect that A1C → Spec C is not proper,

because the complex manifold corresponding to A1C is not compact. However, note

that this map is separated (it is a map of affine schemes), finite type, and closed.So the “universally” is what matters here.

1c:proper

239

240

14.1.A. Exercise. Show that A1C → Spec C is not proper, by finding a base

change that turns this into a non-closed map. (Hint: Consider A1C × P1

C → P1C.) 2

14.1.1. As a first example: closed immersions are proper. They are clearly sepa-rated, as affine morphisms are separated, §12.1.12. They are finite type. After basechange, they remain closed immersions, and closed immersions are always closed. 3

14.1.2. Proposition. — [forme: Check which parts of H.E.II.4.1 are here4 ]

(a) The notion of “proper morphism” is stable under base change.(b) The notion of “proper morphism” is local on the target (i.e. f : X → Y

is proper if and only if for any affine open cover Ui → Y , f−1(Ui) → Ui

is proper). Note that the “only if” direction follows from (a) — considerbase change by Ui → Y .

(c) The notion of “proper morphism” is closed under composition.(d) The product of two proper morphisms is proper (i.e. if f : X → Y and g :

X ′ → Y ′ are proper, where all morphisms are morphisms of Z-schemes)then f × g : X ×Z X ′ → Y ×Z Y ′ is proper.

(e) Suppose 5

(29) Xf //

g

@@@

@@@@

Yh

~~~~

~~~

Z

is a commutative diagram, and g is proper, and h is separated. Then f isproper. [forme: (f) (I don’t know if this is useful, but I may as well say it

anyway.) Suppose (29) is a commutative diagram, and f is surjective, g is

proper, and h is separated and finite type. Then h is proper. [Mumford p.

169]]

A sample application of (e): A morphism (over Spec k) from a proper k-schemeto a separated k-scheme is always proper.

Proof. (a) We have already shown that the notions of separatedness and finite typeare local on the target. The notion of closedness is local on the target, and henceso is the notion of universal closedness.

(b) The notions of separatedness, finite type, and universal closedness are allpreserved by fibered product. (Notice that this is why universal closedness is betterthan closedness — it is automatically preserved by base change!)

(c) The notions of separatedness, finite type, and universal closedness are allpreserved by composition.

(d) By (a) and (c), this follows from Exercise 11.4.E.(e) Closed immersions are proper, so we invoke the Cancellation Theorem 12.1.19

for properties of morphisms.

We now come to the most important example of proper morphisms.

2propercompact

3climmproper, H.II.4.8 I think

4H.E.II.4.1

5commie

241

14.1.3. Theorem. — Projective A-schemes are proper over A.6

It is not easy to come up with an example of an A-scheme that is proper butnot projective! We will see a simple example of a proper but not projective surface,[forme: REF] . Once we discuss blow-ups, I’ll give Hironaka’s example of a properbut not projective nonsingular (“smooth”) threefold over C (§27.5.A).

Proof. The structure morphism of a projective A-scheme X → Spec A factors as aclosed immersion followed by Pn

A. Closed immersions are proper, and compositionsof proper morphisms are proper, so it suffices to show that Pn

A → Spec A is proper.We have already seen that this morphism is finite type (Easy Exercise 10.2.D) andseparated (Prop. 12.1.5), so it suffices to show that Pn

A → Spec A is universallyclosed. As Pn

A = PnZ ×Z Spec A, it suffices to show that Pn

X := PnZ ×Z X → X is

closed for any scheme X . But the property of being closed is local on the target onX , so by covering X with affine open subsets, it suffices to show that Pn

A → Spec Ais closed. This is important enough to merit being stated as a Theorem.

14.1.4. Theorem. — π : PnA → Spec A is a closed morphism. 7

ex.doneThis is sometimes called the fundamental theorem of elimination theory. Here

are some examples to show you that this is a bit subtle.First, let A = k[a, b, c, . . . , i], and consider the closed subscheme of P2

A (takenwith coordinates x, y, z) corresponding to ax + by + cz = 0, dx + ey + fz = 0,gx+hy+ iz = 0. Then we are looking for the locus in Spec A where these equationshave a non-trivial solution. This indeed corresponds to a Zariski-closed set — where

determinant

det

a b cd e fg h i

= 0.

As a second example, let A = k[a0, a1, . . . , am, b0, b1, . . . , bn]. Now considerthe closed subscheme of P1

A (taken with coordinates x and y) corresponding toa0x

m + a1xm−1y + · · · + amym = 0 and b0x

n + b1xm−1y + · · · + bnyn = 0. Then

there is a polynomial on the coefficients a0, . . . , bn (an element of A) which vanishesif and only if these two polynomials have a common non-zero root — this polynomialis called the resultant. resultant

More generally, this question boils down to the following question. Given anumber of homogeneous equations in n+1 variables with indeterminate coefficients,Proposition 14.1.4 implies that one can write down equations in the coefficients thatwill precisely determine when the equations have a nontrivial solution.Proof of Theorem 14.1.4. Suppose Z → Pn

A is a closed subset. We wish to showthat π(Z) is closed.

Suppose y /∈ π(Z) is a closed point of Spec A. We’ll check that there is adistinguished open neighborhood D(f) of y in Spec A such that D(f) doesn’t meetπ(Z). (If we could show this for all points of π(Z), we would be done. But I preferto concentrate on closed points for now.) Suppose y corresponds to the maximalideal m of A. We seek f ∈ A−m such that π∗f vanishes on Z.

Let U0, . . . , Un be the usual affine open cover of PnA. The closed subsets π−1y

and Z do not intersect (see Figure 1). On the affine open set Ui, we have two

6projproperA

7projclosed

242

Figure 1

closed subsets Z ∩ Ui and π−1y ∩ Ui that do not intersect, which means that theideals corresponding to the two closed sets generate the unit ideal, so in the ring offunctions A[x0/i, x1/i, . . . , xn/i]. on Ui, we can write

1 = ai +∑

mijgij

where mij ∈ m, and ai vanishes on Z. Note that ai, gij ∈ A[x0/i, . . . , xn/i], so bymultiplying by a sufficiently high power xn

i of xi, we have an equality

xNi = a′

i +∑

mijg′ij

on Ui, where both sides are expressions in S• = A[x0, . . . , xn]. [forme: Perhaps do

more so that we get something correct on the affine cone.] We may take N largeenough so that it works for all i. Thus for N ′ sufficiently large, we can write anymonomial in x1, . . . , xn of degree N ′ as something vanishing on Z plus a linearcombination of elements of m times other polynomials. Hence

SN ′ = I(Z)N ′ + mSN ′

where I(Z)∗ is the graded ideal of functions vanishing on Z. We now need Nakayama’slemma. If you haven’t seen this result before, we will prove it shortly, in §15.3. WeNakayama

will use the following form of it: if M is a finitely generated module over A suchthat M = mM for some maximal ideal m, then there is some f /∈ m such thatfM = 0 (see Lemma 15.3.1). Applying this in the case where M = SN ′/I(Z)N ′ ,we see that there exists f ∈ A−m such that

fSN ′ ⊂ I(Z)N ′ .

Thus we have found our desired f .We now tackle Theorem 14.1.4 in general. Suppose y = [p] not in the image of

Z. Applying the above argument in Spec Ap, we find SN ′ ⊗ Ap = I(Z)N ′ ⊗ Ap +mSN ′ ⊗ Ap, from which g(SN ′/I(Z)N ′) ⊗ Ap = 0 for some g ∈ Ap − pAp, fromwhich (SN ′/I(Z)N ′) ⊗ Ap = 0. As SN ′ is a finitely generated A-module, there issome f ∈ A− p with fSN ⊂ I(Z) (if the module-generators of SN ′ , and f1, . . . , fa

are annihilate the generators respectively, then take f =∏

fi), so once again wehave found D(f) containing p, with (the pullback of) f vanishing on Z.

Notice that projectivity was essential to the proof: we used graded rings in anessential way.

This also concludes the proof of Theorem 14.1.3.

14.1.5. Corollary. — Finite morphisms are proper.

Proof. Suppose f : X → Y is a finite morphism. As properness is local on thebase, to check properness of f , we may assume Y is affine. But finite morphismsto Spec A are projective [forme: ref ] , and projective morphisms are proper.

In particular, as promised in our initial discussion of finiteness (§9.1.6):

243

14.1.6. Corollary. — Finite morphisms are closed. 8 9 [primordial: Proof by closed morphism, finite

morphism is closedgoing-up: If f∗ : B → A is the ring map, inducing f : SpecA → SpecB, then suppose IsubsetA

cuts out a closed set of SpecA, and J = (f∗)−1(I), then note that B/J ⊂ A/I, and apply the

going-up theorem here. ] [forexperts: The proof by going up takes a littlecare.]

14.1.7. Unproved facts that may help you correctly think about finite-ness.

We conclude with some interesting facts that we will prove later. They mayshed some light on the notion of finiteness.

A morphism is finite if and only if it is proper and affine, if and only if it isproper and quasifinite. We have verify the “only if” parts of this statement; the“if” parts are harder.

As an application: quasifinite morphisms from proper schemes to separatedschemes are finite. Here is why: suppose f : X → Y is a quasifinite morphismover Z, where X is proper over Z. Then by the Cancellation Theorem 12.1.19 forproperties of morphisms, X → Y is proper. Hence as f is quasifinite and proper,f is finite.

[forme: Here is an explicit example: consider a morphism P1 → P1 given by two

distinct sections of OP1 (2). The fibers are finite, hence this is a finite morphism. (This

could also be checked directly.)]

8finiteclosed, M.I.7.P.3.1, M.II.7.P.4, he seemed to use going up

9H.E.II.3.5b, make sure I have the right reference for finite UC

Part IV

Harder properties of schemes

[primordial: [This might not be a “part”.]There are some fundamental notions the reader will want to know. “Let X be a smooth

surface.” Strangely, dimension and “smoothness” will be much harder to deal with than earlierproperties, and we won’t need much about them for a while. So the reader is encouraged to skimthrough the section, or ignore it completely, knowing that he or she can return to finish it at theirleisure. [I should let them know when they should come back to this.]

Not so strange: definition for manifolds is equally problematic.

]

CHAPTER 15

Dimension

1

15.1 Dimension and codimension

[forme: Teaching: The notion of dimension is the first of two algebraically “hard”

properties of schemes, the other being smoothness = nonsingularity.]

15.1.1. Dimension. One rather basic notion we expect to have of geometricobjects is dimension, and our goal in this chapter is to define the dimension ofschemes. This should agree with, and generalize, our geometric intuition. Keep inmind that although we think of this as a basic notion in geometry, it is a slipperyconcept, and has been so for historically. (For example, how do we know that thereisn’t an isomorphism between some 1-dimensional manifold and some 2-dimensionalmanifold?) [forme: If we know what balls are, we can use homotopy theory.]

A caution for those thinking over the complex numbers: our dimensions will bealgebraic, and hence half that of the “real” picture. For example, A1

C, which youmay picture as the complex numbers (plus one generic point), has dimension 1.

Surprisingly, the right definition is purely topological — it just depends on thetopological space, and not on the structure sheaf. We define the dimension ofa topological space X as the supremum of lengths of chains of closed irreduciblesets, starting the indexing with 0. (This dimension may be infinite.) Scholars ofthe empty set can take the dimension of the empty set to be −∞. Define thedimension of a ring as the Krull dimension of its spectrum — the sup of the Krull dimension, dim

lengths of the chains of nested prime ideals (where indexing starts at zero). Thesetwo definitions of dimension are sometimes called Krull dimension. (You mightthink a Noetherian ring has finite dimension because all chains of prime ideals arefinite, but this isn’t necessarily true — see Exercise 15.1.6.)

As we have a natural homeomorphism between Spec A and Spec A/n(A) (§5.4.5:the Zariski topology doesn’t care about nilpotents), we have dim A = dim A/n(A).

Examples. We have identified all the prime ideals of k[t] (they are 0, and(f(t)) for irreducible polynomials f(t)), Z (0 and (p)), k (only 0), and k[x]/(x2)(only 0), so we can quickly check that dim A1

k = dim Spec Z = 1, dim Spec k = 0,dim Spec k[x]/(x2) = 0.

We must be careful with the notion of dimension for reducible spaces. If Z isthe union of two closed subsets X and Y , then dimZ = max(dim X, dim Y ). Inparticular, if Z is the disjoint union of something of dimension 2 and something of

1c:dimension

247

248

dimension 1, then it has dimension 2. Thus dimension is not a “local” characteristicof a space. This sometimes bothers us, so we will often talk about dimensions ofirreducible topological spaces. If a topological space can be expressed as a finiteunion of irreducible subsets, then say that it is equidimensional or pure dimen-sional (resp. equidimensional of dimension n or pure dimension n) if each of itscomponents has the same dimension (resp. they are all of dimension n).equidimensional

pure dimension An equidimensional dimension 1 (resp. 2, n) topological space is said to be acurve (resp. surface, n-fold).curve, surface

15.1.2. Codimension. Because dimension behaves oddly for disjoint unions, weneed some care when defining codimension, and in using the phrase. For example,if Y is a closed subset of X , we might define the codimension to be dim X−dim Y ,but this behaves badly. For example, if X is the disjoint union of a point Y and acurve Z, then dim X − dim Y = 1, but the reason for this has nothing to do withthe local behavior of X near Y .

A better definition is as follows. In order to avoid excessive pathology, wedefine the codimension of Y in X only when Y is irreducible. We define the codi-mension of an irreducible closed subset Y ⊂ X of a topological space as thesupremum of lengths of increasing chains of irreducible closed subsets starting atY (where indexing starts at 0). The codimension of a point is defined to be thecodimension of its closure.codim

We say that a prime ideal p in a ring has codimension equal to the supremumof lengths of the chains of decreasing prime ideals starting at p, with indexingstarting at 0. Thus in an integral domain, the ideal (0) has codimension 0; and inZ, the ideal (23) has codimension 1. Note that the codimension of the prime idealp in A is dim Ap (see §5.3.E). (This notion is often called height.) [forme: Usingcodimension

codimension for height follows [E, Ch. 9].] Thus the codimension of p in A is thecodimension of [p] in Spec A.height

15.1.A. Exercise. Show that if Y is an irreducible subset of a scheme X withgeneric point y, show that the codimension of Y is the dimension of the local ringOX,y. [forme: TO DO: when discussing stalks of schemes earlier, give an exercise

showing that the primes correspond to irreducibles through the point. Then hint

back to that from here.]ex.done

Note that Y is codimension 0 in X if it is an irreducible component of X .Similarly, Y is codimension 1 if it is strictly contained in an irreducible componentY ′, and there is no irreducible subset strictly between Y and Y ′. (See Figure 1 forexamples.) An closed subset all of whose irreducible components are codimension1 in some ambient space X is said to be a hypersurface in X .hypersurface

15.1.B. Easy exercise. Show that 2

(30) codimX Y + dim Y ≤ dim X.

We will soon see that equality always holds if X and Y are varieties [forme:

ref ] , but equality doesn’t always hold (§15.6.4).Warnings. (1) We have only defined codimension for irreducible Y in X . Exer-

cise extreme caution in using this word in any other setting. We may use it in thecase where the irreducible components of Y each have the same codimension.

2codimdim

249

q

p

C

Figure 1. Behavior of codimension

(2) The notion of codimension still can behave slightly oddly. For example,consider Figure 1. (You should think of this as an intuitive sketch, but once we de-fine dimension correctly, this will be precise.) Here the total space X has dimension2, but point p is dimension 0, and codimension 1. We also have an example of acodimension 2 subset q contained in a codimension 0 subset C with no codimension1 subset “in between”.3

Worse things can happen; we will soon see an example of a closed point in anirreducible surface that is nonetheless codimension 1, not 2, §15.6.4. However, forirreducible varieties (finitely generated domains over a field), this can’t happen, andthe inequality (30) must be an inequality. We’ll show this in Proposition 15.4.D.

15.1.3. What will happen in this chapter.In this chapter, we’ll explore the notions of dimension and codimension, and

show that they satisfy properties that we find desirable, and (later) useful. Inparticular, we’ll learn some techniques for computing dimension.

We would certainly want affine n-space to have dimension n. We will indeedshow that dim An

k = n, and show more generally that the dimension of an irreduciblevariety over k is its transcendence degree, in §15.4. En route, we will see someuseful facts, including the Going-Up Theorem, and Noether Normalization. (Whileproving the Going-Up Theorem, we will see a trick that will let us prove manyforms of Nakayama’s Lemma, which will be useful to us in the future.) Related tothe Going-Up Theorem is the fact that certain nice (“integral”) morphisms X → Ywill have the property that dim X = dim Y (Exercise 15.2.H).

Noether Normalization will let us prove Chevalley’s Theorem [forme: ref ] ,stating that the image of a finite type morphism of Noetherian schemes is always

3heightexample

250

constructable §15.5. We will also give a short proof of the Nullstellensatz [forme:

ref ] .We’ll then spend Section 15.6 discussing two useful facts about codimension

one. A linear function on a vector space is either vanishes in codimension 0 (if it isthe 0-function) or else in codimension 1. The same is true much more generally forfunctions on Noetherian schemes. Informally: a function on a Noetherian schemealso vanishes in pure codimension 0 or 1. More precisely, the irreducible componentsof its vanishing locus are all codimension at most 1. This is Krull’s Principal IdealTheorem. A second fact, that we’ll call “Algebraic Hartogs’ Lemma”, informallystates that on a normal scheme, any rational function with no poles is in fact aregular function. These two codimension one facts will come in very handy in thefuture.

We end this introductory section with a first property about codimensions (andhypersurfaces) that we’ll find useful, and a pathology.

15.1.4. Warm-up proposition. — In a unique factorization domain A, allcodimension 1 prime ideals are principal. 4

ex.doneWe will see that the converse (in the case where A is Noetherian domain) holds

as well (Proposition ??). [forme: This is here because we need it in the proof of

the big dimension theorem.]

Proof. Suppose p is a codimension 1 prime. Choose any f 6= 0 in p, and let g beany irreducible/prime factor of f that is in p (there is at least one). Then (g) is aprime ideal contained in p, so (0) ⊂ (g) ⊂ p. As p is codimension 1, we must havep = (g), and thus p is principal.

15.1.5. A fun but unimportant counterexample. As a Noetherian ring has noinfinite chain of prime ideals, you may think that Noetherian rings must have finitedimension. Here is an example of a Noetherian ring with infinite dimension, dueto Nagata, the master of counterexamples. [forme: Placement isn’t great.]infinite-dimensional N

ring; Nagata [primordial:

15.1.C. Less important exercise. Here is an easier example of a similar phenomenon. Show

that the following topological space is Noetherian, but has infinite dimension. Let X = Z≥0, with

topology given by declaring that the closed sets are ∅, X, and sets of the form 0, . . . , n. [forme:

All are irreducible, so we clearly have an infinite chain. It is clearly Noetherian too.

Anssi told me this one.] ]

15.1.6. Exercise ?. Choose an increasing sequence of positive integers m1,m2, . . . whose differences are also increasing (mi+1 −mi > mi −mi−1). Let Pi =(xmi+1, . . . , xmi+1) and S = A − ∪iPi. Show that S is a multiplicative set. Show

that S−1A is Noetherian. Show that each S−1P is the smallest prime ideal in achain of prime ideals of length mi+1 −mi. Hence conclude that dim S−1A = ∞.[forme: See also [E, p. 231]; this was taken from [AM, p. 126, exercise 4]] 5

4ufdht1prA

5infdimNring

251

15.2 Integral ring extensions and the Going-up theorem

[forme: Greg has a note on integral ring extensions, which I have in my 2007

210B directory.] 6A ring homomorphism φ : B → A is integral if every elementof A is integral over φ(B). [forme: Notation from [AM, p. 60].] In other words, ifa is any element of A, then a satisfies some monic polynomial

an+?an−1 + · · ·+? = 0

where all the coefficients lie in φ(B). We call φ an integral extension if φ is aninclusion of rings. [forme: FOAG: integral morphism to integral homomorphism] integral morphism of

rings, [AM, p. 60]

integral extension of

rings

[forme: Example: normalization is an integral morphism.]

15.2.A. Exercise. Show that if f : B → A is a ring homomorphism, and(b1, . . . , bn) = 1 in B, and Bbi → Af(bi) is integral, then f is integral. Thus [forme:

by XXX] we can define the notion of integral morphism of schemes.

15.2.B. Exercise. Show that the notion of integral homomorphism is wellbehaved with respect to localization and quotient of B, and quotient of A, but notlocalization of A. [forme: witness k[t]→ k[t], but k[t]→ k[t](t)] Show that the notionof integral extension is well behaved with respect to localization and quotient of B,but not quotient of A. [forme: same example, k[t] → k[t]/(t)] If possible, drawpictures of your examples.7

15.2.C. Exercise. Show that if B is an integral extension of A, and C is anintegral extension of B, then C is an integral extension of A. 8

15.2.1. Proposition (finite implies integral). — If A is a finite B-algebra,then φ is an integral homomorphism. 9

ex.done

The converse is false: integral does not imply finite, as Q → Q is an integralhomomorphism, but Q is not a finite Q-module.

15.2.D. Unimportant Exercise: finite = integral + finite type. Showthat a morphism is finite if and only if it is integral and finite type.

Proof. [forme: If B is Noetherian, then this has a short argument: suppose a ∈ B.

Then A is a Noetherian B-module, and hence the ascending chain of B-submodules

of A (1) ⊂ (1, a) ⊂ (1, a, a2) ⊂ (1, a, a2, a3) ⊂ · · · eventually stabilizes, say (1, a, . . . , an−1) =

(1, a, . . . , an−1, an). Hence an is a B-linear combination of 1, . . . , an−1, i.e. is integral

over B. So Noetherian-minded readers can stop reading.] The proof involves auseful trick.

6s:integralmorphism

7e:integralnice

8integralintegral; this may not get used, except to motivate Exercise 9.1.L; in that case, make it a side

comment.9finiteintegral

252

Choose a finite generating set m1, . . . , mn of A as a B-module. Then ami =∑

bijmj , for some bij ∈ B. Thus10

(31) (aIn×n − [bij ]ij)

m1

...mn

=

0...0

.

We can’t quite invert this matrix (aIn×n − [bij ]ij), but we almost can. Recallthat any n × n matrix M has an adjoint matrix adj(M) such that adj(M)M =adjoint matrix

det(M)Idn. (The ijth entry of adj(M) is the determinant of the matrix obtainedfrom M by deleting the ith column and jth row, times (−1)i+j [forme: give

book ref] .) The coefficients of adj(M) are polynomials in the coefficients of M .(You’ve likely seen this in the form of a formula for M−1 when there is an inverse.)Multiplying both sides of (32) on the left by adj(Idn −A), we obtain

det(Idn −A)

m1

...mn

= 0.

Multiplying (31) by the adjoint of (aIn×n − [bij ]ij), we get

det(aIn×n − [bij ]ij)

m1

...mn

=

0...0

.

So det(aI−M) annihilates A, i.e. det(aI−M) = 0. But expanding the determinantyields an integral equation for a with coefficients in B.

We now state the Going-up theorem.

15.2.2. The Cohen-Seidenberg Going up theorem. — Suppose φ : B → Ais an integral extension. Then for any prime ideal q ⊂ B, there is a prime idealp ⊂ A such that p ∩ B = q. 11 [forme: We don’t need B Noetherian for any of(C-S) Going-Up theo-

rem this. Also, A needn’t be finitely generated.]

ex.doneAlthough this is a theorem in algebra, the name can be interpreted geometri-

cally: the theorem asserts that the corresponding morphism of schemes is surjective,and that “above” every prime q “downstairs”, there is a prime q “upstairs”, seeFigure 2. (For this reason, it is often said that q is “above” p if p ∩ B = q.) 12

[forme: (Joe points out that my speculation on the origin of the name “going up” is

wrong.)]

15.2.E. Exercise (reality check). The morphism k[t] → k[t](t) is not inte-gral, as 1/t satisfies no monic polynomial with coefficients in k[t]. Show that theconclusion of the Going-up theorem 15.2.2 fails.

ex.doneProof of the Cohen-Seidenberg Going-Up theorem 15.2.2 ?. This proof is eminentlyreadable, but could be skipped on first reading. We start with an exercise. [forme:

Greg Brumfiel has a going-up proof, which I have in my 210B file (2007).]

10thiseq

11goingup

12goinguppic

253

[p]

SpecA

SpecB

[q]

Figure 2. A picture of the Going-up theorem

15.2.F. Exercise. Show that the special case where A is a field translates to: ifB ⊂ A is a subring with A integral over B, then B is a field. Prove this. (Hint: allyou need to do is show that all nonzero elements in B have inverses in B. Here isthe start: If b ∈ B, then 1/b ∈ A, and this satisfies some integral equation over B.)

ex.done

Proof of the Going-Up Theorem 15.2.2. We first make a reduction: by localizingat q, so we can assume that (B, q) is a local ring.

Then let p be any maximal ideal of A. We will see that p ∩ B = q. Considerthe following diagram.

A // // A/p field

B?

OO

// // B/(B ∩ p)?

OO

By the Exercise above, the lower right is a field too, so B∩p is a maximal ideal,hence q.

15.2.G. Important but straightforward exercise (sometimes also calledthe going-up theorem). Show that if q1 ⊂ q2 ⊂ · · · ⊂ qn is a chain of primeideals of B, and p1 ⊂ · · · ⊂ pm is a chain of prime ideals of A such that pi “liesover” qi (and m < n), then the second chain can be extended to p1 ⊂ · · · ⊂ pn sothat this remains true.

ex.done

This version of the Going-up Theorem has an important consequence.

15.2.H. Important exercise. Show that if f : Spec A→ Spec B corresponds toan integral extension of rings, then dim Spec A = dim Spec B.13 (Hint: show thata chain of prime ideals downstairs gives a chain upstairs, by the previous exercise,

13finitedim

254

of the same length. Conversely, a chain upstairs gives a chain downstairs. Weneed to check that no two elements of the chain upstairs goes to the same element[q] ∈ Spec B of the chain downstairs. As integral extensions are well-behaved bylocalization and quotients of Spec B (Exercise 15.2.B), we can replace B by Bq/qBq

(and A by A⊗B (Bq/qBq)). Thus we can assume B is a field. Hence we must showthat if φ : k → A is an integral extension, then dim A = 0. Outline of proof:Suppose p ⊂ m are two prime ideals of p. Mod out by p, so we can assume that Ais a domain. I claim that any non-zero element is invertible: Say x ∈ A, and x 6= 0.Then the minimal monic polynomial for x has non-zero constant term. But then xis invertible — recall the coefficients are in a field.)

15.2.I. Exercise. Show that integral ring extensions induce a surjective map ofspectra.

15.3 As long as we’re in the neighborhood: Nakayama’slemma ?

[forme: I’d like to state Nakayama in plain english.] 14 The trick in the proofof Proposition 15.2.1 is very handy, and can be used to quickly prove Nakayama’slemma. This name is used for several different but related results. Nakayama isn’tespecially closely related to dimension, but we may as well prove it while the trickis fresh in our minds.

15.3.1. Nakayama’s Lemma version 1. — Suppose A is a ring, I an ideal ofA, and M is a finitely-generated A-module. Suppose M = IM . Then there existsan a ∈ A with a ≡ 1 (mod I) with aM = 0.15Nak

Proof. Say M is generated by m1, . . . , mn. Then as M = IM , we have mi =∑

j aijmj for some aij ∈ I . Thus16

(32) (Idn −A)

m1

...mn

= 0

where Idn is the n×n identity matrix in A, and A = (aij). Multiplying both sidesof (32) on the left by adj(Idn −A), we obtain

det(Idn −A)

m1

...mn

= 0.

But when you expand out det(Idn −A), you get something that is 1 (mod I).

Here is why you care: Suppose I is contained in all maximal ideals of A. (Theintersection of all the maximal ideals is called the Jacobson radical, but we won’tuse this phrase. For comparison, recall that the nilradical was the intersection of

14s:nakayama

15nak1 (used)

16nakpfeq

255

the prime ideals of A.) Then I claim that any a ≡ 1 (mod I) is invertible. Forotherwise (a) 6= A, so the ideal (a) is contained in some maximal ideal m — buta ≡ 1 (mod m), contradiction. Then as a is invertible, we have the following. Jacobson radical (don’t

use)15.3.2. Nakayama’s Lemma version 2. — Suppose A is a ring, I an ideal ofA contained in all maximal ideals, and M is a finitely-generated A-module. (Themost interesting case is when A is a local ring, and I is the maximal ideal.) SupposeM = IM . Then M = 0. 17 Nak

15.3.A. Exercise (Nakayama’s lemma version 3). Suppose A is a ring, andI is an ideal of A contained in all maximal ideals. Suppose M is a finitely generatedA-module, and N ⊂M is a submodule. If N/IN →M/IM an isomorphism, thenM = N . (This can be useful, although it won’t come up again for us.)

15.3.B. Important exercise (Nakayama’s lemma version 4). Suppose(A, m) is a local ring. Suppose M is a finitely-generated A-module, and f1, . . . , fn ∈M , with (the images of) f1, . . . , fn generating M/mM . Then f1, . . . , fn generateM . (In particular, taking M = m, if we have generators of m/m2, they also generatem.)18

15.3.C. Unimportant exercise (Nakayama’s lemma version 5). ProveNakayama version 1 (Lemma 15.3.1) without the hypothesis that M is finitelygenerated, but with the hypothesis that In = 0 for some n. (This argument doesnot use the trick.) This result is quite useful, although we won’t use it. 19 [forme:

I think it may come up in §29.2.14. Jack says: the Artin version of Nakayama’s

version is: if A is artin, M is any A-module, I is contained in the Jacobson radical,

then M = IM implies M = 0. Pf: A artin implies Jacobson radical has a power which

is zero and so Ik = 0 hence M = IkM = 0 for some k >> 0. ]

15.3.D. Important exercise that we will use soon. Suppose S is a subringof a ring A, and r ∈ A. Suppose there is a faithful S[r]-module M that is finitelygenerated as an S-module. Show that r is integral over S. (Hint: look carefully atthe proof of Nakayama’s Lemma version 1, and change a few words.) [forme: Be

careful before changing A’s to A’s, etc., because I use this in different ways. From

[AM] prop. 2.4. Also part of [AM] 5.1 iv → i. Jack Hall’s argument, not involving

Nak1: Since M is finitely generated as an S-module, then for some n > 0 we have

xn = anxn−1 + · · ·+ a2x+ a1. Since the module M is faithful, we can pull this relation

back to A to obtain that x is integral over S. Is he assuming Noetherianness though?

] 20

15.4 Dimension and transcendence degree

21

17nak2 (used)

18nak4, used in next section

19NakArt

20nakayamaprime

21dimtrdeg

256

We now prove an alternative interpretation for dimension for irreducible vari-eties.

15.4.1. Theorem (dimension = transcendence degree). — Suppose A isa finitely-generated domain over a field k. Then dim Spec A is the transcendencedegree of the fraction field FF(A) over k. 22

ex.doneBy “finitely generated domain over k”, we mean “a finitely generated k-algebra

that is an integral domain”.In case you haven’t seen the notion of transcendence degree, here is a quick

summary of the relevant facts. Suppose K/k is a finitely generated field extension.Then any two maximal sets of algebraically independent elements of K over k (i.e.any set with no algebraic relation) have the same size (a non-negative integer or∞). If this size is finite, say n, and x1, . . . , xn is such a set, then K/k(x1, . . . , xn)is necessarily a finitely generated algebraic extension, i.e. a finite extension. (Sucha set x1, . . . , xn is called a transcendence basis, and n is called the transcendencedegree.) [forme: Fold in better? State clearly what facts are necessary. A short and

well-written proof of this fact is in [?]. Just prove transcendence theory? Fix a field

k, and a finitely generated extension K. Then there is a well-defined number, called

transcendence degree, so that there are x1, . . . , xn, so that K/k(x1, . . . , xn) is finite. ]

In particular, we see that dim Ank = n. However, our proof of Theorem 15.4.1

will go through this fact, so it isn’t really a Corollary.

15.4.2. Sample consequences. We will prove Theorem 15.4.1 shortly. But wefirst show that it is useful by giving some immediate consequences. We begin witha proof of the Nullstellensatz, promised earlier.

15.4.A. Exercise: Nullstellensatz from dimension theory.(a) Suppose A = k[x1, . . . , xn]/I , where k is an algebraically closed field and Iis some ideal. Then the maximal ideals are precisely those of the form (x1 −a1, . . . , xn − an), where ai ∈ k. This version (the “weak Nullstellensatz”) wasstated in §5.3.1.(b) Suppose A = k[x1, . . . , xn]/I where k is not necessarily algebraically closed.Show that every maximal ideal of A has a residue field that is a finite extensionof k. This version was stated in Exercise 7.3.I. (Hint for both parts: the maximalideals correspond to dimension 0 points, which correspond to transcendence degree0 extensions of k, i.e. finite extensions of k. If k = k, the maximal ideals correspondto surjections f : k[x1, . . . , xn] → k. Fix one such surjection. Let ai = f(xi), andshow that the corresponding maximal ideal is (x1 − a1, . . . , xn − an).) 23

15.4.3. Points of A2k.

24We can now confirm that we have named all the primes ofk[x, y] where k is algebraically closed (as promised in Exercise 5.3.D when k = C).Recall that we have discovered the primes (0), f(x, y) where f is irreducible, and(x−a, y−b) where a, b ∈ k. As A2

k is irreducible, there is only one irreducible subsetof codimension 0. By Proposition 15.1.4, all codimension 1 primes are principal. Byinequality (30), there are no primes of codimension greater than 2, and any prime

22bigdimthm

23Nullpf

24pointsofA2

257

of codimension 2 must be maximal. We have identified all the maximal ideals ofk[x, y] by the Nullstellensatz.

15.4.B. Important exercise. Suppose X is an irreducible variety. Show thatdim X is the transcendence degree of the function field (the stalk at the genericpoint) OX,η over k. Thus (as the generic point lies in all non-empty open sets) thedimension can be computed in any open set of X . (This is not true in general, see function field

§15.6.4.)Here is an application that you might reasonably have wondered about before

thinking about algebraic geometry.

15.4.C. Exercise. Suppose f(x, y) and g(x, y) are two complex polynomials(f, g ∈ C[x, y]). Suppose f and g have no common factors. Show that the systemof equations f(x, y) = g(x, y) = 0 has a finite number of solutions. (This isn’tessential for what follows. But it is a basic fact, and very believable.) [forme: Ken

Chan gave the following machinery-free proof: Consider f, g ∈ C(x)[y]. There exist

a, b ∈ C(x)[y] such that af + bg = 1. Choose h ∈ C[x] such that ha, hb ∈ C[x, y]. then

h = haf + hbg. If (s, t) is a common root of f and g, then s is a root of h. But h has

only a finite number of roos, so there are only a finite number of such s. Similarly

for t.]

15.4.D. Exercise. Suppose X ⊂ Y is an inclusion of irreducible k-varieties, andη is the generic point of X . Show that dim X + dimOY,η = dim Y . Hence showthat dim X + codimY X = dim Y . Thus for varieties, the inequality (30) is alwaysan equality. 25

15.4.E. Exercise. Show that Spec k[w, x, y, z]/(wz − xy, wy − x2, xz − y2) is anintegral surface. You might expect it to be a curve, because it is cut out by threeequations in 4-space. (You may recognize this as the affine cone over the twistedcubic.) [forme: The twisted cubic was defined in Exercise 9.1.T.] It turns out twisted cubic

that you actually need three equations to cut out this surface. The first equationcuts out a threefold in four-space (by Krull’s theorem 15.6.3, see later). The secondequation cuts out a surface: our surface, along with another surface. The thirdequation cuts out our surface, and removes the “extraneous component”. One lastaside: notice once again that the cone over the quadric surface k[w, x, y, z]/(wz−xy)makes an appearance.) [forme: Two proofs: describe it as a subring of k[u, v] — (cone over) quadric sur-

facerequires extra insight. Better: look at Aw, etc. (Hint: you can show directly that it

is a domain. But that requires some cleverness. Instead, you could cut it into pieces.

For example, what is D(w)?)]

15.4.4. Noether Normalization.Hopefully you are now motivated to understand the proof of Theorem 15.4.1

on dimension and transcendence degree. To set up the argument, we introduceanother important and ancient result, Noether’s Normalization Lemma.

15.4.5. Noether Normalization Lemma. — Suppose A is an integral domain,finitely generated over a field k. If tr.deg.kA = n, then there are elements x1, . . . ,

25heightdim

258

xn ∈ A, algebraically independent over k, such that A is a finite (hence integral byProposition 15.2.1) extension of k[x1, . . . , xn].26

ex.doneThe geometric content behind this result is that given any integral affine k-

scheme X , we can find a surjective finite morphism X → Ank , where n is the

transcendence degree of the function field of X (over k). Surjectivity follows fromthe Going-Up Theorem 15.2.2.Nagata’s proof of Noether normalization ?. [forme: Greg has a note on Noether

normalization using transcendence bases, in my 2007 210B directory. Nagata’s proof

[?, §1.1].] 27 Suppose we can write A = k[y1, . . . , ym]/p, i.e. that A can be chosento have m generators. Note that m ≥ n. We show the result by induction on m.The base case m = n is immediate.

Assume now that m > n, and that we have proved the result for smallerm. We will find m − 1 elements z1, . . . , zm−1 of A such that A is finite overA′ := k[z1, . . . , zm−1] (i.e. the subring of A generated by z1, . . . , zm−1). Then bythe inductive hypothesis, A′ is finite over some k[x1, . . . , xn], and A is finite overA′, so by Exercise 9.1.L, A is finite over k[x1, . . . , xn].

A

finite

A′ = k[z1, . . . , zm−1]/p

finite

k[x1, . . . , xn]

As y1, . . . , ym are algebraically dependent, there is some non-zero algebraicrelation f(y1, . . . , ym) = 0 among them (where f is a polynomial in m variables).

Let z1 = y1 − yr1m , z2 = y2 − yr2

m , . . . , zm−1 = ym−1 − yrm−1m , where r1, . . . ,

rm−1 are positive integers to be chosen shortly. Then

f(z1 + yr1m , z2 + yr2

m , . . . , zm−1 + yrm−1m , ym) = 0.

Then upon expanding this out, each monomial in f (as a polynomial in m variables)will yield a single term in that is a constant times a power of ym (with no zi factors).By choosing the ri so that 0 r1 r2 · · · rm−1, we can ensure that thepowers of ym appearing are all distinct, and so that in particular there is a leadingterm yN

m , and all other terms (including those with zi-factors) are of smaller degreein ym. Thus we have described an integral dependence of ym on z1, . . . , zm−1 asdesired.

15.4.6. Aside: the geometric idea behind Nagata’s proof. There is some geometricintuition behind this. Suppose we have an m-dimensional variety in An

k with m < n,for example xy = 1 in A2. One approach is to project it to a hyperplane via a finitemorphism. In the case of xy = 1, if we projected to the x-axis, it wouldn’t be finite,roughly speaking because the asymptote x = 0 prevents the map from being closed.But if we projected to a line, we might hope that we would get rid of this problem,and indeed we usually can: this problem arises for only a finite number of directions.

26nnl

27nnlpf

259

But we might have a problem if the field were finite: perhaps the finite number ofdirections in which to project each have a problem. (The reader may show that ifk is an infinite field, then the substitution in the above proof zi = yi − yri

m can bereplaced by the linear substitution zi = yi−aiym where ai ∈ k, and that for a non-empty Zariski-open choice of ai, we indeed obtain a finite morphism.) Nagata’strick in general is to “jiggle” the variables in a non-linear way, and that this isenough to prevent non-finiteness of the map.Proof of Theorem 15.4.1 on dimension and transcendence degree. Suppose X is anintegral affine k-scheme. We show that dim X equals the transcendence degree nof its function field, by induction on n. Fix X , and assume the result is known forall transcendence degrees less than n.

By Noether normalization, there exists a surjective finite morphism map X →An

k . By Exercise 15.2.H, dim X = dim Ank . If n = 0, we are done, as dim A0

k = 0.We now show that dim An

k = n for n > 0, by induction. Clearly dim Ank ≥ n,

as we can describe a chain of irreducible subsets of length n + 1: if x1, . . . , xn arecoordinates on An, consider the chain of ideals

(0) ⊂ (x1) ⊂ · · · ⊂ (x1, . . . , xn)

in k[x1, . . . , xn]. Suppose we have a chain of prime ideals of length at least n:

(0) = p0 ⊂ · · · ⊂ pm.

where p1 is a codimension 1 prime ideal. Then p1 is principal (as k[x1, . . . , xn] is aunique factorization domain, Proposition 15.1.4) say p1 = (f(x1, . . . , xn)), where fis an irreducible polynomial. Then k[x1, . . . , xn]/(f(x1, . . . , xn)) has transcendencedegree n− 1, so by induction,

dim k[x1, . . . , xn]/(f) = n− 1.

[forme: PERHAPS PROVE PROPERTIES OF TRANSCENDENCE DEGREE

HERE.]

15.5 Images of morphisms, and Chevalley’s theorem

28We can now prove Chevalley’s Theorem 15.5.1, discussed in §9.2.C.

15.5.1. Chevalley’s Theorem. — Suppose f : X → Y is a morphism offinite type of Noetherian schemes. Then the image of any constructable set is con-structable.29 [forme: I am following Mumford p. 70, although I have changed the constructable set,

Chevalley’s theoremexposition. I should compare this to Atiyah-Macdonald, Exercise 7.25. Matsumura

Commutative Algebra p. 42, Theorem 6 in flatness section. Algebraic proof in [AM,

Ex. 7.25]. 30]ex.done

The proof will use Noether normalization. This is remarkable: Noether normal-ization is about finitely generated algebras over a field, but there is no field in thestatement of Chevalley’s theorem. Hence if you prefer to work over arbitrary rings

28chevalleypf

29chevalley

30Chevalley

260

(or schemes), this shows that you still care about facts about finite type schemesover a field. Conversely, even if you are interested in finite type schemes over agiven field (like C), the field that comes up in the proof of Chevalley’s theorem isnot that field, so even if you prefer to work over C, this argument shows that youstill care about working over arbitrary fields, not necessarily algebraically closed.

15.5.A. Hard Exercise. Reduce the proof of Chevalley’s theorem 15.5.1 tothe following statement: suppose f : X = Spec A → Y = Spec B is a domi-nant morphism, where A and B are domains, and f corresponds to φ : B →B[x1, . . . , xn]/I ∼= A. Then the image of f contains a dense open subset of Spec B.(Hint: Make a series of reductions. The notion of constructable is local, so reduceto the case where Y is affine. Then X can be expressed as a finite union of affines;reduce to the case where X is affine. X can be expressed as the finite union of irre-ducible components; reduce to the case where X is irreducible. Reduce to the casewhere X is reduced. By considering the closure of the image of the generic pointof X , reduce to the case where Y also is integral (irreducible and reduced), andX → Y is dominant. Use Noetherian induction in some way on Y .) [forme: This

problem currently requires the induced reduced subscheme structure, Exercise 9.2.6.

This is a tricky problem, so give them an outline.]

Proof. We prove the statement given in the previous exercise. [forme: We no longer

need the Noetherian hypothesis.] Let K := FF(B). Now A⊗B K is a localizationof A with respect to B∗ (interpreted as a subset of A), so it is a domain, and it isfinitely generated over K (by x1, . . . , xn), so it has finite transcendence degree r overK. Thus by Noether normalization, we can find a subring K[y1, . . . , yr] ⊂ A⊗B K,so that A⊗B K is integrally dependent on K[y1, . . . , yr]. We can choose the yi tobe in A: each is in (B∗)−1A to begin with, so we can replace each yi by a suitableK-multiple.

Sadly A is not necessarily integrally dependent on A[y1, . . . , yr] (as this wouldimply that Spec A→ Spec B is surjective by the Going-Up Theorem 15.2.2). How-ever, each xi satisfies some integral equation

xni + f1(y1, . . . , yr)x

n−1i + · · ·+ fn(y1, . . . , yr) = 0

where fj are polynomials with coefficients in K = FF(B). Let g be the productof the denominators of all the coefficients of all these polynomials (a finite set).Then Ag is integral over Bg[y1, . . . , yr], and hence Spec Ag → Spec Bg is surjective;Spec Bg is our open subset.

[forme: Does this work?: We actually get more out of this proof.

15.5.B. Exercise. Suppose X → Y is a dominant morphism of k-varieties. Show thatthere is a dense open subset U of Y such that for each p ∈ U ,

dimY + dimXp = dimX

where Xp is the fiber over p.

15.5.C. Exercise. Suppose X → Y is a finite type morphism of Noetherian schemes.

Consider the function dimXp from Y to Z, where dim ∅ := −1. Show that this function

is constructable, i.e. the locus where dimXp = n is constructable for each n. ]

[primordial:

15.5.2. Elimination of quantifiers and the Lefschetz principle.

261

Greg thinks of this as elimination of quantifiers: this reduces to the fact that the image ofconstructable sets in An+1 in An is constructable. (Translation: the locus of things where variousalgebraic equalities and inequalities is satisfied is constructable.) He thinks of the proof as usingthe Euclidean algorithm. elim of quants

Lefschetz princApparently this implies the Nullstellensatz, but I don’t yet see how. This could be an exercise.Exercise. Suppose k is an algebraically closed subfield of an algebraically closed field k′.

(Think of Q and C.) Then f ∈ k[x1, . . . , xn] is irreducible iff it is irreducible in k′[x1, . . . , xn].Exercise. Two k-schemes are isomorphic over k if and only if they are isomorphic over k′.

This all has something to do with the fact that a bunch of equations has a solution over k iff

it has a solution over k′, basically by the Nullstellensatz. ]

15.6 Fun in codimension one: Krull’s Principal IdealTheorem, Algebraic Hartogs’ Lemma, and more

31 In this section, we’ll explore a number of results related to codimension one.Codimension one primes of Z and k[x, y] correspond to prime numbers and

irreducible polynomials respectively. We will make this link precise for uniquefactorization domains. Then we introduce two results that apply in more generalsituations, and link functions and the codimension one points where they vanish,Krull’s Principal Ideal Theorem 15.6.3, and Algebraic Hartogs’ Lemma 15.6.6. Wewill find these two theorems very useful. For example, Krull’s Principal IdealTheorem will help us compute codimensions, and will show us that codimension canbehave oddly, and Algebraic Hartogs’ Lemma will give us a useful characterizationof Unique Factorization Domains (Proposition 15.6.8).

The results in this section will require (locally) Noetherian hypotheses.

15.6.1. Krull’s Principal Ideal Theorem. [forme: Earlier note: Sort out

Krull including Eisenbud p. 233. Later note: Iitaka proves Krull quickly on p. 116.]

As described earlier in the chapter, in analogy with linear algebra, we have thefollowing.

15.6.2. Krull’s Principal Ideal Theorem (geometric version). — SupposeX is a Noetherian scheme, and f is a function. Then the irreducible componentsof V (f) are codimension 0 or 1.

ex.doneThis is clearly equivalent to the following algebraic statement.

15.6.3. Krull’s Principal Ideal Theorem (algebraic version). — SupposeA is a Noetherian ring, and f ∈ A. Then every minimal prime p containing f hascodimension at most 1. If furthermore f is not a zero-divisor, then every minimalprime p containing f has codimension precisely 1. [forme: [AM, p. 122]. [H,

T.I.1.11A] — He says neither 0-divisor nor a unit. (We could have V (f) = ∅, if f

is a unit — but that doesn’t violate the statement.) (Two toggles: domain or non-

domain. Non-zero-divisors, and zero-divisors. They should reduce all this to domains

and non-zero-divisors.) ] 32 [forme: Transcendence degree version: Mumford Krull

p. 57, in I.1.7. He quotes Zariski-Samuel, vol. 2, ch. 7, §7. Joe Rabinoff says this is

near the end of Atiyah-Macdonald, along with the other version of Krull.] 33 Krull

31codimonefun

32krull

33Krullt

262

The full proof is technical, and postponed to in §15.7.But this immediately useful. For example, consider the scheme Spec k[w, x, y, z]/(wx−

yz). What is its dimension? It is cut out by one non-zero equation wx− yz in A4,so it is a threefold.

15.6.A. Exercise. What is the dimension of Spec k[w, x, y, z]/(wz−xy, y17+z17)?(Be careful to check they hypotheses before invoking Krull!)

15.6.B. Exercise. Show that an irreducible homogeneous polynomial in n + 1variables over a field k describes an integral scheme of dimension n− 1.

15.6.C. Exercise (important for later). (a) (Hypersurfaces meet everythingof dimension at least 1 in projective space — unlike in affine space.) Suppose X isa closed subset of Pn

k of dimension at least 1, and H a nonempty hypersurface inPn

k . Show that H meets X . (Hint: consider the affine cone, and note that the coneover H contains the origin. Use Krull’s Principal Ideal Theorem 15.6.3.)(b) (Definition: Subsets in Pn cut out by linear equations are called linear sub-spaces. Dimension 1, 2 linear subspaces are called lines and planes respectively.)Suppose X → Pn

k is a closed subset of dimension r. Show that any codimension rline, plane

linear space meets X . Hint: Refine your argument in (a). (In fact any two thingsin projective space that you might expect to meet for dimensional reasons do infact meet. We won’t prove that here.) [forme: But I should give a ref]

(c) Show further that there is an intersection of r + 1 hypersurfaces missing X .(The key step: show that there is a hypersurface of sufficiently high degree thatdoesn’t contain every generic point of X . Show this by induction on the numberof generic points. To get from n to n + 1: take a hypersurface not vanishing on p1,. . . , pn. If it doesn’t vanish on pn+1, we’re done. Otherwise, call this hypersurfacefn+1. Do something similar with n + 1 replaced by i (1 ≤ i ≤ n). Then consider∑

i f1 · · · fi · · · fn+1.) [forme: [less important:] If k is infinite, show that there is

a codimension r + 1 linear subspace missing X. (The key step: show that there is

a hyperplane not containing any generic point of a component of X. [This uses the

same trick as Exercise 7.5.F(c), which may be cut.])] 34

[forme: [In the above, showing that the intersection of ≤ n hypersurfaces in Pn

is non-empty, refer forward to Bezout.]]

15.6.4. Pathologies of the notion of “codimension”. 35 We can use Krull’sPrincipal Ideal Theorem to produce the long-promised example of pathology inthe notion of codimension. Let A = k[x](x)[t]. In other words, elements of Aare polynomials in t, whose coefficients are quotients of polynomials in x, whereno factors of x appear in the denominator. (Warning: A is not isomorphic tok[x, t](x).) Clearly, A is a domain, and (xt− 1) is not a zero divisor. You can verifythat A/(xt − 1) ∼= k[x](x)[1/x] ∼= k(x) — “in k[x](x), we may divide by everythingbut x, and now we are allowed to divide by x as well” — so A/(xt − 1) is a field.Thus (xt − 1) is not just prime but also maximal. By Krull’s theorem, (xt − 1) iscodimension 1. Thus (0) ⊂ (xt− 1) is a maximal chain. However, A has dimensionat least 2: (0) ⊂ (t) ⊂ (x, t) is a chain of primes of length 3. (In fact, A hasdimension precisely 2, although we don’t need this fact in order to observe the

34projectivefact

35heightpath

263

pathology.) Thus we have a codimension 1 prime in a dimension 2 ring that isdimension 0. Here is a picture of this lattice of ideals.

(x, t)

(t)

DDDD

DDDD

(xt− 1)

wwww

wwww

w

(0)

This example comes from geometry; it is enlightening to draw a picture see Fig-ure 3. Spec k[x](x) corresponds to a germ of A1

k near the origin, and Spec k[x](x)[t]corresponds to “this × the affine line”. You may be able to see from the picturesome motivation for this pathology — note that V (xt − 1) doesn’t meet V (x), soit can’t have any specialization on V (x), and there nowhere else for V (xt − 1) tospecialize. 36 [forme: Don’t define non-catenary]

Figure 3. Dimension and codimension behave oddly on the sur-face Spec k[x](x)[t]

It is disturbing that this misbehavior turns up even in a relative benign-lookingring.

15.6.D. Unimportant exercise. Show that it is false that if X is an integralscheme, and U is a non-empty open set, then dim U = dim X . [forme: Use the

pathology!] 37

15.6.5. Algebraic Hartogs’ Lemma for Noetherian normal schemes.Hartogs’ Lemma in several complex variables states (informally) that a holo-

morphic function defined away from a codimension two set can be extended overthat. We now describe an algebraic analog, for Noetherian normal schemes.

15.6.6. Algebraic Hartogs’ Lemma. — Suppose A is a Noetherian normaldomain. 38

A = ∩p codimension 1Ap.

The equality takes place inside FF (A); recall that any localization of a do-main A is naturally a subset of FF (A) (§5.3.4). Warning: No one else calls thisAlgebraic Hartogs’ Lemma. I’ve called it this because I find the that it parallelsthe statement in complex geometry. The proof is technical, so we postpone it to§15.6.9. (One can state Algebraic Hartogs’ Lemma more generally in the case that

36noncatenary, H.E.II.3.21ad

37H.E.II.3.21e

38t:hartogs, hartogs, H.P.II.6.3A

264

Spec A is a Noetherian normal scheme, meaning that A is a product of Noether-ian normal domains; the reader may wish to do so. A more general statement isthat if A is a Noetherian domain, then ∩codimP=1AP is the integral closure of A[AM, Cor. 5.22]. We won’t need this. And this “domain” condition can also berelaxed.) [primordial: [forme: [M, III.8 p. 272] states that if A is a Noetherian

integral domain, then it is integrally closed iff (i) A = ∩PAP , and for all minimal

prime ideals P , AP is a principal valuation ring. He doesn’t give a proof, but refers

to Zariski-Samuel, vol. I, Ch. V, §6; or Bourbaki Ch. 7.] ]One might say that if f ∈ FF (A) does not lie in Ap where p has codimension

1, then f has a pole at [p], and if f ∈ FF (A) lies in pAp where p has codimension1, then f has a zero at [p]. It is worth interpreting Algebraic Hartogs’ Lemma assaying that a rational function on a normal scheme with no poles is in fact regular(an element of A). More generally, if X is a Noetherian normal scheme, we candefine zeros and poles of rational functions on X . (We will soon define the order ofa zero or a pole.) [forme: Ref!]

15.6.E. Exercise. Suppose f is an element of a normal domain A, and f iscontained in no codimension 1 primes. Show that f is a unit. 39

15.6.F. Exercise. Suppose f and g are two global sections of a Noetheriannormal scheme, not vanishing at any associated point, with the same poles andzeros. Show that each is a unit times the other.

[forme: DO THIS: Remark on how we need normality, using the nodal cubic.

Explain that we think of the nodal cubic as the affine line with two points glued

together. Draw a picture of course.]

15.6.7. A useful characterization of unique factorization domains.We can use Algebraic Hartogs’ Lemma 15.6.6 to prove one of the four things

you need to know about unique factorization domains, promised in §7.4.5.

15.6.8. Proposition. — Suppose that A is a Noetherian domain. Then A is aUnique Factorization Domain if and only if all codimension 1 primes are principal.40

This contains Proposition 15.1.4 and (in some sense) its converse.

Proof. We have already shown in Proposition 15.1.4 that if A is a Unique Factoriza-tion Domain, then all codimension 1 primes are principal. Assume conversely thatall codimension 1 primes of A are principal. I claim that the generators of theseideals are irreducible, and that we can uniquely factor any element of A into theseirreducibles, and a unit. First, suppose (f) is a codimension 1 prime ideal p. Thenif f = gh, then either g ∈ p or h ∈ p. As codim p > 0, p 6= 0, so by Nakayama’sLemma 15.3.B (as p is finitely generated), p 6= p2. Thus both g and h cannot be inp. Say g /∈ p. Then g is contained in no codimension 1 primes (as f was containedin only one, namely p), and hence is a unit by Exercise 15.6.E.

Finally, we show that any non-zero element f of A can be factored into irre-ducibles. Now V (f) is contained in a finite number of codimension 1 primes, as (f)as a finite number of associated primes (§7.5), and hence a finite number of minimalprimes. We show that any nonzero f can be factored into irreducibles by induction

39unittest, used

40ufdht1pr, H.P.I.1.12A, [M, p. 259], [E, p. 98-99, Prop. 3.11b]

265

on the number of codimension 1 primes containing f . In the base case where thereare none, then f is a unit by Exercise 15.6.E. For the general case where there isat least one, say f ∈ p = (g). Then f = gnh for some h /∈ (g). (Reason: otherwise,we have an ascending chain of ideals (f) ⊂ (f/g) ⊂ (f/g2) ⊂ · · · , contradictingNoetherianness.) Thus f/gn ∈ A, and is contained in one fewer codimension 1primes.

15.6.9. Proof of Algebraic Hartogs’ Lemma 15.6.6 ?. 41This proof does not shedlight on any of the other discussion in this section, and thus should not be read.However, you should sleep soundly at night knowing that the proof is this short.[forme: Kiran wrote: One stylistic quibble: you may as well give a direct proof, which

shows that any minimal ideal Q containing I has codimension 1: given any minimal

Q containing I, your argument shows that the maximal ideal of the local ring at Q

is principal, whence codimension 1. I don’t see how this can work...] Obviouslythe right side is contained in the left. Assume we have some x in all AP but not inA.42 Let I be the “ideal of denominators”: ideal of denominators

I := r ∈ A | rx ∈ A.(The ideal of denominators arose in an earlier discussion about normality, see theproof of Proposition 7.4.2.) We know that I 6= A, so choose q a minimal primecontaining I .

Observe that this construction behaves well with respect to localization (i.e.if p is any prime, then the ideal of denominators x in Ap is the Ip, and it againmeasures the failure of ‘Algebraic Hartogs’ Lemma for x,’ this time in Ap). ButHartogs’ Theorem is vacuously true for dimension 1 rings, so hence no codimension1 prime contains I . Thus q has codimension at least 2. By localizing at q, we canassume that A is a local ring with maximal ideal q, and that q is the only primecontaining I . Thus

√I = q, so there is some n with I ⊂ qn. Take a minimal such

n, so I 6⊂ qn−1, and choose any y ∈ qn−1−qn. Let z = yx. Then z /∈ A (so qz /∈ q),but qz ⊂ A: qz is an ideal of A.

I claim qz is not contained in q. Otherwise, we would have a finitely-generatedA-module (namely q) with a faithful A[z]-action, forcing z to be integral over A(and hence in A) by Exercise 15.3.D.

Thus qz is an ideal of A not contained in q, so it must be A! Thus qz = Afrom which q = A(1/z), from which q is principal. But then codim Q = dim A ≤dimA/Q Q/Q2 ≤ 1 by Nakayama’s lemma 15.3.B, contradicting the fact that q hascodimension at least 2.

[primordial: We will later give a more long-winded, more complicated proof, in §16.6, in

order to introduce Serre’s criterion for normality. We are holding off because we will use a small

fact about discrete valuation rings. ][primordial:

15.6.G. Exercise. Show that the irreducible subsets of dimension n − 1 of Pnk correspond

to homogeneous irreducible polynomials modulo multiplication by non-zero scalars. 43 [This isessentially the same as Exercise 9.1.9, so I should avoid duplication. The first one should trump.]

41hartogspf

42hartogsproof

43secondirrhyp

266

By determining the prime ideals of Z[x], show that dim Z[x] = 2.

15.6.H. Important but straightforward exercise. If A is a finitely generated domain over

k, show that dimA[x] = dimA + 1. If A is any ring, show that dimA[x] ≥ dimA + 1. (Fact,

proved later: if A is a Noetherian ring, then dimA[x] = dimA + 1. We’ll prove this later. You

may use this fact in later exercises.) [forme: I should add this proof soon! [AM, p. 126,

ex. 7]] 44 ]

15.7 Proof of Krull’s Principal Ideal Theorem 15.6.3 ??

The details of this proof won’t matter much to us, so you should probably notread it. It is included so you can glance at it and believe that the proof is fairlyshort, and you could read it if you really wanted to. 45 [forme: I am following

Eisenbud’s proof, [E, Thm. 10.1, p. 232]. Be sure that this isn’t a straight copy.] 46

15.7.1. Lemma. — If A is a Noetherian ring with one prime ideal. Then A isArtinian, i.e., it satisfies the descending chain condition for ideals. 47Artinian, descending

chain condition The notion of Artinian rings is very important, but we will get away withoutdiscussing it much.

Proof. If A is a ring, we define more generally an Artinian A-module, which is anA-module satisfying the descending chain condition for submodules. Thus A is anArtinian module

Artinian ring if it is Artinian over itself as a module.If m is a maximal ideal of R, then any finite-dimensional (R/m)-vector space

(interpreted as an R-module) is clearly Artinian, as any descending chain

M1 ⊃M2 ⊃ · · ·must eventually stabilize (as dimR/m Mi is a non-increasing sequence of non-negativeintegers).

15.7.A. Exercise. Show that for any n, mn/mn+1 is a finitely-dimensional A/m-vector space. (Hint: show it for n = 0 and n = 1. Use the dimension for n = 1 tobound the dimension for general n.) Hence mn/mn+1 is an Artinian A-module.

As√

0 is prime, it must be m.

15.7.B. Exercise. Prove that mn = 0 = 0 for some n. (Hint: suppose m

can be generated by m elements, each of which has kth power 0, and show thatmm(k−1)+1 = 0.)

15.7.C. Exercise. Show that if 0→ M ′ → M → M ′′ → 0 is an exact sequenceof modules. then M is Artinian if and only if M ′ and M ′′ are Artinian. (Hint: seethe solution to Exercise 5.5.K.)

Thus as we have a finite filtration

A ⊃ m ⊃ · · · ⊃ mn = 0

44dimRx

45proofofKrullt

46krullproof

47artinian, part of [E, Thm. 2.14, p. 75]. Here I’m following more of [AM], e.g. Theorem 8.5.

267

all of whose quotients are Artinian, so A is Artinian as well. This completes theproof of the Lemma.

Proof of Krull’s principal ideal theorem 15.6.3. Suppose we are given x ∈ A, with p

a minimal prime containing x. By localizing at p, we may assume that A is a localring, with maximal ideal p. Suppose q is another prime strictly contained in p.

x o

???

????

?

p // A

q/

@@

For the first part of the theorem, we must show that Aq has dimension 0. Thesecond part follows from our earlier work: if any minimal primes are height 0, f isa zero-divisor, by Proposition 7.5.4.

Now p is the only prime ideal containing (x), so A/(x) has one prime ideal. ByLemma 15.7.1, A/(x) is Artinian.

We invoke a useful construction, the nth symbolic power of a prime ideal: if A symbolic power

is a ring, and q is a prime ideal, then define

q(n) := r ∈ A | rs ∈ qn for some s ∈ A− q.We have a descending chain of ideals in A

q(1) ⊃ q(2) ⊃ · · · ,so we have a descending chain of ideals in A/(x)

q(1) + (x) ⊃ q(2) + (x) ⊃ · · ·which stabilizes, as A/(x) is Artinian. Say q(n) + (x) = q(n+1) + (x), so

q(n) ⊂ q(n+1) + (x).

Hence for any f ∈ q(n), we can write f = ax + g with g ∈ q(n+1). Hence ax ∈ q(n).As p is minimal over x, x /∈ q, so a ∈ q(n). Thus

q(n) = (x)q(n) + q(n+1).

As x is in the maximal ideal p, the second version of Nakayama’s lemma 15.3.2gives q(n) = q(n+1).

We now shift attention to the local ring Aq, which we are hoping is dimension 0.

We have q(n)Aq = q(n+1)Aq (the symbolic power construction clearly construction

commutes with respect to localization). For any r ∈ qnAq ⊂ q(n)Aq, there is somes ∈ Aq − qAq such that rs ∈ qn+1Aq. As s is invertible, r ∈ qn+1Aq as well. ThusqnAq ⊂ qn+1Aq, but as qn+1Aq ⊂ qnAq, we have qnAq = qn+1Aq. By Nakayama’sLemma version 4 (Exercise 15.3.B),

qnAq = 0.

Finally, any local ring (R, m) such that mn = 0 has dimension 0, as Spec A consistsof only one point: [m] = V (m) = V (mn) = V (0) = Spec A.

[primordial: Hidden here is a cut proof of Krull from Hartogs. It only works for Noe-therian rings whose normalizations are also Noetherian; this isn’t always the case! Ref: Eisenbud

268

p. 263, who says that examples are given by Nagata, 1962, Appendix, Examples 3–5. It is how-ever true for finite type over a field, Eisenbud p. 293 Cor. 13.13. This argument uses NoetherNormalization, so I could again follow Mumford.

We start by proving it for a Noetherian normal domain A. Suppose f ∈ FF(A). We wish toshow that the minimal primes containing f are all height 1. If there is one which is height greaterthan 1, then localize at it; then we may assume that A is a local ring with maximal ideal m ofheight at least 1, and that the only prime containing f is m. Let g = 1/f ∈ FF(A). Then g ∈ Ap

for all height 1 primes p, so by “Hartogs’ Theorem”, g ∈ A. Thus gf = 1. But g, f ∈ A, andf ∈ m, so we have a contradiction.

Next, we prove it in general. Suppose f ∈ A, where A is a Noetherian ring. We wish to showthat all the minimal primes containing f are height 1 or 0. Reduce to the case where A is anintegral domain (by quotienting by a height 0 prime of A). Then let A be the integral closure of

A (in its fraction field). We know Krull for A, so by the going up theorem 15.2.2, we know it forA. (More discussion needed!)

This is probably a repeat: The following gives Hartogs implying Krull, in a nice case (wherewe have normality).

The first statement implies the second: because A is a domain, the associated primes ofSpecA are precisely the minimal (i.e. height 0) primes. If f is a not a zero-divisor, then f is notan element of any of these primes, by Proposition 7.5.4. So we will now prove the first statement.

Suppose f ∈ FF(A). We wish to show that the minimal primes containing f are all height 1.If there is one which is height greater than 1, then after localizing at this prime, we may assumethat A is a local ring with maximal ideal m of height at least 2, and that the only prime containingf is m. Let g = 1/f ∈ FF(A). Then g ∈ Ap for all height 1 primes p, so by Algebraic Hartogs’Lemma 15.6.6, g ∈ A. Thus gf = 1. But g, f ∈ A, and f ∈ m, so we have a contradiction.

15.8 Dimension 0 rings, including Artinian rings ?

We have seen interesting rings of dimension 1, including notably the polynomial ring k[t]and the integers Z. But there are also interesting rings of dimension 0. We’ve seen k[ε]/ε2 before[REF]. Here are a few more examples.

15.8.A. Exercise. Show that Z/2[x, y]/(x3, y5) and Z/(215) are dimension 0.We might have the intuition that both of these rings have “size 15”, and we will make that

more precise with the notion of length very soon.But first, let’s hunt for more examples.

15.8.B. Exercise. Suppose A is a local k-algebra.

(a) If the dimension of A as a k-vector space is 2, show that A ∼= k[ε]/ε2.(b) Classify all such A whose dimension as a k-vector space is 3. (Hint: there are two.)

In analogy with rings, if M is a module over a ring, we say that M is Noetherian (resp.

Artinian) if it satisfies the ascending (descending) chain condition for submodules. We say that achain of submodules of M

M = M0 ⊃M1 ⊃ · · · ⊃Mn

is a composition series if each Mj/Mj+1 is a nonzero simple module (is nonzero and has nocomp seriesnonzero proper submodules). The length of M is the least length of a composition series, or ∞ ifM has no finite composition series.

15.8.C. Exercise. If M is a non-zero simple A-module, show that M ∼= A/m for some maximalideal m.

15.8.D. Exercise. Show that M has finite length (has a finite composition series) if and only ifM is Artinian and Noetherian. [forme: Part of [E, Thm. 2.13, p. 72]]

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15.8.1. Theorem. — If the A-module M has a finite composition series48

M = M0 ⊃M1 ⊃ · · · ⊃Mn = 0

of length n, then

(a) Every chain of submodules of M can be refined to a composition series, and in partic-ular has length at most n.

(b) The natural map M → ⊕mMm (where the sum is taken over all m such that someMi/Mi+1

∼= A/m) is an isomorphism. The number of Mi/Mi+1 isomorphic to A/mis the length of Mm as a module over Am, and is thus independent of the choice ofcomposition series.

(c) M is annihilated by some power of a maximal ideal m if and only if M = Mm.

[forme: [E, Thm. 2.13, p. 72]]

This theorem generalizes the Jordan-Holder theorem, and the “classical” form of the Chineseremainder theorem (involving integers modulo some number).

[forme: WHAT DO WE NEED?]

Proof. (a) [forme: Distill from Eisenbud p. 73](b) [forme: Distill from Eisenbud p. 73-4](c) [forme: Distill from Eisenbud p. 74]

We now define an important class of ring that we will see has dimension 0. We say a ring Ais Artinian if it satisfies the descending chain condition on ideals: every descending sequence Artinian ring

I1 ⊃ I2 ⊃ · · ·of ideals eventually stabilizes: there is an r such that Ir = Ir+1 = · · · . Although this definitionis similar to the definition of Noetherian ring (§5.5.4), involving an ascending chain condition,Artinian rings behave quite differently. For example, we’ll soon see that all Artinian rings areNoetherian (Theorem 15.8.2), although the converse certainly fails (consider k[t]). If A is anArtinian ring, we say that SpecA is an Artinian scheme. Artinian scheme

15.8.E. Easy Exercise. Suppose A is a local k-algebra of finite dimension. Show that A isArtinian. Suppose that Z/(n) (n 6= 0) is Artinian.

15.8.2. Theorem. — The following are equivalent.49

(1) A is Noetherian, and all prime ideals are maximal.(2) A has finite length as an A-module.(3) A is Artinian.

Proof. (a) implies (b). [forme: from Eisenbud.](b) implies (c) follows immediately from Theorem 15.8.1.(c) implies (a) [forme: from Eisenbud.]

15.8.F. Exercise. Suppose X is an affine finite-type k-scheme (a closed subscheme of some Ank ).Show that the following are equivalent. [forme: Variant of [E, Cor. 2.15, p. 76].]

(a) X contains a finite number of points.(b) Γ(X,OX) is a finite-dimensional k-vector space.(c) X is Artinian.

15.8.3. Exercise. Show that any Artinian ring is a finite direct product of local Artinian rings.[forme: [E, Cor. 2.16, p. 76].]

15.8.G. Important Exercise. Suppose A is a Noetherian ring, and M is a finitely-generatedA-module. The following are equivalent.

(a) M has finite length.(b) Some finite product of maximal ideals

Qni=1 mi annihilates M .

48t:compositionseries

49ArtinianNoetherian, [E, Cor. 2.17, p. 77]

270

(c) All primes containing the annihilator of M are maximal.(d) A/ ann(M) is Artinian.

(Hint: Show the implications (a) ⇒ (b) ⇒ (c) ⇒ (d) ⇒ (a). For (a) ⇒ (b), use Theorem 15.8.1.For (c) ⇒ (d), use Theorem 15.8.2.) [forme: Short corollary from [E, Cor. 2.17, p. 77].]50

15.8.H. Important exercise. Suppose I ⊂ A is an ideal of a Noetherian ring. Show that thefollowing are equivalent for a prime ideal p containing I.

(a) p is minimal among primes containing I.(b) Ap/Ip is Artinian.(c) In Ap, pnp ⊂ Ip for all sufficiently large n.

(Hint: For (a) ⇒ (b), use the previous exercise 15.8.G. For (b) ⇒ (c), use Theorems 15.8.1 and15.8.2.) [forme: Short cor from [E, Cor. 2.19, p. 77-78].]

15.9 Older notes on height, Hartogs, Krull

When giving algebraic Hartogs, mention the valuative criterion: as normal implies R1, allheight 1 primes are regular. Then the regular functions are things with each of these valuationsnonnegative.

[forme: Turn this into a theorem!]

15.9.A. Important exercise. (This will be useful soon.) (a) Suppose X = SpecA where A isa Noetherian domain, and Z is an irreducible component of V (r1, . . . , rn), where r1, . . . , rn ∈ A.Show that the height of (the prime associated to) Z is at most n. Conversely, suppose X = SpecAwhere A is a Noetherian domain, and Z is an irreducible subset of height n. Show that there aref1, . . . , fn ∈ A such that Z is an irreducible component of V (f1, . . . , fn).51

(b) (application to finitely generated k-algebras) Suppose X = SpecA where A is a finitelygenerated domain over k, and Z is an irreducible component of V (r1, . . . , rn), where r1, . . . , rn ∈A. Show that dimZ ≥ dimX − n. Conversely, suppose X = SpecA where A is a Noetheriandomain, and Z is an irreducible subset of codimension n. Show that there are f1, . . . , fn ∈ A suchthat Z is an irreducible component of V (f1, . . . , fr). [forme: Is this really that important?!I gave this in 216, and it was a harder problem than I realized. One possibility isjust to ask it in the case of a k-algebra. Mumford Cor.I.7.3 and 4 p. 62, and thatis much easier. Kirsten proves the following. 1. Let A be Noetherian ring, and〈r1, . . . , rn〉 ⊂ A a proper ideal. Then for any minimal prime q containing 〈r1, . . . , rn〉we have that codim q ≤ n. From [E] Theorem 10.2, p. 235. This takes some time.2. Let X = SpecA with A a finitely generated k-algebra and let Z be an irreduciblecomponent of V (r1, . . . , rn) with 〈r1, . . . , rn〉 6= A. Then dimZ ≥ dimXi − n, where Xiis any irreducible component of X containing Z. 3. Let A be a finitely generatedk-domain, and let q be a prime of height n. Then there exist f1, . . . , fn ⊂ A suchthat q is a minimal prime containing f1, . . . , fn and such that every minimal primecontaining f1, . . . , fn has height n. Eisenbud shows this wihtout the hypothesis thatA is a finitely generated k-domain, p. 235 Cor. 10.5: Let A be a Noetherian ringand let q be a prime of height n. Then, there exist f1, . . . , fn ∈ q such that q is aminimal prime containing 〈f1, . . . , fn〉 and such that every minimal prime containing

〈f1, . . . , fn〉 has height n.]

15.10 Elements of a Noetherian ring that vanish analyticallyvanish

50e:art

51e:htbound

271

[forme: We need this proof!!! At least for a Noetherian dimension 1 local ring.]

15.10.1. Proposition. — If (A,m) is a Noetherian local ring, then ∩imi = 0. 52

It is tempting to argue that m(∩imi) = ∩imi, and then to use Nakayama’s lemma 15.3.B toargue that ∩imi = 0. Unfortunately, it is not obvious that this first equality is true: product doesnot commute with infinite intersections in general. [forme: I heard this argument fromKirsten Wickelgren, who heard it from Greg Brumfiel. This is used in 16.3.1, and in“good facts to know about smoothness”, although it may be cut there..]

It would be nice to give a geometric intuition for this. Any function that isanalytically zero is in fact zero near here.

Proof. Let I = ∩imi. We wish to show that I ⊂ mI; then as mI ⊂ I, we have I = mI, and henceby Nakayama’s Lemma 15.3.B, I = 0. Fix a primary decomposition of mI. It suffices to show thatp contains I for any p in this primary decomposition, as then I is contained in all the primaryideals in the decomposition of mI, and hence mI.

Let q =√

p. If q 6= m, then choose x ∈ m − q. Now x is not nilpotent in R/p, and hence isnot a zero-divisor. But xI ⊂ p, so I ⊂ p.

On the other hand, if q = m, then as m is finitely generated, and each generator is in√

p,there is some a such that ma ⊂ p. But I ⊂ ma, so we are done.

[Earlier notes on this: the plan is apparently to show that m(∩mi) ⊂ ∩mi using primarydecomposition of left hand side.]

Iitaka has an argument on this on p. 111.

15.11 Proof of Krull’s Principal Ideal Theorem ?

[forme: Clean up massively! Compare to the previous exposition too.]

15.11.1. Theorem. — Suppose A is a Noetherian domain. If x ∈ A, and p is minimal amongprimes of A containing x, then ht p ≤ 1. [forme: [E, p. 232]]

The proof will use the notion of the nth symbolic power of a prime ideal: symbolic power of a

prime idealq(n) := r ∈ A | sr ∈ qn for some s ∈ A− q.(The idea behind this: these are the elements of A that, when “restricted to” Aq, lie in (qAq)n.)[forme: Reason to consider this: when you are thinking of height of a prime ideal,you should just localize there — this is a usual trick.]Proof. Suppose p is any minimal prime ideal containing x. Suppose q ( p.

Strategy: We’ll show that q has height 0; the result will follow. To show that q has height0, it suffices to show that qAq has height 0 in Aq, i.e. that there are no smaller prime ideals. It

would suffice to show that (qAq)n = 0 for some n, as then√

0 is qAq; but it is also the intersectionof all prime ideals.

Let’s get to work. By localization at p, we may assume that p is maximal. We have contain-ments of ideal as follows:

A

p

>>>>

>>>>

(x) q

Now A/(x) is Noetherian and all prime ideals are maximal, so A/(x) is Artinian (Theo-rem 15.8.2 above).

52capmi, used in ??

272

By the descending chain condition for

· · · ⊃ (x) + q(n) ⊃ (x) + q(n+1) ⊃ · · ·

we have that for some n

(x) + q(n) = (x) + q(n+1).

Thus q(n) ⊂ (x) + q(n+1), so for any f ∈ q(n), we can write f = ax + b where a ∈ A, b ∈ q(n+1).

Hence ax ∈ q(n+1). But as x /∈ q (as p is minimal among the primes containing x), we have

a ∈ q(n) (by the definition of q(n)). Thus q(n) = (x)q(n) + q(n+1). As x lies in the maximal ideal

q, by Nakayama’s lemma we have q(n) = q(n+1). Localizing at q, we have

(qAq)n = (qAq)n+1.

By Nakayama’s Lemma in this ring (where qAq is maximal), we have (qAq)n = 0, as desired.

[primordial:Older notes. A function is determined by its value at closed points if reduced of finite

type over field. Reason: closed points are dense. If base field is algebraically closed, a function isdetermined by values at k-points.

Can check reducedness at closed points. (We’ve already said it is stalk-local.) I think thismay now be proved. Reason: closed points are dense.

Locally of finite type: every non-empty open set contains a closed point. (Translation: closedpoints are dense.) Sneaky corollary: Hence an infinite number of closed points.

f vanishes on closed points implies f = 0 on a reduced lft/k scheme. Translation: N is theJacobson radical for a finite-type k-algebra.

Mumford p. 123: X finite type / k but Γ(X,OX) not finitely generated. “this is the factthat Hilbert’s 14th problem is false.” Ask Daniel Allcock!

15.11.A. Exercise. If A is a finitely generated k-algebra, show that every nonempty open setof SpecA contains a closed point.

Brian Osserman has some notes on dimension, BOdimension.pdf in “other references” direc-tory.

]

15.12 Older notes start here, on products

IfX and Y are subschemes of Pnk , and Z is a component ofX∩Y , then dimZ ≥ dimX+dimY −n.In fact, this is true if you are in some smooth, dimension n finite type k-scheme. Method:

X ∩ Y ∼= X × Y ∩∆. From dimX × Y = dimX + dimY . We need the fact that the diagonal is alocal complete intersection, which needs a proof.

[forme: Place:

15.12.A. Exercise (Dimensions add nicely when taking products of finite-type k-schemes.).Suppose X and Y are finite-type k-schemes. Show that dimX×SpeckY = dimX×Spec kY .[What should they be allowed to use? Do I want to make X and Y irreducible for

them? Certainly they should be able to use the fact that if you tensor two fields, thenthe transcendence degrees add. But perhaps it would be nicer to have them provethat in the end.] (Caution: show that this need not be true if not taking fiberedproduct over k! [It goes wrong in many many ways.]) ]

Place: Show how dimension behaves in fibered product, over Spec k. Ideally discuss somedimensionbound in general: dimX ×Z Y ≤ dimX + dimY − dimZ. Make sure that is right! On a relatednote: dimensions of fibers.

Fix and place?:

15.12.B. Exercise. If S• is a Noetherian domain over k, and ProjS• is non-empty show thatdimSpec S• = dimProjS• + 1. (Hint: throw out the origin. Look at a distinguished D(f) where

273

deg f = 1. Use the fact mentioned in Exercise 15.6.H. By the previous exercise 19.1.A, you canassume that S• is generated in degree 1 over S0 = A.) 53 [forme: This isn’t quite right.]

]

53e:dimproj

CHAPTER 16

Nonsingularity (or regularity or “Smoothness”)

[primordial: Place:

16.0.1. Morphisms and tangent spaces. Suppose f : X → Y , and f(p) = q. Then if wewere in the category of manifolds, we would expect a tangent map, from the tangent space of pto the tangent space at q. Indeed that is the case; we have a map of stalks OY,q → OX,p, whichsends the maximal ideal of the former n to the maximal ideal of the latter m (we have checkedthat this is a “local morphism” when we briefly discussed local-ringed spaces). Thus n2 → m2,from which n/n2 → m/m2, from which we have a natural map (m/m2)∨ → (n/n2)∨. This is themap from the tangent space of p to the tangent space at q that we sought.

16.0.C.Exercise. SupposeX is a k-scheme. Describe a natural bijection Homk(Spec k[ε]/(ε2), X)

to the data of a k-valued point (a point whose residue field is k, necessarily closed) and a tangent

vector at that point. [forme: It seems that “finite type” is not necessary, as long as

you require the maps to be k-morphisms.] ]One natural notion we expect to see for geometric spaces is the notion of when

an object is “smooth”. In algebraic geometry, this notion, called nonsingularity (orregularity, although we won’t use this term) is easy to define but a bit subtle inpractice. We will soon define what it means for a scheme to be nonsingular (or reg-ular) at a point. A point that is not nonsingular is (not surprisingly) called singular(“not smooth”). A scheme is said nonsingular if all its points are nonsingular, andsingular if one of its points is singular. smooth, regular, nonsin-

gular, singular, Zariski

tangent space

The notion of nonsingularity is surprisingly not as useful as you might think. In-stead, there is a “relative notion” that applies to a morphism of schemes f : X → Ythat is much better-behaved (corresponding to the notion of submersion in differen-tial geometry). For this reason, the word “smooth” is reserved for these morphisms.We will discuss smooth morphisms in Chapter 30. However, nonsingularity is stilluseful, especially in (co)dimension 1.

In this chapter, we’ll say as little as possible to satisfy your curiosity, and toset us up for later work. In Section ????, we mention a few facts without proof,that are interesting to know, but not useful to us in the short term.

The main things you should remember from this chapter are:[forme: ZARISKI TANGENT SPACE TRICK, Zariski-Krull; MAYBE JACO-

BIAN; DIM 1; AFFINE SPACE IS REGULAR IN CODIMENSION 1, and more

generally, facotrial, or locally Noetherian and normal implies R1.]

16.1 The Zariski tangent space

Z tangent spaceWe begin by defining the notion of the tangent space of a scheme at a point. It

will behave like the tangent space you know and love at “smooth” points, but will

275

276

also make sense at other points. In other words, geometric intuition at the smoothpoints guides the definition, and then the definition guides the algebra at all points,which in turn lets us refine our geometric intuition.

This definition is short but surprising. The main difficulty is convincing yourselfthat it deserves to be called the tangent space. I’ve always found this tricky toexplain, and that is because we want to show that it agrees with our intuition; butunfortunately, our intuition is worse than we realize. So I’m just going to define itfor you, and later try to convince you that it is reasonable.

Suppose p is a prime ideal of a ring A, so [p] is a point of Spec A. Then [pAp]is a point of the scheme Spec Ap. For convenience, we let m := pAp ⊂ Ap =: B.Let k = B/m be the residue field. Then m/m2 is a vector space over the residuefield k: it is an B-module, and m acts like 0. This is defined to be the Zariskicotangent space. The dual is the Zariski tangent space. Elements of theZariski cotangent space are called cotangent vectors or differentials; elementsof the tangent space are called tangent vectors. 1differential = cotangent

vector, tangent vector Note that this definition is intrinsic. It doesn’t depend on any specific de-scription of the ring itself (such as the choice of generators over a field k, which isequivalent to the choice of embedding in affine space). Notice that in some sense,the cotangent space is more algebraically natural than the tangent space. There isa moral reason for this: the cotangent space is more naturally determined in termsof functions on a space, and we are very much thinking about schemes in terms of“functions on them”. This will come up later.

I’m now going to give you a bunch of plausibility arguments that this is areasonable definition.

First, I’ll make a moral argument that this definition is plausible for the cotan-gent space of the origin of An. Functions on An should restrict to a linear functionon the tangent space. What function does x2 +xy +x + y restrict to “near the ori-gin”? You will naturally answer: x+y. Thus we “pick off the linear terms”. Hencem/m2 are the linear functionals on the tangent space, so m/m2 is the cotangentspace.

Here is a second argument. In differential geometry, the tangent space at apoint is sometimes defined as the vector space of derivations at that point. Aderivation is a function that takes in functions near the point that vanish at thepoint, and gives elements of the field k, and satisfies the Leibniz rule

(fg)′ = f ′g + g′f.

Translation: a derivation is a map m→ k. But m2 → 0, as if f(p) = g(p) = 0, then

(fg)′(p) = f ′(p)g(p) + g′(p)f(p) = 0.

Thus we have a map m/m2 → k, i.e. an element of (m/m2)∨.

16.1.A. Exercise. Check that this is reversible, i.e. that any map m/m2 → kgives a derivation. In other words, verify that the Leibniz rule holds.

16.1.1. Old-fashioned example. 2Here is an example to help tie this down toearth. Computing the Zariski-tangent space is actually quite hands-on, becauseyou can compute it in a multivariable calculus way. In A3, we have a curve cut

1M.III.4.FD.1

2oldfashdiff

277

out by x + y + z2 + xyz = 0 and x − 2y + z + x2y2z3 = 0. (You know enough tocheck that this is a curve, but it is not important to do so.) What is the tangentline near the origin? (Is it even smooth there?) Answer: the first surface looks likex + y = 0 and the second surface looks like x− 2y + z = 0. The curve has tangentline cut out by x+y = 0 and x−2y+z = 0. It is smooth (in the analytic sense). Inmultivariable calculus, the students do a page of calculus to get the answer, becauseI can’t tell them to just pick out the linear terms.

Let’s make explicit the fact that we are using. If A is a ring, m is a maximalideal, and f ∈ m is a function vanishing at the point [m] ∈ Spec A, then the Zariskitangent space of Spec A/(f) at m is cut out in the Zariski tangent space of Spec A(at m) by the single linear equation f (mod m2). The next exercise will force youthink this through.

16.1.B. Important exercise (“Krull’s theorem for the Zariski tangentspace”). Suppose A is a ring, and m a maximal ideal. If f ∈ m, show thatthe dimension of the Zariski tangent space of Spec A at [m] is the dimension of theZariski tangent space of Spec A/(f) at [m], or one less. 3

Here is another example to see this principle in action: x + y + z2 = 0 andx+y+x2+y4 +z5 = 0 cuts out a curve, which obviously passes through the origin.If I asked my multivariable calculus students to calculate the tangent line to thecurve at the origin, they would do a page of calculus which would boil down topicking off the linear terms. They would end up with the equations x + y = 0 andx+y = 0, which cuts out a plane, not a line. They would be disturbed, and I wouldexplain that this is because the curve isn’t smooth at a point, and their techniquesdon’t work. We on the other hand bravely declare that the cotangent space is cutout by x + y = 0, and (will soon) define this as a singular point. (Intuitively, thecurve near the origin is very close to lying in the plane x + y = 0.) Notice: thecotangent space jumped up in dimension from what it was “supposed to be”, notdown. We’ll see that this is not a coincidence very soon, in Theorem 16.2.1.

Here is a nice consequence of the notion of Zariski tangent space.

16.1.2. Problem. Consider the ring A = k[x, y, z]/(xy− z2). Show that (x, z) isnot a principal ideal.4

As dim A = 2, and A/(x, z) ∼= k[y] has dimension 1, we see that this idealis height 1 (as height behaves as codimension should for finitely generated k-domains!). Our geometric picture is that Spec A is a cone (we can diagonalizethe quadric as xy − z2 = ((x + y)/2)2 − ((x − y)/2)2 − z2, at least if char k 6= 2— see Remark 7.4.6), and that (x, z) is a ruling of the cone. (See Figure 4 for asketch.) This suggests that we look at the cone point.

Solution. Let m = (x, y, z) be the maximal ideal corresponding to the ori-gin. Then Spec A has Zariski tangent space of dimension 3 at the origin, andSpec A/(x, z) has Zariski tangent space of dimension 1 at the origin. But Spec A/(f)must have Zariski tangent space of dimension at least 2 at the origin.

3tangentKrull

4conenonpr

278

16.1.C. Exercise. Show that (x, z) ⊂ k[w, x, y, z]/(wz − xy) is a codimension1 ideal that is not principal. (See Figure 1 for a sketch.) 5 This example waspromised in Exercise 7.4.D. It could be used in Exercise 16.3.I.

Figure 1. The ruling V (x, z) on V (wz − xy) ⊂ P3.

We conclude with some exercises to give you practice with the concept.

16.1.D. Exercise. Find the dimension of the Zariski tangent space at the point[(2, x)] of Z[2i] ∼= Z[x]/(x2 +4). Find the dimension of the Zariski tangent space at

the point [(2, x)] of Z[√

2i] ∼= Z[x]/(x2 + 2).

16.1.E. Exercise (the Jacobian criterion for computing the Zariskitangent space). 6 Suppose k is an algebraically closed field, and X is a finiteJacobian criterion

type k-scheme. Then locally it is of the form Spec k[x1, . . . , xn]/(f1, . . . , fr). Showthat the Zariski tangent space at the closed point p (with residue field k, by theNullstellensatz) is given by the cokernel of the Jacobian map kr → kn given by theJacobian matrix7

(33) J =

∂f1

∂x1(p) · · · ∂fr

∂x1(p)

.... . .

...∂f1

∂xn(p) · · · ∂fr

∂xn(p)

.

(This is just making precise our example of a curve in A3 cut out by a couple ofequations, where we picked off the linear terms, see Example 16.1.1 .) You might

be alarmed: what does ∂f∂x1

mean?! Do you need deltas and epsilons? No! Justdefine derivatives formally, e.g.

∂

∂x1(x2

1 + x1x2 + x22) = 2x1 + x2.

5rulingnotprincipal

6Jacobian, M.III.4.P.1

7Jacobianmatrix

279

(Hint: Do this first when p is the origin, and consider linear terms, just as inExample 16.1.1. Note for future reference that you are not using the fact thatk is algebraically closed. Then in the general case (with k algebraically closed),“translate p to the origin.” [forme: Make very clear to them this problem is about

making that “linear term” idea concrete. People weren’t clear on this. Perhaps for

experts explain where algebraic closure came in, by giving an inseparable example

where this fails. Mention that it holds if the field is perfect.] [forme: In Laszlo’s

course, p. 81, Prop. 6.4.10: If A is a Noetherian regular ring, then AnA is regular at

all closed points. ]

16.1.F. Less important exercise (“higher-order data”). In Exercise 5.6.B,you computed the equations cutting out the three coordinate axes of A3

k. (Call thisscheme X .) Your ideal should have had three generators. Show that the ideal hasat least three generators. (Hint: working modulo m = (x, y, z) won’t give any usefulinformation, so work modulo m2.)

16.2 The local dimension is at most the dimension of thetangent space

16.2.1. Theorem. — Suppose (A, m) is a Noetherian local ring. Then dim A ≤dimk m/m2. 8

If equality holds, we say that A is a regular local ring. If a Noetherian ringA is regular at all of its primes, we say that A is a regular ring. regular local ring, regu-

lar ring, regular schemeA locally Noetherian scheme X is regular or nonsingular at a point p if thelocal ring OX,p is regular. It is singular at the point otherwise. A scheme isregular or nonsingular if it is regular at all points. It is singular otherwise (i.e.if it is singular at at least one point). regular scheme, nonsin-

gular, singularProof of Theorem 16.2.1: Note that m is finitely generated (as R is Noetherian),so m/m2 is a finitely generated (R/m = k)-module, hence finite-dimensional. Saydimk m/m2 = n. Choose n elements of m/m2, and lift them to elements f1, . . . , fn

of m. Then by Nakayama’s lemma (version 4, Exercise 15.3.B), (f1, . . . , fn) = m.Recall Exercise 15.9.A(a). In this case, V ((f1, . . . , fn)) = V (m) is just the

point [m], so the height of m is at most n. Thus the longest chain of prime idealscontaining m is at most n + 1. But this is also the longest chain of prime ideals inX (as m is the unique maximal ideal), so n ≥ dim X .

16.2.A. Exercise. Show that if A is a Noetherian local ring, then A has finitedimension. (Warning: Noetherian rings in general could have infinite dimension,see Remark 15.1.6.)

In the case of finite type schemes over an algebraically closed field k, the Jaco-bian criterion (Exercise 16.1.E) gives a hands-on method for checking for singularityat closed points.

8dimtaninequality

280

16.2.B. Exercise. Suppose k is algebraically closed. Show that the singularclosed points of the hypersurface f(x1, . . . , xn) = 0 in An

k are given by the equations

f = ∂f∂x1

= · · · = ∂f∂xn

= 0.[forme: Now might be a good time to mention k-smooth. At this point this

depends on embedding; we’ll wait until we talk about differentials and smoothness

before showing that this is intrinsic.]

16.2.C. Exercise. Show that A1k and A2

k are nonsingular. (Make sure to checknonsingularity at the non-closed points! Fortunately you know what all the pointsof A2

k are; this is trickier for A3k.)

Let’s apply this technology to an arithmetic situation.

16.2.D. Easy Exercise. Show that Spec Z is a nonsingular curve.Here are some fun comments: What is the derivative of 35 at the prime 5?

Answer: 35 (mod 25), so 35 has the same “slope” as 10. What is the derivativeof 9, which doesn’t vanish at 5? Answer: the notion of derivative doesn’t applythere. You’d think that you’d want to subtract its value at 5, but you can’t subtract“4 (mod 5)” from the integer 9. Also, 35 (mod 2)5 you might think you want torestate as 7 (mod 5), by dividing by 5, but that’s morally wrong — you’re dividingby a particular choice of generator 5 of the maximal ideal of the 5-adics Z5; in thiscase, one appears to be staring you in the face, but in general that won’t be true.Follow-up fun: you can talk about the derivative of a function only for functionsvanishing at a point. And you can talk about the second derivative of a functiononly for functions that vanish, and whose first derivative vanishes. For example, 75has second derivative 75 (mod 125) at 5. It’s pretty flat.

16.2.E. Exercise. (This exercise is for those who know about the primes ofthe Gaussian integers Z[i].) Note that Z[i] is dimension 1, as Z[x] has dimension 2(problem set exercise), and is a domain, and x2 + 1 is not 0, so Z[x]/(x2 + 1) hasdimension 1 by Krull. Show that Spec Z[i] is a nonsingular curve. [Refer experts tothe Jacobian test. Or point out that they can look at it as cut out by linear terms,and use the prime (2, 1 + x) as an example.]Gaussian integers Z[i]

16.2.F. Exercise. Show that there is one singular point of Spec Z[5i], anddescribe it.

16.2.G. Handy Exercise (the Euler test for projective hypersurfaces).There is an analogous Jacobian criterion for hypersurfaces f = 0 in Pn

k . Show that

the singular closed points correspond to the locus f = ∂f∂x1

= · · · = ∂f∂xn

= 0. If thedegree of the hypersurface is not divisible by the characteristic of any of the residuefields (e.g. if we are working over a field of characteristic 0), show that it suffices

to check ∂f∂x1

= · · · = ∂f∂xn

= 0. (Hint: show that f lies in the ideal ( ∂f∂x1

, . . . , ∂f∂xn

)).Euler test

(Fact: this will give the singular points in general. I don’t want to prove this, andI won’t use it.) [forme: We’ll need k to be algebraically closed.]

16.2.H. Exercise. Suppose that k is algebraically closed. Show that y2z =x3 −xz2 in P2

k is an irreducible nonsingular curve. (This is for practice.) Warning:I didn’t say char k = 0, so be careful when using the Euler test. [forme: Warning

because of the Euler test]

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16.2.I. Exercises. Find all the singular closed points of the following planecurves. Here we work over a field of characteristic 0 for convenience.

(a) y2 = x2 + x3. This is called a node.(b) y2 = x3. This is called a cusp.(c) y2 = x4. This is called a tacnode.

(I haven’t given a precise definition of a node, etc.) [forme: Here’s a false def-

inition: the Noetherian local ring (R,m) has Zariski tangent space of dimension 2;

and the natural map Sym2(m/m2) → m2/m3 has a one-dimensional kernel, which can

be interpreted as a quadratic form on m/m2, and this quadratic form should have

distinct zeros. Translation: if x and y give a basis of m/m2, then the quadratic form

Ax2 +Bxy+Cy2 should have distinct roots, i.e. B2 − 4AC 6= 0. You can see that power

series are influencing my thinking. Refer forward to where I discuss power series

later in this chapter. You may want to think about the right definition of cusp and

tacnode.) Add more singularities here One possibility: give the definition in the node, cusp, tacnode

case where you are on a nonsingular surface?]

16.2.J. Exercise. Show that the twisted cubic Projk[w, x, y, z]/(wz − xy, wy −x2, xz− y2) is nonsingular. (You can do this by using the fact that it is isomorphicto P1. I’d prefer you to do this with the explicit equations, for the sake of practice.By the way, The twisted cubic was defined in Exercise 9.1.T.)

16.2.K. Exercise. Show that the only dimension 0 Noetherian regular localrings are fields. (Hint: Nakayama.) 9

16.2.L. Exercise. Suppose (S, n) is a Noetherian local ring of dimension 0. Show

that nn = 0 for some n. Hint:√

0 = n (Spec S has only one point, and hence onlyone radical ideal!), and n is finitely generated. 10 [forme: Where should this be

placed? This gets used in the DVR section at the end. Perhaps I should place it

there. Also I may as well say that the only dimension 0 domains are also fields. ]

The case of dimension 1 is richer, and more important. We address that now.[forme: Place this! It tells us something we expect: A “manifold point” is locally

irreducible. Perhaps here is a good place to mention useful facts about regular local

rings. Other things that we might need: regular local rings have unique factorization,

hence normal. That is fact 3 in my notes.]

16.2.2. Proposition. — Every Noetherian regular local ring is an integraldomain. [forme: [E, Cor. 10.14, p. 241]] 11

Proof. Let A be the ring. We prove the result by induction on dim A. If dim A = 0,then A is a field (Exercise 16.2.K), so the result holds. Assume now that dim A =n > 0, and that the result holds for regular local rings of dimension less than n.Then as dim A > 0, m 6= m2 (or else m = 0, from which dim A = 0). Let p1, . . . , pm

be the minimal primes of A. (There are a finite number, as A is Noetherian.) ByExercise 7.5.F(c), m 6⊂ p1 ∪ · · · pn. Thus m−m2 6⊂ p1 ∪ · · · pn. Choose any elementof m − m2 − p1 − · · · − pn. By Krull’s theorem 15.6.3, dim A/(x) = dim A − 1.[Make sure I have the refined version of Krull’s theorem! We need the

9dim0regfield

10Nlddim0

11rlrdomain

282

dimension version here: If (A, m) is a local ring, then A/(x) ≥ dim A − 1.This is a special case of [E, Cor. 10.9].] Yowch, this is not immediate! Also,the dimension of the Zariski tangent space of A/(x) at the maximal ideal is n− 1[refer back to Zariski tangent space Krull]. Thus A/(x) is a regular local ring, andby the inductive hypothesis, it is an integral domain, so (x) is a prime ideal. (x)is not a minimal prime (as by construction x doesn’t lie in any minimal prime), so(x) contains some minimal prime p. We wish to show that p = 0. Now if y ∈ p

then y ∈ (x), so y = ax, so (as x /∈ p) a ∈ p. Thus xQ = Q. Thus by the secondversion of Nakayama’s lemma 15.3.2, we are done.

16.2.M. Easy Exercise. Show that a nonsingular Noetherian scheme is irre-ducible if and only if it is connected. 12 (Hint: Exercise 7.3.G.)locally integral (temp.)

16.3 Dimension 1 Noetherian regular local rings = discretevaluation rings

The case of dimension 1 is important, because if you understand how primes be-have that are separated by dimension 1, then you can use induction to prove facts inarbitrary dimension. This is one reason why Krull’s Principal Ideal Theorem 15.6.3is so useful.

A dimension 1 Noetherian regular local ring can be thought of as a “germ of acurve” (see Figure 2). Two examples to keep in mind are k[x](x) = f(x)/g(x) | x 6|g(x) and Z(5) = a/b | 5 6 |b.

Figure 2. A germ of a curve

The purpose of this section is to give a long series of equivalent definitions ofthese rings. The statement and proof of this theorem will take us most of the waythrough this section. We will eventually have six equivalent definitions, (a) through(f).

16.3.1. Theorem. — Suppose (A, m) is a Noetherian dimension 1 local ring.The following are equivalent. (a) A is regular. 13

We think of this informally as: A is the germ of a smooth curve.(b) m is principal. If A is regular, then m/m2 is one-dimensional. Choose any

element t ∈ m−m2. Then t generates m/m2, so generates m by Nakayama’s lemma.Such an element is called a uniformizer. (Warning: we needed to know that muniformizer

was finitely generated to invoke Nakayama — but fortunately we do, thanks to theNoetherian hypothesis!)

12nsNirriffconn2

13dvrtfae

283

Conversely, if m is generated by one element t over A, then m/m2 is generatedby one element t over A/m = k.

(c) All ideals are of the form mn or 0. Suppose (A, m, k) is a Noetherianregular local ring of dimension 1. Then I claim that mn 6= mn+1 for any n. Proof:Otherwise, mn = mn+1 = mn+2 = · · · . Then ∩im

i = mn. But ∩imi = (0) by

Proposition 15.10.1. Then as tn ∈ mn, we must have tn = 0. But A is a domain,so t = 0 — but t ∈ m−m2.

I next claim that mn/mn+1 is dimension 1. Reason: mn = (tn). So mn isgenerated as as a A-module by one element, and mn/(mmn) is generated as a(A/m = k)-module by 1 element, so it is a one-dimensional vector space.

So we have a chain of ideals A ⊃ m ⊃ m2 ⊃ m3 ⊃ · · · with ∩mi = (0). Wewant to say that there is no room for any ideal besides these, because “each pairis “separated by dimension 1”, and there is “no room at the end”. Proof: supposeI ⊂ A is an ideal. If I 6= (0), then there is some n such that I ⊂ mn but I 6⊂ mn+1.Choose some u ∈ I − mn+1. Then (u) ⊂ I . But u generates mn/mn+1, hence byNakayama it generates mn, so we have mn ⊂ I ⊂ mn, so we are done. Conclusion:in a Noetherian local ring of dimension 1, regularity implies all ideals are of theform mn or (0).

Conversely, suppose we have a dimension 1 Noetherian local domain that is notregular, so m/m2 has dimension at least 2. Choose any u ∈ m−m2. Then (u, m2)is an ideal, but m ( (u, m2) ( m2. We’ve thus shown that (c) is equivalent to theprevious cases.

(d) A is a principal ideal domain.

16.3.A. Exercise. Show that (d) is equivalent to (a)–(c).(e) A is a discrete valuation ring. I will now define something for you that

will be a very nice way of describing such rings, that will make precise some ofour earlier vague ramblings. We’ll have to show that this definition accords with(a)–(d) of course.

Suppose K is a field. A discrete valuation on K is a surjective homomorphismv : K∗ → Z (homomorphism: v(xy) = v(x) + v(y)) satisfying

v(x + y) ≥ min(v(x), v(y))

except if x + y = 0. discrete valuation

Here are some examples. (i) (the 5-adic valuation) K = Q, v(r) is the “powerof 5 appearing in r”, e.g. v(35/2) = 1, v(27/125) = −3.

(ii) K = k(x), v(f) is the “power of x appearing in f .”(iii) K = k(x), v(f) is the negative of the degree. This is really the same as

(ii), with x replaced by 1/x.Then 0 ∪ x ∈ K∗ | v(x) ≥ 0 is a ring. It is called the valuation ring of v.

16.3.B. Exercise. Describe the valuation rings in those two examples. (No-tice that they are familiar-looking dimension 1 Noetherian local rings. What acoincidence!)

16.3.C. Exercise. Show that 0∪x ∈ K∗ | v(x) ≥ 1 is the unique maximal idealof the valuation ring. (Hint: show that everything in the complement is invertible.)Thus the valuation ring is a local ring.

284

An integral domain A is called a discrete valuation ring if there exists adiscrete valuation v on its fraction field K = FF(A).

Now if A is a Noetherian regular local ring of dimension 1, and t is a uniformizer(generator of m as an ideal = dimension of m/m2 as a k-vector space) then any non-zero element r of A lies in some mn − mn+1, so r = tnu where u is a unit (as tn

generates mn by Nakayama, and so does r), so FF A = At = A[1/t]. So any elementof FF A can be written uniquely as utn where u is a unit and n ∈ Z. Thus we candefine a valuation v(utn) = n.

16.3.D. Exercise. Show that v is a discrete valuation.Thus (a)-(c) implies (d).Conversely, suppose (A, m) is a discrete valuation ring.

16.3.E. Exercise. Show that (A, m) is a Noetherian regular local ring of dimen-sion 1. (Hint: Show that the ideals are all of the form (0) or In = r ∈ A | v(r) ≥n, and I1 is the only prime of the second sort. Then we get Noetherianness, anddimension 1. Show that I1/I2 is generated by any element of I1 − I2.)

As a corollary:

16.3.F. Exercise. Show that there is only one discrete valuation on a discretevaluation ring.

Whenever you see a Noetherian regular local ring of dimension 1, we have avaluation on the fraction field. If the valuation of an element is n > 0, we say thatthe element has a zero of order n. If the valuation is −n < 0, we say that theelement has a pole of order n.pole

For example, consider the function (x − 2)2/(x − 3)4. What are its zeros andpoles?

So we can finally make precise the fact that 75/34 has a double zero at 5,and a single pole at 2! Also, you can easily figure out the zeros and poles ofx3(x + y)/(x2 + xy)3 on A2. Note that we can only make sense of zeros and polesat nonsingular points of codimension 1.

Definition. More generally: suppose X is a locally Noetherian scheme. Thenfor any regular codimension 1 points (i.e. any point p where OX,p is a regular localring of dimension 1), we have a valuation v. If f is any non-zero element of thefraction field of OX,p (e.g. if X is integral, and f is a non-zero element of thefunction field of X), then if v(f) > 0, we say that the element has a zero of orderv(f), and if v(f) < 0, we say that the element has a pole of order −v(f).

We aren’t yet allowed to discuss order of vanishing at a point that is not regularcodimension 1. One can make a definition, but it doesn’t behave as well as it doeswhen have you have a discrete valuation.

16.3.G. Exercise. Suppose X is an integral Noetherian scheme, and f ∈FF(Γ(X,OX))∗ is a non-zero element of its function field. Show that f has afinite number of zeros and poles. (Hint: reduce to X = Spec A. If f = f1/f2, wherefi ∈ A, prove the result for fi.)

14

[[Some notes from Sam Vandervelde. The definition of zeros and poles wasconfusing to people.

14finitezerospoles, H.L.II.6.1. From here: use primary decomposition, or finiteness of associated points.

285

1) (stalk version) the stalk at P has dimension 1, and the value of f at thisstalk is contained in the maximal ideal. (More precisely, there is an inclusion ofthis stalk in the function field, and under this identification f lies in the stalk andin fact inside the maximal ideal.)

2) (open affine version) consider any open affine Spec R in X. Then the functionfield of X can be identified with Frac(R), so write f = g/h for g,h in R via thisidentification. Then the set of minimal primes over (g) which are not minimalprimes over (h) are certainly zeros. We can deal with primes P, minimal overboth (g) and (h), if R is regular at P. After that it gets murkier. Is there a cleandefinition in the open affine version of where the zeros of f are? Maybe via primarydecomposition, I suppose... but I think there are issues here as well since theprimary components are not necessarily unique?

Remarks: at any regular codim 1 point P of X we can further define the orderof zero via the associated DVR. And poles of f correspond to zeros of 1/f , so weget pole theory for free. Finally, there are points P of codim 2 or greater on whichf vanishes when dim X > 1, but we do not speak of these as ”zeros of f” I gather.Is this right? ]]

Finally:(f) (A, m) is a unique factorization domain,(g) A is integrally closed in its fraction field K = FF(A).(a)-(e) clearly imply (f), because we have the following stupid unique factor-

ization: each non-zero element of r can be written uniquely as utn where n ∈ Z≥0

and u is a unit. Also, (f) implies (b), by the easy direction of Proposition 15.6.8.[forme: In fact, we could just have (b) ↔ (f) from Proposition 15.6.8.]

(f) implies (g), because checked earlier that unique factorization domains arealways integrally closed in its fraction field (Exercise 7.4.F).

So it remains to check that (g) implies (a)-(e). This is the only tricky part ofthis entire theorem, but it will take us less than half a page.

Proof that (g) implies (b). Suppose (A, m) is a Noetherian local domain ofdimension 1, that is integrally closed in its fraction field K = FF(A). Choose anyr ∈ A 6= 0. Then S = A/(r) is dimension 0, and is Noetherian and local, so ifn is its maximal ideal, then there is some n such that nn = 0 but nn−1 6= 0 byExercise 16.2.L. Thus mn ⊆ (r) but mn−1 6⊂ (r). Choose s ∈ mn−1 − (r). Considerx = r/s. Then x−1 /∈ A, so as A is integrally closed, x−1 is not integral over A.

Now x−1m 6⊂ m (or else x−1m ⊂ m would imply that m is a faithful A[x−1]-module, contradicting Exercise 15.3.D). But x−1m ⊂ A. Thus x−1m = A, fromwhich m = xA, so m is principal.

16.3.2. Geometry of normal Noetherian schemes. Now I’d like to discussthe geometry of normal Noetherian schemes. Suppose A is an Noetherian integrallyclosed domain. Then it is regular in codimension 1 (translation: all its codimensionat most 1 points are regular). If A is dimension 1, then obviously A is nonsingular.

Example: Spec Z[i] is nonsingular. Reason: it is dimension 1, and Z[i] is aunique factorization domain, hence its Spec is normal. [forme: Give a reference] Gaussian integers Z[i]

Remark: A (Noetherian) scheme can be singular in codimension 2 and still benormal. Example: you have shown that the cone x2 + y2 = z2 in A3 is normal[forme: Ref? PS4, problem B4] , but it is clearly singular at the origin (the Zariskitangent space is visibly three-dimensional).

286

So integral (locally Noetherian) schemes can be singular in codimension 2. Buttheir singularities turn out to be not so bad. I mentioned earlier, before we evenknew what normal schemes were, that they satisfied Hartogs’ Theorem 15.6.6, thatyou could extend functions over codimension 2 sets.

Remark: We know that for Noetherian rings we have inclusions:

regular in codimension 1 ⊃ integrally closed ⊃ unique factorization domain.[Hence for locally Noetherian schemes: R1, normal, factorial.]

Here are two examples to show you that these inclusions are strict.

16.3.H. Exercise. Let A be the subring k[x3, x2, xy, y] ⊂ k[x, y]. (The ideabehind this example: I’m allowing all monomials in k[x, y] except for x.) Show thatit is not integrally closed (easy — consider the “missing x”). Show that it is regularin codimension 1 (hint: show it is dimension 2, and when you throw out the originyou get something nonsingular, by inverting x2 and y respectively, and consideringAx2 and Ay).

16.3.I. Exercise. You have checked that k[w, x, y, z]/(wz − xy) is integrallyclosed (at least if k is algebraically closed of characteristic not 2, Exercise 7.4.I).Show that it is not a unique factorization domain. (One possibility is to do this“directly”. This might be hard — how do you know that x is irreducible ink[w, x, y, z]/(wz − xy)? Another possibility, faster but less intuitive, is to use theintermediate result that in a unique factorization domain, any height 1 prime isprincipal, Exercise 15.6.8 and considering Exercise 16.1.C.) [forme: Think this

through. In particular, there is a partial converse to that commutative algebra fact.

I may want to state this explicitly.] 15

I have earlier defined factorial, so I should use this word here.

16.3.3. Finitely generated modules over discrete valuation rings. Here issomething that will be useful later [forme: Ref!] . Recall the structure theoremfor finitely generated modules over principal ideal domains. We can write them asthe direct sum of A/(a). For DVR’s, this means that the summands are of the formA or A/mk. Hence torsion-free is the same as free. 16finitely generated mod-

ules over discrete valua-

tion rings

[forme: W] hen I note that torsion-free = locally free for finitely generated,I could mention various examples. If you give up finitely generated, it is false:FF (DV R). If I give up dimension 1, take the ideal sheaf of a point (rank jumps).If I give up nonsingularity, take the ideal sheaf of a point in a node (rank jumpsagain). Unimportant exercise: verify.

[forme: When discussing valuation rings: here is an interesting and useful exam-

ple of a valuation ring that is not a discrete valuation ring: the ring of Puisseux series

(over a field k): K = ∪n≥1k((x1/n)). The degree map is K∗ → Q given by deg xq = q.]

[primordial:[Lie algebra as tangent space at origin to a Lie group.][When discussing projective checks of nonsingularity over algebraically closed fields, have

them do the Shafarevich exercise, that the biggest linear space a nonsingular variety of degreegreater than 1 can contain has dimension roughly half.]

[Define smooth k-scheme to be a geometrically nonsingular k-scheme. In other words, itshould be nonsingular upon base-changing to the algebraic closure. “smooth = geometricallynonsingular”. Warn them that this definition will be improved upon later. Another version:

15notufd

16fgoverdvr

287

Define k-smoothness of a locally finite type k-scheme. We will later extend this to the notion togeneral morphisms. Note that the Jacobian criterion can be used here.]

[To place: Any regular local ring is a UFD; reference given in Iitaka p. 120, attributed toAuslander-Buchsbaum. Localization of regular local ring is also regular local. Reference given inIitaka p. 120, attributed to Serre. It should be in Eisenbud’s book.]

[We now know that for UFD’s, all height 1 primes are principal. We should use this here, tosay that factorial implies regular in codimension 1. We’ll say that we’ll soon see that normal andlocally Noetherian implies regular in codimension 1.]

16.4 Good facts to know about regular local rings

As it stands, I really don’t like this section. I would rather say as little as possiblehere: prove that regular local rings are integral domains. State that any localizationof a regular local ring at a prime is also a regular local ring, but make clear thatwe don’t really care. State the important fact that regular local rings are uniquefactorization domains, hence normal. Everything else can go.

Suppose (A,m) is a Noetherian regular local ring.Fact 1. Any localization of A at a prime is also a regular local ring [E, Cor. 19.14, p. 479].17

[forme: Hartshorne Theorem II.8.14A. He quotes Matsumura [2, p. 139]. Is singularlocus closed? That would imply this. True for finite type over algebraically closedfield, Hartshorne Thm 8.15.] [I state this again at 22.4.1.]

Hence to check if SpecA is nonsingular, then it suffices to check at closed points (at maximalideals). For example, to check if A3 is nonsingular, you can check at all closed points, because allother points are obtained by localizing further. (You should think about this — it is confusingbecause of the order reversal between primes and closed subsets.)

16.4.A. Exercise. Assuming fact 1, show that on a locally finite type k-scheme, you can checknonsingularity by checking at closed points. (Caution: a scheme in general needn’t have any closedpoints!) 18 [forme: There are four references to closed points: 7.4.D, 7.3.I, 7.1.E,16.4.A. I should smooth this discussion out.] [forme: Jason asks: does locallyNoetherian suffice?]

16.4.B. Less important exercise. Show that there is a nonsingular hypersurface of degreed. Show that there is a Zariski-open subset of the space of hypersurfaces of degree d. The twoprevious sentences combine to show that the nonsingular hypersurfaces form a Zariski-open set.Translation: almost all hypersurfaces are smooth. [forme: Another proof is given byBertini’s theorem ??. At this point, we need to work over an algebraically closedfield.] 19

Fact 2. (“leading terms”, proved in an important case). [Give a ref?] The naturalmap Symn(m/m2)→ mn/mn+1 is an isomorphism. Even better, the following diagram commutes:

(mi/mi+1)× (mj/mj+1)× //

∼

mi+j/mi+j+1

∼

Symi(m/m2)× Symj(m/m2)

× // Symi+j(m/m2)

Easy Exercise. Suppose (A,m, k) is a regular Noetherian local ring of dimension n. Show

that dimk(mi/mi+1) =

`n+i−1i

´.

Exercise. Show that Fact 2 also implies that (A,m) is a domain. (Hint: show that iff, g 6= 0, then fg 6= 0, by considering the leading terms.) [This is now done in 16.2.2.]

I don’t like facts pulled out of nowhere, so I want to prove it in an important case. Suppose

(A,m) is a Noetherian local ring containing its residue field k: k // A // // A/m = k . (For

17localizationofreg

18nsclosedpoints

19nonsingularhypersurface

288

example, if k is algebraically closed, this is true for all local rings of finite type k-schemes atmaximal ideals, by the Nullstellensatz. But it is not true if (A,m) = (Zp, pZp), as the residue fieldFp is not a subring of Zp.)

Suppose A is a regular of dimension n, with x1, . . . , xn ∈ A generating m/m2 as a vector space(and hence m as an ideal, by Nakayama’s lemma). Then we get a natural map k[t1, . . . , tn]→ A,taking ti to xi.

16.4.1. Theorem. — Suppose (A,m) is a Noetherian regular local ring containing its residue

field k: k // A // // A/m = k . Then k[t1, . . . , tn]/(t1, . . . , tn)m → A/mm is an isomor-

phism for all m. 20

[forme: Remark: without the regular hypothesis, we still have surjectivity. Per-haps this is worth mentioning in the proof too.]

Proof: See Section 16.5.21To interpret this better, and to use it: define the inverse limit A := lim←−A/m

n. This is

the completion of A at m. (We can complete any ring at any ideal of course.) For example, ifcompletion

S = k[x1, . . . , xn], and n = (x1, . . . , xn), then S = k[[x1, . . . , xn]], power series in n variables. Wehave a good intuition for for power series, so we will be very happy with the next result.

16.4.2. Theorem. — Suppose A contains its residue field k: k // A // // A/m = k .

Then the natural map k[[t1, . . . , tn]]→ A taking ti to xi is an isomorphism. [forme: Impliedby Hartshorne Theorem 8.21A (e).]

This follows immediately from the previous theorem, as both sides are inverse limits of thesame things.

[forme: Think of this as: functions have power series, and are determined bytheir power series. Example: In k[x, y](0,0), 1/(1 − xy) = 1 + xy + x2y2 + · · · ....]

I’ll now give some consequences.

Note that A → A. Here’s why. (Recall the interpretation of inverse limit: you can interpret

A as a subring of A/m× A/m2 × A/m3 × · · · such that if j > i, the jth element maps to the ith

factor under the natural quotient map.) What can go to 0 in A? It is something that lies in mn

for all n. But ∩imi = 0 (Proposition 15.10.1), so the map is injective. (Important note: We aren’tassuming regularity of A in this argument!!)

Thus we can think of the map A→ A as a power series expansion.[forme: [Do we care?: Cohen structure theorem. If A is a complete regular local ring of dimension

n containing some field, then A ∼= k[[x1, . . . , xn]], the ring of formal power series over the residue field

k of A. [forme: Hartshorne Thm. I.5.5A. this may follow from previous comment.]]]

This implies the “leading term” fact in this case (where the local ring contains the residuefield). (Exercise: Prove this. This isn’t hard; it’s a matter of making sure you see what thedefinitions are.) Hence in this case we have proved that A is a domain.

We go back to stating important facts that we will try not to use.Fact 3. Not only is (A,m) a domain, it is a unique factorization domain, which we have

shown implies integrally closed in its fraction field. Reference: Eisenbud Theorem 19.19, p. 483.This implies that regular schemes are normal.

[forme: Does this require Noetherianness? He uses Cor. 10.6, a domain isfactorial iff every codim 1 prime is principal.] [forme: If the ring comes from ak-variety, where k is possibly algebraically closed, Mumford shows that the local ringis a UFD in [M, III.7.T.1]. 22 He mentions that the general case, that all regular localrings are UFD’s, is due to Auslander-Buchsbaum, and he refers to Zariski-Samuel,appendix, vol. 2.]

16.4.3. Remark: factoriality is weaker than nonsingularity. There are local rings that aresingular but still factorial, so the implication factorial implies nonsingular is strict. Here are twoexamples, that we will verify later: (i) If k is algebraically closed of characteristic not 2, thenk[x1, . . . , xn]/(x2

1 + · · ·+ x2n) is a unique factorization domain, but clearly singular at the origin,

20dimk

21hat

22M.III.7.T.1

289

where n ≥ 5. [forme: [M, p. 271]] Note that we need n ≥ 5, because of our friend thenonsingular quadric!(ii) If char k 6= 2, and k does not contain a square root of −1, then k[x, y, z]/(x2 + y2 − z2) is aunique factorization, but also clearly singular at the origin. 23

16.5 +Proof of Theorem 16.4.1

Let’s now set up the proof of Theorem 16.4.1, with a series of exercises.24

16.5.A. Exercise. If S is a Noetherian ring, show that S[[t]] is Noetherian. (Hint: SupposeI ⊂ S[[t]] is an ideal. Let In ⊂ S be the coefficients of tn that appear in the elements of I forman ideal. Show that In ⊂ In+1, and that I is determined by (I0, I1, I2, . . . ).) [forme: Langpaperback Algebra, p. 147.]

16.5.1. Corollary. k[[t1, . . . , tn]] is a Noetherian local ring.

16.5.B. Exercise. Show that dimk[[t1, . . . , tn]] is dimension n. (Hint: find a chain of n + 1prime ideals to show that the dimension is at least n. For the other inequality, use Krull.)

16.5.C. Exercise. If A is a Noetherian local ring, show that A := lim←−A/mn is a Noetherian local A

ring. (Hint: Suppose m/m2 is finite-dimensional over k, say generated by x1, . . . , xn. Describe a

surjective map k[[t1, . . . , tn]]→ A.)We now outline the proof of the Theorem, as an extended exercise. (This is hastily and

informally written.)

Suppose p ⊂ A is a prime ideal. Define p ⊂ A by p/mm ⊂ A/mm. Show that p is a prime

ideal of A. (Hint: if f, g /∈ p, then let mf ,mg be the first “level” where they are not in p (i.e. the

smallest m such that f /∈ p/mm+1). Show that fg /∈ p by showing that fg /∈ p/mmf +mg+1.)

Show that if p ⊂ q, then p ⊂ q. Hence show that dim A ≥ dimA. But also dim A ≤dimm/m2 = dimA. Thus dim A = dimA.

We’re now ready to prove the Theorem. We wish to show that k[[t1, . . . , tn]]→ A is injective;we already know it is surjective. Suppose f ∈ k[[t1, . . . , tn]] 7→ 0, so we get a map k[[t1, . . . , tn]/f

surjects onto A. Now f is not a zero-divisor, so by Krull’s Principal Ideal Theorem 15.6.3, the leftside has dimension n−1. But then any quotient of it has dimension at most n−1, contradiction.

16.6 +A second proof of “Hartogs’ Theorem” for Noetheriannormal rings, and Serre’s criterion for normality

25 [forme: Matsumura Commutative Algebra p. 124, Theorem 38. Better is Atiyah-Maconald Cor. 5.22 on p. 56. Let A be a subring of a field K. Then the integralclosure of A in K is the intersection of all the valuation rings of K which contain A.Atiyah-Macdonald’s definition [of what?] is great. See also Eisenbud Cor. 11.4, p.251. The argument below is extracted from Eisenbud.]

The proof uses two lemmas.

16.6.1. Lemma. — Suppose A is a reduced Noetherian ring, and x ∈ FF(A). Then x ∈ A if andonly if the image of x in FF(Ap) belongs to Ap for every prime p associated to a nonzerodivisorin A. [forme: cite[Prop. 11.3, p. 250]e]

23mumfordprepromise

24promisedproofs

25hartogssecondproof

290

By “a prime associated to a non-zero-divisor b in A”, we mean the associated primes of theideal (b) (which are the same as the associated primes to the zero ideal of A/(b), considered asprimes of A).

Proof. Clearly if x ∈ A, then x ∈ Ap for all p, so we show the other direction. Suppose thata/b ∈ FF(A), with a, b ∈ A, b not a zerodivisor, and a/b /∈ A. We will show that x /∈ Ap for somep associated to the non-zero-divisor b. Then a /∈ (b). Let B := A/(b). Then a is not 0 in B, soby Proposition 7.5.5 (important fact (3) about associated primes, §7.5), there is some associatedprime q of B where a is not 0 in Bq. Considering q as a prime in A, we have Bq

∼= Aq/bAq, so inAq, a /∈ bAq, from which a/b /∈ Aq.

16.6.2. Lemma. — If A is a normal Noetherian domain, and b ∈ A, then for every prime p

associated to A/(b) (considered as a prime in A), pAp is a principal ideal. 26

We will prove this in a moment.First note that in Ap, dimAp/p p/p2 is generated by one element, so dimAp = codim p ≤ 1.

Furthermore, if b is not a zero divisor, then p has height precisely 1. Hence we have provedHartogs’ Theorem 15.6.6.

16.6.A. Exercise. Show that the converse to Lemma 16.6.2 is true: suppose that A is aNoetherian domain. Show that if for every prime p associated to A/(b) (considered as a prime inA), pAp is a principal ideal, then A is normal (i.e. integrally closed). (Hint: We have A = ∩Ap

in FFA, where the intersection runs over all primes p associated to various (b). Show that theintersection of normal domains with common quotient fields is normal. Use Theorem 16.3.1 (b)and (g).)

16.6.B. Exercise: Serre’s criterion that “normal = R1+S2”. Prove Serre’s criterionfor normality: 27 suppose A is a reduced domain. Show that A is normal if and only if A isSerre’s criterion for nor-

mality, n=R1+S2 regular in codimension 1, and every associated prime of a principal ideal generated by a non-zero-divisor is of height 1 (i.e. if b is a non-zero-divisor, then SpecA/(b) has no embedded points).The first hypothesis, regularity in codimension 1, is called “R1”, and the second is called “Serre’sS2 criterion”. The S2 criterion says rather precisely what is needed for normality in addition toregularity in codimension 1. [forme: WORK THIS OUT! Allen’s version of S2? Do Ineed a Noetherian hypothesis?]

16.6.C. Exercise. [Show that ’knotted plane’ does not satisfy Serre’s S2 criterion, by exhibitinga nonzerodivisor with an associated point. Add this to the index. Call this the knotted plane.]

16.6.D. Useful exercise. [Two planes meeting at a point. I should have mentioned it earlieras something that is not normal, but certainly R1. Have them show that it is not S2 directly,by producing a function which has an associated point. Tell them the function! Add this to theindex.]Proof of Lemma 16.6.2. Suppose a ∈ A is a non-zero-divisor, and p is an associated prime. Then

by Theorem 7.5.3 on uniqueness of primary decomposition, p =p

((a) : b) for some b ∈ A. AsA is reduced, p = ((a) : b). By localizing at p if necessary, we may assume that p is a maximalideal of A. (All of our constructions behave well with respect to localization. For example, thenotion of associated prime behaves well with respect to localization, by Exercise 7.5.G.) Considerp−1 := r ∈ FF(A) | rp ⊂ A. (This can be interpreted as (A : p) in the ring FF(A).) Clearlyp ⊂ p−1p ⊂ A. As p is maximal, we have only two possibilities: p−1p = p or p−1p = A.

Suppose first that p−1p = p. Choose any r ∈ p−1. Then applying Exercise 15.3.D (withS = A, R = A[r], M = p), we find that r is integral over A, so r ∈ A. As this is true for allr ∈ p−1, we conclude that p−1 = A. But as bp ⊂ (a), we have that b/a ∈ p−1 = A, from whichb ∈ (a), from which A ⊂ ((a) : b) = p, yielding a contradiction.

Thus p−1p = A. We next show that there is some r ∈ p−1 such that rp = A: as p−1p = A,there are some ai ∈ p−1 and bi ∈ p such that a1b1 + · · · + anbn = 1. At least one (ai, bi) (say

26ppp, half of cite[Thm. 11.2, p. 250]e; other half is in subsequent exercise.

27R1S2 H.T.II.8.22A; H ascribes this to Matsumura Th.39 p. 125

291

(a, b)) satisfies ab /∈ p; say ab = c where c is not in the maximal ideal p. Then c is invertible in A,and (c−1a)b = 1, from which A ⊂ (c−1a)p ⊂ A, so (c−1a)p = A. We take r = c−1a.

Thus p = (r−1) is principal.

16.7 Cohen-Macaulay rings and schemes

28

[Auslander-Buchsbaum theorem: depth plus homological dimension = dimension.]This is first used in the blow-up section. Likely first move this into nonsingularity section.

Have one section where I state facts, and a starred section where I prove them.I’ll start with a list of things I want.To figure out soon: Do we need Noetherian hypotheses? Why X Cohen-Macaulay

implies D Cohen-Macaulay?State Noetherian hypotheses at the outset.Motivation.Two definitions: depth (order of regular sequence doesn’t matter), and slicing (associated

points picture)Nonsingular is Cohen-Macaulay. Local complete intersection is Cohen-Macaulay. CM is

stalk-local.More generally, if A is Cohen-Macaulay (recall that I’ve stated that nonsingular schemes

are Cohen-Macaulay), and x1, . . . , xr ∈ m is a regular sequence, with I = (x1, . . . , xr), then thenatural map Symn I/Ir → In/In+1 is an isomorphism. You can read about this at p. 110 ofMatsumura’s Commutative Algebra. from H. This is stated in 16.7.4.

Bezout’s theorem and CM-ness. Slice a CM with an effective Cartier, and degrees add.Slice CM with effective Cartier, get CM.CM implies equidimensional.R1+CM implies normal by Serre’s criterion.Example of things that aren’t CM: non equidimensional, associated points, 2 planes.The Cohen-Macaulay/Flatness theorem: (exercise 29.13.A) Suppose π : X → Y is a

map of locally Noetherian schemes, where both X and Y are equidimensional, and Y is nonsin-gular. Show that if any two of the following hold, then the third does as well: 29

• π is flat.• X is Cohen-Macaulay.• Every fiber Xy is Cohen-Macaulay of the expected dimension.

Perhaps: π : X → Y , where the target is CM. Then f is flat iff X is CM and fibers areequidimensional.

Exercise. Extend the statement to the case where Y is a Cohen-Macaulay finite type k-scheme. (Hint: Locally Y → An with finite fibers, and Y is pure dimension n. Apply the theoremto X → An and Y → An.

Earlier notes:

• motivation: degree of projective scheme• depth: longest regular sequence; ≤ dim; independent of order• local coho criterion• open condition

16.7.1. Start of exposition.We come down to another one of these notions that ends up being very important, even

though a priori it wouldn’t have been very obvious that this is a good topic. Here is a briefindication of why we care.

It generalizes the notion of nonsingular, i.e. we allow some singularities. And it captures auseful fact about nonsingularity (repeated slicing fact), so we may as well define the more general

28s:CM

29cmf

292

kinds of rings/schemes satisfying this, and giving it a name. Then complete intersections satisfymany of the same facts as smooth things.

(Remark that it is not so easy to find non-CM things.)Then we get some nice things out of it, including: (1) Bezout’s theorem (see around 23.5.J).

(2) We’ll later get Serre duality; and our proof even for smooth things will use this nice property.(3) behaves well with flatness.

The notion is also very important in commutative algebra.Here’s basically the idea. Cohen-Macaulay at a point. You have a dimension n scheme. You

can cut down to dimension 0 by a sequence of effective Cartier divisors, i.e. single equations, notzero-divisors, = not vanishing on any associated prime. Let’s now make this precise. There is amore general notion of CM-module.

16.7.A. Exercise to place later. (word correctly) Define complete intersection. Show thata complete intersection is cut out by (f1, . . . , fr) in projective space (r = codim), then in factf1, . . . , fr generate the homogeneous ideal.

(This ends the CM introduction.)CM, depth, Koszul com-

plex Depth of an A-module M , where A is a ring: the length of the longest regular sequence. ByKrull’s Principal Ideal Theorem ??, depthM ≤ dimA. [That’s a bit of crap.]

[Depth: Used to be called homological codimension. Reason: The Auslander-BuchsbaumTheorem: depth + hd = dimension. Fact: homological dimension = 0 implies free (in some goodAuslander-Buchsbaum

theorem case?)]If A is a Noetherian local ring then we say that M is Cohen-Macaulay if depthM = dimA.

A local ring A is Cohen-Macaulay if it is Cohen-Macaulay over itself. A ring is Cohen-Macaulayif all local rings are Cohen-Macaulay.

16.7.B. Exercise. A CM scheme is normal iff it is regular in codimension 1. [forme: Answer:Serre’s criterion, in 16.6.B. Make sure I know this.] 30 Trivial consequence: For a R1surface, CM iff normal.

As promised:

16.7.C. Exercise. Show that a regular local ring is Cohen-Macaulay. [Take an element ofm−m2; show that it is prime, and that R/f is a regular local ring of dimension one less. Finishthinking this through at some point.] 31regular implies CM

Informal Translation of CM-ness: dualizating sheaf exists. (That’s not quite right I think,according to Hartshorne. But it doesn’t satisfy duality otherwise.)

This is a point-local criterion. It is an open condition, hence for finite-type k-schemes, it maybe checked on closed points.

Cohen-Macaulay schemes are locally equidimensional, and (locally) have only equidimen-sional associated primes (all associated points are minimal primes).

Criterion (at least for finite-type k-schemes): 0-dimensional schemes are Cohen-Macaulay. A

scheme is Cohen-Macaulay at a point if all of its associated points are the same dimension (“noembedded points”), and when you slice with a Cartier divisor you get a Cohen-Macaulay scheme.

Example: Curves with no 0-dimensional associated points (no “embedded points”) are Cohen-Macaulay.

16.7.2. Corollary. — Complete intersections in projective space (or indeed local completeintersections in general) are Cohen-Macaulay. 32ci, loc comp int; lci are

CM That’s one way of getting interesting CM things.Let’s now connect this to flatness. The following is extremely handy — both directions are.

16.7.3. Proposition. — Suppose π : X → Y is a morphism where Y is regular. (a) Supposeπ is flat and the fibers are Cohen-Macaulay of fixed dimension. Then X is Cohen-Macaulay. (b)Suppose X is Cohen-Macaulay, and the fibers are equidimensional. Then π is flat. 33

30partH.P.II.8.23

31regularisCM

32lciCM

33CMregularflat, used in Serre duality, H.E.III.10.9

293

Proof. [Think this through, and make sure that I have the hypotheses right. Certainly, for (a) weonly need Y to be CM; what about (b)?] For (a), use the local criterion for CM-ness. (Have anexplicit statement somewhere.) For (b), use the local criterion for flatness.

(It would be fun to use this proposition to give examples of things that aren’t CM, or ofmorphisms that aren’t flat.)

Exercise: verify that determinantal schemes are Cohen-Macaulay.Example of non-CM schemes:

• Something not locally equidimensional.• Two planes meeting at a point. (Refer elsewhere) 2 planes meeting at pt• The example of a surface with no “embedded points” that is still not Cohen-Macaulay,

from [H]; where is it?

Perhaps define Gorenstein. GorensteinNeeded in discussion of blowing up:

16.7.4. Theorem. — If A is Cohen-Macdaulay, and x1, . . . , xr ∈ m is a regular sequence (withI = (x1, . . . , xr)), then the natural map is an isomorphism.34

H quotes Matsumura CA p. 110. PROOF NEEDED.

16.7.5. Koszul notes (not yet mixed in).

16.7.6. Koszul complex. 35 K complexSuppose A is a ring, and f1, . . . , fr ∈ A, and M is an A-module. 36 We say f1, . . . , fr is a regular sequence

regular sequence for M is f1 is not a zero-divisor on M , f2 is not a zero-divisor on M/f1M , f3is not a zero-divisor on M/(f1f2M), etc. If M isn’t mentioned, it is assumed to be A. Geometrictranslation in this situation: f1 doesn’t contain any associated points of SpecA. Then this givesa Cartier divisor. Slice with this Cartier divisor.

Define the Koszul complex K•(f1, . . . , fr) as follows. K1 is a free A-module of rank r withbasis e1, . . . , er, and Kj = ∧jKr . In other words, Kj is a free module with basis

ei1 ∧ · · · ∧ eij .The boundary map d : Kj → Kj−1 is defined by

d(ei1 ∧ · · · ∧ eij ) =X

(−1)j−1fij ei1 ∧ · · · ∧ eij ∧ · · · ∧ eijwhere eij indicates that this term is to be omitted from the wedge product.

16.7.D. Exercise. Verify that K• is indeed a complex, i.e. that d2 = 0.If M is an A-module, define K•(f1, . . . , fr;M) := K•(f1, . . . , fr)⊗AM .

Thus for example, the Koszul complex with r = 1 is 0 // M×f //M // 0 , with

cohomology M/fM in degree 0 and (0 : f)M := m ∈M | fm = 0 in M in degree 1.

16.7.7. Proposition. — If the fi form a regular sequence for M , then

hi(K(f1, . . . , fr;M) = 0

for i > 0 and M/(f1, . . . , fr)M for i = 0. 37

16.7.E. Exercise. Prove this. [Hint: Here is one approach, using an easy spectral sequence.Clearly we will work by induction. Consider the double complex

e1 ∧K•(f2, . . . , fr;M)

K•(f2, . . . , fr ;M)

34cmii2, H.T.II.8.21Ae

35Koszul

36regularsequence

37H.P.III.7.10A; he refers to Matsumura Th. 43 p. 135 or Serre [11, IV.A].

294

where the vertical maps is multiplication by f1. Show that the double complex is isomorphic toK•(f1, . . . , fr;M). Compute the spectral sequence by using the vertical arrow first. The resultingcomplex (using the fact that f1 is not a zero-divisor on M) after the first step lives in one row,and is K•(f2, . . . , fr ;M/f1M).

Another approach is to describe a homotopy.]

16.7.F. Exercise. For convenience, let N = (0 : f1)M . Refine the above argument to show that

0→ HiK(M/f1)→ HiK(M)→ Hi−1K(N)→ 0.

This gives the converse to the proposition: if the higher Koszul homology vanishes, then thef ’s form a regular sequence, and the 0th homology is M modulo the f ’s. Proof by induction.If the higher Koszul homology vanishes, then: the same is true for N , from which H0K(N) =N/(f2, . . . , fr)N . This is 0, so N = (f2, . . . , fr)N . Somehow this will give N = 0. How? Dowe need some Noetherian hypothesis? Thus f1 is not a zero-divisor on M . Furthermore,the higher homology of M/f1M is 0, from which the f2, . . . , fr form a regular sequence there.Finally, by looking at the 0th homology groups, we get the rest of the statement.

Corollary: the order of the fi doesn’t matter.

16.8 Valuative criteria for separatedness

[forme: Some of this discussion should not be starred.

16.8.A. Exercise. [Reword, and move somewhere appropriate.] If X is separated,then each valuation has at most one “center”. This leads to the valuative criterion.]

Much or all of this section should skipped upon first reading.valuative criterion for

separatedness (2 ver-

sions)

Describe fact that some people love. It can be useful. I’ve never used it. But it gives goodintuition.

It is possible to verify separatedness by checking only maps from valuations rings.We begin with a valuative criterion that applies in a case that will suffice for the interests of

most people, that of finite type morphisms of Noetherian schemes. We’ll then give a more generalversion for more general readers. [The fact in the second proof trumps the one from the firstproof. But the first is more elementary. Proof possibly in the notes later.]

16.8.1. Theorem (Valuative criterion for separatedness for morphisms of finite typeof Noetherian schemes). — Suppose f : X → Y is a morphism of finite type of Noetherianschemes. Then f is separated if and only if the following condition holds. For any discretevaluation ring R with function field K, and or any diagram of the form38

(34) SpecK _

// X

f

SpecR // Y

(where the vertical morphism on the left corresponds to the inclusion R → K), there is at mostone morphism SpecR→ X such that the diagram

(35) SpecK // _

X

f

SpecR

<<

// Y

commutes. 39

A useful thing to take away from this statement is the intuition behind it. We think of SpecRas a “germ of a curve”, and SpecK as the “germ minus the origin”. Then this says that if we

38valsephyp

39valcritsepdvr, H.E.II.4.11, except he seems to do it a hard way

295

have a map from a germ of a curve to Y , and have a lift of the map away from the origin to X,then there is at most one way to lift the map from the entire germ. (A picture is helpful here.)

For example, this captures the idea of what is wrong with the map of the line with thedoubled origin over k: we take SpecR to be the germ of the affine line at the origin, and considerthe map of the germ minus the origin to the line with doubled origin. Then we have two choicesfor how the map can extend over the origin.

16.8.B. Exercise. Make this precise: show that the line with the doubled origin fails thevaluative criterion for separatedness. [Link this to 12.1.D and 12.1.K.]

16.8.2. Note on moduli spaces and the valuative criterion of separatedness. I said alittle more about separatedness of moduli spaces, for those familiar such objects. Suppose we areinterested in a moduli space of a certain kind of object. That means that there is a scheme Mwith a “universal family” of such objects over M , such that there is a bijection between familiesof such objects over an arbitrary scheme S, and morphisms S → B. (One direction of this mapis as follows: given a morphism S → B, we get a family of objects over S by pulling back theuniversal family over B.) The separatedness of the moduli space (over the base field, for example,if there is one) can be interpreted as follows. Fix a valuation ring A (or even discrete valuationring, if our moduli space of of finite type) with fraction field K. We interpret Spec intuitively as agerm of a curve, and we interpret SpecK as the germ minus the “origin” (an analogue of a smallpunctured disk). Then we have a family of objects over SpecK (or over the punctured disk), orequivalently a map SpecK → M , and the moduli space is separated if there is at most one wayto fill in the family over the origin, i.e. a family over SpecA.

? The rest of this section should be ignored upon first reading.

Proof. (This proof is more telegraphic than I’d like. I may fill it out more later. Because wewon’t be using this result later in the course, you should feel free to skip it, but you may wantto skim it.) One direction is fairly straightforward. Suppose f : X → Y is separated, and sucha diagram (34) were given. suppose g1 and g2 were two morphisms SpecR → X making (35)commute. Then g = (g1, g2) : SpecR→ X ×Y X is a morphism, with g(SpecK) contained in thediagonal. Hence as SpecK is dense in SpecR, and g is continuous, g(SpecR) is contained in theclosure of the diagonal. As the diagonal is closed (the separated hypotheses), g(SpecR) is alsocontained set-theoretically in the diagonal. As SpecR is reduced, g factors through the inducedreduced subscheme structure (§9.2.6) of the diagonal. Hence g factors through the diagonal:

SpecR // X δ // X ×Y X,

which means g1 = g2 by Exercise 12.1.J.Suppose conversely that f is not separated, i.e. that the diagonal ∆ ⊂ X×Y X is not closed.

As X ×Y X is Noetherian (X is Noetherian, and X ×Y X → X is finite type as it is obtainedby base change from the finite type X → Y ) (Exercise 9.3.D), we have a well-defined notion of

dimension [ref!]of all irreducible closed subsets, and it is bounded. Let P be a point in ∆−∆ of

largest dimension. Let Q be a point in ∆ such that P ∈ Q. (A picture is handy here.) Let Z

be the scheme obtained by giving the induced reduced subscheme structure to Q. Then P is acodimension 1 point on Z; let R′ = OZ,P be the local ring of Z at P . Then R′ is a Noetherianlocal domain of dimension 1. Let R′′ be the normalization of R. Choose any point P ′′ of SpecR′′

mapping to P ; such a point exists because the normalization morphism SpecR → SpecR′ is

surjective (normalization is an integral extension, hence surjective by the Going-up theorem,lecture 21 theorem 1.5). Give better ref in FOAG! Let R be the localization of R′′ at P ′′.Then R is a normal Noetherian local domain of dimension 1, and hence a discrete valuation ring(Section A.2.7). Let K be its fraction field. Then SpecR→ X ×Y X does not factor through thediagonal, but SpecK → X ×Y X does, and we are done. [A little more discussion needed!]

With a more powerful invocation of commutative algebra, we can prove a valuative criterionwith much less restrictive hypotheses.

16.8.3. Theorem: Valuative criterion of separatedness. — Suppose f : X → Y is aquasiseparated morphism. Then f is separated if and only if the following condition holds. For

296

any valuation ring R with function field K, and any diagram of the form (34), there is at mostone morphism SpecR→ X such that the diagram (35) commutes. 40

[I think the proof given in Hartshorne Theorem II.4.3 applies (even though the hypothesesare more restrictive).] [Hartshorne gives a ref for the general statement as EGA I, new edition,5.5.4. He also suggests EGA IV.8 for those wishing to avoid Noetherian hypotheses.]

In order to prove this, we will need a preliminary result

16.8.4. Lemma. — Hartshorne quasicompact result. Suppose X → W is a quasicompactmorphism. Then f(X) is a closed subset if and only if it is closed under specialization. 41

Observe that this is false without the quasicompact hypothesis. Example: let W = A1, andX be the disjoint union of its closed points. Then f(X) is closed under specialization, but it isnot a closed set.Proof of the valuative criterion, Theorem 16.8.3. The proof of the simpler direction is the sameas before: if f is separated, then there is at most one g making (35) commute.

In the converse direction, we also proceed similarly. Suppose conversely that f is not sep-arated, i.e. that the diagaonal ∆ ⊂ X ×Y X is not closed. Let P be a point in ∆ − ∆. Thequasiseparated hypothesis implies that δ is quasicompact, so by Lemma 16.8.4, there is someQ ∈ ∆ with P ∈ Q. Let Z be the scheme obtained by giving the induced reduced subscheme(9.2.6) structure to Q.

By Lemma A.2.3, there is a discrete valuation ring R with quotient field K dominating thelocal ring OZ,P . Precisely, this means that there is a discrete valuation on the fraction field Kof P , with discrete valuation ring R, such that the induced morphism SpecR→ SpecOZ,P sendsthe closed point of SpecR to P and the open point SpecK to Q. Then SpecR→ X ×Y X doesnot factor through the diagonal, but SpecK → X ×Y X does, and we are done as before.

16.8.C. Less important exercise (Max Lieblich): “Sections are universally closed.”.Show that X → Y is separated if and only if for any base change X ′ → Y ′, any section Y ′ → X′

is a closed immersion. [Check that this requires the valuative criterion.]

16.9 Aside: Valuative criteria of propertness

This section should be skipped on a first reading, and in a first course! [Maybe not!]Discussion with Martin Olsson Oct. 17 2007.Martin: I was looking at the section on valuative criterion for separatedness/properness

in your course notes, and I was struck by the statement that you have never usedthe valuative criterion for properness! Perhaps I misunderstood this. For example,how do you prove that Mg is proper or many other moduli spaces for that matter?Or checking that a stack has proper diagonal.

My response: I should be more specific here. It clearly gives the only reason-able *definition* of properness for stacks, and it has a very nice moduli-theoreticinterpretation. But I’ve never used it for schemes, and don’t know of a case where Iwould want to use it. You mention one thing below, that it is easy to show that P n

is proper using the valuative criterion. But I found this a hard proof (given that thisrequires the proof of the valuative criterion), while directly showing that projectivemorphisms are proper using the definition of proper (finite type: trivial. separated;trivial. universally closed: reduce to closed, and then that is just a little bit unpleas-ant) is not bad at all. It isn’t even useful for showing that the Hilbert scheme (of aprojective scheme, with given hilbert polynomial) is proper, because the construction(necessary to show finite-type-ness anyway) immediately gives projectivity.

If you could give students the valuative criterion without proof, it would lead

to easy proofs. But when reading Hartshorne, I found II.4 difficult, because I foundthe proofs of the val. criteria (of both separatedness and properness) difficult (to

40valcritsep, H.T.II.4.3

41qclosurecriterion, H.L.II.4.5

297

read and remember). And when I read (part of EGA), I had this amazing realizationthat these notions were easy, and Grothendieck’s proofs were really simple, and we’dsomehow gotten away from that... — Ravi

Martin: Probably somewhere you should note that it is very easy to prove thatPn is proper using the valuative criterion.

There is a valuative criterion for properness too. I’ve never used it personally, but it is useful,both directly, and also philosophically. I’ll make statements, and then discuss some philosophy.

[[H, Thm. II.4.7]]

16.9.1. Theorem (Valuative criterion for properness for morphisms of finite typeof Noetherian schemes). — Suppose f : X → Y is a morphism of finite type of locallyNoetherian schemes. Then f is proper if and only if the following condition holds. For anydiscrete valuation ring R with function field K, and or any diagram of the form42

(36) SpecK _

// X

f

SpecR // Y

(where the vertical morphism on the left corresponds to the inclusion R → K), there is exactlyone morphism SpecR→ X such that the diagram 43

(37) SpecK // _

X

f

SpecR

<<

// Ycommutes.

Recall that the valuative criterion for properness was the same, except that exact was replacedby at most.

In the case where Y is a field, you can think of this as saying that limits of one-parametersalways exist, and are unique. [Make a similar comment in the separated bit.]

16.9.2. Theorem (Valuative criterion of properness). — Suppose f : X → Y is aquasiseparated, finite type (hence quasicompact) morphism. Then f is proper if and only if thefollowing condition holds. For any valuation ring R with function field K, and or any diagram ofthe form (36), there is exactly one morphism SpecR→ X such that the diagram (37) commutes.

Uses: (1) intuition. (2) moduli idea: exactly one way to fill it in (stable curves). (3) motivatesthe definition of properness for stacks.

[Argument is like this, in both cases. Suppose proper. Then we get at most one by thevaluative criterion for separatedness. We need to show there is at least one.

Conversely: suppose it satisfies the criterion. Then finite type is free, and separatednessfollows by that valuative criterion. We need to show that it is universally closed. We show thatit is closed.]

16.9.3. Older notes. Valuative criterion of properness [H, Thm. II.4.7]. The hypothesesinclude finite type. We’ve proved the valuative criterion of separatedness. So we need a valuativecriterion for universal closedness.

Use the valuative criterion to prove stuff.To show scheme-theoretic image is everything, do we use valuative criterion?Chow’s Lemma could be discussed here, but it will be mentioned in Section 21.6.

]

42valprophyp

43valpropconc

Part V

Quasicoherent sheaves

[primordial:

CHAPTER 17

Quasicoherent and coherent sheaves

1

[forme: Mention that if f∗OX is of constant rank, then we get a well-defineddegree, preserved under base change. For finite morphisms.]

[forme: Here is a third fact: If π : X → Y is proper, and F is a coherent sheaf onX, then π∗F is coherent.

In particular, if X is proper over k, H0(X,F) is finite-dimensional. (This is justthe special case of the morphism X → k.) ]

[primordial: If f is finite, then f∗OX is a finite type quasicoherent sheaf of algebras (=coherent if X is locally Noetherian). ]

[primordial: We remark here that we have shown that if f : X → Y is an affine morphism,then f∗OX is a quasicoherent sheaf of algebras (a quasicoherent sheaf with the structure of analgebra “over OX”). We’ll soon reverse this process to obtain Spec of a quasicoherent sheaf ofalgebras. ]

[Place: Quasicoherent sheaves are modules over the structure sheaf over any open set. (Thisis ready to go in.)]

17.1 Toward quasicoherent sheaves: the distinguished affinebase

[forme: Done earlier: OX-modules. It is immediate to verify that they form a OX -modulecategory. In particular, we have the notion of isomorphisms.]

Schemes generalize and geometrize the notion of “ring”. It is now time to define the corre-sponding analogue of “module”, which is a quasicoherent sheaf.

One version of this notion is that of an OX -module. They form an abelian category, withtensor products (cf. Exercise 4.5.H). (That might be called a tensor category — I should check.)

We want a better one — a subcategory of OX -modules. Because these are the analogues ofmodules, we’re going to define them in terms of affine open sets of the scheme. So let’s think a bitharder about the structure of affine open sets on a general scheme X. I’m going to define what I’llcall the distinguished affine base of the Zariski topology. This won’t be a base in the sense thatyou’re used to. (For experts: it is a first example of a Grothendieck topology.) It is more akin toa base.

The open sets are the affine open subsets of X. We’ve already observed that this forms abase. But forget about that.

We like distinguished open sets SpecRf → SpecR, and we don’t really understand openimmersions of one random affine in another. So we just remember the “nice” inclusions.

Definition. The distinguished affine base of a scheme X is the data of the affine open setsand the distinguished inclusions. 2 [sheaf argument] sheaf on affine base

distinguished affine base(Remark we won’t need, but is rather fundamental: what we are using here is that we havea collection of open subsets, and some subsets, such that if we have any x ∈ U, V where U and Vare in our collection of open sets, there is some W containing x, and contained in U and V suchthat W → U and W → V are both in our collection of inclusions. In the case we are considering

1chapter:qcs

2x:sheafonaffinebase

301

302

here, this is the key Proposition in Class 9 that given any two affine open sets SpecA, SpecB inX, SpecA ∩ SpecB could be covered by affine open sets that were simultaneously distinguishedin SpecA and SpecB. This is a cofinal condition.) [forme: This is a cofinal sub-directedset of this directed set.]cofinal

We can define a sheaf on the distinguished affine base in the obvious way: we have a set (orabelian group, or ring) for each affine open set, and we know how to restrict to distinguished opensets.

Given a sheaf F on X, we get a sheaf on the distinguished affine base. You can guess wherewe’re going: we’ll show that all the information of the sheaf is contained in the information of thesheaf on the distinguished affine base.

As a warm-up: We can recover stalks. Here’s why. Fx is the direct limit lim−→(f ∈ F(U))

where the limit is over all open sets contained in U . We compare this to lim−→(f ∈ F(U)) where thelimit is over all affine open sets, and all distinguished inclusions. You can check that the elementsof one correspond to elements of the other. (Think carefully about this! It corresponds to the factthat the basic elements are cofinal in this directed system.)

17.1.A. Exercise. Show that a section of a sheaf on the distinguished affine base is determinedby the section’s germs.

17.1.1. Theorem. — 3sheaf determined by its

behavior on a nice base (a) A sheaf on the distinguished affine base F b determines a unique sheaf F , which whenrestricted to the affine base is Fb. (Hence if you start with a sheaf, and take the sheafon the distinguished affine base, and then take the induced sheaf, you get the sheaf youstarted with.)

(b) A morphism of sheaves on an affine base determines a morphism of sheaves.(c) An OX -module “on the distinguished affine base” yields an OX -module.

Proof of (a). (Two comments: this is very reminiscent of our sheafification argument. It alsotrumps our earlier theorem on sheaves on a nice base.) [forme: Cut the old proof, but keepit hidden.]

Suppose Fb is a sheaf on the distinguished affine base. Then we can define stalks.For any open set U of X, define

F(U) := (fx ∈ Fbx)x∈U : ∀x ∈ U, ∃UX with x ⊂ Ux ⊂ U, Fx ∈ Fb(Ux) | Fxy = fy∀y ∈ Uxwhere each Ux is in our base, and Fxy means “the germ of F x at y”. (As usual, those who want

to worry about the empty set are welcome to.)This is a sheaf: convince yourself that we have restriction maps, identity, and gluability,

really quite easily.I next claim that if U is in our base, that F(U) = F b(U). We clearly have a map Fb(U) →

F(U). For the map F(U)→ Fb(U): gluability exercise (a bit subtle).

These are isomorphisms, because elements of F(U) are determined by stalks, as are elementsof Fb(U).

(b) Follows as before.(c) Exercise.

17.2 Quasicoherent sheaves

4 We now define a quasicoherent sheaf. In the same way that a scheme is defined byquasicoherent sheaf“gluing together rings”, a quasicoherent sheaf over that scheme is obtained by “gluing togethermodules over those rings”. We will give two equivalent definitions; each definition is useful indifferent circumstances. The first just involves the distinguished topology.

[forme: In retrospect the order of these definitions should be reversed. But thenbe careful — I refer back to this a lot.]

3t:sheafonaffinebase

4s:qcs

303

Definition 1. An OX -module F is a quasicoherent sheaf if for every affine open SpecRand distinguished affine open SpecRf thereof, the restriction map φ : Γ(SpecR,F)→ Γ(SpecRf ,F)factors as: [forme: Make this a triangle.]

φ : Γ(SpecR,F)→ Γ(SpecR,F)f ∼= Γ(SpecRf ,F).

The second definition is more directly “sheafy”. Given a ring R and a module M , we defineda sheaf M on SpecR long ago — the sections over D(f) were Mf . [forme: Exercise 6.1.C]

Definition 2. An OX -module F is a quasicoherent sheaf if for every affine open SpecR,

F|SpecR∼= ˜Γ(SpecR,F).

(The “wide tilde” is supposed to cover the entire right side Γ(SpecR,F).) This isomorphism isas sheaves of OX -modules.

Hence by this definition, the sheaves on SpecR correspond to R-modules. Given an R-moduleM , we get a sheaf M . Given a sheaf F on SpecR, we get an R-module Γ(X,F). These operationsare inverse to each other. So in the same way as schemes are obtained by gluing together rings,quasicoherent sheaves are obtained by gluing together modules over those rings.

[forme: Fix this.] By Theorem 17.1.1, we have:Definition 2’. An OX-module on the distinguished affine base yields an OX -module.

17.2.1. Proposition. — Definitions 1 and 2 are the same. 5

Proof. Clearly Definition 2 implies Definition 1. (Recall that the definition of M [forme: justbefore exercise 6.1.C] was in terms of the distinguished topology on SpecR.) We now showthat Definition 1 implies Definition 2. We use Theorem 17.1.1. By Definition 1, the sections overany distinguished open SpecRf of M on SpecR is precisely Γ(SpecR,M)f , i.e. the sections of

˜Γ(SpecR,M) over SpecRf , and the restriction maps agree. Thus the two sheaves agree. [forme:Say this better. We don’t need Theorem 4.2, I think. Just use sheaf on distinguishedbase — not DAbase— of topology of SpecR.]

We like Definition 1 because it says that to define a quasicoherent OX -module is that wejust need to know what it is on all affine open sets, and that it behaves well under inverting singleelements.

One reason we like Definition 2 is that it glues well.

17.2.2. Proposition (quasicoherence is affine-local). — Let X be a scheme, and M an

OX-module. Then let P be the property of affine open sets thatM|SpecR∼= ˜Γ(SpecR,M). Then

P is an affine-local property.

Proof. By the Affine Communication Lemma 7.3.3, we must check two things. Clearly if SpecRhas property P , then so does the distinguished open SpecRf : if M is an R-module, then

M |SpecRf∼= Mf as sheaves of OSpecRf

-modules (both sides agree on the level of distinguished

open sets and their restriction maps).We next show the second hypothesis of the Affine Communication Lemma. Suppose we

have modules M1, . . . , Mn, where Mi is an Rfi-module, along with isomorphisms φij : (Mi)fj

→(Mj)fi

of Rfifj-modules (i 6= j; where φij = φ−1

ji ). We want to construct an M such that M gives

us Mi on D(fi) = SpecRfi, or equivalently, isomorphisms Γ(D(fi), M) ∼= Mi, with restriction

maps

Γ(D(fi), M)

Γ(D(fj), M)

Γ(D(fi), M)fj

oo ∼= // Γ(D(fj), M)fi

that agree with φij .

5quasicoherent2defs

304

We already what M should be, thanks to Theorem 6.1.1 (or more precisely Exercise 6.1.C).Consider elements of M1 × · · · ×Mn that “agree on overlaps”; let this set be M . Then

0→M →M1 × · · · ×Mn →M12 ×M13 × · · · ×M(n−1)n

is an exact sequence (where Mij = (Mi)fj∼= (Mj)fi

, and the latter morphism is the “difference”

morphism). So M is a kernel of a morphism of R-modules, hence an R-module. We show thatMi∼= Mfi

; for convenience we assume i = 1. Localization is exact, so 6

(38) 0→Mf1 →M1 × (M2)f1 × · · · × (Mn)f1 →M12 × · · · × (M23)f1 × · · · × (M(n−1)n)f1 .

[Do you see why (M1)f1 = M1?] Then by interpreting this exact sequence, you can verify thatthe kernel is M1. I gave one proof in class, and I’d like to give two proofs here. We know that∪ni=2D(f1fi) is a distinguished cover of D(f1) = SpecR1. So we have an exact sequence

0→M1 → (M1)f2 × · · · × (M1)fn → (M1)f2f3 × · · · × (M1)fn−1fn .

Put two copies on top of each other, and add vertical isomorphisms, alternating between identityand the negative of the identity:

0 //M1//

id

(M1)f2 × · · · × (M1)fn//

−id

(M1)f2f3 × · · · × (M1)fn−1fn

0 //M1

// (M1)f2 × · · · × (M1)fn// (M1)f2f3 × · · · × (M1)fn−1fn

Then the total complex of this double complex is exact as well (exercise). (The total complexis obtained as follows. The terms are obtained by taking the direct sum in each southwest-to-northeast diagonal. This is a baby case of something essential so check it, if you’ve never seen itbefore!). But this is the same sequence as (38), except Mf1 replaces M1, so we have our desiredisomorphism. [forme: This should be cut. But in any case, in this exercise, we onlyuse the fact that the rows form a complex, not that they are exact.]

Here is a second proof that the sequence7

(39) 0→M1 →M1 × (M2)f1 × · · · × (Mn)f1 →M12 × · · · × (M23)f1 × · · · × (M(n−1)n)f1

is exact. To check exactness of a complex of R-modules, it suffices to check exactness “at eachprime p”. In other words, if a complex is exact once tensored with Rp for all p, then it was exactto begin with. Now note that if N is an R-module, then (Nfi

)p is 0 if fi ∈ p, and Np otherwise.Hence after tensoring with Rp, each term in (39) is either 0 or Np, and the reader will quicklyverify that the resulting complex is exact. (If any reader thinks I should say a few words as towhy this is true, they should let me know, and I’ll add a bit to these notes. I’m beginning tothink that I should re-work some of my earlier arguments, including for example base gluabilityand base identity of the structure sheaf, in this way.)

At this point, you probably want an example. I’ll give you a boring example, and save amore interesting one for the end of the class.

Example: OX is a quasicoherent sheaf. Over each affine open SpecR, it is isomorphic themodule M = R. This is not yet enough to specify what the sheaf is! We need also to describe thedistinguished restriction maps, which are given by R → Rf , where these are the “natural” ones.

(This is confusing because this sheaf is too simple!) A variation on this theme is O⊕nX (interpretedin the obvious way). This is called a rank n free sheaf. It corresponds to a rank n trivial vectorbundle.free sheaf

Not every OX-module is a quasicoherent sheaf. One example is in the following exercise. Wewill see more examples in Exercises 9.1.R and ??; both are examples of pathologies though.

17.2.A. Exercise. (a) Suppose X = Spec k[t]. Let F be the skyscraper sheaf supported at theorigin [(t)], with group k(t). Give this the structure of an OX-module. Show that this is not aquasicoherent sheaf. (More generally, if X is an integral scheme, and p ∈ X that is not the genericpoint, we could take the skyscraper sheaf at p with group the function field of X. Except in a sillycircumstances, this sheaf won’t be quasicoherent.) [forme: Jarod warns to be careful withthis: you could give this a different qcs structure. By saying “skyscraper sheaf”, I

6earlierthing

7nextthing

305

need to endow it with an OX-module structure. You could take the sheaf on Spec k,and push it forward.](b) Suppose X = Spec k[t]. Let F be the skyscraper sheaf supported at the generic point [(0)],with group k(t). Give this the structure of an OX-module. Show that this is a quasicoherentsheaf. Describe the restriction maps in the distinguished topology of X. 8

17.2.B. Important exercise for later. Suppose X is a Noetherian scheme. Suppose F is aquasicoherent sheaf on X, and let f ∈ Γ(X,OX) be a function on X. Show that the restrictionmap resXf⊂X : Γ(X,F) → Γ(Xf ,F) (here Xf is the open subset of X where f doesn’t vanish)

is precisely localization. In other words show that there is an isomorphism Γ(X,F)f → Γ(Xf ,F)making the following diagram commute.

Γ(X,F)resXf ⊂X //

⊗RRf %%KKKKKKKKKKΓ(Xf ,F)

Γ(X,F)f

∼

88rrrrrrrrrr

All that you should need in your argument is that X admits a cover by a finite number of opensets, and that their pairwise intersections are each quasicompact. We will later rephrase this assaying that X is quasicompact and quasiseparated (Section 12.1.9). (Hint: cover by affine opensets. Use the sheaf property. A nice way to formalize this is the following. Apply the exact functor⊗RRf to the exact sequence

0→ Γ(X,F)→ ⊕iΓ(Ui,F)→ ⊕Γ(Uijk,F)

where the Ui form a finite cover of X and Uijk form an affine cover of Ui ∩ Uj .)9

17.2.C. Less important exercise. Give a counterexample to show that the above statementneed not hold if X is not quasicompact. (Possible hint: take an infinite disjoint union of affineschemes. Answer: see the hint to Ex. 20.2.B. The key idea is that infinite direct sums do notcommute with localization.) 10

For the experts: I don’t know a counterexample to this when the quasiseparated hypothesis isremoved. Using the exact sequence above, I can show that there is a map Γ(Xf ,F)→ Γ(X,F)f .

[forme: Something about visualizing quasicoherent sheaves.]

17.3 Locally free sheaves

[Place here: Hom(F ,G ⊗ L) ∼= Hom(F ⊗ L∨,G). Presumably this works with L replaced byE . Make sure they know that this works for Hom replaced by Hom. Make sure they know howtransition functions play with inverses.]

I want to show you how that quasicoherent sheaves somehow generalize the notion of vectorbundles.

(For arithmetic people: don’t tune out! Fractional ideals of the ring of integers in a numberfield will turn out to be an example of a “line bundle on a smooth curve”.)

Since this is motivation, I won’t make this precise, so you should feel free to think of this inthe differentiable category (i.e. the category of differentiable manifolds). A rank n vector bundleon a manifold M is a fibration π : V →M that locally looks like the product with n-space: everypoint of M has a neighborhood U such that π−1(U) ∼= U × Rn, where the projection map is the

8oddqcs

9e:localizeqcs, used in Ex. 20.2.1

10qccounterexample

306

obvious one, i.e. the following diagram commutes.

π−1(U)

π|π−1(U) ##F

FFFF

FFFFoo ∼= // U × Rn

projection to first factorxx

xxxx

xxx

U

This is called a trivialization over U . We also want a “consistent vector space structure”. Thusgiven trivializations over U1 and U2, over their intersection, the two trivializations should berelated by an element of GL(n) with entries consisting of functions on U1 ∩ U2.

Examples of this include for example the tangent bundle on a sphere, and the moebius stripover R1.

Pick your favorite vector bundle, and consider its sheaf of sections F . Then the sections overany open set form a real vector space. Moreover, given a U and a trivialization, the sections arenaturally n-tuples of functions of U .

U × Rn

π

U

f1,...,fn

VV

The open sets over which V is trivial forms a nice base of the topology.Motivated by this, we define a locally free sheaf of rank n on a scheme X as follows. It is a

quasicoherent sheaf that is locally, well, free of rank n. A locally free sheaf is determined by thefollowing data: a cover Ui of X, and for each i, j transition functions Tij lying in GL(n,Γ(Ui ∩Uj ,OX)) satisfying

Tii = Idn, TijTjk = Tik

(which implies Tij = T−1ji ). Given this data, we can find the sections over any open set U as

follows. Informally, they are sections of the free sheaves over each U ∩ Ui that agree on overlaps.

More formally, for each i, they are ~si =

0BB@

si1...sin

1CCA ∈ Γ(U ∩ Ui,OX)n, satisfying Tij~s

i = ~sj on

U ∩ Ui ∩ Uj .In the differentiable category, locally free sheaves correspond precisely to vector bundles (for

example, you can describe them with the same transition functions). So you should really thinkof these “as” vector bundles, but just keep in mind that they are not the “same”, just equivalentnotions.

A rank 1 vector bundle is called a line bundle. Similarly, a rank 1 locally free sheaf is calledan invertible sheaf. I’ll later explain why it is called invertible; but it is still a somewhat heinousterm for something so fundamental.

Caution: Not every quasicoherent sheaf is locally free. Easy exercise: give an example.In a few sections, we will define some operations on quasicoherent sheaves that generate

natural operations on vector bundles (such as dual, Hom, tensor product, etc.). The constructionswill behave particularly well for locally free sheaves. We will see that the invertible sheaves on Xwill form a group under tensor product, called the Picard group of X.PicX

We first make precise our discussion of transition functions. Given a rank n locally free sheafF on a scheme X, we get transition functions as follows. Choose an open cover Ui of X so that Fis a free rank n sheaf on each Ui. Choose a basis ei,1, . . . , ei,n of F over Ui. Then over Ui ∩ Uj ,for each k, ei,k can be written as a Γ(Ui ∩ Uj ,OX)-linear combination of the ej,l (1 ≤ l ≤ n), sowe get an n × n “transition matrix” Tji with entries in Γ(Ui ∩ Uj ,OX). Similarly, we get Tij ,and TijTji = TjiTij = In, so Tij and Tji are invertible. Also, on Ui ∩ Uj ∩ Uk, we readily haveTik = TijTjk: both give the matrix that expresses the basis vectors of ei,q in terms of ek,q . [Makesure this is right!]

17.3.A. Exercise. Conversely, given transition functions Tij ∈ GL(n,Γ(Ui∩Uj ,OX)) satisfyingthe cocycle condition TijTjk = Tik “on Ui ∩ Uj ∩ Uk”, describe the corresponding rank n locallyfree sheaf.

We end this section with a few stray comments.

307

[forme: Move forward to when we define invertible sheaf! We similarly definerational and regular sections of an invertible sheaf L on a scheme X. ] rational and regular sec-

tions of an invertible

sheaf

Caution: there are new morphisms between locally free sheaves, compared with what peopleusually say for vector bundles. Here is an example on A1:

0→ tk[t]→ k[t]→ k[t]/(t)→ 0.

For vector bundle people: the thing on the left isn’t a morphism of vector bundles (at least locally free sheaf, invert-

ible sheaf, rankaccording to some definitions). (If you think it is a morphism of vector bundles, then you shouldstill be disturbed, because its cokernel is not a vector bundle!)

17.3.1. Remark. Based on your intuition for line bundles on manifolds, you might hopethat every point has a “small” open neighborhood on which all invertible sheaves (or locally freesheaves) are trivial. Sadly, this is not the case. We will eventually see (XXXX) that for the curvey2 − x3 − x = 0 in A2

C, every nonempty open set has nontrivial invertible sheaves. (This will usethe fact that it is an open subset of an elliptic curve.) elliptic curve

17.3.B. Exercise (for arithmetically-minded people only — I won’t define my terms).Prove that a fractional ideal on a ring of integers in a number field yields an invertible sheaf. Showthat any two that differ by a principal ideal yield the same invertible sheaf. [Later, I should spellout what the definitions are. It is something like this. K is a finite extension of Q. OK is calledthe ring of integers of K, which I can define. If a is a subset closed under multiplication by OK ,then we say that it is a fractional ideal. We require that there is another b such that ab = OK .We can multiply these. Principal fractional ideals are of the form (r) where r ∈ K. The classgroup is the group of fractional ideals modulo K∗.] class group in number

theoryThus we have described a map from the class group of the number field to the Picard groupof its ring of integers. It turns out that this is an isomorphism. So strangely the number theoristsin this class are the first to have an example of a nontrivial line bundle.

17.3.C. Less impmortant exercise. Show that locally free sheaves on Noetherian normalschemes satisfy “Hartogs’ theorem”: sections defined away from a set of codimension at least 2extend over that set. (Hartogs’ theorem for Noetherian normal schemes is Theorem 15.6.6.) Hartogs

17.4 Quasicoherent sheaves form an abelian category

The category of R-modules is an abelian category. (Indeed, this is our motivating exampleof our notion of abelian category.) Similarly, quasicoherent sheaves form an abelian category. I’llexplain how.

When you show that something is an abelian category, you have to check many things,because the definition has many parts. However, if the objects you are considering lie in someambient abelian category, then it is much easier. As a metaphor, there are several things you haveto do to check that something is a group. But if you have a subset of group elements, it is mucheasier to check that it is a subgroup.

You can look at back at the definition of an abelian category, and you’ll see that in order tocheck that a subcategory is an abelian subcategory, you need to check only the following things:

(i) 0 is in your subcategory(ii) your subcategory is closed under finite sums(iii) your subcategory is closed under kernels and cokernels

In our case of quasicoherent sheaves ⊂ OX -modules, the first two are cheap: 0 is certainlyquasicoherent, and the subcategory is closed under finite sums: if F and G are sheaves on X, and

over SpecR, F ∼= M and G ∼= N , then F ⊕ G = M ⊕N , so F ⊕ G is a quasicoherent sheaf.We now check (iii). Suppose α : F → G is a morphism of quasicoherent sheaves. Then on

any affine open set U , where the morphism is given by β : M → N , define (kerα)(U) = ker β and(cokerα)(U) = coker β. Then these behave well under inversion of a single element: if

0→ K →M → N → P → 0

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is exact, then so is

0→ Kf →Mf → Nf → Pf → 0,

from which (ker β)f ∼= ker(βf ) and (coker β)f ∼= coker(βf ). Thus both of these define quasicoher-ent sheaves. Moreover, by checking stalks, they are indeed the kernel and cokernel of α. Thus thequasicoherent sheaves indeed form an abelian category.

As a side benefit, we see that we may check injectivity, surjectivity, or exactness of a morphismof quasicoherent sheaves by checking on an affine cover.

Warning: If 0 → F → G → H → 0 is an exact sequence of quasicoherent sheaves, then forany open set

0→ F(U)→ G(U)→H(U)

is exact, and we have exactness on the right is guaranteed to hold only if U is affine. (To set youup for cohomology: whenever you see left-exactness, you expect to eventually interpret this as astart of a long exact sequence. So we are expecting H1’s on the right, and now we expect thatH1(SpecR,F) = 0. This will indeed be the case.)

17.4.A. Exercise. You can check exactness of a sequence of quasicoherent sheaves on an affinecover. (In particular, taking sections over an affine open SpecR is an exact functor from thecategory of quasicoherent sheaves on X to the category of R-modules. Recall that taking sectionsis only left-exact in general, see Section 4.5.C.) Similarly, you can check surjectivity on an affinecover (unlike sheaves in general). 11

[This exercise slightly repeats the previous couple of paragraphs, so smooth this out.]

17.4.B. Exercise. Suppose 0 → F ′ → F → F ′′ → 0 is a short exact sequence of locally freesheaves onX. Suppose U = SpecA is an affine open set where F ′, F ′′ are free, say F ′|SpecA = Ra,

F ′′|SpecA = Rb. Show that F is also free, and that 0→ F ′ → F → F ′′ → 0 can be interprted as

coming from the tautological exact sequence 0 → Ra → Ra+b → Rb → 0 Show that given suchan open cover, the transition matrices of F may be interpreted as block upperdiagonal matrices,where thet top a×a blok are transition matrices for F ′, and the bottom b×b blocks are transitionmatrices for F ′′. [Perhaps link this to the Sym problem.] [Mention that if the last two are locallyfree and finite rank, over a Noetherian ring, then so is the first [ref?].]

17.5 Module-like constructions on quasicoherent sheaves

In a similar way, basically any nice construction involving modules extends to quasicoherentsheaves.

As an important example, we consider tensor products. Exercise. If F and G are qua-sicoherent sheaves, show that F ⊗ G is given by the following information: If SpecR is anaffine open, and Γ(SpecR,F) = M and Γ(SpecR,G) = N , then Γ(SpecR,F ⊗ G) = M ⊗ N ,and the restriction map Γ(SpecR,F ⊗ G) → Γ(SpecRf ,F ⊗ G) is precisely the localizationmap M ⊗R N → (M ⊗R N)f ∼= Mf ⊗Rf

Nf . (We are using the algebraic fact that that

(M ⊗R N)f ∼= Mf ⊗RfNf . You can prove this by universal property if you want, or by us-

ing the explicit construction.) [forme: algebraic glossary?]Note that thanks to the machinery behind the distinguished affine base, sheafification is taken

care of.For category-lovers: this makes the category of quasicoherent sheaves into a monoid. [Tech-

nically no! Choose a skeleton.]Re-arrange this an the next section. I made a special section for locally free

sheaves in response to a comment from Justin, but I now regret it.

17.5.A. Exercise. If F and G are locally free sheaves, show that F ⊗G is locally free. [forme:H Ex. II.5.16(a); add to refs] (Possible hint for this, and later exercises: check on sufficientlysmall affine open sets.)

11surjectivityonaffine

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17.5.B. Exercise. (a) Tensoring by a quasicoherent sheaf is right-exact. More precisely, if F isa quasicoherent sheaf, and G′ → G → G′′ → 0 is an exact sequence of quasicoherent sheaves, thenso is G′ ⊗ F → G ⊗F → G′′ ⊗ F → 0 is exact.(b) Tensoring by a locally free sheaf is exact. More precisely, if F is a locally free sheaf, and G ′ →G → G′′ is an exact sequence of quasicoherent sheaves, then then so is G ′⊗F → G⊗F → G′′⊗F .(c) The stalk of the tensor product of quasicoherent sheaves at a point is the tensor product ofthe stalks.

Note: if you have a section s of F and a section t of G, you get a section s⊗ t of F ⊗ G. Ifeither F or G is an invertible sheaf, this section is denoted st.

We now describe other constructions.

17.5.C. Exercise. Sheaf Hom Hom is quasicoherent, and is what you think it might be.(Describe it on affine open sets, and show that it behaves well with respect to localization withrespect to f . To show that HomA(M,N)f ∼= HomAf

(Mf ,Nf ), take a “partial resolution” Aq →Ap → M → 0, and apply Hom(·,N) and localize.) (Hom was defined earlier in Section 4.3.B.) sheaf Hom (Hom) of qcsShow that Hom is a left-exact functor in both variables.

Definition. Hom(F ,OX) is called the dual of F , and is denoted F∨. dual

17.5.D. Exercise. The direct sum of quasicoherent sheaves is what you think it is.

17.6 Some notions especially relevant for locally free sheaves

Exercise. Show that if F is locally free then F∨ is locally free, and that there is a canonicalisomorphism (F∨)∨ ∼= F . (Caution: your argument showing that if there is a canonical isomor-phism (F∨)∨ ∼= F better not also show that there is a canonical isomorphism F∨ ∼= F ! We’ll seean example soon of a locally free F that is not isomorphic to its dual. The example will be the linebundle O(1) on P1.) [forme: One possibility: look at transition functions. Another:show that the distinguished restriction maps behave the same. I think I should adda finite rank hypothesis, and mention that it is unnecessary.]

Remark. This is not true for quasicoherent sheaves in general, although your argument willimply that there is always a natural morphism F → (F∨)∨. Quasicoherent sheaves for which thisis true are called reflexive sheaves. We will not be using this notion. Your argument may also reflexive sheaflead to a canonical map F ⊗F∨ →OX . This could be called the trace map — can you see why?

17.6.A. Exercise. Given transition functions for the locally free sheaf F , describe the transitionfunctions for the locally free sheaf F∨. Note that if F is rank 1 (i.e. locally free), the transitionfunctions of the dual are the inverse of the transition functions of the original; in this case,F ⊗ F∨ ∼= OX .

17.6.B. Exercise. If F and G are locally free sheaves, show that F ⊗G and Hom(F ,G) are bothlocally free.

17.6.C. Exercise. Show that the invertible sheaves on X, up to isomorphism, form an abeliangroup under tensor product. This is called the Picard group of X, and is denoted PicX. (For PicXarithmetic people: this group, for the Spec of the ring of integers R in a number field, is the classgroup of R.) class group

For the next exercises, recall the following. If M is an A-module, then the tensor algebraT ∗(M) is a non-commutative algebra, graded by Z≥0, defined as follows. T 0(M) = A, Tn(M) =M ⊗A · · · ⊗A M (where n terms appear in the product), and multiplication is what you expect.The symmetric algebra Sym∗M is a symmetric algebra, graded by Z≥0, defined as the quotient ofT ∗(M) by the (two-sided) ideal generated by all elements of the form x⊗y−y⊗x for all x, y ∈M .

Thus SymnM is the quotient of M ⊗ · · · ⊗M by the relations of the form m1 ⊗ · · · ⊗mn −m′1 ⊗· · · ⊗m′n where (m′1, . . . ,m

′n) is a rearrangement of (m1, . . . ,mn). The exterior algebra ∧∗M is

defined to be the quotient of T ∗M by the (two-sided) ideal generated by all elements of the formx⊗y+y⊗x for all x, y ∈M . Thus ∧nM is the quotient of M⊗· · ·⊗M by the relations of the form

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m1 ⊗ · · · ⊗mn − (−1)sgnm′1 ⊗ · · · ⊗m′n where (m′1, . . . ,m′n) is a rearrangement of (m1 , . . . ,mn),

and the sgn is even if the rearrangement is an even permutation, and odd if the rearrangement isan odd permutation. (It is a “skew-commutative” A-algebra.) It is most correct to write T ∗A(M),Sym∗A(M), and ∧∗A(M), but the “base ring” A is usually omitted for convenience.tensor algebra, symm al-

gebra, exterior algebra17.6.D. Exercise. If F is a quasicoherent sheaf, then define the quasicoherent sheaves T nF ,Symn F , and ∧nF . If F is locally free of rank m, show that TnF , Symn F , and ∧nF are locallyfree, and find their ranks.

You can also define the sheaf of non-commutative algebras T ∗F , the sheaf of algebras Sym∗ F ,and the sheaf of skew-commutative algebras ∧∗F .cokernels, images, ten-

sor products of quasico-

herent sheaves17.6.E. Important exercise. If 0 → F ′ → F → F ′′ → 0 is an exact sequence of locally freesheaves, then for any r, there is a filtration of Symr F :

Symr F = F 0 ⊇ F 1 ⊇ · · · ⊇ F r ⊃ F r+1 = 0

with quotients

F p/F p+1 ∼= (Symp F ′)⊗ (Symr−p F ′′)for each p. [forme: Ex. II.5.16c. I’ve copied the wording.] (Possible hint for this, andthe corresponding wedge problem. It suffices to consider small enough affine open sets, where F ′,F , F ′′ are free, adn to show that your construction behaves well with respect to localization atan element f ∈ R. In such an open set, the sequence is 0→ Rp → Rp+q → Rq → 0. Let e1, . . . ,en be the standard basis of Rn, and f1, . . . , fq be the the standard basis of Rq . Let e′1, . . . , e′pbe denote the images of e1, . . . , ep in Rp+q . Let f ′1, . . . , f ′q be any lifts of f1, . . . , fq to Rp+q .

Note that f ′i is well-defined modulo e′1, . . . , e′p. Note that

Syms F|SpecR∼= ⊕si=0 Symi F ′|SpecR ⊗OSpec R

Syms−i F ′′|SpecR.

Show that Fp := ⊕si=p Symi F ′|SpecR ⊗OSpec RSyms−i F ′′|SpecR gives a well-defined (locally

free) subsheaf that is independent of the choices made, e.g. of the basis e1, . . . , ep (this is inGLp(R)), f1, . . . , fq (this is in GLq(R)), and the lifts f ′1, . . . , f ′q .)

17.6.F. Exercise. Suppose F is locally free of rank n. Then ∧nF is called the determinant(line) bundle. Show that ∧rF × ∧n−rF → ∧nF is a perfect pairing for all r. [forme: Ex.determinant bundleII.5.16(b)]

17.6.G. Exercise. If 0→ F ′ → F → F ′′ → 0 is an exact sequence of locally free sheaves, thenfor any r, there is a filtration of ∧rF :

∧rF = F 0 ⊇ F 1 ⊇ · · · ⊇ F r ⊃ F r+1 = 0

with quotients

F p/F p+1 ∼= (∧pF ′)⊗ (∧r−pF ′′)for each p. In particular, detF = (detF ′)⊗ (det F ′′). [forme: Ex. II.5.16d. I’ve copied thewording. Kirsten notes that we need only F ′′ is locally free; the other two can bemerely quasicoherent. The uses ta lemma from Lang’s 3rd edition, Prop. 1.4, p. 736:M and N are R-modules. then the natural homomorphisms ∧iM,∧iN → ∧i(M ⊕ N)and ∧i(M ⊕ N) ⊗ ∧r−i(M ⊕ N) → ∧r(M ⊕ N) combine to form a natural isomorphism∧r(M ⊕N) ∼= ⊕ri=0 ∧iM ⊗ ∧r−iN . Presumably the same is true with Sym.]

17.6.H. Exercise. Determinant line bundles behave well in exact sequences for finite rankcoherent sheaves. Remark: we can extend this to coherent sheaves; make a link to that.12

17.6.I. Exercise (torsion-free sheaves). An R-module M is torsion-free if rm = 0 impliesr = 0 or m = 0. Show that this satisfies the hypotheses of the affine communication lemma.Hence we make a definition: a quasicoherent sheaf is torsion-free if for one (or by the affinecommunication lemma, for any) affine cover, the sections over each affine open are torsion-free.By definition, “torsion-freeness is affine-local”. Show that a quasicoherent sheaf is torsion-free if allits stalks are torsion-free. Hence “torsion-freeness” is “stalk-local.” (“Stalk-locality” was definedtorsion-free quasicoher-

ent sheaf in Section 7.4.1.) [This exercise is wrong! “Torsion-freeness” is should be defined as “torsion-free

12detexact

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stalks” — it is (defined as) a “stalk-local” condition. Here is a better exercise. Show that if Mis torsion-free, then so is any localization of M . In particular, Mf is torsion-free, so this notionsatisfies half the hypotheses of the affine communication lemma. Also, Mp is torsion-free, so this

implies that M is torsion-free. Find an example on a two-point space showing that R might not

be torsion-free even though OSpecR = R is torsion-free.][I still haven’t defined torsion sheaf and torsion section of a sheaf. I should do this around

here, for quasicoherent sheaves, and make a comment about this for OX -modules in general.] torsion sheaf, torsion

section of a sheaf17.6.J. Exercise. Show that torsion-free coherent sheaves on a smooth curve are locally free.This uses stuff from 16.3.3. Show that any coherent sheaf is a direct sum of a torsion free sheafand a torsion sheaf in a unique way. (Is that right? Does it work in higher-dimension?) Show

that any subsheaf of a locally free sheaf is also locally free (over a nonsingular curve).

17.7 Finiteness conditions on quasicoherent sheaves: finitelygenerated quasicoherent sheaves, and coherent sheaves

17.7.1. [Place: At least if A is Noetherian, HomA(M,N) is coherent if M and N are coherent.Used. 13 In particular, the dual of a coherent sheaf is coherent. Why is this true? ]

There are some natural finiteness conditions on an A-module M . I will tell you three. In thecase when A is a Noetherian ring, which is the case that almost all of you will ever care about,they are all the same.14

The first is the most naive: a module could be finitely generated. In other words, there is asurjection Ap →M → 0. finitely generated

The second is reasonable too: it could be finitely presented. In other words, it could have afinite number of generators with a finite number of relations: there exists a finite presentation finite presentation

Aq → Ap →M → 0.

The third is frankly a bit surprising, and I’ll justify it soon. We say that an A-module M iscoherent if (i) it is finitely generated, and (ii) whenenver we have a map Ap →M (not necessarilysurjective!), the kernel is finitely generated. coherent sheaf

Clearly coherent implies finitely presented, which in turn implies finitely generated.

17.7.2. Proposition. — If A is Noetherian, then these three definitions are the same.15

Preparatory facts. [Perhaps move this back with facts to know about Noetherian rings, near5.5.K.] If R is any ring, not necessarily Noetherian, we say an R-module is Noetherian if it satisfiesthe ascending chain condition for submodules. Thus for example A is a Noetherian ring if and onlyif it is a Noetherian A-module. Exercise. Show that M Noetherian implies that any submoduleof M is a finitely generated R-module. Exercise. If 0→M ′ →M →M ′′ → 0 is exact, then M ′

and M ′′ are Noetherian if and only if M is Noetherian. (Hint: Given an ascending chain in M , weget two simultaneous ascending chains in M ′ and M ′′.) [This is Exercise 5.5.K. People sometimes

did it in a complicated way. They should use the following lemma: if M ′ // Mφ // M ′′

is exact, and N,N ′ ⊂M , and “N ∩M ′ = N ′∩M ′” and φ(N) = φ(N ′), then N = N ′.] Exercise.Show that if A is a Noetherian ring, then An is a Noetherian A-module. Exercise. Show that ifA is a Noetherian ring and M is a finitely generated A-module, then M is a Noetherian module.

Proof. Clearly both finitely presented and coherent imply finitely generated.

Suppose M is finitely generated. Then take any Apα // M . kerα is a submodule of a

finitely generated module over A, and is thus finitely generated. (Here’s why submodules of finitelygenerated modules over Noetherian rings are also finitely generated: Show it is true for M = Rn

— this takes some inspiration. Then given N ⊂ N , consider Rn M , and take the submodule

13homcoh

14s:finitetype

15Ncohfg

312

corresponding to N .) Thus we have shown coherence. By choosing a surjective Ap →M , we getfinite presentation.

Hence almost all of you can think of these three notions as the same thing.

17.7.3. Lemma. — The coherent A-modules form an abelian subcategory of the category ofA-modules.

I will prove this in the case where A is Noetherian, but I’ll include a series of short exercisesin the notes that will show it in general.Proof if A is Noetherian. Recall that we have four things to check (see our discussion earliertoday). We quickly check that 0 is finitely generated (=coherent), and that if M and N arefinitely generated, then M ⊕ N is finitely generated. Suppose now that f : M → N is a map offinitely generated modules. Then coker f is finitely generated (it is the image of N), and ker f istoo (it is a submodule of a finitely generated module over a Noetherian ring).

Easy Exercise (only important for non-Noetherian people). Show A is coherent (asan A-module) if and only if the notion of finitely presented agrees with the notion of coherent.

I want to say a few words on the notion of coherence. There is a good reason for thisdefinition — because of this lemma. There are two sorts of people who should care. Complexgeometers should care. They consider complex-analytic spaces with the classical topology. Onecan define the notion of coherent OX -module in a way analogous to this. You can then show thatthe structure sheaf is coherent, and this is very hard. (It is called Oka’s theorem, and takes alot of work to prove.) I believe the notion of coherence may have come originally from complexOka’s theoremgeometry.

The second sort of people who should care are the sort of arithmetic people who sometimesare forced to consider non-Noetherian rings. (For example, for people who know what they are,

the ring of adeles is non-Noetherian.)Warning: it is common in the later literature to define coherent as finitely generated. It’s

possible that Hartshorne does this. Please don’t do this, as it will only cause confusion. (In fact,if you google the notion of coherent sheaf, you’ll get this faulty definition repeatedly.) I will tryto be scrupulous about this. Besides doing this for the reason of honesty, it will also help yousee what hypotheses are actually necessary to prove things. And that always helps you rememberwhat the proofs are —- in other words, why things are true.

17.7.A. Exercise. If f ∈ A, show that if M is a finitely generated (resp. finitely presented,coherent) A-module, thenMf is a finitely generated (resp. finitely presented, coherent) Af -module.16 [This problem also appears below in the “coherence” section.] [First two are easy. The third iscomplicated by the “for al” in the definition. Possible hint: Hom of quasicoherent is quasicoherent.]

Exercise. If (f1, . . . , fn) = A, and Mfiis finitely generated (resp. coherent) Afi

-module forall i, then M is a finitely generated (resp. coherent) A-module.

I’m not sure if that exercise is even true for finitely presented. That’s one of several reasonswhy I think that “finitely presented” is a worse notion than coherence.

Definition. A quasicoherent sheaf F is finite type (resp. coherent) if for every affine openSpecR, Γ(SpecR,F) is a finitely generated (resp. coherent) R-module.finite type, coherent

Thanks to the affine communication lemma, and the two previous exercises, it suffices tocheck this on the open sets in a single affine cover. [Side remark: finite type does not implyO⊕n → F → 0. Examples? Rason: quasicompactness; also lack of any global sections. This is agood excuse to mention “generated by global sections”.]

17.8 +Coherent modules over non-Noetherian rings

[Remark: stalks of coherent etc. are coherent. But this isn’t used, so don’t mention it.]Here are some notes on coherent modules over a general ring. Read this only if you really

want to!

16localizefg

313

[Show that the hom’s from one to another are also coherent? Is that even true?]Suppose A is a ring. We say an A-module M is finitely generated if there is a surjection

An → M → 0. We say it is finitely presented if there is a presentation Am → An → M → 0.We say M is coherent if (i) M is finitely generated, and (ii) every map An → M has a finitelygenerated kernel. The reason we like this third definition is that coherent modules form an abeliancategory.

Here are some quite accessible problems working out why these notions behave well.1. Show that coherent implies finitely presented implies finitely generated.2. Show that 0 is coherent.

Suppose for problems 3–9 that

(40) 0→M → N → P → 0

is an exact sequence of A-modules. In thise series of problems, we will show that if two of (40)are coherent, the third is as well, whiich will prove very useful.Hint ?. Here is a hint which applies to several of the problems: try to write

0 // Ap //

Ap+q //

Aq //

0

0 //M // N // P // 0

and possibly use the Snake Lemma 3.8.5.3. Show that N finitely generated implies P finitely generated. (You will only need right-exactnessof (40).)4. Show that M , P finitely generated implies N finitely generated. (Possible hint: ?.) (You willonly need right-exactness of (40).)5. Show that N,P finitely generated need not imply M finitely generated. (Hint: if I is an ideal,we have 0→ I → A→ A/I → 0.)6. Show that N coherent, M finitely generated implies M coherent. (You will only need left-exactness of (40).)7. Show that N , P coherent implies M coherent. Hint for (i):

Aq

!!CCC

CCCC

C

Ap

BBB

BBBB

B

0 // M

// N

// P //

???

????

0

0 0 0

(You will only need left-exactness of (40).)8. Show that M finitely generated and N coherent implies P coherent. (Hint for (ii): ?.) [forme:Joe Rabinoff: you will only need right-exactness of (40).]9. Show that M , P coherent implies N coherent. (Hint: ?.) [forme: I had earlier written:“You will only need right-exactness of (40),” but I think I should have written it forthe previous problem; Joe R thinks it is essential.]10. Show that a finite direct sum of coherent modules is coherent.11. Suppose M is finitely generated, N coherent. Then if φ : M → N is any map, then show thatImφ is coherent.12. Show that the kernel and cokernel of maps of coherent modules are coherent.

At this point, we have verified that coherent A-modules form an abelian subcategory of the

category of A-modules. (Things you have to check: 0 should be in this set; it should be closedunder finite sums; and it should be closed under taking kernels and cokernels.)13. Suppose M and N are coherent submodules of the coherent module P . Show that M + Nand M ∩N are coherent. (Hint: consider the right map M ⊕N → P .)

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14. Show that if A is coherent (as an A-module) then finitely presented modules are coherent. (Ofcourse, if finitely presented modules are coherent, then A is coherent, as A is finitely presented!)15. If M is finitely presented and N is coherent, show that Hom(M,N) is coherent. (Hint: Homis left-exact in its first entry.)16. If M is finitely presented, and N is coherent, show that M ⊗N is coherent.17. If f ∈ A, show that if M is a finitely generated (resp. finitely presented, coherent) A-module,then Mf is a finitely generated (resp. finitely presented, coherent) Af -module. (Hint: localizationis exact.) [This problem appears earlier as well, as Exercise 17.7.A.]18. Suppose (f1, . . . , fn) = A. Show that if Mfi

is finitely generated for all i, then M is too.(Hint: Say Mfi

is generated by mij ∈ M as an Afi-module. Show that the mij generate M . To

check surjectivity ⊕i,jA→M , it suffices to check “on D(fi)” for all i.)19. Suppose (f1, . . . , fn) = A. Show that if Mfi

is coherent for all i, then M is too. (Hint:

if φ : A2 → M , then (kerφ)fi= ker(φfi

), which is finitely generated for all i. Then apply theprevious exercise.)

[forme: Old hint: (Hint for (ii): say Aa → M . Then for each i we have

A⊕rifi

// A⊕afi// Mfi

// 0 . We can arrange this so that the generators of

A⊕rifi

map to (images of) elements of A⊕a in A⊕afi. Then consider A⊕

P

riα // Aa

β // M .

To check if the image of α surject onto the kernel of β, it suffices to check in D(fi)for all i.)]20. Show that the ring A := k[x1, x2, . . . ] is not coherent over itself. (Hint: consider A→ A withx1, x2, . . . 7→ 0.) Thus we have an example of a finitely presented module that is not coherent;a surjection of finitely presented modules whose kernel is not even finitely generated; hence anexample showing that finitely presented modules don’t form an abelian category.Questions I still want to figure out, but that I haven’t thought about much.

A moral reason why finitely presented is worse than coherent: “there exists” is worse than“for all”.

[What is an easy example of a module that is finitely generated but not finitely presented?The obvious candidate example is R = k[x1, . . . ], I = (x1, . . . ), M = R/I. Max says that this isdiscussed in Matsumura.

Suppose (f1, . . . , fn) = A. If Mfiis finitely presented for all i, is M too? (If not, then the

right notion is of a locally finitely presented sheaf. This is equivalent to the stalks being finitelypresented.)]

17.9 Support of a sheaf

[Support behaves well with respect to localization?]Suppose F is a sheaf of abelian groups (resp. OX -module) on a topological space X (resp.

ringed space (X,OX)). Define the support of a section s of F to be

Supp s = p ∈ X | sp 6= 0 in Fp.I think of this as saying where s “lives”. Define the support of F as

SuppF = p ∈ X | Fp 6= 0.It is the union of “all the supports of sections on various open sets”. I think of this as sayingwhere F “lives”. [forexperts: In analysis, would you take the closure of this locus?]

17.9.A. Exercise. 17 The support of a finite type quasicoherent sheaf on a scheme is a closedsubset. (Hint: Reduce to an affine open set. Choose a finite set of generators of the correspondingmodule.) [forme: Check H ref: Ex. II.5.6(c) and/or II.5.8(a)?] Show that the supportof a quasicoherent sheaf need not be closed. (Hint: If A = C[t], then C[t]/(t − a) is an A-modulesupported at a. Consider ⊕a∈CC[t]/(t−a). Warning: this example won’t work if ⊕ is replaced byQ

, so be careful!) [Do I just need locally finitely generated? Yes.] [Support was defined in 4.6.G.Move this stuff further back.]

17supportcoherentclosed

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[Caution. where the germs are nonzero, not where the value is nonzero (in which case onlymakes sense sometimes).]

17.10 Rank of a finite type sheaf at a point

The rank F of a finite type sheaf at a point p is dimFp/mFp where m is the maximal ideal rankcorresponding to p. [Say why “finite type” is required.] More explicitly, on any affine set SpecAwhere p = [p] and F(SpecA) = M , then the rank is dimA/p Mp/pMp. By Nakayama’s lemma,this is the minimal number of generators of Mp as an Ap-module.

If F is quasicoherent (not necessarily finite type), then Fp/mFp can be interpreted as thefiber of the sheaf at the point. A section of F over an open set containing p can be said to takeon a value at that point, which is an element of Fp/mFp. value of a quasicoherent

sheaf at a point17.10.A. Exercise. 18

(a) If m1, . . . , mn are generators at P , they are generators in an open neighborhood of P .

(Hint: Consider cokerAn(f1,...,fn) //M and Exercise 17.9.A.)

(b) Show that at any point, rank(F ⊕ G) = rank(F) + rank(G) and rank(F ⊗ G) =rankF rankG at any point. (Hint: Show that direct sums and tensor products com-mute with ring quotients and localizations, i.e. (M ⊕N)⊗R (R/I) ∼= M/IM ⊕N/IN ,(M ⊗R N)⊗R (R/I) ∼= (M ⊗R R/I) ⊗R/I (N ⊗R R/I) ∼= M/IM ⊗R/I N/IM , etc.)

(c) Show that rank is an upper semicontinuous function on X. (Hint: Generators at P are rank of cs is uscgenerators nearby.)19

Note that this definition is consistent with the notion of rank of a locally free sheaf. In thatcase, the rank is a (locally) constant function of the point. The converse is sometimes true, as isshown in Exercise 17.10.B below.

[Should I say somewhere that a coherent sheaf is locally free if all of its stalks are free?Possibly even finite type?]

17.10.B. Important Hard Exercise. (a) If X is reduced, F is coherent, and the rank isconstant, show that F is locally free. (Hint: choose a point p ∈ X, and choose generatorsof the stalk Fp. Let U be an open set where the generators are sections, so we have a map

φ : O⊕nU → F|U . The cokernel and kernel of φ are supported on closed sets by Exercise 17.9.A.Show that these closed subsets don’t include p. Make sure you use the reduced hypothesis!) Thuscoherent sheaves are locally free on a dense open set. [forme: Note that there are noNoetherian hypotheses! This is a bit harder! This is used in the proof of Grauert’stheorem 29.8.1. Here is the argument. We seem to need only that F is finite type.Preliminaries (including 17.10.A(a): we can assume X is affine, say SpecA, so theclosed points are dense. We will show it in a neighborhood of a closed point [m].Choose m1, . . . , mn generators of M/mM , and lift them to elements of M . Then theygenerate M , by Nakayama. Let φ : An → M with (r1, . . . , rn) 7→P

rimi. Let K be thecokernel, which is finitely generated. Then Km = 0 (because ⊗Am is right-exact), sothere is an f ∈ A such that Kf = 0 (take the product of the annihilators of a finitegenerating set of K). Replace A by Af . We now have that coker φ = 0, and we wantto prove kerφ = 0. Otherwise, say (r1, . . . , rn) is in the kernel, with r1 6= 0. As r1 6= 0,there is some p where r1 /∈ p — here we use the reduced hypothesis. Then r1 isinvertible in Ap, so Mp has fewer than n generators, contradicting the constancy ofrank.](b) Show that this can be false if X is not reduced. (Hint: Spec k[x]/x2, M = k.) 20

You can use the notion of rank to help visualize finite type sheaves, or even quasicoherentsheaves. (We discussed first finite type sheaves on reduced schemes. We then generalized toquasicoherent sheaves, and to nonreduced schemes.) [Give pictures.]

18generationex

19uscftsheaf; this is used, so be careful if I change this from (c)

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17.10.C. Useful Exercise: Geometric Nakayama. Suppose X is a scheme, and F is a finitetype quasicoherent sheaf. Show that if Fx ⊗ k(x) = 0, then there exists V such that F|V = 0.Better: if I have a set that generates the fiber, it generates the stalk. [forme: State thiscorrectly! Perhaps: If x ∈ U ⊂ X is a neighborhood of x in X and a1, . . . , an ∈ F(U)so that the images a1, . . . , an ∈ Fx generate F ⊗ k(x), then there is a neighborhoodx ⊂ V ⊂ U of x so that a1|V , . . . , an|V generate F|V .] [forme: [E, p. 124]]

17.10.D. Less important Exercise. Suppose F and G are finite type sheaves such thatF ⊗ G ∼= OX . Then F and G are both invertible (Hint: Nakayama.) [forme: H. Ex. 5.7(c)]21 This is the reason for the adjective “invertible” these sheaves are the invertible elements of the“monoid of finite type sheaves”.

17.11 Quasicoherent sheaves of ideals, and closed subschemes

22 23 [With the usual definition, you have to work harder to show closed subschemes of SpecRcorrespond to ideals! ]

Older: Now might be a good time to mention (in an exercise) that pulling back and effec-tive Cartier divisor, if it pulls back to an effective Cartier divisor induces the pullback of thecorresponding invertible sheaf. But be caution if I move the previous discussion earlier!

[forme: When pulling back closed immersions, The ideal sheaf of W is sometimescalled the inverse image (quasicoherent) ideal sheaf.

Note for experts: It is not necessarily the quasicoherent pullback (f∗) of theideal sheaf, as the following example shows. (Thanks Joe!)

Spec k[x]/(x) //

Spec k[x]/(x)

Spec k[x]/(x) // Spec k[x]

Instead, the correct thing to pullback (the thing that “pulls back well”) is the sur-jection OZ → OY → 0, which pulls back to OX → OW → 0. The key issue is thatpullback of quasicoherent sheaves is right-exact, so we shouldn’t expect the pullback

of 0→ IY/Z →OZ →OY → 0 to be exact, only right-exact. (Thus for example we get

a natural map f∗IY/Z → IW/X .)]

[Student confusion: quotient corresponds to subojbect. Forshadows: morphism of schemes.]Suppose I →OX is a quasicoherent sheaf of ideals. (Quasicoherent sheaves of ideals are, not

suprisingly, sheaves of ideals that are quasicoherent.) Not all sheaves of ideals are quasicoherent.The cokernel of I → OX is also quasicoherent, so we have an exact sequence of quasicoherent

sheaves 24closed subscheme exact

sequence (41) 0→ I → OX →OX/I → 0.

(This exact sequence will come up repeatedly. We could call it the closed subscheme exact se-quence.) Now OX/I is finite type (as over any affine open set, the corresponding module isgenerated by a single element), so SuppOX/I is a closed subset. Also, OX/I is a sheaf of rings.Thus we have a topological space SuppOX/I with a sheaf of rings. I claim this is a scheme.To see this, we look over an affine open set SpecR. Here Γ(SpecR,I) is an ideal I of R. ThenΓ(SpecR,OX/I) = R/I (because quotients behave well on affine open sets).

I claim that on this open set, SuppOX/I is the closed subset V (I), which I can identify withthe topological space SpecR/I. Reason: [p] ∈ Supp(OX/I) if and only if (R/I)p 6= 0 if and onlyif p contains I if and only if [p] ∈ SpecR/I.

(Remark for experts: when you have a sheaf supported in a closed subset, you can interpretit as a sheaf on that closed subset. More precisely, suppose X is a topological space, i : Z → X is

21e57c

22qcsideals

23H.E.II.3.11

24subschemeES

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an inclusion of a closed subset, and F is a sheaf on X with SuppF ⊂ Z. Then we have a naturalmap F → i∗i−1F (corresponding to the map i−1F → i−1F , using the adjointness of i−1 andi∗). You can check that this is an isomorphism on stalks, and hence an isomorphism, so F can beinterpreted as the pushforward of a sheaf on the closed subset. Thanks to Jarod and Joe for thiscomment.)

I next claim that on the distinguished open set D(f) of SpecR, the sections of OX/I areprecisely (R/I)f ∼= Rf/If . (Reason that (R/I)f ∼= Rf/If : take the exact sequence 0 → I →R→ R/I → 0 and tensor with Rf , which preserves exactness.) Reason: On SpecR, the sectionsof OX/I are R/I, and OX/I is quasicoherent, hence the sections over D(f) are (R/I)f .

That’s it!We say that a closed subscheme of X is anything arising in this way from a quasicoherent

sheaf of ideals. In other words, there is a tautological correspondence between quasicoherentsheaves of ideals and closed subschemes.

Important remark. Note that closed subschemes of affine schemes are affine. (This istautological using our definition, but trickier using other definitions.)

Exercise (sections of locally free sheaves cut out closed subschemes). Suppose F isa locally free sheaf on a scheme X, and s is a section of F . Describe how s = 0 “cuts out” a closedsubscheme. [forme: This is kind of a weird problem. but it may be worth sayingexplicitly. Kirsten makes a definition if F is quasicoherent: O → F gives F∨ → O,

and the image is a quasicoherent sheaf. However, this behaves badly. On Spec Z, let

F = gZ/p, and s = 1, then Y = Spec Z but s does not vanish at [(p)].]

17.11.1. From closed subschemes to effective Cartier divisors. [I could move thislater.] There is a special name for a closed subscheme locally cut out by one equation that isnot a zero-divisor. More precisely, it is a closed subscheme such that there exists an affine coversuch that on each one it is cut out by a single equation, not a zero-divisor. (This does not meanthat on any affine it is cut out by a single equation — this notion doesn’t satisfy the “gluability”hypothesis of the Affine Communication Theorem. If I ⊂ R is generated by a non-zero divisor,then If ⊂ Rf is too. But “not conversely”. I might give an example later, involving ellipticcurves.) We call this an effective Cartier divisor. (This admittedly unwieldy terminology! But effective Cartier divisorthere is a reason for it.) By Krull’s Principal Ideal Theorem, it is pure codimension 1.

Remark: if I = (u) = (v), and u is not a zero-divisor, then u and v differ by a unit in R.Proof: u ∈ (v) implies u = av. Similarly v = bu. Thus u = abu, from which u(1 − ab) = 0. As uis not a zero-divisor, 1 = ab, so a and b are units.

Reason we care: effective Cartier divisors give invertible sheaves. If I is an effective Cartierdivisor on X, then I is an invertible sheaf. Reason: locally, sections are multiples of a singlegenerator u, and there are no “relations”.

The invertible sheaf corresponding to an effective Cartier divisor is for various reasons definedto be the dual of the ideal sheaf. This line bundle has a canonical section: Dualizing 0→ I → Ogives us O → I∗. [Or even not dualizing... instead twisting...]

17.11.A. Exercise. This section vanishes along our actual effective Cartier divisor. (This is agood exercise to make sure that you are clear on the the construction of this canonical section!)

17.11.B. Exercise. Describe the invertible sheaf in terms of transition functions. More precisely,on any affine open set where the effective Cartier divisor is cut out by a single equation, theinvertible sheaf is trivial. Determine the transition functions between two such overlapping affineopen sets. Verify that there is indeed a canonical section of this invertible sheaf, by describing it.

To describe the tensor product of such invertible sheaves: if I = (u) (locally) and J = (v),then the tensor product corresponds to (uv). Motivated by this, we define the sum of two effectiveCartier divisors in this way.

We get a monoid of effective Cartier divisors, with unit I = O. Notation: D is an effectiveCartier divisor. O(D) is the corresponding line bundle. O(−D) is the ideal sheaf.

0→O(−D)→O → OD → 0.

D is associated to the closed subscheme.Hence we can get a bunch of invertible sheaves, by taking differences of these two. Surprising

fact: we almost get them all! (True for all nonsingular schemes. This is true for all projectiveschemes. It is very hard to describe an invertible sheaf that is not describable in such a way.)

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17.12 ++Quasicoherent sheaves on ringed spaces

(This section should certainly be omitted on first reading. It is included only for completeness.)Many of the definitions we have made for quasicoherent sheaves work just fine forOX -modules

in general, where (X,OX) is a ringed space. If you wish, you can figure out what the definitionsof the following should be in this generality, and verify that all the same results about them hold:

quasicoherent sheaf, finite type sheaf, coherent sheaf, locally free sheaf, tensor products (this onewe did earlier), Hom (done earlier), T ∗F , Sym∗ F , ∧∗F , torsion-free sheaves.

17.13 Pre-216 notes

[Add later: for a finite type k-scheme, a morphism of quasicoherent sheaves is surjective iffit is at the closed points.]

Stalk information: a sheaf is 0 if it is 0 at all stalks. Support (need not be closed), see [H]exercise [which one? when is it closed?]. Support was defined in 4.6.G.

17.13.1. More on invertible sheaves.Parametrized by H1(X,O∗X). [not described yet]Sections, and rational sections. Multiplying sections of different invertible sheaves. Quotient

of 2 sections is a rational function.

17.13.A. Exercise. Coherent sheaves (or locally frees) form a monoid. Grothendieck group.Associated points. If two modules have disjoint sets of associated points, then their tensor

product is 0.

][primordial:

CHAPTER 18

Invertible sheaves and divisors

18.1 Invertible sheaves and divisors

We next develop some mechanism of understanding invertible sheaves (line bundles) on agiven scheme X. Define PicX to be the group of invertible sheaves on X. How can we describemany of them? How can we describe them all? Our goal for the first part of today will be topartially address this question. As an important example, we’ll show that we have already foundall the invertible sheaves on projective space Pnk — they are the O(m).

One moral of this story will be that invertible sheaves will correspond to “codimension 1information”.

Recall one way of getting invertible sheaves, by way of effective Cartier divisors. Recall that effective Cartier divisoran effective Cartier divisor is a closed subscheme such that there exists an affine cover such thaton each one it is cut out by a single equation, not a zero-divisor. (This does not mean that on anyaffine it is cut out by a single equation — this notion doesn’t satisfy the “gluability” hypothesis ofthe Affine Communication Lemma. If I ⊂ R is generated by a non-zero divisor, then If ⊂ Rf istoo. But “not conversely”, as we will see in Exercise 18.1.F. [forme: I could refer to 25.5.2.]) By Krull’s Principal Ideal Theorem 15.6.3, it is pure codimension 1.

Remark: if I = (u) = (v), and u is not a zero-divisor, then u and v differ multiplicativelyby a unit in R. Proof: u ∈ (v) implies u = av. Similarly v = bu. Thus u = abu, from whichu(1 − ab) = 0. As u is not a zero-divisor, 1 = ab, so a and b are units. In other words, thegenerator of such an ideal is well-defined up to a unit.

The reason we care: effective Cartier divisors give invertible sheaves. If I is an effectiveCartier divisor on X, then I is an invertible sheaf. Reason: locally, sections are multiples of asingle generator u, and there are no “relations”.

Recall that the invertible sheaf O(D) corresponding to an effective Cartier divisor is definedto be the dual of the ideal sheaf ID . The ideal sheaf itself is sometimes denoted O(−D). We havean exact sequence

0→O(−D)→O → OD → 0.

The invertible sheaf O(D) has a canonical section: Dualizing 0→ I → O gives us O → I∗.Exercise. This section vanishes along our actual effective Cartier divisor.Exercise. Conversely, if L is an invertible sheaf, and s is a section that is not locally a zero

divisor (make sense of this!), then s = 0 cuts out an effective Cartier divisor D, and O(D) ∼= L. (IfX is locally Noetherian, “not locally a zero divisor” translate to “does not vanish at an associatedpoint”.)

Define the sum of two effective Cartier divisors as follows: if I = (u) (locally) and J = (v),then the sum corresponds to (uv) locally. (Verify that this is well-defined!) [Better: IJ .]

Exercise. Show that O(D + E) ∼= O(D)⊗O(E).Thus we have a map of semigroups, from effective Cartier divisors to invertible sheaves with

sections not locally zero-divisors (and hence also to the Picard group of invertible sheaves).Hence we can get a bunch of invertible sheaves, by taking differences of these two. The

surprising fact: we “usually get them all”! In fact it is very hard to describe an invertible sheafon a finite type k-scheme that is not describable in such a way (we will see later today that thereare none if the scheme is nonsingular or even factorial; and we might see later in the year thatthere are none if the scheme is quasiprojective).

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[forme: Here is a failed exposition: I could define a virtual Cartier divisor to bea difference of two effective Cartier divisors. Then each virtual Cartier divisor gives aninvertible sheaf, with a rational section not vanishing at the associated points. The downside is that given an invertible sheaf with a rational section not vanishing at the associatedpoints, it is not clear that we can get an associated virtual Cartier divisor. to patch this, wecould define some version of a Cartier divisor as the following data (modulo an equivalencerelation): a cover Ui of X; on each Ui, a virtual Cartier divisor Di −Ei, and on each Ui ∩Uj ,and equality of virtual Cartier divisors Di − Ei = Dj − Ej . Two such objects are declaredthe same if they are the same after a common refinement of their open sets. But then ona triangle of P1’s, we have something weird — a nontrivial bundle would give me a zero“pseudo-Cartier” divisor.

Finally, we could also define K, the sheaf of total fractions: K(U) is the ring of total

fractions of U. Then we have an inclusion O∗X → K of sheaves of abelian groups. This is

just unpleasant.]Instead, I want to take another tack. Some of what we do will generalize to the non-normal

case, which is certainly important, and experts are invited to think about this.Define a Weil divisor as a formal sum of height 1 irreducible closed subsets of X. (ThisWeil divisor

makes sense more generally on any pure dimensional, or even locally equidimensional, scheme.)In other words, a Weil divisor is defined to be an object of the form

X

Y ⊂ X height 1

nY [Y ]

the nY are integers, all but a finite number of which are zero. Weil divisors obviously form anabelian group, denoted WeilX.

A Weil divisor is said to be effective if nY ≥ 0 for all Y . In this case we say D ≥ 0, andby D1 ≥ D2 we mean D1 −D2 ≥ 0. The support of a Weil divisor D is the subset ∪nY 6=0Y . IfU ⊂ X is an open set, there is a natural restriction map WeilX → WeilU , where

PnY [Y ] 7→P

Y ∩U 6=∅ nY [Y ∩ U ].effective; support

Suppose now thatX is a Noetherian scheme, regular in codimension 1. We add this hypothesisbecause we will use properties of discrete valuation rings. Suppose that L is an invertible sheaf,and s a rational section not vanishing on any irreducible component of X. Then s determines aWeil divisor

div(s) :=X

Y

valY (s)[Y ].

(Recall that valY (s) = 0 for all but finitely many Y , by Exercise 16.3.G.) This is the “divisorof poles and zeros of s”. (To determine the valuation valY (s) of s along Y , take any openset U containing the generic point of Y where L is trivializable, along with any trivializationover U ; under this trivialization, s is a function on U , which thus has a valuation. Any twosuch trivializations differ by a unit, so this valuation is well-defined.) This map gives a grouphomomorphism1

(42)div : (invertible sheaf L, rational section s not vanishing at any minimal prime)/Γ(X,O∗X)→WeilX.

18.1.A. Exercise. (a) (divisors of rational functions) Verify that on A1k, div(x3/(x + 1)) =

3[(x)]− [(x+ 1)] = 3[0]− [−1].(b) (divisor of a rational sections of a nontrivial invertible sheaf) Verify that on P1

k, there is a

rational section of O(1) “corresponding to” x2/y. Calculate div(x2/y).We want to classify all invertible sheaves on X, and this homomorphism (42) will be the key.

Note that any invertible sheaf will have such a rational section (for each irreducible component,take a non-empty open set not meeting any other irreducible component; then shrink it so thatL is trivial; choose a trivialization; then take the union of all these open sets, and choose thesection on this union corresponding to 1 under the trivialization). We will see that in reasonablesituations, this map div will be injective, and often even an isomorphism. Thus by forgetting therational section (taking an appropriate quotient), we will have described the Picard group. Let’sput this strategy into action. Suppose from now on that X is normal.

18.1.1. Proposition. — The map div is injective.

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Proof. Suppose div(L, s) = 0. Then s has no poles. Hence by Hartogs’ theorem 15.6.6, s is aregular section. Now s vanishes nowhere, so s gives an isomorphismOX → L (given by 1 7→ s).

Motivated by this, we try to find the inverse map to div.Definition. Suppose D is a Weil divisor. If U ⊂ X is an open subscheme, define FF(U) to

be the field of total fractions of U , i.e. the product of the stalks at the minimal primes of U . (Asdescribed earlier, if U is irreducible, this is the function field.) Define FF(U)∗ to be those rationalfunctions not vanishing at any generic point of U (i.e. not vanishing on any irreducible componentof U). Define the sheaf O(D) by

Γ(U,O(D)) := s ∈ FF(U)∗ | div s+D|U ≥ 0.Note that the sheaf O(D) has a canonical rational section, corresponding to 1 ∈ FF(U)∗.

18.1.2. Proposition. — Suppose L is an invertible sheaf, and s is a rational section notvanishing on any irreducible component of X. Then there is an isomorphism (L, s) ∼= (O(div s), t),where t is the canonical section described above.

Proof. We first describe the isomorphism O(div s) ∼= L. Over open subscheme U ⊂ X, we have abijection Γ(U,L) → Γ(U,O(div s)) given by s′ 7→ s′/s, with inverse obviously given by t′ 7→ st′.Clearly under this bijection, s corresponds to the section 1 in FF(U)∗; this is the section we arecalling t.

We denote the subgroup of WeilX corresponding to divisors of rational functions the sub-group of principal divisors, which we denote PrinX. Define the class group of X, ClX, byWeilX/PrinX. By taking the quotient of the inclusion (42) by PrinX, we have the inclusion

PicX → ClX.

We’re now ready to get a hold of PicX rather explicitly!

18.1.B. Exercise. Suppose that A is a Noetherian domain. Show that A is a Unique Factor-ization Domain if and only if A is integrally closed and Cl SpecA = 0. (One direction is easy:

we have already shown that Unique Factorization Domains are integrally closed in their fractionfields. Also, Exercise 15.6.8 shows that all height 1 primes are principal, so that implies thatCl SpecA = 0. It remains to show that if A is integrally closed and ClX = 0, then all height 1prime ideals are principal. Hartogs’ theorem 15.6.6 may arise in your argument.) This is the thirdimportant characterization of unique factorization domains promised in §7.4.5. 2 [Given p height1. There exists f ∈ FFA with div f = [p], so by Hartogs’ Theorem 15.6.6 f ∈ A, and f ∈ p. Weshow that f generates p. If g ∈ p, then g/f ∈ A by Hartogs.]

Hence Cl(Ank ) = 0, so Pic(Ank ) = 0 . (Geometers will find this believable: “Cn is a con-

tractible manifold, and hence should have no nontrivial line bundles”.)Another handy trick is the following. Suppose Z is an irreducible codimension 1 subset of

X. Then we clearly have an exact sequence:

0 // Z17→[Z]// WeilX // Weil(X − Z) // 0.

When we take the quotient by principal divisors, we lose exactless on the left, and get:

(43) Z17→[Z]// ClX // Cl(X − Z) // 0.

[Easier example: An minus any union of Weil divisors. Then this has no class group. Whathappens if you take out something of higher codimension? Discuss that case too!]

For example, let X = Pnk , and Z be the hyperplane x0 = 0. We have

Z→ Cl Pnk → Cl Ank → 0

from which Cl Pnk = Z[Z] (which is Z or 0), and Pic Pnk is a subgroup of this.

18.1.C. Important exercise. Verify that [Z] → O(1). In other words, let Z be the Cartierdivisor x0 = 0. Show that O(Z) ∼= O(1). (For this reason, people sometimes call O(1) thehyperplane class in PicX.) hyperplane class

2ufdcl

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Hence Pic Pnk → Cl Pnk is an isomorphism, and Pic Pnk ∼= Z , with generator O(1). The degree

of an invertible sheaf on Pn is defined using this: the degree of O(d) is of course d.degree of invertible sheaf

on PnkMore generally:

18.1.3. Proposition. — If X is factorial — all stalks are unique factorization domains — thenfactorialfor any Weil divisor D, O(D) is invertible, and hence the map PicX → ClX is an isomorphism.

Proof. It will suffice to show that [Y ] is Cartier if Y is any irreducible codimension 1 set. Ourgoal is to cover X by open sets so that on each open set U there is a function whose divisor is[Y ∩ U ]. One open set will be X − Y , where we take the function 1. Next, suppose x ∈ Y ; wewill find an open set U ⊂ X containing x, and a function on it. As OX,x is a unique factorizationdomain, the prime corresponding to 1 is height 1 and hence principal (by Exercise 15.6.8). Letf ∈ FFA be a generator. Then f is regular at x. f has a finite number of zeros and poles, andthrough x there is only one 0, notably [Y ]. Let U be X minus all the others zeros and poles.

I will now mention a bunch of other examples of class groups and Picard groups you cancalculate.

For the first, I want to note that you can restrict invertible sheaves on X to any subschemeY , and this can be a handy way of checking that an invertible sheaf is not trivial. For example,if X is something crazy, and Y ∼= P1, then we’re happy, because we understand invertible sheaves

on P1. Effective Cartier divisors sometimes restrict too: if you have effective Cartier divisor onX, then it restricts to a closed subscheme on Y , locally cut out by one equation. If you arefortunate that this equation doesn’t vanish on any associated point of Y , then you get an effectiveCartier divisor on Y . You can check that the restriction of effective Cartier divisors correspondsto restriction of invertible sheaves.

(The following discussion comes in several clumps. We have 3 problems relatedto projective space minus a hypersurface. We have the smooth quadric surface, andthe cone. We have Nagata’s Lemma and two consequences. We have generic finaldiscussions.

18.1.D. Exercise: a torsion Picard group. Show that Y is an irreducible degree d hypersur-face of Pn. Show that Pic(Pn − Y ) ∼= Z/d. (For differential geometers: this is related to the factthat π1(Pn − Y ) ∼= Z/d.) [Give example of plane minus conic.]

The previous exercise provides us with some examples tht we have been waiting for. Thenext two exercises explore its consequences.

18.1.E. Exercise. Keeping the same notation, assume d > 1, and let H0, . . . , Hn be the n+ 1coordinate hyperplanes on Pn. Show that Pn − Y is affine, and Pn − Y − Hi is a distinguishedopen subset of it. Show that the Pn − Y − Hi form an open cover of Pn − Y . Show thatPic(Pn−Y −Hi) = 0. Then by Exercise 18.2, each Pn−Y −Hi is the Spec of a unique factorizationdomain, but Pn − Y is not. Thus the property of being a unique factorization domain is not anaffine-local property — it satisfies only one of the two hypotheses of the affine communicationlemma 7.3.3. 3

18.1.F. Exercise. Keeping the same notation as the previous exercise, show that on Pn−Y , Hi(restricted to this open set) is an effective Cartier divisor that is not cut out by a single equation.(Hint: Otherwise it would give a trivial element of the class group.) 4

18.1.G. Exercise. Let X = Projk[w,x, y, z]/(wz−xy), a smooth quadric surface. [Make figureand refer to it!!] Show that PicX ∼= Z ⊕ Z as follows: Show that if L and M are two lines indifferent rulings (e.g. L = V (w,x) and M = V (w, y)), then X −L−M ∼= A2. This will give you asurjection Z⊕ Z ClX. Show that O(L) restricts to O on L and O(1) on M . Show that O(M)restricts to O on M and O(1) on L. (This is a bit longer to do, but enlightening.)

18.1.H. Exercise. Let X = Spec k[w,x, y, z]/(xy − z2), a cone. show that PicX = 1, andClX ∼= Z/2. (Hint: show that the ruling Z = x = z = 0 generates ClX by showing that its

3ufdnotaffloc

4locprnotpr

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complement is isomorphic to A2k. Show that 2[Z] = div(x) (and hence principal), and that Z is not

principal (an example we did when learning how to use the Zariski tangent space, Problem 16.1.2).

18.1.I. Exercise. Prove Nagata’s Lemma: Suppose A is a Noetherian domain, x ∈ A anelement such that (x) is prime and A[1/x] is a unique factorization domain. Then A is a uniquefactorization domain.5 (Hint: Exercise 18.2. Use the short exact sequence [(x)] → Cl SpecA →ClA[1/x]→ 0 (43) to show that Cl SpecA = 0. Show that A[1/x] is integrally closed, then showthat A is integrally closed as follows. Suppose Tn + an−1Tn−1 + · · ·+ a0 = 0, where ai ∈ A, andT ∈ FF(A). Then by integral closure of A[1/x], we have that T = r/xm, where if m > 0, thenr /∈ x. Then we quickly get a contradiction if m > 0.) Nagata, N’s lemma

The next two results were promised in §16.4.3.

18.1.J. Exercise. (a) Suppose that k is algebraically closed of characteristic not 2. Show thatif m ≥ 3, then A = k[a, b, x1, . . . , xm]/(ab − x2

1 − · · · − x2m) is a unique factorization domain, by

using the Nagata’s Lemma with x = a.(b) Suppose that char k 6= 2 and k does not contain a square root of −1, then A = k[x, y, z]/(x2 +y2 − z2) is a unique factorization domain, by using Nagata’s Lemma.

[(a) A[1/a] ∼= k[a, 1/a, x1, . . . , xm] is a ufd. A/(a) ∼= k[b, x1, . . . , xm]/Pmi=1 x

2i is an integral

domain, hence (a) is prime (here we use that n ≥ 5).]Final random remarks:Note: on curves, the invertible sheaves correspond to formal sums of points, modulo equiva-

lence relation.Number theorists should note that we have recovered a common description of the class

group: formal sums of primes, modulo an equivalence relation. (I would like to discuss ringsof integers in number fields here. I should look up the correct definition.) Fractionalideals give you an invertible sheaf. Each invertible sheaf gives you a fractional ideal. Principalfractional ideals give you an the trivial bundle.

Remark: Much of this discussion works without the hypothesis of normality, and indeedbecause non-normal schemes come up all the time, we need this additional generality. Thinkthrough this if you like.

18.2 Earlier notes

Lefschetz? We can’t yet say that PicX × A1 ∼= PicX.This one needs to be kept! I refer to it! [H, Prop. 6.2]: Suppose A is a Noetherian

domain. Then A is a unique factorization domain if and only if SpecA is normal and Cl SpecA = 0.6

Later: Show that on a projective scheme, any divisor is a difference of 2 effective Cartierdivisors?

Example. We showed earlier that Proj(N ⊕ 1) ∼= SpecN‘

ProjN , and that the latter is cut

out scheme-theoretically by a section of O(1).

][primordial:

5nagataslemma

6ufdcl, one of the main criteria to know for ufd’s.

CHAPTER 19

Quasicoherent sheaves on projective A-schemes

and graded modules

[This was originally before the invertible sheaf chapter.]We now describe quasicoherent sheaves on a projective A-scheme. Recall that a projective A-

scheme is produced from the data of Z≥0-graded ring S•, with S0 = A, and S+ finitely generatedas an A-module. The resulting scheme is denoted ProjS•.

Let X = ProjS•. Suppose M∗ is a graded S• module, graded by Z. We define the quasi-

coherent sheaf gM∗ as follows. (I will avoid calling it M , as this might cause confusion with theaffine case.) On the distinguished open D(f), we let

gM∗(D(f)) ∼= (Mf )0.

(More correctly: we define a sheaf M∗(f) on D(f) for each f . We give identifications of the

restriction of M∗(f) and M∗(g) to D(fg). Then by an earlier problem set problem telling how

to glue sheaves, these sheaves glue together to a single sheaf on gM∗ on X. We then discard the

temporary notation M∗(f).)This is clearly quasicoherent, because it is quasicoherent on each D(f). If M∗ is finitely

generated over S•, then so M∗ is a finite type sheaf.[forme: I haven’t checked finite type / coherence things.] I will now give some

straightforward facts.

If M∗ and M ′∗ agree in high enough degrees, then gM∗ ∼= gM ′∗. Thus the map from gradedS•-modules to quasicoherent sheaves on ProjS• is not a bijection.

Given a map of graded modules φ : M∗ → N∗, we we get an induced map of sheaves gM∗ →fN∗. Explicitly, over D(f), the map M∗ → N∗ induces M∗[1/f ] → N∗[1/f ] which induces φf :(M∗[1/f ])0 → (N∗[1/f ])0; and this behaves well with respect to restriction to smaller distinguishedopen sets, i.e. the following diagram commutes.

(M∗[1/f ])0φf //

(N∗[1/f ])0

(M∗[1/(fg)])0

φfg // (N∗[1/(fg)])0 .

In fact ∼ is a functor from the category of graded S•-modules to the category of quasicoherentsheaves on ProjS•. This isn’t quite an isomorphism, but it is close. The relationship is akin tothat between presheaves and sheaves, and the sheafification functor, as we will see before long.

19.0.A. Exercise. Show that gM∗ ⊗ fN∗ ∼= ˜M∗ ⊗S• N∗. (Hint: describe the isomorphism ofsections over each D(f), and show that this isomorphism behaves well with respect to smallerdistinguished open sets.)1

19.0.1. Graded ideals of S• give closed subschemes of ProjS•. If I∗ ⊂ S• is a gradedideal, we get a closed subscheme.

Example: Suppose S• = k[w,x, y, z], so ProjS• ∼= P3. Suppose I∗ = (wz− xy, x2 −wy, y2 −xz). Then we get the exact sequence of graded S•-modules

0→ I∗ → S• → S•/I∗ → 0.

1tensorproj

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Which closed subscheme of P3 do we get? The twisted cubic! [Refer to Exercise 9.1.T, where thetwisted cubic was defined.]

19.0.B. Exercise. Show that if I∗ is a graded ideal of S•, show that we get a closed subschemeProjS•/I∗ → ProjS•.

2

We will soon see (§19.3.2) that all closed subschemes of ProjS• arise in this way.

19.1 Invertible sheaves (line bundles) on projectiveA-schemes

We now come to one of the most fundamental concepts in projective geometry.

19.1.A.Exercise. Suppose S• is generated over S0 by f1, . . . , fn. Suppose d = lcm(deg f1, . . . , deg fn).Show that Sd∗ is generated in “new” degree 1 (= “old” degree d). (Hint: I like to show this byinduction on the size of the set deg f1, . . . ,deg fn.) This is handy, because we can stick everyProj in some projective space via the construction of Exercise 19.0.B. [forme: I earlier re-ferred to 10.4.D] [forme: I’ve already incorporated this into FOAG. In FOAG, inthe next paragraph, refer to the correct Exercise number.] 3

Suppose that S• is generated in degree 1. By Exercise 19.1.A, this is not a huge assumption.Suppose M∗ is a graded S•-module. Define the graded module M(n)∗ so that M(n)m := Mn+m.

Thus the quasicoherent sheaf M(n)∗ is given by

Γ(D(f), M(n)∗) = (Mf )n

where here the subscript means we take the nth graded piece. (These subscripts are admittedlyconfusing!)

As an incredibly important special case, define OProj S•(n) := S(n)∗. When the space isimplicit, it can be omitted from the notation: O(n) (pronounced “oh of n”).

19.1.B. Important exercise. If S• is generated in degree 1, show that OProjS•(n) is aninvertible sheaf.

(We will soon see that on Pmk , these are the only invertible sheaves. [Give ref.])

19.1.C. Essential exercise. Show that dimk Γ(Pmk ,OPmk

(n)) =`m+n

n

´. 4 [Refer back to P1

case earlier.][Mention: H0(P1,OP1 (n)) realizes the irreducible representations of SL2(C) with highest

weight n. This is the beginning of the Borel-Weil story.]I’ll get you started on this. As always, consider the “usual” affine cover. Consider the

n = 1 case. Say S• = k[x0, . . . , xm]. Suppose we have a global section of O(1). On D(x0),

the sections are of the form f(x0, . . . , xn)/xdeg f−10 . On D(x1), the sections are of the form

g(x0, . . . , xn)/xdeg g−11 . They are supposed to agree on the overlap, so

xdeg f−10 g(x0, . . . , xn) = xdeg g−1

1 f(x0, . . . , xn).

How is this possible? Well, we must have that g = xdeg g−11 × some linear factor, and f =

xdeg f−10 × the same linear factor. Thus on D(x0), this section must be some linear form. OnD(x1), this section must be the same linear form. By the same argument, on each D(xi), thesection must be the same linear form. Hence (with some argumentation), the global sections ofO(1) correspond to the linear forms in x0, . . . , xm, of which there are m+ 1.

Thus x+y+2z is a section of O(1) on P2. It isn’t a function, but I can say where this sectionvanishes — precisely where x+ y + 2z = 0.

We make some follow-up remarks about this exercise. Note that sections of O(m) on Pnk arenaturally identified with forms degree m polynomials in our n+ 1 variables. Also, notice that for

2projclimm

3generateddeg1

4dimohn

327

fixed n,`m+n

n

´is a polynomial in m of degree m for m ≥ 0 (or better: for m ≥ −n − 1). This

should be telling you that this function “wants to be a polynomial” but has not succeeded. We willlater define h0(Pnk ,O(m)) := Γ(Pnk ,O(m)), and later still we will define higher cohomology groups,and we will define the Euler characteristic χ(Pnk ,O(m)) :=

P∞i=0 h

n(Pnk ,O(m)) (cohomology willvanish in degree higher than n). We will discover the moral that the Euler characteristic is better-behaved than h0, and so we should now suspect (and later prove) that this polynomial is in factthe Euler characteristic, and the reason that it agrees with h0 for m ≥ 0 because all the othercohomology groups should vanish. Euler characteristic

19.1.D. Exercise. Show that F(n) ∼= F ⊗O(n).

19.1.E. Exercise. Show that O(m+ n) ∼= O(m)⊗O(n). [forme: Use Exercise 19.0.A.]

19.1.F. Exercise. Show that if m 6= n, then OPlk(m) is not isomorphic to OPl

k(n) if l > 0.

(Hence we have described a countable number of invertible sheaves (line bundles) that are non-

isomorphic. We will see later that these are all the line bundles on projective space Pnk .)

19.2 Generation by global sections, and Serre’s Theorem

Replace “generated by global sections” with “globally generated”. Mention that earlier term.Then define “relatively generated” or possibly “relatively globally generated”.

19.2.1. Generated by global sections. Suppose F is a sheaf on a scheme X. We say that

F is generated by global sections at a point p if we can find φ : O⊕I → F that is surjective at

the stalk of p: φp : O⊕Ip → Fp is surjective. (Some what more precisely, the stalk of F at p is

generated by global sections of F . The global sections in question are the images of the 1’s in |I|factors of O⊕I

p .) We say that F is generated by global sections if it is generated by global sections

at all p, or equivalently, if we can we can find O⊕I → F that is surjective. (By our earlier resultthat we can check surjectivity at stalks, so this is the same as saying that it is surjective at allstalks.) [forme: Give ref to that earlier result. Mention also that we can take a singleindex set for all of X, by taking the union of all the index sets for each p.] If I canbe taken to be finite, we say that F is generated by a finite number of global sections. We’ll see generated by global sec-

tionssoon why we care.

19.2.A. Exercise. If quasicoherent sheaves L andM are generated by global sections at a pointp, then so is L⊗M. (This exercise is less important, but is good practice for the concept.)

19.2.B. Easy exercise. An invertible sheaf L on X is generated by global sections if andonly if for any point x ∈ X, there is a section of L not vanishing at x. (Hint: Nakayama.) SeeTheorem 20.4.1 and Exercise 20.4.E for why we care. [forme: Or perhaps a motivation iscoming up soon?] 5

19.2.2. Lemma. — Suppose F is a finite type sheaf on X. Then the set of points where F isgenerated by global sections is an open set.

Proof. Suppose F is generated by global sections at a point p. Then it is generated by a finitenumber of global sections, say m. This gives a morphism φ : O⊕m → F , hence imφ → F . Thesupport of the (finite type) cokernel sheaf is a closed subset not containing p.

19.2.C. Important exercise (an important theorem of Serre). Suppose S0 is a Noetherianring, and S• is generated in degree 1. Let F be any finite type sheaf on ProjS•. Then for someinteger n0, for all n ≥ n0, F(n) can be generated by a finite number of global sections. [forme:H.T.II.5.17] [forme: coherence?!! or finite type?!] 6

5invertiblesheafgenerated

6serreexercise

328

I’m going to sketch how you should tackle this exercise, after first telling you the reason wewill care.

19.2.3. Corollary. — Thus any coherent sheaf F on ProjS• can be presented as:

⊕finiteO(−n)→ F → 0.

[forme: Hartshorne C.II.5.18]We’re going to use this a lot!

Proof. Suppose we have m global sections s1, . . . , sm of F(n) that generate F(n). This gives amap

⊕mO // F(n)

given by (f1, . . . , fm) 7→ f1s1+· · ·+fmsm on any open set. Because these global sections generateF , this is a surjection. Tensoring with O(−n) (which is exact, as tensoring with any locally freeis exact) gives the desired result. [forme: FOAG: give ref]

Here is now a hint/sketch for the Serre exercise 19.2.C.We can assume that S• is generated in degree 1 [already assumed in the hypothesis]; we can

do this thanks to Exercise 19.1.A. Suppose deg f = 1. Say F|D(f) = M , where M is a (S•[1/f ])0-module, generated by m1, . . . , mn. As these elements generate the module, they clearly generateall the stalks over all the points of D(f). They are sections over this big distinguished open set. Itwould be wonderful if we knew that they had to be restrictions of global sections, i.e. that therewas a global section m′i that restricted to mi on D(f). If that were always true, then we would

cover X with a finite number of each of these D(f)’s, and for each of them, we would take thefinite number of generators of the corresponding module. Sadly this is not true.

However, we will see that fNm “extends”, where m is any of the mi’s, and N is sufficientlylarge. We will see this by (easily) checking first that fNm extends over another distinguished openD(g) (i.e. that there is a section of F(N) over D(g) that restricts to fnm onD(g)∩D(f) = D(fg)).[forme: Two solvers got stuck here — they thought we were done.]

So we’re done, right? Wrong — we still don’t that these extensions on various open sets gluetogether, and in fact they might not! More precisely: we don’t know that the extension over D(g)and over some other D(g′) agree on the overlap D(g) ∩ D(g′) = D(gg′). But after multiplying

both extensions by fN′

for large enough N ′, we will see that they agree on the overlap. Byquasicompactness, we need to to extend over only a finite number of D(g)’s, and to make sureextensions agree over the finite number of pairs of such D(g)’s, so we will be done.

Great, let’s make this work. Let’s investigate this on D(g) = SpecA, where the degree of g

is also 1. Say F|D(g) ∼= N . Let f ′ = f/g be “the function corresponding to f on D(g)”. We havea section over D(f ′) on the affine scheme D(g), i.e. an element of Nf ′ , i.e. something of the form

n/(f ′)N for some n ∈ N . So then if we multiply it by f ′N , we can certainly extend it! So if wemultiply by a big enough power of f , m certainly extends over any D(g).

As described earlier, the only problem is, we can’t guarantee that the extensions over D(g)and D(g′) agree on the overlap (and hence glue to a single extensions). Let’s check on the

intersection D(g) ∩D(g′) = D(gg′). Say m = n/(f ′)N = n′/(f ′)N′

where we can take N = N ′

(by increasing N or N ′ if necessary). We certainly may not have n = n′, but by the (concrete)definition of localization, after multiplying with enough f ′’s, they become the same.

In conclusion after multiplying with enough f ’s, our sections overD(f) extend over each D(g).After multiplying by even more, they will all agree on the overlaps of any two such distinguishedaffine. Exercise 19.2.C is to make this precise.

19.3 Every quasicoherent sheaf on a projective A-schemearises from a graded module

We have gotten lots of quasicoherent sheaves on ProjS• from graded S•-modules. We’ll nowsee that we can get them all in this way.

329

We want to figure out how to “undo” the M construction. When you do the essentialexercise 19.1.C, you’ll suspect that in good situations,

Mn∼= Γ(ProjS•, M(n)).

Motivated by this, we define

Γn(F) := Γ(ProjS•,Fn).

Then Γ∗(F) is a graded S•-module, and we can dream that Γ∗(F)∼ ∼= F . We will see that this

is indeed the case! Γ∗(F)

19.3.A. Exercise. Show that Γ∗ gives a functor from the category of quasicoherent sheaves onProjS• to the category of graded S•-modules. (In other words, show that if F → G is a morphismof quasicoherent sheaves on ProjS•, describe the natural map Γ∗(F) → Γ∗(G), and show thatsuch natural maps respect the identity and composition.)

Note that ∼ and Γ∗ cannot be inverses, as ∼ can turn two different graded modules into the

same quasicoherent sheaf.

Our initial goal is to show that there is a natural isomorphism Γ∗(F) → F , and that there

is a natural map M∗ → Γ∗(gM∗). We will show something better: that ∼ and Γ∗ are adjoint. adjoint

We start by describing the natural map M∗ → Γ∗gM∗. We describe it in degree n. Given

an element mn, we seek an element of Γ(ProjS•,gM∗(n)) = Γ(ProjS•, M(n+∗)). By shifting the

grading of M∗ by n, we can assume n = 0. For each D(f), we certainly have an element of(M [1/f ])0 (namely m), and they agree on overlaps, so the map is clear.

19.3.B. Exercise. Show that this canonical map need not be injective, nor need it be surjective.(Hint: S• = k[x], M∗ = k[x]/x2 or M∗ = polynomials with no constant terms .)

The natural map gΓ∗F → F is more subtle (although it will have the advantage of being anisomorphism).

19.3.C. Exercise. Describe the natural map gΓ∗F → F as follows. First describe it over D(f).

Note that sections of the left side are of the formm/fn where m ∈ Γn deg fF , and m/fn = m′/fn′

if there is some N with fN (fn′m − fnm′) = 0. Show that your map behaves well on overlaps

D(f) ∩D(g) = D(fg).

19.3.D. Longer Exercise. Show that the natural map gΓ∗F → F is an isomorphism, by showingthat it is an isomorphism over D(f) for any f . Do this by first showing that it is surjective. Thiswill require following some of the steps of the proof of Serre’s theorem (Exercise 19.2.C). Thenshow that it is injective. [forme: Hartshorne Proposition II.5.15. Add to notes. Havethem do Lemma 5.14 as exercises.]

19.3.1. Corollary. — Every quasicoherent sheaf arises from this tilde construction.

19.3.2. Exercise. — Each closed subscheme of ProjS• arises from a graded ideal I• ⊂ S•.7 (Hint: Suppose Z is a closed subscheme of ProjS•. Consider the exact sequence 0 → IZ →OProjS• →OZ → 0. Apply Γ∗, and then ∼.) [forme: After applying Γ∗, we get somethingonly left-exact. Applying ∼, we get something into something that is still left-exact,but I think it becomes exact again.]

In particular, let x0, . . . , xn be generators of S1 as an A-module. Then we have a surjectionof graded rings

A[t0, . . . , tn]→ S•

where ti 7→ xi. Then this describes ProjS• as a closed subscheme of PnA.

19.3.E. Exercise (Γ∗ and ∼ are adjoint functors). Prove part of the statement that Γ∗

and ∼ are adjoint functors, by describing a natural bijection Hom(M∗,Γ∗(F)) ∼= Hom(gM∗,F).For the map from left to right, start with a morphism M∗ → Γ∗(F). Apply ∼, and postcompose

with the isomorphism gΓ∗F → F , to obtain

gM∗ → gΓ∗F → F .

7gradedideals

330

Do something similar to get from right to left. Show that “both compositions are the identity inthe appropriate category”. (Is there a clever way to do that?)

19.3.3. Saturated S•-modules. We end with a remark: different graded S•-modules give thesame quasicoherent sheaf on ProjS•, but the results of this section show that there is a “best”graded module for each quasicoherent sheaf, and there is a map form each graded module to its

“best” version, M∗ → Γ∗(gM∗). A module for which this is an isomorphism (a “best” module) iscalled saturated. I don’t think we’ll use this notation, but other people do.saturated module

This “saturation” map M∗ → Γ∗(gM∗) should be seen as analogous to the sheafification map,taking presheaves to sheaves.8

[forme: Presumably prove that if F is coherent, then Γ∗(M) is coherent.]

19.4 Pre-216 left-overs

9

19.4.A. Exercise. More generally, show that there is a natural map of k-modules

φ : S1 → h0(X,OX(1)).

(In fact 1 isn’t special here.)(a) Show that this need not be injective. (Hint: S• = k[x, y, z]/x2.)(b) Let S• be the polynomials in x and y with no monomial x in it. Show that φ is not surjectivein this example.

19.4.B. Exercise. Hence we have an example promised earlier (Section 4.5.H): why you needto sheafify when tensoring. (Hint: O(1)⊗O(−1) ∼= O on projective space. Look at global sectionsof O.) 10

19.4.C. Exercise. O(m) ∼= O(1)⊗m . Hence O(m) ∼= Symn(O(1)).

19.4.D. Exercise. Promised earlier 4.5.C: Consider 0→O(−2)→O → O(2P )→ 0. Applyingthe global section function, we get something left exact. Show that it is not right exact. (Thissuggests long exact sequence...) Notice that we get a subscheme. “Fat point.” [What is this anexample of? Non-right-exactness? I guess so.]

Description of rational sections in this natural way. Multiplication of these rational sections.Notation: “twisting” by O(n) (or with a line bundle).Do I need this?:

19.4.1. Corollary. — [State better!] If F is coherent on PnR, then we can construct

0→ G → O(m)⊕j → F → 0.

11

This is false for quasicoherent sheaves in general; consider ⊕m≤0O(m).

19.4.E. Exercise. Quasicoherent sheaves of ideals on Proj correspond to homogeneous ideals.[H, Cor. II.5.16]

Saturated ideals [H, Ex. II.5.10].

19.4.2. Lemma [H, Prop. II.5.13]. — If S is a polynomial ring, then Γ∗OX = S•.This is quite simple.Warning [H, II.5.13.1]: this need not be true if S is not a polynomial ring. Give example:

rational space quartic. We’ll discuss this later.Worked example: Rational space quartic in P3. Do this example after introducing O(1).

8saturatedtemp

9s:ohm

10tensorsheaf2

11Pgen

331

Projective normality discussion earlier: S• → Γ∗(ProjS•).

][primordial:

CHAPTER 20

Pushforwards and pullbacks of quasicoherent

sheaves

20.1 Introduction

There are two things you can do with modules and a ring homomorphism B → A. If M isan A-module, you can create an B-module MB by simply treating it as an B-module. If N is anB-module, you can create an A-module N ⊗B A.

These notions behave well with respect to localization (in a way that we will soon makeprecise), and hence work (often) in the category of quasicoherent sheaves (and indeed always inthe category of modules over ringed spaces, see Remark XXXX, although this will not concern ushere). The two functors are adjoint:

HomA(N ⊗B A,M) ∼= HomB(N,MB)

(where this isomorphism of groups is functorial in both arguments), and we will see that thisremains true on the scheme level.

One of these constructions will turn into our old friend pushforward. The other will be arelative of pullback, whom I’m reluctant to call an “old friend”.

20.2 Pushforwards of quasicoherent sheaves

[forme: Not worth saying: For schemes over k, the pushforward to a point Spec kis same as global section functor. Pushforwards are left exact; expect there to besome higher pushforwards. Indeed this is true (in reasonable situations); see later inthis section. ] pushforward of quasico-

herent sheavesThe main message of this section is that in “reasonable” situations, the pushforward of aquasicoherent sheaf is quasicoherent, and that this can be understood in terms of one of themodule constructions defined above. We begin with a motivating example:

20.2.A. Exercise. Let f : SpecA → SpecB be a morphism of affine schemes, and suppose M

is an A-module, so M is a quasicoherent sheaf on SpecA. Show that f∗M ∼= gMB . [forme: H.Prop. II.5.2(d)] 1 (Hint: There is only one reasonable way to proceed: look at distinguishedopen sets!)

In particular, f∗M is quasicoherent. Perhaps more important, this implies that the push-forward of a quasicoherent sheaf under an affine morphism is also quasicoherent. The followingresult doesn’t quite generalize this statement, but the argument does.

20.2.1. Theorem. — Suppose f : X → Y is a morphism, and X is a Noetherian scheme.Suppose F is a quasicoherent sheaf on X. Then f∗F is a quasicoherent sheaf on Y . 2

[forme: We will soon see that the hypotheses can be weakened, to quasicompactand quasiseparated (Prop. ??). [forme: (This will then generalize the preceding

1pushforwardaffine

2t:qcspushforward

333

334

exercise as well.)] More precisely, once we define quasiseparated, the followingproof will follow without change.]

[forme: qs is now done!!!]The fact about f that we will use is that the preimage of any affine open subset of Y is

a finite union of affine sets (f is quasicompact), and the intersection of any two of these affinesets is also a finite union of affine sets. (This is a definition of the notion of a quasiseparatedmorphism. We will meet this notion properly in §12.1.9.) Thus the “correct” hypothesis here isquasiseparatedthat f is quasicompact and quasiseparated. [We’ll give a better definition of “quasiseparated”later, in XXXX.]

Proof. By the first definition of quasicoherent sheaves (Section 17), it suffices to show the following:if F is a quasicoherent sheaf on X, and f : X → SpecR, then the following diagram commutes:

Γ(X,F)resD(g)⊂Spec R //

⊗RRg

%%KKKKKKKKKKΓ(Xg ,F)

Γ(X,F)g

∼

88rrrrrrrrrr

This is Exercise 17.2.B. [forme: Careful: In that exercise, the element is f , but that’s

used for the morphism now.] [forme: H. Prop. II.5.8]

20.2.B. Exercise. Give an example of a morphism of schemes π : X → Y and a quasicoherentsheaf F on X such that π∗F is not quasicoherent. (Answer: Y = A1, X = countably many copiesof A1. Then let f = t. Xt has a global section (1/t, 1/t2 , 1/t3 , . . . ). The key point here is thatinfinite direct sums do not commute with localization.) 3

Coherent sheaves don’t always push forward to coherent sheaves. For example, consider the

structure morphism f : A1k → Spec k, given by k 7→ k[t]. Then f∗OA1

kis the k[t], which is not a

finitely generated k-module. Under especially good situations, coherent sheaves do push forward.For example:

20.2.C. Exercise. Suppose f : X → Y is a finite morphism of Noetherian schemes. If F is acoherent sheaf on X, show that f∗F is a coherent sheaf. (Hint: Show first that f∗OX is finite type= locally finitely generated.) [Is there a straightforward counterexample if we drop the Noetherianhypothesis?] [forme: This uses Exercise 9.1.L, which is currently not yet pasted in.]

Once we define cohomology of quasicoherent sheaves, we will quickly prove that if F is acoherent sheaf on Pnk , then Γ(Pnk ) is a finite-dimensional k-module, and more generally if F isa coherent sheaf on ProjS•, then Γ(ProjS•) is a coherent A-module (where S0 = A). This isa special case of the fact the “pushforwards of coherent sheaves by projective morphisms arealso coherent sheaves”. We will first need to define “projective morphism”! This notion is ageneralization of ProjS• → SpecA.

[forme: Unimportant exercise: pushforward induces morphisms of stalks, whichis in general neither injective nor surjective.]

20.3 Pullback of quasicoherent sheaves

[Fold in near start: I will try to reserve the phrase “pullback of a sheaf” for pullbacks ofquasicoherent sheaves f∗, and “inverse image sheaf” for f−1, which applies in a more generalsituation, in the category of sheaves on topological spaces. Give a reference back to that.

Also: give an exercise that says for f : pt → X, f∗F is the stalk of F . This is just formorphisms of rather general things — how general?]

I will give four definitions of the pullback of a quasicoherent sheaf. The first one is the mostuseful in practice, and is in keeping with our emphasis of quasicoherent sheaves as just “modulesglued together”. The second is the “correct” definition, as an adjoint of pushforward. The third,

3qcspushforwardcounterexample, cf. 17.2.C

335

which we mention only briefly, is more correct, as adjoint in the category of OX -modules. Andwe end with a fourth definition.

We note here that pullback to a closed subscheme or an open subscheme is often calledrestriction. restriction of a quasico-

herent sheaf20.3.1. Construction/description of the pullback. Let us now define the pullback functorprecisely. Suppose X → Y is a morphism of schemes, and G is a quasicoherent sheaf on Y .We will describe the pullback quasicoherent sheaf f∗G on X by describing it as a sheaf on thedistinguished affine base. In our base, we will permit only those affine open sets U ⊂ X such thatf(U) is contained in an affine open set of Y . The distinguished restriction map will force thissheaf to be quasicoherent. [forme: (refer back).]

Suppose U ⊂ X, V ⊂ Y are affine open sets, with f(U) ⊂ V , U ∼= SpecA, V ∼= SpecB.

Suppose F|V ∼= N . Then define Γ(f∗V F , U) := N ⊗B A. Our main goal will be to show that thisis independent of our choice of V .

We begin as follows: we fix an affine open subset V ⊂ Y , and use it to define sections overany affine open subset U ⊂ f−1(V ). We show that this gives us a quasicoherent sheaf f∗V G on

f−1(V ), by showing that these sections behave well with respect to distinguished restrictions.First, note that if D(f) ⊂ U is a distinguished open set, then

Γ(f∗V F ,D(f)) = N ⊗B Af ∼= (N ⊗B A)⊗A Af = Γ(f∗V F , U)⊗A Af .Define the restriction map Γ(f∗V F , U)→ Γ(f∗V F , D(f)) by4

(44) Γ(f∗V F , U)→ Γ(f∗V F , U)⊗A Af(with α 7→ α ⊗ 1 of course). Thus on the distinguished affine topology of SpecA [forme: ref ]we have defined a quasicoherent sheaf. (We have checked that this describes a sheaf in XXXX,and the restriction map (44) ensures that it is quasicoherent by XXXX.) [forme: Ref!]

Finally, we show that if f(U) is contained in two affine open sets V1 and V2, then the allegedsections of the pullback we have described do not depend on whether we use V1 or V2. Moreprecisely, we wish to show that

Γ(f∗V1F , U) and Γ(f∗V2

F , U)

have a canonical isomorphism, which commutes with the restriction map (44).Let Wii∈I be an affine cover of V1 ∩ V2 by sets that are distinguished in both V1 and V2

(possible by the Proposition 7.3.2 we used in the proof of the Affine Communication Lemma).Then by the previous paragraph, as f∗V1

F is a sheaf on the distinguished base of V1,

Γ(f∗V1F , U) = ker

“⊕iΓ(f∗V1

F , f−1(Wi))→ ⊕i,jΓ(f∗V1F , f−1(Wi ∩Wj))

”.

If V1 = SpecB1 and Wi = D(gi), then

Γ(f∗V1F , f−1(Wi)) = N ⊗B1

Af#gi∼= N ⊗(B1)gi

Af#gi= Γ(f∗Wi

F , f−1(Wi)),

so 5

(45) Γ(f∗V1F , U) = ker

“⊕iΓ(f∗Wi

F , f−1(Wi))→ ⊕i,jΓ(f∗WiF , f−1(Wi ∩Wj))

”.

The same argument for V2 yields6

(46) Γ(f∗V2F , U) = ker

“⊕iΓ(f∗Wi

F , f−1(Wi))→ ⊕i,jΓ(f∗WiF , f−1(Wi ∩Wj))

”.

But the right sides of (45) and (46) are the same, so the left sides are too. Moreover, (45) and(46) behave well with respect to restricting to a distinguished open D(g) of SpecA (just apply⊗AAg to the the right side) so we are done.

Hence we have described a quasicoherent sheaf f∗G on X whose behavior on affines mappingto affines was as promised.

20.3.2. Theorem. —

(1) The pullback of the structure sheaf is the structure sheaf.(2) The pullback of a finite type (=locally finitely generated) sheaf is finite type.

4pullbackdist

5partone

6parttwo

336

(3) The pullback of a finitely presented sheaf is finitely presented. Hence if f : X → Yis a morphism of locally Noetherian schemes, then the pullback of a coherent sheaf iscoherent. (It is not always true that the pullback of a coherent sheaf is coherent, and theinterested reader can think of a counterexample.) [forme: Spec k[x1, . . . ]→ Spec k]

(4) The pullback of a locally free sheaf of rank r is another such. (In particular, thepullback of an invertible sheaf is invertible.)

(5) (functoriality in the morphism) π∗1π∗2F ∼= (π2 π1)∗F

(6) (functoriality in the quasicoherent sheaf) F1 → F2 induces π∗F1 → π∗F2

(7) If s is a section of F then there is a natural section π∗s that is a section of π∗F .(8) (stalks) If π : X → Y , π(x) = y, then (π∗F)x ∼= Fy ⊗OY,y

OX,x. The previous map,

restricted to the stalks, is f 7→ f ⊗ 1. (In particular, the locus where the section onthe target vanishes pulls back to the locus on the source where the pulled back sectionvanishes.)

(9) (fibers) Pullbacks of fibers are given as follows: if π : X → Y , where π(x) = y, thenπ∗F/mX,xπ∗F ∼= (F/mY,yF)⊗OY,y/mY,y

OX,x/mX,x(10) (tensor product) π∗(F ⊗ G) = π∗F ⊗ π∗G(11) pullback is a right-exact functor

[Somewhere, have the exercise that the stalk is the pullback by x→ X.]All of the above are interconnected in obvious ways. For example, given F1 → F2 and a

section s of F1, then we can pull back the section and then send it to π∗F2, or vice versa, and weget the same thing.

[Mention that much more is true, that the reader should be able to prove on a moment’snotice, such as for example that the pullback of the symmetric power of a locally free sheaf isnaturally isomorphic to the symmetric power of the pullback, and similarly for wedge powers andtensor powers.]

I used some of these results to help give an intuitive picture of the pullback.

Proof. Most of these are left to the reader. It is convenient to do right-exactness early (e.g. beforeshowing that finitely presented sheaves pull back to finitely presented sheaves). For the tensorproduct fact, show that (M ⊗S R)⊗ (N ⊗S R) ∼= (M ⊗N)⊗S R, and that this behaves well withrespect to localization. The proof of the fiber fact is as follows. (S, n)→ (R,m).

S //

R

S/n // R/m

(N⊗SR)⊗R (R/m) ∼= (N⊗S (S/n))⊗S/n (R/m) as both sides are isomorphic to N⊗S (R/m).

20.3.A. Unimportant Exercise. Verify that the following is a example showing that pullbackis not left-exact: consider the exact sequence of sheaves on A1, where p is the origin:

0→OA1 (−p)→OA1 →Op → 0.

(This is a closed subscheme exact sequence; also an effective Cartier exact sequence. Algebraically,we have k[t]-modules 0→ tk[t]→ k[t]→ k→ 0.) Restrict to p.

20.3.3. Pulling back closed subschemes. Suppose Z → Y is a closed immersion, andX → Y is any morphism. Then we define the pullback of the closed subscheme Z to X as follows.We pullback the quasicoherent sheaf of ideals on Y defining Z to get a quasicoherent sheaf of

ideals on X (which we take to define W ). Equivalently, on any affine open Y , Z is cut out bysome functions; we pull back those functions to X, and denote the scheme cut out by them by W .

Exercise. Let W be the pullback of the closed subscheme Z to X. Show that W ∼= Z ×Y X.In other words, the fibered product with a closed immersion always exists, and closed immersions

337

are preserved by fibered product (or by pullback), i.e. if

Wg′ //

X

Z

g // Y

is a fiber diagram, and g is a closed immersion, then so is g′.Perhaps move this to fibered products. 7 Another version: Suppose Y → Z is

a closed immersion, and X → Z is any morphism. Show that the fibered productX ×Z Y exists, by explicitly describing it. Show that the projection X ×Z Y → Y isa closed immersion. We say that “closed immersions are preserved by base change”or “closed immersions are preserved by fibered product”. (Base change is anotherword for fibered products.)

20.3.4. Three more “definitions”. Pullback is left-adjoint of the pushforward. This is atheorem (which we’ll soon prove), but it is actually a pretty good definition. If it exists, then itis unique up to unique isomorphism by Yoneda nonsense.

The problem is this: pushforwards don’t always exist (in the category of quasicoherentsheaves); we need the quasicompact and quasiseparated hypotheses. However, pullbacks alwaysexist. So we need to motivate our definition of pullback even without the quasicompact and

quasiseparated hypothesis. (One possible motivation will be given in Remark 20.3.6.)

20.3.5. Theorem. — Suppose π : X → Y is a quasicompact, quasiseparated morphism. Thenpullback is left-adjoint to pushforward. More precisely, Hom(f∗G,F) ∼= Hom(G, f∗F). 8

(The quasicompact and quasiseparated hypothesis is required to ensure that the pushforwardexists, not because it is needed in the proof.)

More precisely still, we describe natural homomorphisms that are functorial in both argu-

ments. We show that it is a bijection of sets, but it is fairly straightforward to verify that it is anisomorphism of groups. Not surprisingly, we will use adjointness for modules.

Proof. Let’s unpack the right side. What’s an element of Hom(G, f∗F)? For every affine V in Y ,we get an element of Hom(G(V ),F(f−1(V ))), and this behaves well with respect to distinguishedopen sets. Equivalently, for every affine V in Y and U in f−1(V ) ⊂ X, we have an elementHom(G(V ),F(U)), that behaves well with respect to localization to distinguished open sets on both

affines. By the adjoint property, this corresponds to elements of Hom(G(V )⊗OY (V )OX(U),F(U)),which behave well with respect to localization. And that’s the left side.

20.3.6. +Pullback for ringed spaces. 9 (This is actually conceptually important but distracting pullback for [locally?]

ringed spacesfor our exposition; we encourage the reader to skip this, at least on the first reading.) Pullbacksand pushforwards may be defined in the category of modules over ringed spaces. We define

pushforward in the usual way, and then define the pullback of an OY -module using the adjointproperty. Then one must show that (i) it exists, and (ii) the pullback of a quasicoherent sheaf isquasicoherent. The fourth definition is as follows: suppose we have a morphism of ringed spacesπ : X → Y , and an OY -module G. Then we define f∗G = f−1G ⊗f−1OY

OX . We will not show

that this definition is equivalent to ours, but the interested reader is welcome to try this as anexercise. There is probably a proof in Hartshorne. The statements of Theorem 20.3.5 apply inthis more general setting. (Really the third definition “requires” the fourth.)

Here is a hint as to why this definition is equivalent to ours (a hint for the exercise if youwill). We need to show that f−1F ⊗f−1OY

OX (“definition 4”) and f∗F (“definition 1”) are

isomorphic. You should (1) find a natural morphism from one to the other, and (2) show thatit is an isomorphism at the level of stalks. The difficulty of (1) shows the disadvantages of ourdefinition of quasicoherent sheaves.

7fiberclosed1

8pullbacktheorem

9pullbackasadjoint

338

[forme: (1) Which direction is the map? Presumably we need a theorem sayingthat a map from a quasicoherent sheaves on an affine, to any other OX-module, canbe determined by map on each affine, with a compatibility relation.

(2) boils down to an easy fact. Suppose we have a ring map R → S, with M anR-module. Then

(M ⊗R S)⊗S Sq ∼= (M ⊗R Rp)⊗S Sqwhich is true because both are isomorphic to M ⊗R Sq .]

20.4 Important example: Morphisms to projective (andquasiprojective) schemes, and invertible sheaves

[Max made an interesting point: This can be taken as the definition of projective space!]This will tell us why invertible sheaves are crucially important: they tell us about maps to

projective space, or more generally to quasiprojective schemes. (And given that we have hada hard time naming any non-quasiprojective schemes, they tell us about maps to essentially allschemes that are interesting to us.)

20.4.1. Important theorem. — Maps to Pn correspond to n+1 sections of a line bundle, notall vanishing at any point (= generated by global sections, Ex. 19.2.B), modulo global sections ofO∗X . 10

Notes from teaching 2007. Emphasize how important this is. Every time they see a map toprojective space, they should immediately think of an invertible sheaf and some sections. Mentionmapping to a projective variety. For example, what if you wanted to map to y2z = x3 − xz2?You have three sections satisfying the same equation — as sections of the third tensor power ofthat line bundle.

First give An example. Then Pn with functions nowhere zero. Then generalize to thisstatement.

At some point, give the following “reality check” exercise. A degree e rational curve in Pm

is actually contained in a Pe.Here more precisely is the correspondence. If you have n + 1 sections, then away from the

intersection of their zero-sets, we have a morphism. Conversely, if you have a map to projectivespace f : X → Pn, then we have n+1 sections of OPn(1), corresponding to the hyperplane sections,x0, . . . , xn+1. then f∗x0, . . . , f∗xn+1 are sections of f∗OPn(1), and they have no common zero.

So to prove this, we just need to show that these two constructions compose to give theidentity in either direction.

Given n+1 sections s0, . . . , sn of an invertible sheaf. We get trivializations on the open setswhere each one vanishes. The transition functions are precisely si/sj on Ui ∩ Uj . We pull backO(1) by this map to projective space, This is trivial on the distinguished open sets. Furthermore,f∗D(xi) = D(si). Moreover, si/sj = f∗(xi/xj). Thus starting with the n + 1 sections, takingthe map to the projective space, and pulling back O(1) and taking the sections x0, . . . , xn, werecover the si’s. That’s one of the two directions.

Correspondingly, given a map f : X → Pn, let si = f∗xi. The map [s0; · · · ; sn] is preciselythe map f . We see this as follows. The preimage of Ui is D(si) = D(f∗xi) = f∗D(xi). So theright open sets go to the right open sets. And D(si)→ D(xi) is precisely by sj/si = f∗xj/xi.

Earlier: Given n+ 1 sections s0, . . . , sn of an invertible sheaf L. We get trivializations onthe open sets where each one vanishes. Map to Ui given by xj/i = sj/si. These glue. Transition

functions of O(1) is xi/xj on Ui ∩Uj (= xi/j in terms of Uj-coordinates, 1/xj/i in terms of Ui-

coordinates). Pull this back to X. Trivialized open sets are f∗Ui = D(si). Transition functionson f∗Ui ∩ f∗Uj = D(si) ∩ D(sj). Get si/sj . These are transition functions for L! Also, thesections xi pullback to each of these.

For the opposite direction: given a map f : X → Pn, let si = f∗xi. The map [s0; · · · ; sn] isprecisely the map f . We see this as follows. The preimage of Ui is D(si) = D(f∗xi) = f∗D(xi).

10H.T.II.7.1, mapstoPn

339

So the right open sets go to the right open sets. And D(si) → D(xi) is precisely by sj/si =f∗xj/xi.

Here are some examples.Example 1. Consider the n + 1 functions x0, . . . , xn on An+1 (otherwise known as n + 1

sections of the trivial bundle). They have no common zeros on An − 0. Hence they determine amorphism An+1 − 0→ Pn. (We’ve talked about this morphism before. But now we don’t have toworry about gluing.)

Example 2: the Veronese morphism. Consider the line bundle OPn(m) on Pn. We’ve checked

that the number of sections of this line bundle are`n+mm

´, and they correspond to homogeneous

degree m polynomials in the projective coordinates for Pn. Also, they have no common zeros (asfor example the subset of sections xm0 , xm1 , . . . , xmn have no common zeros). Thus these determine

a morphism Pn → P“

n+mm

”

−1. This is called the Veronese morphism. For example, if n = 2 and

m = 2, we get a map P2 → P5. 11 d-tuple embedding,

Veronese embeddingThis is in fact a closed immersion. Reason: This map corresponds to a surjective map ofgraded rings. The first ring R1 has one generator for each of degree m monomial in the xi.The second ring is not k[x0, . . . , xn], as R1 does not surject onto it. Instead, we take R2 =k[x0, . . . , xn](m), i.e. we consider only those polynomials all of whose terms have degree divisibleby m. Then the natural map R1 → R2 is fairly clearly a surjection. Thus the corresponding mapof projective schemes is a closed immersion by an earlier exercise.

How can you tell in general if something is a closed immersion, and not just a map? Here is

one way.

20.4.A. Exercise. Let f : X → PnA be a morphism of A-schemes, corresponding to an invertiblesheaf L on X and sections s0, . . . , sn ∈ Γ(X,L) as above. Then φ is a closed immersion iff (1)each open set Xi = Xsi is affine, and (2) for each i, the map of rings A[y0, . . . , yn]→ Γ(Xi,OXi

)

given by yj 7→ sj/si is surjective. [forme: Prop. H. 7.2. Reword!] 12

We’ll give another method of detecting closed immersions later. The intuition for this willcome from differential geometry: the morphism should separate points, and also separate tangentvectors.

Example 3. The rational normal curve. The image of the Veronese morphism when n =1 is called a rational normal curve of degree m. Our map is P1 → Pm given by [x; y] →[xm;xm−1y; · · · ; xym−1; ym]. When m = 3, we get our old friend the twisted cubic. [Referto Exercise 9.1.T, where the twisted cubic was defined.] When m = 2, we get a smooth conic.What happens when m = 1? rational normal curve

20.4.B. Exercise (Automorphisms of projective space). Show that all the automorphismsof projective space Pnk correspond to (n+ 1) × (n + 1) invertible matrices over k, modulo scalars(also known as PGLn+1(k)). (Hint: Suppose f : Pnk → Pnk is an automorphism. Show thatf∗O(1) ∼= O(1). Show that f∗ : Γ(Pn,O(1))→ Γ(Pn,O(1)) is an isomorphism.) [H, E.g. 7.1.1]

This exercise will be useful later, especially for the case n = 1.(A question for experts: why did I not state that previous exercise over an arbitrary base

ring A? Where does the argument go wrong in that case?)

20.4.C. Neat Exercise. Show that any map from projective space to a smaller projective spaceis constant.

Cut: [forme: what does this mean?:

20.4.D. Exercise. Any two rational normal curves are projectively equivalent.]

20.4.E. Less important exercise. (a) Suppose X is a quasicompact scheme, and L an in-vertible sheaf on it, that is generated by global sections. Show that there are a finite number ofsections of L that give a map to projective space. (b) Give an example to show that the statementis wrong without the quasicompact hypothesis. [Stupid example showing need for quasicompact-ness, X =

‘n Pn, L = O(1).] 13

11d:Veronese

12referforward

13generatedmapstoPn

340

Here are some useful phrases to know.A linear series or linear system on a scheme X over a field k is an invertible sheaf L and a

finite-dimensional k-vector space V of sections. (We will not require that this vector space be asubspace of Γ(X,L); in general, we just have a map V → Γ(X,L).) If the linear series is Γ(X,L),we call it a complete linear series, and is often written |L|. Given a linear series, any point x ∈ Xon which all elements of the linear series V vanish, we say that x is a base-point of V . If Vhas no base-points, we say that it is base-point-free. The union of base-points is called the baselocus. In fact, it naturally has a scheme-structure — it is the (scheme-theoretic) intersection ofthe vanishing loci of the elements of V (or equivalently, of a basis of V ). In this incarnation, it iscalled the base scheme of the linear series. [forme: Check out definition of linear series(complete) linear series,

base-point(-free), base

locus, base scheme, lin-

ear system, |L|

vs. linear system.]Then Theorem 20.4.1 says that each base-point-free linear series gives a morphism to projec-

tive space X → PV ∗ = Proj⊕nL⊗n. The resulting morphism is often written X|V | // Pn . (I

may not have this notation quite standard; I should check with someone. I always forget whetherI should use “linear system” or “linear series”.)

20.4.F. Exercise. If the image scheme-theoretically lies in a hyperplane of projective space, wesay that it is degenerate (and otherwise, non-degenerate). Show that a base-point-free linear seriesV with invertible sheaf L is non-degenerate if and only if the map V → Γ(X,L) is an inclusion.Hence in particular a complete linear series is always non-degenerate.degenerate

non-degenerate

linear series, complete

linear series, base-point

(convention: hyphen-

ated), |L|, base scheme,

base locus

Example: The Veronese and Segre morphisms. The Veronese morphism can be inter-preted in this way. The dth Veronese morphism on Pn corresponds to the complete linear series|OPn(d)|. [Describe ideal.]

Veronese

The Segre morphism can also be interpreted in this way. In case I haven’t defined it yet,suppose F is a quasicoherent sheaf on a Z-scheme X, and G is a quasicoherent sheaf on a Z-schemeY . Let πX , πY be the projections from X ×Z Y to X and Y respectively. Then F G is definedto be π∗XF ⊗ π∗Y G. In particular, OPm×Pn(a, b) is defined to be OPm (a) OPn(b) (over any base

Z). The Segre morphism Pm × Pn → Pmn+m+n corresponds to the complete linear system forthe invertible sheaf O(1, 1).O(a, b), ; this should

be done earlier

d-tuple embedding,

Veronese embedding

Both of these complete linear systems are easily seen to be base-point-free (exercise). Westill have to check by hand that they are closed immersions. (We will later see, in class 34 [addref!], a criterion for linear series to be a closed immersion, at least in the special case where weare working over an algebraically closed field.)

[forme: Whoops! We haven’t done fibered products yet. Segre is done againlater, so move these notes later.]

Older notes.Examples: maps of graded rings. Maps of affines to graded rings missing the origin. Maps

of graded rings where you don’t worry about the locus not being defined outside of your set ofinterest.

Example of a morphism that doesn’t come from a graded ring.S surjects onto T means ProjT → ProjS.Maps of (quasi) projective morphisms to affine spaces.

20.4.G. Exercise. If S is generated in degree 1, scheme is closed immersion in Pn. [I think weshould already have this!]

Criterion for map to projective space to be a closed immersion. I don’t want to assume thebase field is algebraically closed. Can I put this result here, given that we don’t yet understanddifferentials? Probably not.

Idea to place: Here’s an example from Joe R of f : X → Y surjective morphismof Noetherian schemes, F on Y , such that F 6= 0 but f∗F = 0.

A = k[[x]], Y = SpecA, k = A/m, K = FFA, X = Spec k‘

SpecK. Let M = K/(A · 1) =⊕n≤0k · xn

20.5 +Rapid introduction to Geometric Invariant Theory

341

Suppose G acts on L over X (a linearization). Suppose X is affine or projective. ThenX//G := Proj⊕nΓ(X,L⊗n). On the right, we get a finite-generated algebra, because G is areductive group. Suppose for a moment that it is generated in degree 1 upon n-scaling. Then weget

XΓ(L⊗n)//___ PN

and we get a map from the semistable locus (defined as the domain of definition of this rationalmap). [We don’t yet have rational maps.]

This morphism contracts G-orbits! But it is better. (Possibly explain.)Exercises: C∗, with the tautological bundle. λ. Get Pn.P(a, b, c).hardest example: 0

BB@

x1 y1...

...xn yn

1CCA //SL2.

We get the Grassmannian.There are certain obvious equations that it satisfies. Fact: this is it. This is a moduli space!

20.5.1. Brief remark on reductive groups (from Jarod).A linear algebraic group G is linearly reductive if and only if for every action of G on SpecA,

where A is a finitely generated k-algebra, and every invariant ideal J , (A/J)G = AG/(J ∩ AG).“The direction which implies that G is linear reductive is easy, if not trivial.”

20.6 Pre 216 notes

Coherent sheaves push forward to coherent sheaves when the morphism is projective.

20.6.1. Theorem. — If f : X → Y is projective, and F is a coherent sheaf on X, then f∗Fis coherent on Y . 14 pushforward of coherent

sheavesThis is currently mentioned twice, see also Theorem 23.6.2. Is this best left forafter discussing higher pushforwards? In particular, the ring of global sections of a projectivescheme is finitely generated.

We may see later that this is true in more general circumstances, when f is proper (Theo-rem 21.6.3).

Proof. It suffices to prove this over an affine SpecR ⊂ Y . For convenience, we may take Y =SpecR. Factor X → Y into X → PnR → SpecR. It suffices to prove the result i) for closedimmersiona, and ii) for PnR → SpecR.

i) As usual, we’ll assume our original Y was locally Noetherian, so R is Noetherian. If M isa finitely generated (R/I)-module, then it is obviously a finitely generated R-module.

ii) We’ll use cohomology facts (Section 23.11). We’ll use the fact that H i(PnR,O(m)) is afinite number of copies of R. We also use the fact (XXXX) that F is surjected onto by someO(n)⊕m, and that cohomology vanishes once you get high enough. Then say

0→ K→ O(n)⊕m → F → 0.

where K is coherent. Induct down from the top.

20.6.2. Remarks. If OX are Y -flat (define here or later, using affine communication theo-rem 7.3.3) then pullback is exact. Also, if H is a flat quasicoherent sheaf on Y , and 0 → F →G → H → 0 is an exact sequence of coherent sheaves on Y , then 0→ f∗F → f∗G → f∗H → 0 isa short exact sequence of quasicoherent sheaves on X. [Again, I need to define flat.]

In theorem on pullbacks with many parts: Place further ahead!: Pullback of cohomologyclasses, if X → Y is a morphism of quasicompact separated schemes. Precisely: suppose π : X → pullback of H∗(F)

14t:projpushforwardcoherent

342

Y , and F is a sheaf on Y . Show that there is a natural map Hi(Y,F)→ Hi(X, π∗F).Define . [Does this have a

name?]20.6.A. Less important exercise. show that if π : Y × Z → Y is the projection, thenHi(Y,F) → Hi(Y × Z, f∗(F)) is injective. [Reason: consider Y → Y × Z → Y .] Show thatthis isn’t true for general morphisms. [Counterexample: A1

‘A1 → P1; such an example can be

produced from any affine cover!.]

20.6.3. Unimportant remark. We give the third definition, which we will not use hereafter. Wework in the category of OX -modules and OY -modules. First I have to define f−1. This is theleft-adjoint of f∗ in categories of sheaves of sets. Then f∗G = f−1F ⊗f∗OY

OX . We show thatthis satisfies the adjoint property. By the definition of morphism of schemes (and indeed ringedspaces), we are given an element of Hom(OY , f∗OX) = Hom(f−1OY ,OX). The left (resp. right)side of equation is in the category of sheaves of rings on Y (resp. X). Then

HomOX(f−1G ⊗f−1OY

OX ,F) = Homf−1OY(f−1G,F)

= HomOY(G, f∗F)

In that last statement, we need that f−1 and f∗ are adjoint also in the category of ringed spaces,where the ringed structures are compatible!

][primordial:

CHAPTER 21

Constructing schemes by universal properties II:

Sheafy spec and proj, and projective morphisms

We now use universal properties to define two more useful constructions, Spec of a sheafof algebras A, and Proj of a sheaf of graded algebras A∗. We’ll see that affine morphisms areprecisely those of the form SpecA → X, and so we’ll define another very useful class of morphismsto be those of the form ProjA∗ → X.

21.1 Sheafy Spec of a (quasicoherent) sheaf of algebras

Spec[forme: I’m a bit nervous about the quasiaffine case. Actually, it isn’t bad I

think. Define it properly: something which extends to some affine.][Say: we now relativize the notion of Spec and Proj. Earlier, we did them over a base field,

and more generally over a base ring. Now we will define them over a general scheme.]Given an A-algebra, B, we can take its Spec to get an affine scheme over SpecA: SpecB →

SpecA. I’ll now give a universal property description of a globalization of that notation. We willtake an arbitrary scheme X, and a quasicoherent sheaf of algebras A on it. We will define how totake Spec of this sheaf of algebras, and we will get a scheme SpecA → X that is “affine over X”,i.e. the structure morphism is an affine morphism.

We will [actually, not!] do this as you might by now expect: for each affine on X, we useour affine construction, and show that everything glues together nicely. [Problem: the target

isn’t necessarily quasicompact!] We do this instead by describing SpecA → X in terms of a gooduniversal property: given any morphism π : Y → X along with a morphism of OX -modules

α : A → π∗OY ,

there is a unique map Y → SpecA factoring π, i.e. so that the following diagram commutes, Spec

Y

π

???

????

?∃!f // SpecA

β||yy

yyyy

yy

X

and an isomorphism φ : A → β∗OSpecA inducing α. [Say more: f : OSpecA → f∗OY and

β∗OSpecA →A.]

[Ben makes a good point: rings ← schemes via Spec and Γ, and these functors are adjoints.Say this back at an earlier spot!]

(For experts: we need OX -modules, and to leave our category of quasicoherent sheaves onX, because we only showed that the pushforward of quasicoherent sheaves are quasicoherent forcertain morphisms, where the preimage of each affine was a finite union of affines, the pairwiseintersection of which were also finite unions. This notion will soon be formalized as quasicompact

and quasiseparated.)At this point we’re getting to be experts on this, so let’s show that this SpecA exists. In the

case where X is affine, we are done by our affine discussion. In the case where X is quasiaffine,[Caution here! we may have a problem here; i : U → X need not be quasicompact, so we may

343

344

have a definitional problem with i∗A. Say more!] we are done for the same reason as before. Andfinally, in the case where X is general, we are done once again!

In particular, note that SpecA → X is an affine morphism.

21.1.A. Exercise. Show that if f : Z → X is an affine morphism, then we have a naturalisomorphism Z ∼= Specf∗OZ of X-schemes.

Hence we can recover any affine morphism in this way. More precisely, a morphism is affineif and only if it is of the form SpecA → X.affine morphisms as

Spec21.1.B. Exercise (Spec behaves well with respect to base change). Suppose f : Z → Xis any morphism, and A is a quasicoherent sheaf of algebras on X. Show that there is a naturalisomorphism Z ×X SpecA ∼= Specf∗A.

An important example of this Spec construction is the total space of a finite rank locally free

sheaf F , which is a vector bundle. It is Spec Sym∗ F∨.vector bundle

21.1.C. Exercise. Show that this is a vector bundle, i.e. that given any point p ∈ X, there isa neighborhood p ∈ U ⊂ X such that Spec Sym∗ F∨|U ∼= AnU . Show that F is isomorphic to thesheaf of sections of it.

As an easy example: if F is a free sheaf of rank n, then Spec Sym∗ F∨ is called AnX , gener-alizing our earlier notions of AnA. As the notion of a free sheave behaves well with respect to baseAnXchange, so does the notion of AnX , i.e. given X → Y , AnY ×Y X ∼= AnX .

Here is one last fact that might come in handy.

21.1.D. Exercise. Suppose f : SpecA → X is a morphism. Show that the category of

quasicoherent sheaves on SpecA is “essentially the same as” (=equivalent to) the category of

quasicoherent sheaves on X with the structure of A-modules (quasicoherent A-modules on X).The reason you could imagine caring is when X is quite simple, and SpecA is complicated.

We’ll use this before long when X ∼= P1, and SpecA is a more complicated curve. (I drew a picture

of this.)[[H] Ex. II.5.17(e)][forme:

21.1.E. Important exercise. The tautological map An+1 − (0, . . . , 0) → P1 is a linebundle. [That’s not quite right.] Show that it is O(−1). (For this reason, O(−1) isoften called the tautological bundle of Pn.) 1 ]tautological bundle

21.1.1. Older notes: Sheafy Spec of a (quasicoherent) sheaf of algebras. F is aquasicoherent sheaf of algebras. Idea: do it affine by affine and glue together.

Instead: we construct the category. Y → X with F → π∗OY .

21.1.F. Exercise. Suppose A is a sheaf of graded algebras on a scheme Y , with S1 coherent,

and π : X → Y is any morphism, then show that Specπ∗A ∼=“SpecA

”YX. (More precisely,

there is a natural isomorphism between the two schemes. Less precisely, it behaves the wayyou want it to, functorially in both π and A.) Show the analogous result for sheaves of gradedalgebras (generated in degree 1, which is coherent) and Proj. [I think this is needed to show thatProj-projective morphisms behave well under base change.

21.2 Sheafy Proj of a sheaf of graded algebras

We’ll now do a global (or “sheafy”) version of Proj, which we’ll denote Proj.Suppose now that S• is a quasicoherent sheaf of graded algebras of X. To be safe, let me

assume that S• is locally generated in degree 1 (i.e. there is a cover by small affine open sets,where for each affine open set, the corresponding algebra is generated in degree 1), and S1 is finitetype. We will define ProjS•.Proj

1tautbundle

345

The essential ideal is that we do this affine by affine, and then glue the result together.[forme: Hartshorne ducks this!] But as before, this is tricky to do, but easier if you statethe right universal property.

As a preliminary, let me re-examine our earlier theorem [ref], that “Maps to Pn correspondto n + 1 sections of an invertible sheaf, not all vanishing at any point (= generated by globalsections, 19.2.B) modulo sections of O∗X .”

I will now describe this in a more “relative” setting, where relative means that we do this withmorphisms of schemes. We begin with a relative notion of base-point free. Suppose f : Y → X isa morphism, and L is an invertible sheaf on Y . We say that L is relatively base point free if forevery point p ∈ X, q ∈ Y , with f(q) = p, there is a neighborhood U for which there is a sectionof L over f−1(U) not vanishing at q. Similarly, we define relatively generated by global sections ifthere is a neighborhood U for which there are sections of L over f−1(U) generating every stalkof f−1(U). This is admittedly hideous terminology. (One can also define relatively generated byglobal sections at a point p ∈ Y . [better ref!:] See class 16 where we defined these notions in anon-relative setting. In class 32, this will come up again.) Explain the idea? Exercise: make themverify that this is the same as f∗f∗F → F being surjective. Direct them to lecture 32. More relatively base point

free, generatedgenerally, we can define the notion of “relatively generated by global sections by a subsheaf off∗L”. [forme: Awk! We only know f∗L is quasicoherent when f is qcqs.]

Definition. (ProjS•,OProjS•(1)) → X satisfies the following universal property. Given any

other X-scheme Y with an invertible sheaf L, and a map of graded OX -algebras

α : S• → ⊕n=0π∗L⊗n,

[Should this be swapped, as in Spec? Or swap Spec back?] Note that there is such a graded

algebra! such that L is relatively generated by the global sections of α(S1) [awkward!], there isa unique factorization

Y

π

???

????

?∃!f // ProjS•

β

||xxxxxxxx

X

and a canonical isomorphism L ∼= f∗OProjS•(1) and a morphism S• → ⊕nβ∗O(n) inducing α.

In particular, ProjS• comes with an invertible sheaf OProjS•(1), and this O(1) should be

seen as part of the data.This definition takes some getting used to.[Remark: Where does L come from? Answer: think about the big theorem of maps to

projective space.]But we prove this as usual!We first deal with the case where X is affine, say X = SpecA, S• = S•. You won’t be

surprised to hear that in this case, (ProjS•,O(1)) satisfies the universal property.We outline why. Clearly, given a map Y → ProjS•, we get a pullback map α. Conversely,

given such a pullback map, we want to show that this induces a (unique) map Y → ProjS•. Nowbecause S• is generated in degree 1, we have a closed immersion ProjS• → Proj Sym∗ S1. Themap in degree 1, S1 → π∗L, gives a map Y → Proj Sym∗ S1 by our magic theorem “Maps to Pn

correspond to n+ 1 sections of an invertible sheaf, not all vanishing at any point (= generated byglobal sections, 19.2.B) modulo sections of O∗X .”

[Jason points out: S1 may not be locally free!][Because the map is actually in this ring with these relations, it satisfies this closed subscheme

fact.]

21.2.A. Exercise. Complete this argument that if X = SpecA, then (ProjS•,O(1)) satisfiesthe universal property.

21.2.B. Exercise. Show that (ProjS•,O(1)) exists in general, by following the analogousuniversal property argument: show that it exists for X quasiaffine, then in general.

21.2.C. Exercise (Proj behaves well with respect to base change). Suppose S• is aquasicoherent sheaf of graded algebras on X satisfying the required hypotheses above for ProjS•

346

to exist. Let f : Y → X be any morphism. Give a natural isomorphism

(Projf∗S•,OProjf∗S•(1)) ∼= (Y ×X ProjS•, g∗OProjS•(1)) ∼=

where g is the natural morphism in the base change diagram

Y ×X ProjS• g //

ProjS•

Y // X.

2

21.2.1. Definition. If F is a finite rank locally free sheaf on X. Then Proj Sym∗ F is called

its projectivization. If F is a free sheaf of rank n+ 1, then we define PnX := Proj Sym∗ F . (Then

PnSpecA agrees with our earlier definition of PnA.) Clearly this notion behaves well with respect to

base change.projectivization of a lo-

cally free sheaf, PnX This “relative O(1)” we have constructed is a little subtle. Here are couple of exercises togive you practice with the concept.

21.2.D. Exercise. Proj(S•[t]) ∼= SpecS•‘

ProjS•, where SpecS• is an open subscheme, and

ProjS• is a closed subscheme. [Say more about this disjoint union: As a set: Spec open, Proj

is closed, both with the subspace topology. As schemes,...] Show that ProjS∗ is an effective

Cartier divisor, corresponding to the invertible sheaf OProjN (1). (This is the generalization of the

projective and affine cone. At some point I should give an explicit reference to our earlier exerciseon this.)

21.2.E. Exercise. Suppose L is an invertible sheaf on X, and S• is a quasicoherent sheafof graded algebras on X satisfying the required hypotheses above for ProjS• to exist. Define

S′∗ = ⊕n=0Sn ⊗Ln. Give a natural isomorphism of X-schemes

(ProjS′∗,OProjS′∗(1)) ∼= (ProjS•,OProjS•(1) ⊗ π∗L),

where π : ProjS• → X is the structure morphism. In other words, informally speaking, the Proj

is the same, but the O(1) is twisted by L.

21.3 Projective morphisms

[forme: to place: Use the Cancellation Theorem 12.1.19 for properties of mor-phisms to prove: A morphism (over Spec k) from a projective k-scheme to a separatedk-scheme is always projective.

finite morphisms are projective. closed immersions are projective.Show that Any quasiprojective morphism is separated., by referring to Proposi-

tion 12.1.14.[Make sure I state and prove: Composition of projective morphisms should be

projective.]]If you are tuning out because of these technicalities, please tune back in! I now want to define

an essential notion.Recall that we have recast affine morphisms in the following way: X → Y is an affine

morphism if X ∼= SpecA for some quasicoherent sheaf of algebras A on Y .I will now define the notion of a projective morphism similarly.

2sProjbasechange

347

21.3.1. Definition. A morphism X → Y is projective if there is an isomorphism

X∼ //

???

????

?ProjS•

||yyyyyyyy

Y

for a quasicoherent sheaf of algebras S• on Y satisfying the required hypothesis for Proj to exist. projective morphism

Two warnings! 1. Notice that I didn’t say anything about the O(1), which is an importantdefinition. The notion of affine morphism is affine-local on the target, but this notion is not affine-local on the target! (In nice circumstances it is, as we’ll see later. We’ll also see an example wherethis is not.) 2. Hartshorne gives a different definition; I’m following the more general definitionof Grothendieck. But again, these definitions turn out to be the same in nice circumstances.

This is the “relative version” of ProjS• → SpecA.

21.3.A. Exercise. Show that closed immersions are projective morphisms. (Hint: Suppose theclosed immersion X → Y corresponds to OY →OX . Consider S0 = OX , Si = OY for i > 1.)

21.3.B. Exercise (suggested by Kirsten). Suppose f : X → PnS [is a closed immersion?]where S is some scheme. Show that the structure morphism π : X → S is a projective morphismas follows: let L = f∗OPn

S(1), and show that X = Projπ∗L⊗n.3

Proj

21.3.2. Older notes. 21.3.C. Exercise. π∗OSpecA = A where π is the structure mor-

phism.Left over notes to place: Do some examples! Twisted cubic, Segre, Veronese. Also, when

you twist A by L, you get the same thing, but O(1) changes. [Refer to Exercise 9.1.T, where thetwisted cubic was defined.]

21.4 Older version: Projective morphisms

I have a newer outline, plus some older notes.I haven’t decide what definition to make for projective morphisms. I will likely put some of

these thoughts in a remark at the end of the final section.1. Hartshone’s definition: X → Y is projective if there is a closed immersion X → PnY . I will

call this Hartshorne-projective. It is not local on the base.2. Proj-projective. We say π : (X,L)→ Y is Proj-projective if π∗L is a coherent sheaf, and

⊕nπ∗L⊗n is generated in degree 1, and the induced map X → P ⊕n π∗L⊗n is an isomorphism.This is local on the base.

3. Weakly Proj-projective. Same thing, except “there exists an L”. This is not local on thebase.

4. Locally # 3.Then Hartshorne projective implies Proj-projective implies weakly Proj-projective implies

locally weakly Proj-projective. I want my criterion to be local on the base. All are preserved bybase change, thanks to Exercise 21.2.C.

Finite morphisms are Proj-projective, where L = OX , hence satisfy 2–4, so long as mydefinition of finite is that the pushforward is coherent, and not just finitely generated. This mayforce me to change my definition of finite morphism. [Does finite imply Hartshorne-projective? Iwould guess not, but don’t know of a counterexample.] 4 finite morphisms are

projectiveI also want projective morphisms to be closed under composition [H, Ex. II.4.9]. Thenproducts of projective morphisms are projective 11.4.E. Also, the Cancellation Theorem 12.1.19

applies.In all cases, we will get that they are separated and finite type and universally closed hence

proper. Also that pushward of coherent is coherent, quasi-finite and projective implies finite,

3recoverproj

4finiteproj

348

21.4.1. Older notes.We can now define projective morphism, tentatively, as f : X → Y with invertible sheafprojective morphism

L on X such that the natural map X → P⊕n≥0 f∗Ln is an isomorphism. (Fix this: we also needthis map! For this we need that L generate fibers. But then this need not be generated in degree1. Ack.)

Note that this generalizes the notion of a projective k-scheme; so we redefine projectivek-scheme.

Now come my earlier notes.Define PnR. Show that projective morphisms locally factor through this.

Exercise: xy − wz in P3 is isomorphic to P1 × P1. By cutting and pasting. See later forcategorical slick proof.

Segre embedding. Product of projective schemes over k are projective, using Segre.Define weakly projective morphisms (in sense of Hartshorne). (Caution: this is unusual

notation.)Using Segre: preserved under base change, composition. Not clearly local on base.Automatically finite type. They are separated, see EGA I.5.5.6 to show this. Mumford Red

book p. 39 exercise.Useful properties of weakly projective morphisms: closed immersions are projective. Com-

position of weakly projective morphisms is projective. Stable under base extension. Products ofweakly projective morphisms [H, Ex. 4.9] via Segre embedding.

Projective bundle. (Can I say what this means yet?)Projective morphisms. (Following EGA) X → Y is projective if X is isomorphic to

ProjY

(S), where S is a quasicoherent graded sheaf generated in degree 1 such that S1 is a finite-

type OY -module.Also preserved under base change, composition etc. Separated. Added bonus: Now local on

the base.

21.4.A. Exercise. Show that finite morphisms are projective. 5 [Caution: not obviously so forfinite implies projectiveHartshorne-projective.]

Corollary: Every curve has a birational model which is smooth and projective: C 99K A1.Use normalization in the function field. C is projective. It is smooth.

Define quasiprojective morphism. Exercise: nice properties.Pushforward of a coherent sheaf under projective morphisms is coherent. This generalizes

(“in families”) the fact that a coherent sheaf on a projective scheme has a finite-dimensional spaceof global sections.

The following theorem is remarkably elementary. We will use this later.

21.4.2. Theorem. — PnR → SpecR is a closed morphism. 6

In other words, if Z is a closed subset of PnR then the projection to SpecR is a closed subset.We use the fundmental theorem in elimination theory, Theorem A.2.2. [FINISH THIS.] THIS

IS NOW 14.1.4.Better: follow Mumford [M, II.7 Thm. 2].

Here is a sketch of an argument.Suppose Z → PnA is a closed immersion, cut out by the graded ideal I∗ ⊂ S• := A[x0, . . . , xn],

and π is the projection to PnA → SpecA, and by abuse of notation, Z → SpecA. We prove theresult by showing that for any prime ideal p ⊂ A such that [p] /∈ π(Z), there is an open subset ofZ containing [p] and not meeting π(Z). Figure 1 may be helpful. We will do this by producinga function f ∈ Z such that f /∈ p, and π∗f vanishes on Z. Then D(f) will be our desired opensubset. 7

A picture of a rectangle (PnA) with an arrow to a line (SpecA) with a vertical line in the

rectangle (π−1[p]) and another closed subset of the rectangle (Z).

Figure 1. Caption needed

5finitemorphismsprojective, H.E.II.4.1 is that finite morphisms are proper, which I state is 14.1.2

6M.II.7.T.2, also M.I.9.T.1

7projclosedfig

349

[Assume p is maximal for now.]Let x0, . . . , xn be the projective coordinates on PnA. On the patch D(xi), take coordinates

xj/i := xj/xi (where j 6= i). Then in A[x0/i, . . . , xn/i], Z∩Ui and V (p) are disjoint closed subsets,

so there is a function on the scheme Z ∩Ui ∪ V (p) that is 1 on Z ∩Ui and vanishing on V (p). Allfunctions on this closed subscheme of AnA are restrictions of functions on the total space, so if thisfunction is ai, we can write

ai +X

mijgij = 1 in A[x0/i, . . . , xn/i]

where ai ∈ I(Z ∩Ui), mij ∈ p, and gij ∈ A[x0/i, . . . , xn/i]. As xj/i = xj/xi, we can multiply thisequation by a high enough power of xi so that we have an equation

a′i +X

mijg′ij = xNi

where a′i ∈ IN and g′ij ∈ SN . Choose one N large enough so that this is true for all i. Then

if N ′ > n(N − 1), any monomial of degree N ′ in A[x0, . . . , xn] has a factor of the form xNi forsome i. Then we have IN′ + pSN′ = SN′ . By Nakayama’s lemma version 1 15.3.1, there exists anf ∈ 1 + p with fSN′/IN′ = 0, i.e. we have found an f vanishing on Z, with value 1 on p. ThenD(f) on SpecA does the trick.

Let’s now patch this. We do the entire thing in Ap, and we obtain that IN′⊗AAp +pSN′⊗AAp = SN′ ⊗A Ap. Thus there is an f ∈ 1 + pAp such that f(SN′/IN′ ) ⊗A Ap = 0. We mayas well multiply by an element of A − p to clear the denominator of f , so f ∈ A − p, and stillf(SN′/IN′ ) ⊗A Ap = 0. By multiplying by another element of A − p, we get f(SN′/IN′ ) = 0.Then we are done! I obviously need to think more about this!

21.4.B. Important exercise. Show that projective morphisms are universally closed. (Defineuniversally closed.) 8 universally closed; pro-

jective implies UCRemark: In fact the scheme-theoretic pushforward [not defined yet] of a closed subscheme ofPnR is the closed subscheme of SpecR we are describing below.

Proof. Let x0, . . . , xn be projective coordinates on PnR. Z is cut out by some hyomogeneouspolynomials f1, . . . , fr ∈ R[x0, . . . , xn]. We want the image of the common zeros, i.e. we want toknow that, given a field Spec k→ R, whether k[x0, . . . , xn]/(f1, . . . , fr) has a point. (Here we areusing the projective Nullstellensatz.)

FINISH THIS! Possible motivation: if all are degree 1, we use linear algebra. If 2 polynomialsin 2 variables, use the resultant.

21.4.3. Put somewhere: A projective [proper] morphism f : X → Y of finite type k-schemes is aclosed immersion iff it separates points and tangent vectors. 9 [Repeated at 23.17.C.] Note thatwe need proper. Example (Brian Osserman): consider P1 to the nodal cubic, where x, y are thepreimages of the node. Let X = P1 − x, and Y the nodal cubic.

You now know enough to define the blow-up. We’ll hold off discussing it until later (Chap-ter 27), because we have no pressing need for it. But feel free to read ahead if you want.

Put somewhere: Suppose X → Y is locally projective. Is it necessarily projective? No:Hironaka’s example 29.3.7 [really I should give the blow-up statement 27.5.A] gives a locallyprojective morphism X → P3 that is not projective. Reason: because then X would be projective.State that this is false here, and refer forward. (Suppose f is proper and the fibers are projective.Then f is not necessarily projective — same example. Put this after describing properness.) Hironaka’s example

8projectiveUC

9separatepointstangentvectors1, H.P.II.7.3

350

21.5 The categorical perspective (Place later?)

Morphisms of schemes give natural transformations of functors.The functor of points: R-valued points of a scheme X are morphisms SpecR→ X. We can

do the same thing for Y -valued points, where Y is a scheme. (Caution about notation!)Brian Osserman notes that this is a good motivation for schemes. The points of a scheme

are weird, but the R-valued points are less weird.Earlier notes: Fibered products behave well from the functor of points point of view. For

example, the wx− yz = 0 example becomes trivial.Put somewhere: Morphisms from a DVR to a scheme (see my notes on p. 97 of my annotated

Hartshorne, and [H, Lemma II.4.4]). Morphisms from Spec of the dual numbers (??) to a k-scheme. Morphisms to A1.

21.6 ?? Chow’s Lemma

[There were some facts about proper things that are now moved to a section on “Chow’slemma”, 21.6, subsection 23.17.1.]

[Very likely this should be a section on “proper facts”, with Chow’s lemma as the heart. Butcurrently the “proper facts” are in Subsection 23.17.1.]

10Do Chow’s Lemma, following Mumford. (Or EGA II.5.6 p. 107)?) [I mention Chow’sLemma earlier.] Use this to prove useful things about proper morphisms. Chow’s lemma is [H,Ex. II.4.10].Chow’s Lemma [M, p. 85-89]; see if he uses the fact that you’re working over a field.Mumford’s statement: If X is complete variety over k, then there is a closed subvariety Y ⊂ Pnkfor some n and a surjective birational morphism π : Y → X.Chow’s lemma

Here is an outline of Dung’s explanation from June 11, 2007.Cover X with affines X = ∪Ui. Let U∗ = ∩Ui, which is a subvariety of U1 × · · · × Uk. Let

Y be the closure of U∗ in this space. It is clearly projective.Required: the existence of a morphism Y → X.Now U∗ → X × Y ; let Y be the closure of the image. We quickly show that Y → X is

surjective and birational.Required: q : Y → Y is an isomorphism.Now things get tricky.Consider Y → X × U1 × · · · × Uk. Let si : Y → X × U i, which takes U∗ to U∗ (put into a

square diagram).

Thus si(Y ) is the closure of U∗ in X × U i.Lemma: si(Y )× (Ui × U i) = si(Y )× (X × Ui).Mumford has a picture here, and states the lemma more nicely as: U ⊂ S, T . U → S × T ,

and U is the closure. Then lemma: U ∩ U × T = U × (S × V ) = ∆U (where ∆ is the “diagonalmorphism”). Proof: RTP ∆U is closed in U × T . But this is the graph of the map U → T , andwe use the separatedness of T . End lemma.

Hence we have the key equality

(47) Y ∩ (Ui × U1 × · · · × Un) = Y ∩ (X × U1 × · · · × Ui × · · · × Un).

As i runs from 1 thorugh n, the left hand side of (47) cover Y .

Define Yi := Y ∩ (U1 × · · · × Ui × · · · × Un). q−1(Yi) = Yi because of the right hand side of(47) (tricky!). As q is surjective, the Yi cover Y .

Required: Yi ∼= Yi. (Better: name the morphism, and say it is an isomorphism.)Define

σi : U1 × · · · × Ui × · · · × Un → X × U i × · · · × Ui × · · · × Unby (u1, . . . , un) 7→ (ui, u1, . . . , un). Now σi : Yi → Yi. Clearly q σi = id.

Required: σi q (“on Yi”) = id. (This requires thought, but is okay.

Finally, the σi agree on the overlap. (Basically, we have covered Y with the Yi, and we have

covered Y with the Yi, and we have q : Y → Y , and we have inverses σ : Yi → Yi.)

10chowslemma

351

21.6.1. Chow’s Lemma (finite type version). — Given finite type X → S where baseis locally Noetherian, then there is X ′ → X which is surjective, projective, and quasiprojectiveover S. There is also an isomorphic open set. Obviously if X is reduced, we may take X ′ to bereduced.

I have no idea why we like this version.

21.6.2. Chow’s Lemma (proper version). — If further X is proper over S, then X ′ → Sis quasiprojective and proper, hence projective. 11

Things we want:

21.6.3. Theorem. — Suppose π : X → Y is a proper morphism of locally Noetherian schemes.Higher pushforwards of coherent sheaves are coherent. 12 pushforward of coherent

sheaves[forme: Weirdness: If you want to extend classical notions, you can go fromproper to projective, or from lN to general schemes, but you can’t do both withoutpossibly losing this. Is it known that you can drop this?]

In particular, if X is proper/k, H0(X,F) is finite-dimensional. And if π is proper, then π∗sends coherents to coherents.

This will be used in the cohomology and base change discussion.

Proof. This is local on Y so we may assume Y = SpecB, where B is a Noetherian ring. We dothis by induction of the dimension of X. We then do it by induction on the number of irreduciblecomponents of X. We then do it by the amount of fuzz on X (the smallest power of the sheaf ofnilpotents that is 0). Note that if it is true for any two in an exact sequence, then it is true forthe third. Using this, we can reduce the fuzz by one. And if there is no fuzz, we can reduce thenumber of components by one. Finally, we have the case where X is integral. Finally: Lemma:H0(X,F) → H0(X′, π∗F) is injective.

Remark: that lemma then implies that if X is a geometrically integral proper scheme overk, then it has only constant global sections.

Remark: this also shows dimensional vanishing.I think these are the only facts we will need about proper morphisms in the future!(One of the few reasons we like proper things: this notion is local on the base.

][primordial:

11Chowslemma

12t:properpushforwardcoherent, EGA.III.3.2, uses devissage EGA.III.3.1

CHAPTER 22

Differentials

1

[Point out that our two nice exact sequences suggest that there should be a long exact se-quence. In fact, the cohomology groups are called Andre-Quillen homology modules. See Eisenbudp. 386. It may be useful to mention that the first one vanishes when the morphism is smooth.Also, we should expect the next term to depend only on two of the schemes, and indeed this givesleft-exactness there. Should I mention the cotangent complex?]

[A third nice exact sequence, to give as an exercise. Suppose X → Y → Z is a completeintersection in a complete intersection. Show that 0 → NX/Y → NX/Z → f∗NY/Z → 0. What

are the right hypotheses? We have f∗N∨Y/Z

→ N∨X/Z

→ N∨X/Y

. We need X → Y is a local

complete intersection, to get exactness on the left. We need Y → Z to be unramified, so that weget exactness on the right. We want NY/Z to be locally free, so that we can dualize this sequence

and retain exactness.][Somewhere here, a remark: define unramified, as the analogue of immersion, but state that

we’ll talk about it in detail later, because it is naturally in a family of three notions, and the othertwo require flatness.] [Caution when globalizing differential exact sequences from affines to thegeneral case: argue that the morphisms localize well.] [Define Ωf , but avoid using it.]

I prefer to start any topic with a number of examples, but in this case I’m going to spend afair amount of time discussing technicalities, and then get to a number of examples. Here is themain message I want you to get. Differentials are an intuitive geometric notion, and we’re goingto figure out the right description of them algebraically. I find the algebraic manifestation a littlenon-intuitive, so I always like to tie it to the geometry. So please don’t tune out of the statements.Also, I want you to notice that although the algebraic statements are odd, none of the proofs arehard or long.

This topic could have been done as soon as we knew about morphisms and quasicoherentsheaves.

22.1 Motivation and game plan

Suppose X is a “smooth” k-variety. We hope to define a tangent bundle. We’ll see that theright way to do this will easily apply in much more general circumstances.• We’ll see that cotangent is more “natural” for schemes than tangent bundle. This is similarto the fact that the Zariski cotangent space is more natural than the tangent space (i.e. if A is aring and m is a maximal ideal, then m/m2 is “more natural” than (m/m2)∨). So we’ll define thecotangent sheaf first. cotangent sheaf• Our construction will work for general X, even if X is not “smooth” (or even at all nice, e.g.finite type). The cotangent sheaf won’t be locally free, but it will still be a quasicoherent sheaf.• Better yet, this construction will work “relatively”. For any X → Y , we’ll define ΩX/Y , aquasicoherent sheaf on X, the sheaf of relative differentials. This will specialize to the earlier sheaf of relative differen-

tialscase by taking Y = Spec k. The idea is that this glues together the cotangent sheaves of the fibersof the family. (I drew an intuitive picture in the “smooth” case. I introduced the phrase “vertical(co)tangent vectors”.) vertical (co)tangent vec-

tors[Another comment to include: functions on a space are contravariant.]

1c:differentials

353

354

22.2 The affine case: three definitions

We’ll first study the affine case. Suppose A is a B-algebra, so we have a morphism of ringsφ : B → A and a morphism of schemes SpecA → SpecB. I will define an A-module ΩA/B inthree ways. This is called the module of relative differentials or the module of Kahler differentials.The module of differentials will be defined to be this module, as well as a map d : A → ΩA/Bsatisfying three properties.Leibniz rule, module

of relative differentials,

module of Kahler differ-

entials

(i) additivity. da + da′ = d(a + a′)(ii) Leibniz. d(aa′) = a da′ + a′da

(iii) triviality on pullbacks. db = 0 for b ∈ φ(B).Trivial exercise. Show that d is B-linear. (In general it will not be A-linear.)Exercise. Prove the quotient rule. Prove the chain rule.As motivation, think of the case B = k. So for example, dan = nan−1da, and moreΩ

generally, if f is a polynomial in one variable, df(a) = f ′(a) da (where f ′ is defined formally: iff =

Pcix

i then f ′ =Pciix

i−1).I’ll give you three definitions of this sheaf in the affine case (i.e. this module). The first is a

concrete hands-on definition. The second is by universal property. And the third will globalizewell, and will allow us to define ΩX/Y conveniently in general.

The first two definitions are analogous to what we have seen for tensor product. Recallthat there are two common definitions of ⊗. The first is in terms of formal symbols satisfyingsome rules. This is handy for showing certain things, e.g. if M → M ′ is surjective, then so isM ⊗N →M ′ ⊗N . The second is by universal property.

22.2.1. First definition of differentials: explicit description. We define ΩA/B to be finite

A-linear combinations of symbols “da” for a ∈ A, subject to the three rules (i)–(iii) above. Forexample, take A = k[x, y], B = k. Then a sample differential is 3x2 dy + 4dx ∈ ΩA/B . We have

identities such as d(3xy2) = 3y2 dx+ 6xy dy.Key fact. Note that if A is generated over B (as an algebra) by xi ∈ A (where i lies in

some index set, possibly infinite), subject to some relations rj (where j lies in some index set,and each is a polynomial in some finite number of the xi), then the A-module ΩA/B is generated

by the dxi, subject to the relations (i)—(iii) and drj = 0. In short, we needn’t take every singleelement of A; we can take a generating set. And we needn’t take every single relation among thesegenerating elements; we can take generators of the relations.

22.2.A. Exercise. Verify the above key fact. [Ziyu used the conormal exact sequence.]In particular:

22.2.2. Proposition. — If A is a finitely generated B-algebra, then ΩA/B is a finite type

(=finitely generated) A-module. If A is a finitely presented B-algebra, then ΩA/B is a finitelypresented A-module.

[Ben says: Consider the diagram

B[x] //

##FFFFFFFFF FB[X]dX

// FAdx

A

d

<<

ΩX/R := FAdx/ im θ|R. Show that it satisfies the universal property.]

(“Finitely presented” algebra means finite number of generators (=finite type) and finitenumber of relations. If A is Noetherian, then the two hypotheses are the same, so most of youwill not care.)

Let’s now see some examples. Among these examples are three particularly important kindsof ring maps that we often consider: adding free variables; localizing; and taking quotients. If weknow how to deal with these, we know (at least in theory) how to deal with any ring map.

22.2.3. Example: taking a quotient. If A = B/I, then ΩA/B = 0 basically immediately:da = 0 for all a ∈ A, as each such a is the image of an element of B. This should be believable;in this case, there are no “vertical tangent vectors”.

355

22.2.4. Example: adding variables. If A = B[x1, . . . , xn], then ΩA/B = Adx1⊕· · ·⊕Adxn. 2

(Note that this argument applies even if we add an arbitrarily infinite number of indeterminates.)The intuitive geometry behind this makes the answer very reasonable. The cotangent bundleshould indeed be trivial of rank n.

22.2.5. Example: two variables and one relation. If B = C, and A = C[x, y]/(y2 − x3),then ΩA/B = C dx⊕ C dy/(2y dy − 3x2 dx).

22.2.6. Example: localization. If S is a multiplicative set of B, and A = S−1B, thenΩA/B = 0. Reason: Note that the quotient rule holds. (If b = as, then db = a ds + s da, which

can be rearranged to give da = (s db− b ds)/s2.) Thus if a = b/s, then da = (s db− b ds)/s2 = 0.(If A = Bf for example, this is intuitively believable; then SpecA is an open subset of SpecB, sothere should be no “vertical cotangent vectors”.)

22.2.B. Exercise: localization (stronger form). If S is a multiplicative set of A, show that

there is a natural isomorphism ΩS−1A/B∼= S−1ΩA/B . (Again, this should be believable from

the intuitive picture of “vertical cotangent vectors”.) If T is a multiplicative set of B, show thatthere is a natural isomorphism ΩS−1A/T−1B

∼= S−1ΩA/B where S is the multiplicative set of A

that is the image of the multiplicative set T ⊂ B. 3 4 [Ziyu used the relative cotangent sequence.][Statement from Kirsten: let T be a multiplicative set of B and let A be a B-algebra. Let S bethe multiplicative set of A that is the image of T ⊂ B. Then there is a natural isomorphismΩS−1A/T−1B

∼= S−1ΩA/B . Proof: Apply part (b) of the next problem to the Cartesian square

S−1A Aoo

T−1B

OO

Boo

OO .]

22.2.C. Exercise. (a) (pullback of differentials) If

A′ Aoo

B′

OO

B

OO

oo

is a commutative diagram, show that there is a natural homomorphism of A′-modules A′ ⊗AΩA/B → ΩA′/B′ . [forme: [Matsumura p. 185 Exercise 1]] An important special case

is B = B′. [One can describe the map on the level of “da’s”. Yu-jong uses universal property:if f : A → A′, then δ : A → ΩA′/B′ given by a → df(a) is a derivation, and hence we get a

natural A-module homomorphism ΩA/B → ΩA′/B′ . Tensor the left with A′ to get an A′-module

homomorphism.][A universal property proof is nice. Ben: ΩA/B → ΩA′/B′ of A-modules. By adjointness of

⊗ and Hom, this gives A′ ⊗A ΩA/B → ΩA′/B′ .]

(b) (differentials behave well with respect to base extension, affine case) If furthermore the abovediagram is a tensor diagram (i.e. A′ ∼= B′ ⊗B A) then show that A′ ⊗A ΩA/B → ΩA′/B′ is an

isomorphism. 5

22.2.D. Exercise. Suppose k is a field, and K is a separable algebraic extension of k. Showthat ΩK/k = 0. (Warning: do not assume that K/k is a finite extension!) [Easy.] [forme:

[Proof: for any α ∈ K, there is a polynomial such that f(α) = 0 and f ′(α) 6= 0. Sinced : K → ΩK/k is a derivation, we have 0 = d(f(α)) = f ′(α)dα, from which dα = 0.

Matsumura p. 185 example 3.] ] 6

2H.E.II.8.2.1

3H.P.II.8.2A2; H cites Matsumura p. 186

4Omegalocalization

5pullbackdiff, H.P.II.8.2A1; H cites Matsumura p. 186. Also H Prop. II.8.10.

6omegasep, used in RH

356

22.2.7. Exercise (Jacobian description of ΩA/B). — Suppose A = B[x1, . . . , xn]/(f1, . . . , fr).

Then ΩA/B = ⊕iBdxi/dfj = 0 maybe interpreted as the cokernel of the Jacobian matrix

(33) J : A⊕r → A⊕n. 7JacobianI now want to tell you two handy (geometrically motivated) exact sequences. The arguments

are a bit tricky. They are useful, but a little less useful than the foundation facts above.

22.2.8. Theorem (the relative cotangent sequence, affine version). — Suppose C →B → A are ring homomorphisms. Then there is a natural exact sequence of A-modules

A⊗B ΩB/C → ΩA/C → ΩA/B → 0.

8

Before proving this, I drew a picture motivating the statement. I drew pictures of two

maps of schemes, SpecAf // SpecB

g // SpecC , where SpecC was a point, SpecB was

A1 (or a “smooth curve”), and SpecA was A2 (or a “smooth surface”). The tangent space toa point upstairs has a subspace that is the tangent space to the vertical fiber. The cokernel isthe pullback of the tangent space to the image point in SpecB. Thus we have an exact sequence0 → TSpecA/ SpecB → TSpecA/ SpecC → f∗TSpecB/ SpecC → 0. We want the correspondingsequence of cotangent vectors, so we dualize. We end up with precisely the statement of theTheorem, except we also have left-exactness. This discrepancy is because the statement of thetheorem is more general; we’ll see that in the “smooth” case, we’ll indeed have left-exactness.[add ref later.]

Proof. (Before we start, note that surjectivity is clear, from da 7→ da. The composition overthe middle term is clearly 0: db → db → 0.) We wish to identify ΩA/B as the cokernel ofA⊗B ΩB/C → ΩA/C . Now ΩA/B is exactly the same as ΩA/C , except we have extra relations:db = 0 for b ∈ B. These are precisely the images of 1⊗ db on the left.

[MAKE SURE to call the map from I/I2 with the name d, as this is what it is.]

22.2.9. Theorem (Conormal exact sequence, affine version). — Suppose B is a C-algebra, I is an ideal of B, and A = B/I. Then there is a natural exact sequence of A-modules9

I/I2δ:i→=1⊗di // A⊗B ΩB/C

a⊗db7→a db // ΩA/C // 0.

Before getting to the proof, some discussion is necessary. (The discussion is trickier than theproof itself!)

[Ben: B/I ⊗ I → B/I ⊗ ΩB/C .] The map δ is a bit subtle, so I’ll get into its details before

discussing the geometry. For any i ∈ I, δi = 1⊗ di. Note first that this is well-defined: If i, i′ ∈ I,i ≡ i′ (mod I2), say i− i′ = i′′i′′′ where i′′, i′′′ ∈ I, then δi−δi′ = 1⊗(i′′ di′′′+ i′′′ di′′) ∈ IΩB/Cis 0 in A ⊗B ΩB/C = (B/I) ⊗B ΩB/C . Next note that I/I2 indeed is an A = (B/I)-module.

Finally, note that the map I/I2 → A⊗BΩB/C is indeed a homomorphism of A-modules: If a ∈ A,

b ∈ I, then ab 7→ 1⊗ d(ab) = 1⊗ (a db+ b da) = 1⊗ (a db) = a(1 ⊗ db).Having dispatched that formalism, let me get back to the geometry. I drew a picture where

SpecC is a point, SpecB is a plane, and SpecA is something smooth in it. Let j be the inclusion.Then we have 0 → TSpecA/SpecC → j∗TSpecB/ SpecC → NSpecB/ SpecC → 0. Dualizing it,

we get 0 → N∨A/B

→ A ⊗ ΩB/C → ΩA/C → 0. This exact sequence reminds me of several

things above and beyond the theorem. First of all, I/I2 will later be the conormal bundle —hence the name of the theorem. Second, in good circumstances, the conormal exact sequence ofTheorem 22.2.9 will be injective on the left. [Give ref.]

22.2.10. Aside: Why should I/I2 be the conormal bundle?. We’ll define I/I2 to be the conormalbundle later, so I’ll try to give you an idea as to why this is reasonable. You believe now thatm/m2 should be the cotangent space to a point in An. In other words, (x1, . . . , xn)/(x1, . . . , xn)2

7JacOmega

8H.P.II.8.3A

9conormalES, H.P.II.8.4A

357

is the cotangent space to ~0 in An. Translation: it is the conormal space to the point ~0 ∈ An.Then you might believe that in An+m, (x1, . . . , xn)/(x1 , . . . , xn)2 is the conormal bundle to thecoordinate n-plane Am ⊂ An+m.10

Let’s finally prove the conormal exact sequence.Proof of the conormal exact sequence (affine version) 22.2.9. We need to identify the cokernel ofδ : I/I2 → A ⊗B ΩB/C with ΩA/C . Consider A⊗B ΩB/C . As an A-module, it is generated by

db (b ∈ B), subject to three relations: dc = 0 for c ∈ φ(C) (where φ : C → B describes B as aC-algebra), additivity, and the Leibniz rule. Given any relation in B, d of that relation is 0.

Now ΩA/C is defined similarly, except there are more relations in A; these are precisely theelements of i ∈ B. Thus we obtain ΩA/C by starting out with A ⊗B ΩB/C , and adding theadditional relations di where i ∈ I. But this is precisely the image of δ!

22.2.11. Second definition: universal property. Here is a second definition that we’ll useat least once, and is certainly important philosophically. Suppose A is a B-algebra, and M is aA-module. An B-linear derivation of A into M is a map d : A→M of B-modules (not necessarilyA-modules) satisfying the Leibniz rule: d(fg) = f dg + g df . As an example, suppose B = k, and derivationA = k[x], and M = A. Then an example of a k-linear derivation is d/dx. As a second example, ifB = k, A = k[x], and M = k. Then an example of a k-linear derivation is d/dx|0.

Then d : A → ΩA/B is defined by the following universal property: any other B-linear

derivation d′ : A→M factors uniquely through d:

A

d

""DDD

DDDD

Dd′ // M

ΩA/B

f

<<yyyyyyyy

Here f is a map of A-modules. (Note again that d and d′ are not! They are only B-linear.) Byuniversal property nonsense, if it exists, it is unique up to unique isomorphism. The candidate Idescribed earlier clearly satisfies this universal property (in particular, it is a derivation!), hencethis is it. [Thus Ω is the “unversal derivation”. I should rewrite this paragraph at some point.Justin points out: the map defined earlier is a derivation, but I never really say that; thus theoriginal map, together with Ω, is a universal derivation.]

The next result will give you more evidence that this deserves to be called the (relative)cotangent bundle.

22.2.12. Proposition. Suppose B is a k-algebra, with residue field k. Then the natural mapδ : m/m2 → ΩB/k ⊗B k is an isomorphism. 11

Proof. By the conormal exact sequence 22.2.9 with I = m and A = C = k, δ is a surjection (as

Ωk/k = 0), so we need to show that it is injection, or equivalently that Homk(ΩB/k ⊗B k, k) →Homk(m/m

2, k) is a surjection. But any element on the right is indeed a derivation from B to k(an earlier exercise from back in the dark ages on the Zariski tangent space), which is precisely anelement of HomB(ΩB/k , k) (by the universal property of ΩB/k), which is canonically isomorphic

to Homk(ΩB/k ⊗B k, k) as desired.

Remark. As a corollary, this (in combination with the Jacobian exercise 22.2.7 above) gives asecond proof of an exercise from the first quarter, showing the Jacobian criterion for nonsingular

varieties over an algebraically closed field.+Aside. If you wish, you can use the universal property to show that ΩA/B behaves well with

respect to localization. For example, if S is a multiplicative set of A, then there is a natural iso-morphism ΩS−1A/B

∼= S−1ΩA/B . This can be used to give a different solution to Exercise 22.2.B

[check if this ref is correct, and if this is true]. It can also be used to give a second definition ofΩX/Y for a morphism of schemes X → Y (different from the one given below): we define it as aquasicoherent sheaf, by describing how it behaves on affine open sets, and showing that it behaveswell with respect to distinguished localization.

10iisquared

11H.P.II.8.7, except he just does it in the local case

358

22.2.13. Third definition. We now want to globalize this definition for an arbitrary morphismof schemes f : X → Y . We could do this “affine by affine”; we just need to make sure that theabove notion behaves well with respect to “change of affine sets”. Thus a relative differential onX would be the data of, for every affine U ⊂ X, a differential of the form

Pai dbi, and on the

intersection of two affine open sets U∩U ′, with representativesPai dbi on U and

Pa′i db

′i on the

second, an equality on the overlap. Instead, we’ll take a different tack. We’ll get what intuitivelyseems to be a very weird definition! I’ll give the definition, then give you some intuition, and thenget back to the definition.

Suppose f : X → Y be any morphism of schemes. Recall that δ : X → X ×Y X is alocally closed immersion [ref — class 9 page 5 in math 216]. Thus there is an open subschemeU ⊂ X ×Y X for which δ : X → U is a closed immersion, cut out by a quasicoherent sheafof ideals I. Then I/I2 is a quasicoherent sheaf naturally supported on X (affine-locally this isthe statement that I/I2 is naturally an A/I-module). We call this the conormal sheaf to X (orsomewhat more precisely,to the locally closed immersion). (For the motivation for this name, see22.2.10.) We denote it by N∨

X/X×Y X. Then we will define ΩX/Y as this conormal sheaf.

(Small technical point for pedants: what does I2 mean? In general, if I and J are quasico-herent ideal sheaves on a scheme Z, what does IJ mean? Of course it means that on each affine,we take the product of the two corresponding ideals. To make sure this is well-defined, we needonly check that if A is a ring, and f ∈ A, and I, J ⊂ A are two ideals, then (IJ)f = IfJf in Af .)

Brief aside on (co)normal sheaves to locally closed immersions. For any locally closed im-mersion W → Z, we can define the conormal sheaf N∨

W/Z, a quasicoherent sheaf on W , similarly,

and the normal sheaf as its dual NW/Z := Hom(N∨,OW ). This is somewhat imperfect notation,

as it suggests that the dual of N is always N∨. This is not always true, as for A-modules, thenatural morphism from a module to its double-dual is not always an isomorphism. (Modules forwhich this is true are called reflexive, but we won’t use this notion.)reflexive sheaf

(co)normal sheaf22.2.E. Exercise: normal bundles to effective Cartier divisors. Suppose D ⊂ X is aneffective Cartier divisor. Show that the conormal sheaf N∨

D/Xis O(−D)|D (and in particular is an

invertible sheaf), and hence that the normal sheaf is O(D)|D . It may be surprising that the normalsheaf should be locally free ifX ∼= A2 andD is the union of the two axes (and more generally ifX isnonsingular butD is singular), because you may be used to thinking that a “tubular neighborhood”being isomorphic to the normal bundle. 12 [0→ O(−2D)→O(−D)→OD(−D)→ 0.]

Let’s get back to talking about differentials.We now define the d operator d : OX → ΩX/Y . Let π1, π2 : X ×Y X → X be the two

projections. Then define d : OX → ΩX/Y on the open set U as follows: df = π∗2f − π∗1f .

(Warning: this is not a morphism of quasicoherent sheaves, although it is OY -linear.) We’ll soonsee that this is indeed a derivation, and at the same time see that our new notion of differentialsagrees with our old definition on affine open sets, and hence globalizes the definition.

][primordial: Before we do, let me try to convince you that this is a reasonable definition

to make. (This paragraph is informal, and is in no way mathematically rigorous.) Say for examplethat Y is a point, and X is something smooth. Then the tangent space to X ×X is TX ⊕ TX :TX×X = TX⊕TX . Restrict this to the diagonal ∆, and look at the normal bundle exact sequence:

0→ T∆ → TX×X |∆ → N∆,X → 0.

Now the left morphism sends v to (v, v), so the cokernel can be interpreted as (v,−v). Thus N∆,X

is isomorphic to TX . Thus we can turn this on its head: we know how to find the normal bundle(or more precisely the conormal sheaf), and we can use this to define the tangent bundle (or moreprecisely the cotangent sheaf). (Experts may want to ponder the above paragraph when Y ismore general, but where X → Y is “nice”. You may wish to think in the category of manifolds,and let X → Y be a submersion.)

Let’s now see how this works for the special case SpecA → SpecB. Then the diagonalSpecA → SpecA⊗B A corresponds to the ideal I of A⊗B A that is the cokernel of the ring map

Xxi ⊗ yi →

Xxiyi.

12nbeffCart

359

The derivation is d : A→ A⊗B A, a 7→ da := 1⊗ a− a⊗ 1 (taken modulo I2). (I shouldn’treally call this “d” until I’ve verified that it agrees with our earlier definition, but bear with me.)

Let’s check that this satisfies the 3 conditions, i.e. that it is a derivation. Two are immediate:it is linear, vanishes on elements of b. Let’s check the Leibniz rule:

d(aa′)− a da′ − a′ da = 1⊗ aa′ − aa′ ⊗ 1− a⊗ a′ + aa′ ⊗ 1− a′ ⊗ a+ a′a⊗ 1

= −a⊗ a′ − a′ ⊗ a+ a′a⊗ 1 + 1⊗ aa′

= (1 ⊗ a− a⊗ 1)(1 ⊗ a′ − a′ ⊗ 1)

∈ I2.

Thus by the universal property of ΩA/B , we get a natural morphism ΩA/B → I/I2 of A-modules.

22.2.14. Theorem. — The natural morphism f : ΩA/B → I/I2 induced by the universal

property of ΩA/B is an isomorphism. 13

Proof. We’ll show this as follows. (i) We’ll show that f is surjective, and (ii) we will describeg : I/I2 → ΩA/B such that g f : ΩA/B → ΩA/B is the identity. [Better: this latter map

g f being the identity shows that f is injective.] Both of these steps will be very short. Thenwe’ll be done, as to show f g is the identity, we need only show (by surjectivity of g) that

(f g)(f(a)) = f(a), which is true (by (ii) g f = id).(i) For surjectivity, we wish to show that I is generated (modulo I2) by a⊗1−1⊗a as a runs

over the elements of A. This has a one sentence explanation: IfPxi ⊗ yi ∈ I, i.e.

Pxiyi = 0 in

A, thenPi xi ⊗ yi =

Pi xi(1 ⊗ yi − yi ⊗ 1).

(ii) Define g : I/I2 → ΩA/B by x⊗ y 7→ x dy. We need to check that this is well-defined, i.e.

that elements of I2 are sent to 0, i.e. we need that“X

xi ⊗ yi”“X

x′j ⊗ y′j”

=X

i,j

xix′j ⊗ yiy′j 7→ 0

wherePi xiyi =

Px′jy′j = 0. But by the Leibniz rule,

X

i,j

xix′j d(yiy

′j) =

X

i,j

xix′jyi dy

′j +

X

i,j

xix′jy′j dyi

=

X

i

xiyi

!0@X

j

x′j dy′j

1A+

X

i

xi dyi

!0@X

j

x′jy′j

1A

= 0.

Then f g is indeed the identity, as

dag // 1⊗ a− a⊗ 1

f // 1 da− a d1 = da

as desired.

We can now use our understanding of how Ω works on affine open sets to state some global

results.

22.2.F. Exercise. Suppose f : X → Y is locally of finite type, and X is locally Noetherian.Show that ΩX/Y is a coherent sheaf on X. 14 [Jarod extends this to Deligne-Mumford stacks:he shows that ΩX/S is well-defined and coherent. Here is what he does. Lemma 1. Suppose

g : U → U ′ is a morphism of etale atlases, so we have

Ug //

f

@@@

@@@@

@ U ′

f ′

~~

X

13H.P.II.8.1A, he refers to Matsumura p. 182. I think my argument is different from Eisenbud and

Matsumura.14

H.C.II.8.5, roughly

360

with f, f ′ etale, surjective, representable. Then g is etale. The proof uses Lemma 2: Suppose

Xf //

g @

@@@@

@@Y

h~~~~

~~~

Z

with g, h formally etale. Then f is formally etale too. He goes on from there.]The relative cotangent exact sequence and the conormal exact sequence for schemes now

directly follow.

22.2.15. Theorem. — (Relative cotangent exact sequence) Suppose Xf // Y

g // Z be

morphisms of schemes. Then there is an exact sequence of quasicoherent sheaves on X 15

f∗ΩY/Z → ΩX/Z → ΩX/Y → 0.

(Conormal exact sequence) Suppose f : X → Y morphism of schemes, Z → X closed subschemeof X, with ideal sheaf I. Then there is an exact sequence of sheaves on Z: 16

I/I2 δ // ΩX/Y ⊗OZ // ΩZ/Y // 0.

Similarly, the sheaf of relative differentials pull back, and behave well under base change.

22.2.16. Theorem (pullback of differentials). — (a) If

X′g //

X

Y ′ // Y

is a commutative diagram of schemes, there is a natural homomorphism of quasicoherent sheaveson X′ g∗ΩX/Y → ΩX′/Y ′ . An important special case is Y = Y ′.

(b) (Ω behaves well under base change) If furthermore the above diagram is a tensor diagram(i.e. X′ ∼= X ⊗Y Y ′) then g∗ΩX/Y → ΩX′/Y ′ is an isomorphism. 17

This follows immediately from Exercise 22.2.C last day. Part (a) implicitly came up in ourearlier discussion of the Riemann-Hurwitz formula.

As a particular case of part (b), the fiber of the sheaf of relative differentials is indeed thesheaf of differentials of the fiber. Thus this notion indeed glues together the differentials on eachfiber.

22.3 Examples

22.3.1. The projective line. As an important first example, let’s consider P1k, with the usual

projective coordinates x0 and x1. As usual, the first patch corresponds to x0 6= 0, and is of theform Spec k[x1/0] where x1/0 = x1/x0. The second patch corresponds to x1 6= 0, and is of the

form Spec k[x0/1] where x0/1 = x0/x1.

Both patches are isomorphic to A1k, and ΩA1

k= OA1

k. (More precisely, Ωk[x]/k = k[x] dx.)

Thus ΩP1k

is an invertible sheaf (a line bundle). Now we have classified the invertible sheaves on

P1k — they are each of the form O(m). So which invertible sheaf is ΩP1/k?

15t:rcES

16H.P.II.8.11 and 12

17H.P.II.8.10

361

Let’s take a section, dx1/0 on the first patch. It has no zeros or poles there, so let’s check

what happens on the other patch. As x1/0 = 1/x0/1, we have dx1/0 = −(1/x20/1

) dx0/1. Thus

this section has a double pole where x0/1 = 0. Hence ΩP1k/k∼= O(−2).

Note that the above argument did not depend on k being a field, and indeed we could replacek with any ring A (or indeed with any base scheme).

22.3.2. A plane curve. Consider next the plane curve y2 = x3 − x in A2k , where the

characteristic of k is not 2. Then the differentials are generated by dx and dy, subject to theconstraint that

2y dy = (3x2 − 1) dx.

Thus in the locus where y 6= 0, dx is a generator (as dy can be expressed in terms of dx). Similarly,in the locus where 3x2− 1 6= 0, dy is a generator. These two loci cover the entire curve, as solvingy = 0 gives x3 − x = 0, i.e. x = 0 or ±1, and in each of these cases 3x2 − 1 6= 0.

Now consider the differential dx. Where does it vanish? Answer: precisely where y = 0. Youshould find this believable from the picture. [give it]

22.3.A. Exercise: differentials on hyperelliptic curves. Consider the double cover f :C → P1

k branched over 2g + 2 distinct points. (We saw earlier that this curve has genus g [ref],modulo the Riemann-Hurwitz formula, which we have yet to prove.) Then ΩC/k is again an

invertible sheaf. What is its degree? (Hint: let x be a coordinate on one of the coordinate patches

of P1k. Consider f∗dx on C, and count poles and zeros.) [Give a picture for a reality check. They

should count 2g + 2 zeros, and 4 poles, for a total 2g − 2.]

22.3.B. Exercise: differentials on nonsingular plane curves. Suppose C is a nonsingularplane curve of degree d in P2

k, where k is algebraically closed. By considering coordinate patches,find the degree of ΩC/k. Make any reasonable simplifying assumption (so that you believe that

your result still holds for “most” curves).Because Ω behaves well under pullback, note that the assumption that k is algebraically

closed may be quickly excised:

22.3.C. Exercise. Suppose that C is a nonsingular projective curve over k such that ΩC/k is an

invertible sheaf. (We’ll see that for nonsingular curves, the sheaf of differentials is always locally

free. But we don’t yet know that.) Let Ck = C ×Speck Spec k. Show that ΩCk/k is locally free,

and that

deg ΩCk/k = deg ΩC/k.

[Many people used Riemann-Roch. This is a little unsatisfying, because it requires Serre duality.]

22.3.3. Projective space and the Euler exact sequence.We next examine the differentials of projective space Pnk , or more generally PnA where A is

an arbitrary ring. As projective space is covered by affine open sets of the form An, on which thedifferentials form a rank n locally free sheaf, ΩPn

A/A is also a rank n locally free sheaf.

22.3.4. Important Theorem (the Euler exact sequence). — The sheaf of differentialsΩPn

A/A satisfies the following exact sequence Euler e.s.

0→ ΩPnA→ OPn

A(−1)⊕(n+1) → OPn

A→ 0.

This is handy, because you can get a hold of Ω in a concrete way. Here is an explicit example,to give you practice.

22.3.D. Exercise. Show that H1(PnA, TnPn

A) = 0. (This later turns out to be an important

calculation for the following reason. If X is a nonsingular variety, H1(X, TX) parametrizes defor-

mations of the variety. [Add ref later.] Thus projective space can’t deform, and is “rigid”.) [Shortcheck using LES.]

Let’s prove the Euler exact sequence. I find this an amazing fact, and while I can prove it, Idon’t understand in my bones why this is true.

362

Proof. (What’s really going on in this proof is that we consider those differentials on An+1A \ 0

that are pullbacks of differentials on PnA.) [check surjectivity!!]

I’ll describe a map O(−1)⊕(n+1) →O, and later identify the kernel with ΩX/Y . The map isgiven by

(s0, s1, . . . , sn) 7→ x0s0 + x1s1 + · · ·+ xnsn.

Note that this is a degree 1 map.Now I have to identify the kernel of this map with differentials, and I can do this on each

open set (so long as I do it in a way that works simultaneously for each open set). So let’s considerthe open set U0, where x0 6= 0, and we have coordinates xj/0 = xj/x0 (1 ≤ j ≤ n). Given adifferential

f1(x1/0, . . . , xn/0) dx1/0 + · · ·+ fn(x1/0, . . . , xn/0) dxn/0

we must produce n+ 1 sections of O(−1). As motivation, let me just look at the first term, andpretend that the projective coordinates are actual coordinates.

f1 dx1/0 = f1 d(x1/x0)

= f1x0 dx1 − x1 dx0

x20

= −x1

x20

f1 dx0 +f1

x0dx1

Note that x0 times the “coefficient of dx0” plus x1 times the “coefficient of dx1” is 0, and alsoboth coefficients are of homogeneous degree −1. Motivated by this, we take:18

(48) f1 dx1/0 + · · ·+ fn dxn/0 7→„−x1

x20

f1 − · · · −xn

x20

fn,f1

x0,f2

x0, · · · , fn

x0

«

Note that over U0, this indeed gives an injection of ΩPnA

to O(−1)⊕(n+1) that surjects onto the

kernel of O(−1)⊕(n+1) →OX (if (g0, . . . , gn) is in the kernel, take fi = x0gi for i > 0).Let’s make sure this construction, applied to two different coordinate patches (say U0 and

U1) gives the same answer. (This verification is best ignored on a first reading.) Note that

f1 dx1/0 + f2 dx2/0 + · · · = f1 d1

x0/1

+ f2 dx2/1

x0/1

+ · · ·

= − f1

x20/1

dx0/1 +f2

x0/1

dx2/1 −f2x2/1

x20/1

dx0/1 + · · ·

= −f1 + f2x2/1 + · · ·

x20/1

dx0/1 +f2x1

x0dx2/1 + · · · .

Under this map, the dx2/1 term goes to the second factor (where the factors are indexed 0 through

n) in O(−1)⊕(n+1), and yields f2/x0 as desired (and similarly for dxj/1 for j > 2). Also, thedx0/1 term goes to the “zero” factor, and yields

0@

nX

j=1

fi(xi/x1)/(x0/x1)2

1A /x1 = fixi/x

20

as desired. Finally, the “first” factor must be correct because the sum over i of xi times the ithfactor is 0.

Generalizations of the Euler exact sequence are quite useful. We won’t use them later, so I’llstate them without proof. Note that the argument applies without change if SpecA is replacedby an arbitrary base scheme.[Make sure base scheme is defined earlier!] The Euler exact sequencefurther generalizes in a number of ways. As a first step, suppose V is a rank n + 1 locally freesheaf (or vector bundle) on a scheme X. Then ΩPV/X sits in an Euler exact sequence:

0→ ΩPV/X →O(−1) ⊗ V ∨ → OX → 0

If π : PV → X, the map O(−1)⊗V ∨ →OX is induced by V ∨⊗π∗O(1) ∼= (V ∨⊗V )⊗OX →OX ,where V ∨ ⊗ V → A is the trace map.

18expliciteuler

363

For another generalization, fix a base field, and let G(m,n+ 1) be the space of vector spacesof dimension m in an (n+ 1)-dimensional vector space V . (This is called the Grassmannian. Wehave not shown that this is actually a variety in any natural way, but it is. The case m = 1 isPn.) Then over G(m, n+ 1) we have a short exact sequence of locally free sheaves Grassmannian

0→ S → V ⊗OG(m,n+1) → Q→ 0

where V ⊗ OG(m,n+1) is a trivial bundle, and S is the “universal subbundle” (such that over a

point [V ′ ⊂ V ] of the Grassmannian G(m,n+1), S|[V ′⊂V ] is V , if you can see what that means).

Then 19

(49) ΩG(m,n+1)/k∼= Hom(Q,S).

22.3.E. Exercise. In the case of projective space, m = 1, S = O(−1). Verify (49) in this case.This Grassmannian fact generalizes further to Grassmannian bundles.

22.4 Varieties over algebraically closed fields

We’ll now discuss differentials in the case of interest to most people: varieties over alge-braically closed fields. I’d like to begin with a couple of remarks.

22.4.1. Remark: nonsingularity may be checked at closed points. Recall from 16.4 a deep factabout regular local rings that we haven’t proved: Any localization of a regular local ring at aprime is again regular local ring. (For a reference, see Matsumura’s Commutative Algebra, p.139.)20 I’m going to continue to use this without proof. [Give proof later?] But in any case, ifthis bothers you, you could re-define nonsingularity of locally finite type schemes over fields tobe what other people call “nonsingularity at closed points”, and the results of this section willhold.21 [f is a closed immersion iff f ⊗k k is.]

22.4.2. Remark for non-algebraically closed people. Even if you are interested in non-algebraicallyclosed fields, this section should still be of interest to you. In particular, if X is a variety over afield k, and Xk = X×Spec k Spec k, then Xk nonsingular implies that X is nonsingular. (You maywish to prove this yourself. By Remark 22.4.1, it suffices to check at closed points.) Possibleexercise. In fact if k is separably closed, then Xk is nonsingular if and only if X is nonsingular,but this is a little bit harder.

Suppose for the rest of this section that X is a pure n-dimensional locally finite type schemeover an algebraically closed field k (e.g. a k-variety).

22.4.3. Proposition. — ΩX/k is locally free of rank n if and only if X is nonsingular. 2223

Proof. By Remark 22.4.1, it suffices to prove that ΩX/k is locally free of rank n if and only ifthe closed points of X is nonsingular. Now ΩX/k is locally free of rank n if and only if its fibers

at all the closed points are rank n (recall that fibers jump in closed subsets). As the fiber of thecotangent sheaf is canonically isomorphic to the Zariski tangent space at closed points [ref], theZariski tangent space at every closed point must have dimension n, i.e. the closed points are allnonsingular.

Using this Proposition, we can get a new result using a neat trick. [Good example of whygeneric points are useful. Add a ref later!]

22.4.4. Theorem. — If X is integral, there is an dense open subset U of X which is nonsin-gular. 24

19OmegaGrass

20H.T.II.8.14A

21hardfact

22H.T.II.8.15

23Omegans

24H.C.II.8.16

364

Proof. The n = 0 case is immediate, so we assume n > 0.We will show that the rank at the generic point is n. Then by uppersemicontinuity of the

rank of a coherent sheaf (Exercise 17.10.A(c)), it must be n in an open neighborhood of the genericpoint, and we are done by Proposition 22.4.3.

We thus have to check that if K is the fraction field of a dimension n integral finite-typek-scheme, i.e. if K is a transcendence degree n extension of k, then ΩK/k is an n-dimensionalvector space. But any transcendence degree n > 1 extension is separably generated: we can findn algebraically independent elements of K over k, say x1, . . . , xn, such that K/k(x1, . . . , xn) isseparable. (This is a fact about transcendence theory, Theorem A.2.1.) Then ΩK/k is generated

by dx1, . . . , dxn (as dx1, . . . , dxn generate Ωk(x1,...,xn)/k, and any element of K is separable

over k(x1, . . . , xn) — this is summarized most compactly using the relative cotangent sequence inthe affine form of Theorem 22.2.8). [No! We need more!]

Curves in P3 example. Exercise? [What did I mean by this comment to myself?]

22.4.5. Bertini’s Theorem. — Suppose X is a nonsingular closed subvariety of Pnk (where thestanding hypothesis for this section, that k is algebraically closed, holds). Then there is an opensubset of hyperplanes H of Pnk such that H doesn’t contain any component of X, and the schemeH∩X is a nonsingular variety. More precisely, this is an open subset of the dual projective spacePnk∨. In particular, there exists a hyperplane H in Pnk not containing any component of X such

that the scheme H ∩X is also a nonsingular variety.25Bertini(We’ve already shown in our section on cohomology that if X is connected, then H ∩ X is

connected.) [give better ref.]We may have used this before to show the existence of nonsingular curves of any genus, for

example. [I don’t think we did.]Note that this implies that a general degree d > 0 hypersurface in Pnk also intersects X in a

nonsingular subvariety of codimension 1 in X: replace X → Pn with the composition X → Pn →PN where the latter morphism is the dth Veronese map (20.4.1).d-tuple embedding,

Veronese embeddingProof. In order to keep the language of the proof as clean as possible, I’ll assume X is irreducible,but essentially the same proof applies in general.

The central idea of the proof is quite naive and straightforward. We’ll describe the hyper-planes that are “bad”, and show that they form a closed subset of dimension at most n−1 of Pnk

∨,and hence that the complement is a dense open subset. More precisely, we will define a projectivevariety Y ⊂ X × Pnk

∨ that will be:

Y = (p ∈ X,H ⊂ Pnk ) | p ∈ H,p is a singular point of H ∩X, or X ⊂ H

We will see that dimY ≤ n−1. Thus the image of Y in Pnk∨ will be a closed subset (the image of a

closed subset by a projective hence closed morphism!), of dimension of n− 1, and its complementis open.

We’ll show that Y has dimension n− 1 as follows. Consider the map Y → X, sending (p,H)

to p. Then a little thought will convince you that there is a (n − dimX − 1)-dimensional familyof hyperplanes through p ∈ X such that X ∩ H is singular at p, or such that X is containedin H. (Those two conditions can be summarized quickly as: H contains the “first-order formalneighborhood of p in X”, Spec(OX,p/m2) where m is the maximal ideal of OX,p.) Hence weexpect Y to be a projective bundle, whose fibers are dimension n− dimX − 1, and hence that Yhas dimension at most dimX + (n− dimX − 1) = n− 1. In fact this is the case, but we’ll show alittle less (e.g. we won’t show that Y → X is a projective bundle) because we don’t need to provethis full statement to complete our proof of Bertini’s theorem.

Let’s put this strategy into action. We first define Y more precisely, in terms of equationson Pn × Pn∨, where the coordinates on Pn are x0, . . .xn, and the dual coordinates on Pn∨ area0, . . . , an. Suppose X is cut out by f1, . . . , fr. (We will soon verify that this definition ofY is independent of these equations.) Then we take these equations as some of the definingequations of Y . (So far we have defined the subscheme X × Pn∨.) We also add the equationa0x0 + · · · + anxn = 0. (So far we have described the subscheme of Pn × Pn∨ corresponding to

25Bertini

365

points (p,H) where p ∈ X and p ∈ H.) Note that the Jacobian matrix

0BB@

∂f1∂x1

(p) · · · ∂fr∂x1

(p)

.... . .

...∂f1∂xn

(p) · · · ∂fr∂xn

(p)

1CCA

has corank equal to dimX at all closed points of X — this is precisely the Jacobian conditionfor nonsingularity [ref ]. (Although we won’t use this fact, in fact it has that corank dimXeverywhere on X. Reason: the locus where the corank jumps is a closed locus, as this is describedby equations, namely determinants of minors. Thus as the corank is constant at all closed points,it is constant everywhere.) We then require that the Jacobian matrix with a new row (a0, · · · , an)has corank ≥ dimX (hence = dimX). This is cut out by equations (determinants of minors). Bythe Jacobian description of the Zariski tangent space, this condition encodes the requirement thatthe Zariski tangent space of H ∩X at p has dimension precisely dimX, which is dimH ∩X + 1(i.e. H ∩ X is singular at p) if H does not contain X, or if H contains X. This is precisely thenotion that we hoped to capture.

] [primordial:Before getting on with our proof, let’s do an example to convince ourselves that this algebra is

describing the geometry we desire. Consider the plane conic x20−x2

1−x22 = 0 over a field of charac-

teristic not 2, which I picture as the circle x2+y2 = 1 from the real picture in the chart U0. (At thispoint I drew a picture.) Consider the point (1, 1, 0), corresponding to (1, 0) on the circle. We ex-pect the tangent line in the affine plane to be x = 1, which should correspond to x0−x1 = 0. Let’ssee what the algebra gives us. The Jacobian matrix is

`2x0 −2x1 −2x2

´=`

2 −2 0´,

which indeed has rank 1 as expected. Our recipe asks that the matrix

„2 −2 0a0 a1 a2

«have

rank 1, which means that (a0 , a1, a2) = (a0 ,−a0, 0), and also that a0x0 +a1x1 +a2x2 = 0, whichis precisely what we wanted!

Returning to our construction, we can see that the Y just described is independent of thechoice of f1, . . . , fr (although we won’t need this fact).

Here’s why. It suffices to show that if we add in a redundant equation (some homogeneousf0 that is a k[x0, . . . , xn]-linear combination of the fi), we get the same Y (as then if we had acompletely different set of f ’s, we could add them in one at a time, and then remove the old f ’sone at a time). If we add in a redundant equation, then that row in the Jacobian matrix will be ak[x0, . . . , xn]-linear combination of other rows, and thus the rank remains unchanged. (There is aslight issue I am glossing over here — f0 may vanish on Y despite not being a linear combinationof f1, . . . , fn.) [Perhaps fix this later.]

We’ll next show that dimY = n − 1. For each p ∈ X, let Zp be the locus of hyperplanescontaining p, such that H ∩X is singular at p, or else contains all of X; what is the dimension ofZp? (For those who have heard of these words: what is the dimension of the locus of hyperplanescontaining a first-order formal neighborhood of p in X?) Suppose dimX = d. Then this shouldimpose d+ 1 conditions on hyperplanes. This means that it is a codimension d+ 1, or dimensionn− d − 1, projective space. Thus we should expect Y → X to be a projective bundle of relativedimension n−d−1 over a variety of dimension d, and hence that dimY = n−1. For convenience,I’ll verify a little less: that dimY ≤ n− 1.

Suppose Y has dimension N . Let H1, . . . , Hd be general hyperplanes such that H1 ∩ · · · ∩Hd ∩X is a finite set of points (this was an exercise from long ago [ref ]). Then if π : Y → X isthe projection to X, then (using Krull’s Principal Ideal Theorem 15.6.3)

n− d− 1 = dimY ∩ π∗H1 ∩ · · · ∩ π∗Hd ≥ dimY − d

from which dimY ≤ n− 1.

22.4.A. Exercise-. Show that Bertini’s theorem still holds even if X is singular in dimension0. (This isn’t that important.)

22.4.6. Remark.. The image in Pn tends to be a divisor. This is classically called the dualvariety. The following exercise will give you some sense of it. dual variety

366

22.4.B. Exercise. Suppose C ⊂ P2 is a nonsingular conic over a field of characteristic not 2.Show that the dual variety is also a nonsingular conic. (More precisely, suppose C is cut out byf(x0, x1, x2) = 0. Show that (a0, a1, a2) | a0x0 + a1x1 + a2x2 = 0 is cut out by a quadraticequation, and that this curve is nonsingular.) Thus for example, through a general point in theplane, there are two tangents to C. (The points on a line in the dual plane corresponds to thoselines through a point of the original plane.) [By choosing coordinates wisely, you can assume theequation is of the form α0x2

0 + α1x21 + α2x2

2 = 0, where αi ∈ k∗.] [Minor issue: nonsingularityover nonalgebraically closed fields. By now they should know that they can just base change tothe algebraic closure.]

22.4.C. Exercise. Investigate what happens in characteristic 2!We’ll soon find the degree of the dual to a degree d curve (after we discuss the Riemann-

Hurwitz formula), at least modulo some assumptions.[State Kleiman-Bertini.]

22.4.7. The Riemann-Hurwitz formula.We’re now ready to discuss and prove the Riemann-Hurwitz formula. (We continue to work

over an algebraically closed field k. Everything below can be mildly modified to work for aperfect field, e.g. any field of characteristic 0, and I’ll describe this at the end of the discussion(Remark 22.4.13).)

Definition (separable morphisms). A finite morphism between integral schemes f : X → Y issaid to be separable if it is dominant, and the induced extension of function fields FF (X)/FF (Y )is a separable extension. (Similarly, a generically finite morphism is generically separable if it isdominant, and the induced extension of function fields is a separable extension. We may not usethis notion.) Note that this comes for free in characteristic 0.(gen.) sep. morphism

22.4.8. Proposition. — If f : X → Y is a finite separable morphism of nonsingular integralcurves, then we have an exact sequence 26

0→ f∗ΩY/k → ΩX/k → ΩX/Y → 0.

Proof. We have right-exactness by the relative cotangent sequence 22.2.15, so we need to checkonly that φ : f∗ΩY/k → ΩX/k is injective. Now ΩY/k is an invertible sheaf on Y , so f∗ΩY/k is

an invertible sheaf on X. Thus it has no torsion subsheaf [Ben says this needs more explanation.],so we need only check that φ is an inclusion at the generic point. We thus tensor with Oη whereη is the generic point of X. This is an exact functor (it is localization), and Oη ⊗ ΩX/Y = 0

(as FF (X)/FF (Y ) is a separable by hypothesis, and Ω for separable field extensions is 0 byExercise 22.2.D). Also, Oη ⊗ f∗ΩY/k and Oη ⊗ΩX/k are both one-dimensional Oη-vector spaces

(they are the stalks of invertible sheaves at the generic point). Thus by considering

Oη ⊗ f∗ΩY/k →Oη ⊗ ΩX/k → Oη ⊗ ΩX/Y → 0

(which is Oη →Oη → 0→ 0) we see that Oη ⊗ f∗ΩY/k → Oη ⊗ΩX/k is injective, and thus thatf∗ΩY/k → ΩX/k is injective.

22.4.9. It is worth noting what goes wrong for non-separable morphisms. For example, supposek is a field of characteristic p, consider the map f : A1

k = Spec k[t] → A1k = Spec k[u] given by

u = tp. Then Ωf [is this the first time I’ve used this notation?] is the trivial invertiblesheaf generated by dt. As another (similar but different) example, if K = k(x) and K ′ = K(xp),then the inclusion K ′ → K induces f : SpecK[t] → SpecK ′[t]. Once again, Ωf is an invertiblesheaf, generated by dx (which in this case is pulled back from ΩK/K′ on SpecK). In both ofthese cases, we have maps from one affine line to another, and there are vertical tangent vectors.

22.4.10. The sheaf ΩX/Y on the right side of Proposition 22.4.8 is a coherent sheaf not supportedat the generic point. Hence it is supported at a finite number of points. These are called theramification points, and the images downstairs are called the branch points. [Add a picture here!!]

ram and branch points

26H.P.IV.1.1

367

Let’s check out what happens at closed points. We have two discrete valuation rings, saySpecA → SpecB. I’ve assumed that we are working over an algebraically closed field k, so thismorphism B → A induces an isomorphism of residue fields (with k). Suppose their uniformizersare s and t respectively, with t 7→ usn where u is a unit of A. Then

dt = d(usn) = unsn−1 ds+ sn du.

This vanishes to order at least n − 1, and precisely n − 1 if n doesn’t divide the characteristic.The former case is called tame ramification, and the latter is called wild ramification. We call thisorder the ramification order at this point of X. 27 tame and wild ram

Define the ramification divisor on X as the sum of all points with their corresponding rami-fications (only finitely many of which are non-zero). The image of this divisor on Y is called thebranch divisor. ram and branch divisor

22.4.D. Straightforward exercise: interpreting the ramification divisor in terms ofnumber of preimages. Suppose all the ramification above y ∈ Y is tame. Show that the degreeof the branch divisor at y is deg(f : X → Y ) − #f−1(y). Thus the multiplicity of the branchdivisor counts the extent to which the number of preimages is less than the degree.28

22.4.11. Proposition. — Suppose R is the ramification divisor of f : X → Y . ThenΩX(−R) ∼= f∗ΩY . 29

Note that we are making no assumption that X or Y is projective. [I should be a bit carefulto make my hypotheses clear!]

Proof. This says that we can interpret the invertible sheaf f∗ΩY over an open set of X asthose differentials on X vanishing along the ramification divisor. But that is the content ofProposition 22.4.8. [Ben says: say more here! Key fact: If you have a dvr, and you knowquotient, that determines the exact sequence.]

Then the Riemann-Hurwitz formula follows!

22.4.12. Theorem (Riemann-Hurwitz). — Suppose f : X → Y is a finite separablemorphism of projective nonsingular curves. Let n = deg f . Then 2g(X)−2 = n(2g(Y )−2)+degR.

Note that we now need the projective hypotheses in order to take degrees of invertible sheaves.[Awkward: Have I defined the degree of a torsion sheaf?]

Proof. This follows by taking the degree of both sides of Proposition 22.4.11 (and using the factthat the pullback of a degree d line bundle by a finite degree n morphism is dn, which was anearlier exercise). [Give ref. — Ex. 3.1 class 29 p. 3, Ex. 2, ps13]

22.4.E. Exercise: degree of dual curves. Describe the degree of the dual to a nonsingulardegree d plane curve C as follows. Pick a general point p ∈ P2. Find the number of tangents toC through p, by noting that projection from p gives a degree d map to P1 (why?) by a curve ofknown genus (you’ve calculated this before), and that ramification of this cover of P1 correspondsto a tangents through p. (Feel free to make assumptions, e.g. that for a general p this branchedcover has the simplest possible branching — this should be a back-of-an-envelope calculation.)

22.4.13. Remark: Riemann-Hurwitz over perfect fields. 30This discussion can be extended towork when the base field is not algebraically closed; perfect will suffice. The place we assumed thatthe base field was algebraically closed was after we reduced to understanding the ramification ofthe morphism of the spectrum of one discrete valuation ring over our base field k to the spectrumof another, and we assumed that this map induced an isomorphism of residue fields. In general,it can be a finite extension. Let’s analyze this case explicitly. Consider a map SpecA → SpecBof spectra of discrete valuation rings, corresponding to a ring extension B → A. Let s be theuniformizer of A, and t the uniformizer of B. Let m be the maximal ideal of A, and n the maximal

27H.P.IV.2.2

28e:ramnum

29H.P.IV.2.3

30rhperfect

368

ideal of A. Then as A/m is a finite extension of B/n, it is generated over B/n by a single element(we’re invoking here the theorem of the primitive element, and we use the “perfect” assumptionhere). Let s′ be any lift of this element of A/m to A. Then A is generated over B by s and s′,so ΩA/B is generated by ds and ds′. The contribution of ds is as described above. You can show

that ds′ = 0. Thus all calculations above carry without change, except for the following.(i) We have to compute the degree of the ramification divisor appropriately: we need to

include as a factor the degree of the field extension of the residue field of the point on the source(over k).

(ii) Exercise 22.4.D doesn’t work, but can be patched by replacing #f−1(y) with the numberof geometric preimages.

As an example of what happens differently in (ii), consider the degree 2 finite morphismX = Spec Z[i]→ Y = Spec Z. We can compute ΩZ[i]/Z directly, as Z[i] ∼= Z[x]/(x2 + 1): ΩZ[i]/Z

∼=Z[i]dx/(2dx). In other words, it is supported at the prime (1 + i) (the unique prime above[(2)] ∈ Spec Z). However, the number of preimages of points in Spec Z is not always 2 away fromthe point [(2)]; half the time (including, for example, over [(3)]) there is one point, but the fieldextension is separable.Gaussian integers Z[i]

22.4.F. Exercise (aside): Artin-Schreier covers. In characteristic 0, the only connected

unbranched cover of A1 is the isomorphism A1∼ // A1 ; that was an earlier example/exercise

[ref, when we discussed RH the first time]. In positive characteristic, this needn’t be true, becauseof wild ramification. Show that the morphism corresponding to k[x] → k[x, y]/(yp − xp − y) issuch a map. (Once the theory of the algebraic fundamental group is developed, this translates to:“A1 is not simply connected in characteristic p.”) Perhaps give some discussion of Artin-Schreier covers. [Jarod: Irreducibility comes from the Eisenstein criterion. Later have themcheck that this is a nontrivial finite etale cover of A1

k. Kirsten: this cover is not even abstractloy

isomorphic to A1, as the ring is not a UFD: x is irreducible but not prime.]

22.4.14. The conormal exact sequence for nonsingular varieties.Recall the conormal exact sequence. Suppose f : X → Y morphism of schemes, Z → X

closed subscheme of X, with ideal sheaf I. Then there is an exact sequence of sheaves on Z:

I/I2 δ // ΩX/Y ⊗OZ // ΩZ/Y // 0.

I promised you that in good situations this is exact on the left as well, as our geometric intuitionpredicts. Now let Z = Spec k (where k = k), and Y a nonsingular k-variety, and X ⊂ Y anirreducible closed subscheme cut out by the quasicoherent sheaf of ideals I ⊂ OY .

22.4.15. Theorem (conormal exact sequence for nonsingular varieties). — X isnonsingular if and only if (i) ΩX/k is locally free, and (ii) the conormal exact sequence is exacton the left also:

0 // I/I2 δ // ΩX/Y ⊗OZ // ΩZ/Y // 0.

Moreover, if Y is nonsingular, then I is locally generated by codim(X, Y ) elements, and I/I2 isa locally free of rank codim(X, Y ). 31

This latter condition is the definition of something being a local complete intersection in anonsingular scheme.local complete intersec-

tion You can read a proof of this in Hartshorne II.8.17. I’m not going to present it in class, as we’llnever use it. The only case I’ve ever seen used is the implication that if X is nonsingular, then (i)and (ii) hold; and we’ve already checked (i). This implication (that in the case of a nonsingularsubvariety of a nonsingular variety, the conormal and hence normal exact sequence is exact) is veryuseful for relating the differentials on a nonsingular subvariety to the normal bundle. [Relatedfact that came up: exactness on the left if the thing on the right is smooth.]

The real content is that in the case of a nonsingular subvariety of a nonsingular variety,

the conormal exact sequence is exact on the left as well, and in this nice case we have a shortexact sequence of locally free sheaves (vector bundles). By dualizing, i.e. applying Hom(·,OX),we obtain the normal exact sequencenormal ES

31t:nsconormal, H.T.II.8.17

369

0→ TX/k → TY/k →NX/Y → 0

which is very handy. Note that dualizing an exact sequence will give you a left-exact sequencein general, but dualizing an exact sequence of locally free sheaves will always be locally free. (Infact, all you need is that the third term is locally free; maybe I should state this fact back at thatpart of the text.) [Make sure to do this when discussing Ext!]

[This proof is a little rough.]I think I need a lemma. If 0 → F → G → H → 0 is an exact sequence of coherent sheaves,

then if G and H are locally free, then so is F . I think we need X to be Noetherian. We need Torfor this: F is flat, hence locally free. Possible kluge: in the case we need it, where X is reduced,we can use the fiber-dimension criterion. But then we still need that if 0 → M → N → An → 0is exact, then it remains so upon tensoring with A/m, for which we (at this point) still need Tor.

Proof. Suppose first that these two conditions (i) and (ii) hold. then we wish to show thatrank ΩX/k = dimX. Let n = dimY and q = rank ΩX/k. Then from (ii), I/I2 is locally free of

rank n− q. [Why? I need a tor fact, see above.] Thus I/I2 is locally generated by n− q elements,from which dimX ≥ n − (n − dim q) by Krull’s principal ideal theorem. On the other hand, byconsidering any closed point x ∈ X, we have q = dimk(mx/m

2x), so q ≥ dimX. Thus we have

equality, and that at every closed point x ∈ X, dimX = q. Thus (i) and (ii) implies that X isnonsingular, and is a local complete intersection.

Conversely, suppose that X is nonsingular. Then (i) holds immediately. Consider ker(ΩY/k⊗OX → ΩX/k). It is locally free of rank r = n− rank ΩX/k, so in some neighborhood [say better],

choose r sections x1, . . . , xr ∈ I such that dx1, . . . , dxr generate ker(ΩY/k ⊗ OX → ΩX/k).

Suppose this cuts out a schemeX ′. [Then show that X ′ is nonsingular and irreducible of dimensionn − r (in this neighborhood). But it contains X, and both are integral schemes of the samedimension.]

22.5 Older notes

[Remark: Zariski’s “Purity of branch locus” theorem. What’s the statement? If X and Yare smooth over k, then the branch locus of X → Y is pure codimension 1. Can I prove this partdirectly?]

Other things I want to include:adjunction formula.• generic smoothness! And give example of failure earlier; even for points.

Some discussion that the first is an absolute notion, while the second is relative. Reservesmooth for relative notion.

Is y2 = x3 +x2 (i.e., Spec k[x, y]/(y2 −x3−x2), where for the sake of simplicity we take k tobe an algebraically closed field) isomorphic to A1? It is certainly affine. We could try to computethe ring of global sections, but this might be hard. There is an easier trick, and it is motivatedby a picture. (Draw a picture of the first curve.) This is a cartoon, or a picture of the real points.There is a point that is not a manifold point (a “singular” point). Hence we should see somethingdifferent happening at the point (0, 0).

Here is an algebraic way of distinguishing them. (This won’t yet have an obvious connectionto the geometry, but it certainly will soon.) (0, 0) is a maximal ideal m of the ring A = k[x, y]/(y2−x3 − x2).

For any closed point of Spec k[t], m/m2 has dimension 1. (That’s because we know thatm = (x − a) for some a, by the Nullstellensatz.) Let’s calculate m/m2. m/m2 in k[x, y] is 2-dimensional. This equation cuts out by ...

The fact that this is 2 has to do with the “local embedding dimension”, in some rough sense.Perhaps make this an exercise. Second example: We work over C. We’re going to glue 4

A1’s together along a point, so that they lie in a plane. xy(y− x)(y − 2x) and xy(y − x)(y − πx).Are these 2 schemes the same? In fact they aren’t; and the “reason” is that, roughly speaking,the cross-ratio of the 4 slopes (considered as points in P1) are the different. Here is a sketch cross-ratioof an argument. First of all, both of these have only one “singular” closed point, where m/m2

370

has dimension greater than 1; it is dimension 2 in both cases. We can recover the informationabout the cross ratio as follows. We have a natural map Sym4 m/m2 → m4/m5. (Think of it asfollows: m/m2 is a 2-dimensional vector space generated by x and y. Sym4 of this vector spacemay be interpreted as 5-dimensional vector space consisting of homogeneous quartics in x andy. m4/m5 actually 4-dimensional; we have the quotient by the defining equation of the quartic.Hence the kernel of this map defines the following quartic. we needn’t have x and y on the leftside corresponding to x and y on the right. (Likely say this in terms of a and b. But we’ll certainlyhave x and y as given by αa + βb and γa + δb modulo m2, where we need

„α βγ δ

«

is not invertible (as we want an isomorphic between both m/m2). Thus we just need to make surethat no element of GL(2) can turn the first quartic into the second.))

22.5.A. Unimportant exercise. 3 lines together: any two are the same. Show it.Related fact: y4 = x4 +x5 has 4 branches through the origin. Think in the analytic topology,

with C. What are their slopes? Intuitive answer: Near the origin (in the analytic topology), thex5 term is trumped. So it looks like y4 = x4, i.e. the slopes are ±1 and ±i. Indeed this is thecase. (You may want to make this precise later.) This suggests that “in general”, something whoseleading (i.e. smallest-degree) term is degree d will have d different branches, and the slopes of thebranches will be the roots of the leading term. (This is indeed the definition of the multiplicity ofa point on a plane curve, or indeed the multiplicity of a point on a hypersurface in An.)

Interesting open-ended questions: y2 = x3; here the leading term has a double root. Whatdoes x2 + y2 = z2 + xyz look like near the origin (in the analytic topology)?

Another example: go back and look at the discussion of the dual numbers in Section ??.dual numbersHopefully these examples are tantalizing to you.

22.5.1. The Zariski tangent space at a point.THE ZARISKI TANGENT SPACE WAS DONE EARLIER. THESE THINGS

SHOULD ALL MIGRATE ELSEWHERE.

22.5.B. Exercise. Suppose X is a finite type k-scheme or Z-scheme. Show that the singularlocus is a Zariski-closed subset. (Hint: the locus where the rank of the Jacobian matrix drops iscut out by determinants of minors. This is a bogus hint! This exercise was stated earlier.)Note that your argument gives the singular locus the structure of a closed subscheme, not just aclosed subset. [Refer to Fitting ideals?]

22.5.2. First definition of the sheaf of relative differentials. I’M NOT SURE IFI LIKE THIS HERE; MAYBE MOVE QUICKLY TO d, and later give the first asan exercise.

22.5.3. Proposition. — Let’s first check that the fibers of this sheaf at x are canonicallyidentified with m/m2 in the case Y = Spec k. [Wait, I’m confused... what happens at non-closedpoints?]32

Proof. Consider the fiber diagram

X //

X ×Spec k X

pr1

x // X

.

Along with the closed subschemes I∆ and I2∆ on X ×Speck X. These pull back to the maximal

ideal m on X [affine?], and m2 respectively. (This requires an exercise... when you pullback anideal, you get another ideal — we’ve checked this earlier when showing that closed immersionsbehave well under base change. The same is true of the square of the ideal. The pullback of thesquare of the ideal is the square of the pullback of the ideal. Perhaps stick this exercise in earlier.)Draw a corresponding pictures too.

32II2mm2

371

Perhaps draw the picture with a bigger fiber diagram, informally given below.

SpecOx/m //

I∆

SpecOx/m2 //

I2∆

X //

X ×Spec k X

pr1

x // X

.

This completes the argument.

22.5.4. The differential d, and an algebraic description.We next describe the differential d. 33 For any open subset U ⊂ X, we define a morphism

d : Γ(U,OX)→ Γ(U,ΩX/Y )

as follows. Let pr1 and pr2 be the projections from X ×Y X onto its two factors. Define

df := pr∗1f |∆2 − pr∗2f |∆2 .

This vanishes on ∆, and hence gives an element of Γ(U,ΩX/Y ).Cautions: this d map is not determined by “the value of f at points”; indeed the entire

point of this map is that it captures (first-order) differential information! Also, although both areΓ(U,OX) and Γ(U,ΩX/Y ) are Γ(U,OX)-modules, d is not a map of Γ(U,OX)-modules, but it is

a map of Γ(U, f−1OY )-modules! (This shouldn’t be suprising: the differential you know about:d : k[x]→ k[x] given by f(x) 7→ f ′(x) is certainly not a map of k[x]-modules, but it is a map ofk-modules. This will become clearer once we delve into the algebra)

This differential behaves the way we would want it to. In particular, it satisfies the followingtwo properties.

22.5.5. Proposition. — 34

(a) If f is the pullback of a function on Y , then df = 0.(b) (Leibniz rule) d(fg) = fdg + gdf . Leibniz

(It will turn out that (d,ΩX/Y ) are universal with respect to these properties, Proposition ??.)Hence we can manipulate these differentials in a way we are accustomed to. For example, ifA = k[x, y] and B = k, then

d(3x2y3) = 6xy3dx+ 9x2y2dy.

If A = k[x, y]/(xy) and B = k, then xdy = −ydx, as d(xy) = d0 = 0.In order to understand the differential map better (for example, to prove the preceding

proposition), we will examine how this works on affines.Suppose now that X = SpecA and Y = SpecB, so π corresponds to π] : B → A. Then the

diagonal morphism δ corresponds to A⊗B A→ A, f ⊗ g 7→ fg, with kernel I, and

df = f ⊗ 1− 1⊗ f (mod I2).

Then I/I2 is denoted ΩA/B . Hence da = 0 if a ∈ π](B), proving Proposition 22.5.5(a) (which ΩA/Bcould also have been shown directly geometrically).

A word about manipulating the A-module I/I2 = ΩB/A, you may wonder whether x(y⊗1−1⊗ y) should be xy ⊗ 1− x⊗ y or y ⊗ x− 1⊗ xy, i.e. whether ΩB/A should be considered as anA-module “on the left or on the right”. In fact both are the same, as can be seen in two ways:algebraically, their difference = (x⊗ 1 − 1 ⊗ x)(y ⊗ 1 − 1 ⊗ y) lies in I2; geometrically, the sheaf(corresponding to) I/I2 is supported on the diagonal, and hence pulling back from either factoramounts to the same thing.

33H.P.II.8.1A, which H refers to Matsumura p. 182; we prove it.

34Leibniz

372

22.5.C. Exercise. Suppose xi is a generating set for A as a B-algebra. Show that ΩA/B =

I/I2 is generated by dxi = xi ⊗ 1 − 1 ⊗ xi. [Answer: use previous exercise, and Proposi-tion 22.5.5(b).]

Hence ΩA/B is generated by da (as a ∈ A) subject to the relations given by Proposition 22.5.5,

i.e. db = 0 for b ∈ π]B and the Leibniz rule. There are no other relations, as I2 is generated by

(x⊗ 1− 1⊗ x)(y ⊗ 1− 1⊗ y)= xy ⊗ 1− x⊗ y − y ⊗ x+ 1⊗ xy= (xy ⊗ 1− x⊗ y + xy ⊗ 1− y ⊗ x)− (xy ⊗ 1− 1⊗ xy)= (x(y ⊗ 1− 1⊗ y) + y(x⊗ 1− 1⊗ x))− dxy= xdy + ydx− dxy

as x, y run over A.

22.6 Examples

Now is a good time to get into some examples to get a feel for how this works.

22.6.A. Exercise. Suppose X = A1k, Y = Spec k. Check that d is

d :X

n≥0

anxn 7→

X

n≥1

nanxn−1dx.

22.6.B. Exercise. Show that there are no global differentials on P1k, i.e. that H0(C,ΩP1

k/k) = 0.

(Imprecise follow-up: Think about dx on A1 ⊂ P1, and how it doesn’t extend over ∞. Intuitively:it has a pole of order 2. We’ll be able to make this precise later (XXXX).) 35

22.6.1. Example. Affine plane curve y2 = x3 − x. Check that it is nonsingular (char not 2?).Notice that Ω is generated by dx and dy. Where does dx vanish? What does it mean when

dx = 0? Notice that dx is pulled back from A1. Investigate: Where is this branched? (We haven’tdefined this yet, but try the problem anyway.)

Notice that where dx 6= 0, dy/dx is a section of the trivial bundle. Interpret this section.Notice: Ω is rank 1, because it is locally generated by dx and dy, and one is always a

multiple of the other, because fortunately df/dx and df/dy are never zero at the same time; nota coincidence.

[Now would be a good time to talk about nonsingularity of things are higherdimension. This could even be an exercise for them. They would have a matrixwhose rank would be...]

22.6.C. Extended exercises: hyperelliptic curve. 36 Fix a field k of characteristic not 2.Let f(x) ∈ k[x] be a polynomial of degree 2g + 2 with distinct roots, and assume f(0) 6= 0. (Thislast hypotheses is unnecessary, and is made for simplicity. MENTION HOW MANY OF THESEHYPOTHESES ARE UNNECESSARY.) Consider the curve C obtained by gluing the affine openpatch U1 given by y2 = f(x) with the affine open patch U2 given by v2 = u2g+2f(1/u), whereU12 = x 6= 0 ⊂ U1 is glued to U21 = u 6= 0 ⊂ U2 via v = y/xg+1, u = 1/x (and symmetricallyy = v/ug+1, x = 1/u).

(a) Show that C is nonsinsingular.(b) Show that the projection of U1 to the x-line and U2 to the u-line glues to give a degree

2 finite morphism C → P1. Hence C is projective (as P1 is projective, and C → P1 is finite henceprojective, and the composition of projective morphisms is projective).

Definition: A projective curve admitting a degree 2 finite morphism to P1 is said to behyperelliptic.hyperelliptic

(c) Show that ΩC/k is an invertible sheaf, by showing that it is actually trivial on U1 and

U2, by showing for example that dx/y is a non-vanishing section over U1. [ydy = f ′(x)dx. We

35diffP1

36hyperelliptic

373

have 2 generators, but we have only 1, dx, where y 6= 0, and 1, dy, where f ′(x) 6= 0. These coverU1, so Ω is invertible. Now dx/y is regular where y 6= 0, and dy/f ′(x) is regular where f ′(x) 6= 0,so dx/y is regular everywhere.]

(d) Show that H0(C,ΩC/k) ∼= kg by explicitly giving a basis of differentials. [Answer: the

differentials on U1 are given by g(x, y)dx/y = (g0(x) + yg1(x))dx/y. On U2 this turns into(g0(1/u) + vg1(1/u)/ug+1)

Hence we have shown that if g > 0, C is not isomorphic to P1, as H0(C,ΩP1k/k) = 0 by

Exercise 22.6.B: we have constructed a new invariant of curves!Definition. A nonsingular projective curve has geometric genus h0(C,ΩC/k). Note that

this is always finite, as ΩC/k is coherent [XXXX], and coherent sheaves on projective kschemes

have finite-dimensional spaces of sections [XXXX]. Sadly, this isn’t really a new invariant. We geometric genuswill see below (in 22.6.E) that this agrees with our earlier definition of genus, i.e. h0(C,ΩC/k) =

h1(C,OC). More precisely, we will see that cup product gives a perfect pairing

H0(C,ΩC/k)×H1(C,OC)→ H1(C,ΩC/k)

onto a one-dimensional vector space H1(C,ΩC/k).

22.6.D. Exercise. (e) This shows that if g > 1, there is only one way of expressing such a curveas hyperelliptic: we get the description by taking global sections of ΩC/k, and out pops a double

cover of a rational normal curve. See also 25.10.A. 37 rational normal curve

22.6.E. Exercise. (f) (Toward Serre duality; this will parallel our proof of Serre duality inTheorem 31.1.1.)38 Compute h1(C,ΩC/k) = 1. Verify that the pullback map H1(P1,ΩP1/k) →H1(C,ΩC/k) is an isomorphism. [Elements of h1(C,Ω), considerd as differentials on U12: 〈xidx/y, yxidx/y〉 =

〈−ug−1−i duv,−u−i−2vdu/v〉. The only term not killed by differentials on U1 or U2 is du/u. Notice

that it is pulled back from P1!]Check that h1(C,OC) = g, and check that there is a natural perfect pairing

H0(C,ΩC/k) ×H1(C,OC)→ H1(C,ΩC/k).

[First, h1(C,OC) = g: sections over U12: 〈xi, xiy〉 = 〈u−i, ug+1−iv〉, so the only terms not killedby U1 or U2 the following:

H1(C,OC) = 〈 yx,y

x2, . . . ,

y

xg〉.

Then the pairing is:

〈dxy, . . . , xg−1 dx

y〉 × 〈 y

x, . . . ,

y

xg〉 → 〈x−1dx〉.

]In particular, we have now produced curves of every arithmetic genus 0, 1, . . . , and we have

seen our first curves of genus 2. (We had earlier seen curves only of genus equal to a triangularnumbers, ??.) Are all curves hyperelliptic? Have we seen all curves? It turns out that the answeris no. Assume first that the field is algebraically closed field. Then all curves of genus up to2 are hyperelliptic, but it isn’t true of genus 3. In fact the nonsingular curves of degree 3 that curves of low genusare plane quartics, described in ??, are not hyperelliptic. But then all genus 3 curves are eitherhyperelliptic, or quartic plane curves. In genus 4 the situation is more complicated, and it getsworse from there. If the base field is not algebraically closed, then it is even worse; all genus 2curves are hyperelliptic, but not all genus 1 curves are hyperelliptic. (There are genus 1 curveswith no points defined over any quadratic extensions of k. But any hyperelliptic curve will havesuch points.) Most of these comments will be illumninated in 25.10.A.

Show that if g > 1, then the hyperelliptic map is unique, by considering the canonicalembedding. If g = 0, is it isomorphic to P1? If f(x) has a root in k, then yes: give that as anexercise. More generally, if the curve has a point, then yes; perhaps we know enough about that?

22.6.F. Exercise: Field extensions. This notion of relative differentials is interesting even forfinite field extensions. In other words, even when you map a reduced point to a reduced point,there is interesting differential information going on. Compute ΩL/K in the following cases.

(a) L = C, K = R,

37hyperellipticcanonical

38towardserreduality

374

(b) More generally, L/K a separable algebraic extension, not necessarily finite(c) Let k be a field of characteristic p, K = k(tp), L = k(t).

22.6.2. Older stuff. Consequence [of Jacobian description]:

22.6.G. Exercise. Is A is a regular ring of dimension n (finite type over k), then ΩA/k is locally

free of rank n. (Hint: Use Exercise 22.2.B and Nakayama’s lemma.)(Foreshadowing: we would like to do this in families, and say that X/Y is well-behaved in

some way, then it should be considered a family of regular varieties, and ΩX/Y should be a locallyfree sheaf of rank equal to the relative dimension. This is the notion of smooth morphism, whichwe will soon discuss (??).)smooth morphism

If X is a regular k-scheme, then we define the canonical (line) bundle (really the canonicalinvertible sheaf) KX to be the top wedge power of ΩX/k. Note that the canonical sheaf, like thesheaf of differentials, is completely intrinsic.canonical bundle

Same as ΩX/k if X is a curve; part of our motivation is to lift these advantages to otherdimensional smooth varieties. The total space of sections is called the geometric genus. Perhapspoint out that E × C is not isomorphic to P2 using this invariant. [arithmetic genus should havegeometric genusbeen described earlier!]

22.6.3. The sheaf of relative differentials for projective space. 22.6.H. Easy butimportant exercise. Show that KPn ∼= O(−(n+ 1)). [Two methods: use wedge power fact. Ortake explicit sections.]

Try to get them to have a geometric picture of this. Their picture should suggest that it isexact on the left too, and indeed it is, in particularly good situations, which is basically TheoremH.II.8.17, discussed in 22.4.15.

An important consequence:

22.6.4. Proposition (the Adjunction formula, take 1). —adjunction formula[Later, we’ll prove the adjunction formula for dualizing sheaves presumably.]39

22.6.5. Bertini’s Theorem. This is now done in Theorem ??

22.6.6. Kleiman-Bertini theorem. — State Kleiman-Bertini theorem too. 40Kleiman-Bertini theo-

rem Question: Are there curves of genus 2? Of any genus? Yes: P1 × P1. Use Bertini. We needvery ampleness. (Do we have all the curves of genus 2? genus g? Answer: yes for 2, no for 3.Reason later.

22.7 Old notes

Things to check:Define the relative canonical bundle. Example: ΩPn and KPn .(relative) canonical bun-

dle K Example: divisor: multiples of f / f× multiples of f . So N∗D = O(−D)|D .

22.7.A. Exercise. Find singular points of weighted projective space (definded in 10.5.7)? Mayask them for some specific examples.

22.7.B. Exercise. if nonsingular, then mk/mk+1 ∼= Symk(m/m2).Remark: nonsingularity is an open condition in good situations.

22.7.1. Example. cannot cut out singular point scheme theoretically with the right number ofequations. Example: nodal curve, and cone point.

22.7.C. Exercise. k[a, b, c, d]/ac = bd. Show that (c, d) is not a principal ideal in this ring, eventhough it is codimension 1. (This answers an earlier question XXXX.)

39adjunction

40kleimanbertini

375

22.7.2. Examples/exercises. 1) Show Spec k[x] → Spec k[y], y → x2 at 0, 1. (Characteristicissue?) 2) Spec Z[i] → Spec Z at 5, 3, 2, 0. 3) Spec k[x, y] → Spec k[x], x → (x, 0) at 2 points ofthe source.

Find a section for a cubic. Hence have example of a nonrational curve! Possible exercise: dothis in degree d. Define genus.

Later:Criterion for locally closed immersion (of finite type schemes over k?): separate points and

tangent vectors.Application: criterion for very ampleness.

22.7.D. Less important exercise (Anders Buch). Define Ω for a ring. Show that the sheafis the sheafification of what you get by doing this on all open sets. (Hint: comes largely for freeby construction. restrict to our base.)

22.7.E. Exercise. Ω for projective bundles.The normal exact sequence. The adjunction formula. Example of a curve not isomorphic to

P1: plane cubic, where K = O. A new curve: plane quartic, where K has (at least) 4 sections.Define genus of a curve.

Adjunction formula for smooth things. (When I introduced locally free sheaves, I included adjunction formulathe symmetric and alternating stuff.) Can I move this back to the differential section?

22.8 Running example: curves

Application. A cubic plane curve is not P1. Differentials have a section.

22.8.A. Exercise. Do the same thing for curves of higher degree.Remark: cubic plane curve has one-dimensional space of sections of tangent sheaf. Infinites-

imal automorphism. Do the same with P1. We do Pn elsewhere in these notes.Definition. Genus of a curve.

22.9 +A bit of Hodge theory, from an algebraic perspectiveΩi, Hodge theory,

Hodge diamond, Lef-

schetz hyperplane the-

orem, Hard Lefschetz

theorem

It would be nice to tell them something about Hodge theory, although I’m not sure what,and I’m not sure where. Likely after Serre duality.

A good deal of the story can described algebraically, but it really is an analytic story. Im-portant because it tells interesting things about the algebraic side of the story.

Suppose X is a nonsingular projective C-variety. Define ΩiX := ∧iΩX , so ΩdimX = KX .

Fact. Old-fashioned cohomology = ⊕Hj(ΩiX). (i, j)-classes. Hodge numbers. Hodgediamond. Serre duality means the bottom is 1, and there is also one symmetry of the Hodgediamond (that last may require an exercise).

Cycle classes: suppose Z → X is also nonsingular. Then use Serre duality to get a(codimZ, codimZ)-class. (In fact: don’t need nonsingularity.)

This is nonzero if Z 6= 0. Here’s why: slice with an ample hyperplane.

Get primitive form of intersection theory: slice in general.Exercise: compute the Hodge diamond of a genus g curve. Notice that you get 1, two g’s, 1.Example/exercise: compute the Hodge diamond of P2. Find the class of a line. Compute the

multiplication H1 ×H1 → H2 explicitly. [I haven’t done this.]

22.9.A. Exercise. Hodge diamond of projective space.Hint: Use nothing more than the knowledge of the cohomology of O(1) and O (Section

XXXX) and Exercise ??.Other interesting facts: Lefschetz hyperplane theorem. Other symmetry in characteristic 0,

which comes from complex conjugation (false in char p, but true if you can “lift to 0”). HardLefschetz; primitive classes.

376

Refer them to [GH] for the analytic setting, maybe Voisin too.What to do if your space isn’t nonsingular projective? Mixed Hodge theory.[Why do Hodge numbers not jump in smooth families? One possible answer: use the fact

that old-fashioned cohomology doesn’t jump; but is there an algebraic reason?]

]

Part VI

Cohomology

[primordial:

CHAPTER 23

Cech cohomology of quasicoherent sheaves

23.1 Cech cohomology of quasicoherent sheaves

[forme: bill fulton’s rec: look at quillen’s article on “higher K-theory”.] [Coho-mology is something that “linearizes nonlinear data”.]

One idea behind the cohomology of quasicoherent sheaves is as follows. If 0 → F → G →H→ 0 is a short exact sequence of sheaves on X, we know that

0→ F(X)→ G(X)→H(X).

In other words, Γ(X, ·) is a left-exact functor. We dream that this is something called H0, andthat this sequence continues off to the right, giving a long exact sequence in cohomology. (Ingeneral, whenever we see a left-exact or right-exact functor, we should hope for this, and in mostgood cases our dreams are fulfilled. The machinery behind this is sometimes called derived functorcohomology, which we may discuss in the third quarter.) derived functor coho-

mologyWe’ll show that these cohomology groups exist. Before defining them explicitly, we firstdescribe their important properties.

Suppose X is an R-scheme. Assume throughout that X is separated and quasicompact.[(Caution: Hartshorne seems to require Noetherian.) Give this as a feature. “Note that wenowhere make use of Noetherian hypotheses.”] Then for each quasicoherent sheaf F on X,we’ll define R-modules Hi(X,F). (In particular, if R = k, they are k-vector spaces.) First,

H0(X,F) = Γ(X,F) . EachHi will be a contravariant functor in the spaceX, and a covariant functor in the sheaf F .

The functor Hi behaves well under direct sums: Hi(X,⊕jFj) = ⊕jHi(X,Fj) . (We will need

infinite sums, not just finite sums.) If 0 → F → G → H → 0 is a short exact sequence of

quasicoherent sheaves on X, then we have a long exact sequence

0→ H0(X,F)→ H0(X,G)→ H0(X,H)

→ H1(X,F)→ H1(X,G)→ H1(X,H)→ · · · .(The maps H)(X, ?) → Hi(X, ??) will be those coming from covariance; the connecting homo-morphisms Hi(x,H) → Hi+1(X,F) will have to be defined.) We’ll see that if X can be covered

by n affines, then Hi(X,F) = 0 for i ≥ n for all F , i. (In particular, all higher quasicoherent

cohomology groups on affine schemes vanish.) If X → Y is a closed immersion, and F is a qua-

sicoherent sheaf on X, then Hi(X,F) = Hi(Y, f∗F) . (We’ll care about this particularly in the

case when X ⊂ Y = PNR , which will let us reduce calculations on arbitrary projective R-schemesto calculations on PnR.)

We will also identify the cohomology of all the invertible sheaves on PnR:

23.1.1. Proposition. — 1

• H0(PnR,OPnR

(m)) is a free R-module of rank`n+m

n

´if i = 0 and m ≥ 0, and 0 other-

wise.• Hn(PnR,OPn

R(m)) is a free R-module of rank

` −m−1−n−m−1

´if m ≤ −n−1, and 0 otherwise.

1projcoh

379

380

• Hi(PnR,OPnR

(m)) = 0 if 0 < i < n.

It is more helpful to say the following imprecise statement: H0(PnR,OPnR

(m)) should be

interpreted as the homogeneous degree m polynomials in x0, . . . , xn (with R-coefficients), andHn(PnR,OPn

R(m)) should be interpreted as the homogeneous degree m Laurent polynomials in x0,

. . . , xn, where in each monomial, each xi appears with degree at most −1.We’ll prove this next day.Here are some features of this Proposition that I wish to point out, that will be the first

appearances of things that we’ll prove later.

• The cohomology of these bundles vanish above the dimension of the space if R = k;we’ll generalize this for SpecR, and even more, in before long.

• These cohomology groups are always finitely-generated R modules.• The top cohomology group vanishes for m > −n − 1. (This is a first appearance of

“Kodaira vanishing”.)• The top cohomology group is “1-dimensional” for m = −n − 1 if R = k. This is the

first appearance of a dualizing sheaf.• We have a natural duality Hi(X,O(m))Hn−i(X,O(−n−1−m)) → Hn(X,O(−n−1)).

This is the first appearance of Serre duality.

I’d like to use all these properties to prove things, so you’ll see how handy they are. We’llworry later about defining cohomology, and proving these properties.

[By Theorem 19.4.1 for any coherent sheaf F on PnR we get...] When we discussed globalsections, we worked hard to show that for any coherent sheaf F on PnR we could find a surjection

O(m)⊕j → F , which yields the exact sequence 2

(50) 0→ G → O(m)⊕j → F → 0

for some coherent sheaf G. We can use this to prove the following.

23.1.2. Theorem. — (i) For any coherent sheaf F on a projective R-scheme where R isNoetherian, hi(X,F) is a finitely generated R-module. (ii) (Serre vanishing) Furthermore, form 0, Hi(X,F(m)) = 0 for all i, even without Noetherian hypotheses. 3 Later say this forSerre vanishinghigher pushforwards! We just need a quasicompact target.

A slightly fancier version will be given in Theorem 26.1.2.

Proof. Because cohomology of a closed scheme can be computed on the ambient space, we mayreduce to the case X = PnR.

(i) Consider the long exact sequence:

0 // H0(PnR,G) // H0(PnR,O(m)⊕j ) // H0(PnR,F) //

H1(PnR,G) // H1(PnR,O(m)⊕j ) // H1(PnR,F) // · · ·

· · · // Hn−1(PnR,G) // Hn−1(PnR,O(m)⊕j) // Hn−1(PnR,F) //

Hn(PnR,G) // Hn(PnR,O(m)⊕j ) // Hn(PnR,F) // 0

The exact sequence ends here because PnR is covered by n + 1 affines. Then Hn(PnR,O(m)⊕j )is finitely generated by Proposition 23.1.1, hence Hn(PnR,F) is finitely generated for all coherent

sheaves F . Hence in particular, Hn(PnR,G) is finitely generated. As Hn−1(PnR,O(m)⊕j) is finitely

generated, and Hn(PnR,G) is too, we have that Hn−1(PnR,F) is finitely generated for all coherentsheaves F . We continue inductively downwards.

2shorty

3Serrevanishing

381

(ii) Twist (50) by O(N) for N 0. Then Hn(PnR,O(m+N)⊕j) = 0, so Hn(PnR,F(N)) = 0.Translation: for any coherent sheaf, its top cohomology vanishes once you twist by O(N) forN sufficiently large. Hence this is true for G as well. Hence from the long exact sequence,Hn−1(PnR,F(N)) = 0 forN 0. As in (i), we induct downwards, until we get thatH1(PnR,F(N)) =

0. (The induction proceeds no further, as it is not true that H0(PnR,O(m + N)⊕j) = 0 for largeN — quite the opposite.)

Exercise for those who like working with non-Noetherian rings: Prove part (i) in the above resultwithout the Noetherian hypotheses, assuming only that R is a coherent R-module (it is “coherentover itself”). (Hint: induct downwards as before. The order is as follows: Hn(PnR,F) finitely

generated, Hn(PnR,G) finitely generated, Hn(PnR,F) coherent, Hn(PnR,G) coherent, Hn−1(PnR,F)

finitely generated, Hn−1(PnR,G) finitely generated, etc.)In particular, we have proved the following, that we would have cared about even before we

knew about cohomology.

23.1.3. Corollary. — Any projective k-scheme has a finite-dimensional space of global sections.More generally, if F is a coherent sheaf on a projective R-scheme, then h0(X,F) is a finitelygenerated R-module. 4

(We will generalize this in Theorem 23.6.2.)This is true more generally for proper k-schemes, not just projective k-schemes, but I won’t

give the argument here.[I could show that h0(X,OX) is one-dimensional. Proper reduced, get one-dimensional family.

The only functions are constants.]Here is another a priori interesting consequence:

23.1.4. Corollary. — If 0→ F → G → H → 0 is an exact sequence of coherent sheaves on pro-jective X with F coherent, then for n 0, 0→ H0(X,F(n))→ H0(X,G(n))→ H0(X,H(n))→0 is also exact.

(Proof: for n 0, H1(X,F(n)) = 0.)This result can also be shown directly, without the use of cohomology. [Maybe give ref.]

23.2 +Other vanishing theorems

(I’m not sure where this should be placed!) We want various cohomology groups to vanish.Sometimes we want to know that a particular group vanish. Mention some vanishing theorems.

Kodaira vanishing. Use this to show Lefschetz hyperplane theorem for Hodge diamonds?One example: cohomology and base change.Another example: cohomology of line bundles of a curve.

23.3 +Proving the things you need to know

You can read this section later on. Although it really isn’t so bad. 5

As you read this, you should go back and check off all the facts, to make sure that I’ve shown

all that I’ve promised.

23.3.1. Cech cohomology. 6 Works nicely here. In general: take finer and finer covers. Here Cech cohomologywe take a single cover.

Suppose X is quasicompact and separated, e.g. X is quasiprojective over A. In particular, Xmay be covered by a finite number of affine open sets, and the intersection of any two affine opensets is also an affine open set; these are the properties we will use. Suppose F is a quasicoherentsheaf, and U = Uini=1 is a finite set of affine open sets of X whose union is U . For I ⊂ 1, . . . , n

4c:cohpushforward, caution because we have only finite generation

5H.III.4

6cech

382

define Ui = ∩i∈IUi. It is affine by the separated hypothesis. Define HiU (U,F) to be the ith

cohomology group of the complex7 [I should probably state explicitly what the boundary map is!Also, have them verify that this is a complex if they’ve never seen anything like this before.]

(51) 0 →M

|I| = 1I ⊂ 1, . . . , n

F(UI)→ · · · →M

|I| = iI ⊂ 1, . . . , n

F(UI)→M

|I| = i + 1I ⊂ 1, . . . , n

F(UI)→ · · · .

Note that if X is an R-scheme, then HiU (X,F) is an R-module. Also H0

U (X,F) = Γ(X,F).

23.3.A. Exercise. Suppose 0→ F1 → F2 → F3 → 0 is a short exact sequence of sheaves on atopological space, and U is an open cover such that on any intersection the sections of F2 surjectonto F3. Show that we get a long exact sequence of cohomology. [Lev’s previous exercise: Showthat the Cech cohomology of an affine scheme with respect to a distinguished cover is “trivial”.)](Note that this applies in our case!)

I ended by stating the following result, which we will prove next day.

23.3.2. Theorem/Definition. — 8 Recall that X is quasicompact and separated. HiU (U,F)

is independent of the choice of (finite) cover Ui. More precisely,

(*) for all k, for any two covers Ui ⊂ Vi of size at most k, the maps HiVi

(X,F)→HiUi

(X,F) induced by the natural maps of complex (51) are isomorphisms.

Define the Cech cohomology group Hi(X,F) to be this group.

I needn’t have stated in terms of some k; I’ve stated it in this way so I can prove it byinduction.

(For experts: we’ll get natural quasiisomorphisms of Cech complexes for various U .)

Proof. We prove this by induction on k. The base case is trivial. We need only prove the result forUini=1 ⊂ Uini=0, where the case k = n is known. Consider the exact sequence of complexes

0

0

0

· · · //L

|I| = i − 10 ∈ I ⊂ 0, . . . , n

F(UI) //

L

|I| = i0 ∈ I ⊂ 0, . . . , n

F(UI) //

L

|I| = i + 10 ∈ I ⊂ 0, . . . , n

F(UI) //

· · ·

· · · //L

|I| = i − 1I ⊂ 0, . . . , n

F(UI) //

L

|I| = iI ⊂ 0, . . . , n

F(UI) //

L

|I| = i + 1I ⊂ 0, . . . , n

F(UI) //

· · ·

· · · //L

|I| = i − 1I ⊂ 1, . . . , n

F(UI) //

L

|I| = iI ⊂ 1, . . . , n

F(UI) //

L

|I| = i + 1I ⊂ 1, . . . , n

F(UI) //

· · ·

0 0 0

We get a long exact sequence of cohomology from this. Thus by Exercise 5.2 of last day, wewish to show that the top row is exact. But the ith cohomology of the top row is preciselyHiUi∩U0i>0

(Ui,F) except at step 0, where we get 0 (because the complex starts off 0→ F(U0)→⊕nj=1F(U0∩Uj)). So we just need to show that higher Cech groups of affine schemes are 0. Hence

we are done by the following result.

7cechcomplex

8t:cechindependence

383

23.3.3. Theorem. — The higher Cech cohomology HiU (X,F) of an affine R-scheme X vanishes

(for any affine cover U , i > 0, and quasicoherent F). 9

Serre describes this as a partition of unity argument. partition of unityA spectral sequence argument can make quick work of this, but I’d like to avoid introducing

spectral sequences until I have to.[Hartshorne does the Noetherian case. [Eisenbud suggests Chapter IX of his syzygies book.]]

Proof. We want to show that the “extended” complex (where you tack on global sections to thefront) has no cohomology, i.e. that 10

(52) 0→ F(X)→ ⊕|I|=1F(UI )→ ⊕|I|=2F(UI)→ · · ·is exact. We do this with a trick.

Suppose first that some Ui (say U0) is X. Then the complex can be described as the middlerow of the following short exact sequence of complexes

0 // 0 //

⊕|I|=1,0∈IF(UI ) //

⊕|I|=2,0∈IF(UI) //

· · ·

0 // F(X) //

⊕|I|=1F(UI ) //

⊕|I|=2F(UI) //

· · ·

0 // F(X) // ⊕|I|=1,0/∈IF(UI ) // ⊕|I|=2,0/∈IF(UI) // · · ·

The top row is the same as the bottom row, slid over by 1. The corresponding long exact sequenceof cohomology shows that the central row has vanishing cohomology. (Topological experts willrecognize a mapping cone in the above construction.) [primordial: Refer to other mentions mapping coneof mapping cone, refmappingcone2, mappingcone3, mappingcone4. ] 11

We next prove the general case by sleight of hand. Say X = Spec S. We wish to show thatthe complex of R-modules (52) is exact. It is also a complex of S-modules, so we wish to showthat the complex of S-modules (52) is exact. To show that it is exact, it suffices to show thatfor a cover of Spec S by distinguished open sets D(fi) (1 ≤ i ≤ s) (i.e. (f1, . . . , fs) = 1 in S)the complex is exact. (Translation: exactness of a sequence of sheaves may be checked locally.)We choose a cover so that each D(fi) is contained in some Uj = SpecRj . Consider the complexlocalized at fi. As

Γ(SpecR,F)f = Γ(Spec(Rj)f ,F)

(as this is one of the definitions of a quasicoherent sheaf), as Uj ∩D(fi) = D(fi), we are in thesituation where one of the Ui’s is X, so we are done.

Later, I was thinking of another proof, using spectral sequences. It was motivatedby my Leray argument. It is sloppily written, because I have only a few minutes;write up better later.

We have our cover Ui of X. Take another cover Vj , that is distinguished, and alsosuch that each Vj is contained in some Ui. Then make a double complex where the(i, j) entry is ⊕|I|=i,|J|=jUI ∩ VJ . If you take the first arrow to be in the U-direction,

then for j > 0, you get 0, because you have a cover of Vj by fewer U ’s. (We areinducting on the number of U ’s; here we still need that trick of removing redundantelements.) So we’re left with j = 1, where we get F(Vj). So by E2 we get F(X).

Next, if we take the first arrow to be in the V -direction, we see a distinguishedcech cover of each UI , and hence get 0 except for j = 0. For j = 0, we get the coverwe’re interested in. The next step consists of the arrows of the Cech cover we’reinterested in, so the result is the cohomology we seek, and we’ve already convergedat step 2.

9H.T.III.3.5 — he does just Noetherian case, but in derived functor setting. I do general case; there refers

to [EGA, III.1.3.1]. [Later add to H that Noetherian case is unnecessary.]10

Cechtoshow11

mappingcone1

384

23.3.B. Exercise. Suppose V ⊂ U are open subsets of X. Show that we have restrictionmorphisms Hi(U,F) → Hi(V,F) (if U and V are quasicompact, and U hence V is separated).Show that restrictions commute. Hence if X is a Noetherian space, H i(,F) this is a contravariantfunctor from the category Top(X) to abelian groups. (For experts: this means that it is a presheaf.But this is not a good way to think about it, as its sheafification is 0, as it vanishes on the affinebase by Remark 4.7.3.) The same argument will show more generally that for any map f : X → Y ,there exist natural maps Hi(X,F)→ Hi(X, f∗F); I should have asked this instead.

23.3.C. Exercise. Show that if F → G is a morphism of quasicoherent sheaves on separated andquasicompact X then we have natural maps Hi(X,F)→ Hi(X,G). Hence Hi(X, ·) is a covariantfunctor from quasicoherent sheaves on X to abelian groups (or even R-modules).

In particular, we get the following facts.1. If X → Y is a closed subscheme then Hi(X,F) = Hi(Y, f∗F), as promised at start of our

discussion on cohomology.2. Also, if X can be covered by n affine open sets, then Hi(X,F) = 0 for all quasicoherent

F , and i ≥ n. In particular, Hi(SpecR,F) = 0 for i > 0.3. Cohomology behaves well for arbitrary direct sums of quasicoherent sheaves.

23.3.4. Dimensional vanishing for projective k-schemes.

23.3.5. Theorem. — Suppose X is a projective k-scheme, and F is a quasicoherent sheaf onX. Then Hi(X,F) = 0 for i > dimX.

In other words, cohomology vanishes above the dimension of X. We will later show that thisis true when X is a quasiprojective k-scheme.

Proof. Suppose X → PN , and let n = dimX. We show that X may be covered by n affine opensets. Long ago, we had an exercise saying that we could find n Cartier divisors on PN such thattheir complements U0, . . . , Un covered X (Exercise 15.6.C). [We did this as follows. Lemma:Suppose Y → PN is a projective scheme. Then Y is Noetherian, and hence has a finite number ofcomponents. We can find a hypersurface H containing none of their associated points. Then Hcontains no component of Y , the dimension of H ∩Y is strictly smaller than Y , and if dimY = 0,then H ∩Y = ∅.] Then Ui is affine, so Ui ∩X is affine, and thus we have covered X with n affineopen sets.

Remark. We actually need n affine open sets to cover X, but I don’t see an easy way toprove it. One way of proving it is by showing that the complement of an affine set is always purecodimension 1. [I wanted to tell the experts something here, but I’m not sure what.]

I’m a little confused about this, so I won’t say it in class.Here is another exercise regarding pushforwards in more generality. It will be part of the

story behind the Leray spectral sequence. [This is used in Exercise 23.6.A.] [This is part of thestory behind the Leray spectral sequence (Section 23.9).]Leray

23.3.D. Exercise. Suppose π : X → Y is a morphism, and F is a quasicoherent sheaf. Assumethat π is quasicompact and quasiseparated, so that we have pushforwards. (For example, we cantake X to be Noetherian.) See Theorem 20.2.1, which is unfortunately in the future. Describe anatural morphism Hi(X,F)→ Hi(Y, f∗F). 12

23.4 Cohomology of line bundles on projective space

I’ll now pay off that last IOU. [Likely this will be moved earlier.]

23.4.1. Proposition. —

• H0(PnR,OPnR

(m)) is a free R-module of rank`n+m

n

´if i = 0 and m ≥ 0, and 0 other-

wise.• Hn(PnR,OPn

R(m)) is a free R-module of rank

` −m−1−n−m−1

´if m ≤ −n−1, and 0 otherwise.

12cohpushforward

385

• Hi(PnR,OPnR

(m)) = 0 if 0 < i < n.

It is more helpful to say the following imprecise statement: H0(PnR,OPnR

(m)) should be

interpreted as the homogeneous degree m polynomials in x0, . . . , xn (with R-coefficients), andHn(PnR,OPn

R(m)) should be interpreted as the homogeneous degree m Laurent polynomials in x0,

. . . , xn, where in each monomial, each xi appears with degree at most −1. [“can be canonicallyidentified with”]

Proof. The H0 statement was an (important) exercise last quarter.Rather than consider O(m) for various m, we consider them all at once, by considering

F = ⊕mO(m).Of course we take the standard cover U0 = D(x0), . . . , Un = D(xn) of PnR. Notice that if

I ⊂ 1, . . . , n, then F(UI) corresponds to the Laurent monomials where each xi for i /∈ I appearswith non-negative degree.

We consider the Hn statement. Hn(PnR,F) is the cokernel of the following surjection

⊕ni=0F(U1,...,n−i)→ FU1,...,n

i.e.

⊕ni=0R[x0, . . . , xn, x−10 , . . . , ˆx−1

i , . . . x−1n ]→ R[x0, . . . , xn, x

−10 , . . . , x−1

n ].

This cokernel is precisely as described.We last consider the Hi statement (0 < i < n). We prove this by induction on n. The cases

n = 0 and 1 are trivial. Consider the exact sequence of quasicoherent sheaves:

0 // F ×xn // F // F ′ // 0

where F ′ is analogous sheaf on the hyperplane xn = 0 (isomorphic to Pn−1R ). (This exact sequence

is just the direct sum over all m of the exact sequence

0 // OPnR

(m − 1)×xn // OPn

R(m) // OPn−1

R(m) // 0 ,

which in turn is obtained by twisting the closed subscheme exact sequence

0 // OPnR

(m − 1)×xn // OPn

R(m) // OPn−1

R(m) // 0

by OPnR

(m).)

The long exact sequence in cohomology gives us:

0 // H0(PnR,F)×xn // H0(PnR,F) // H0(Pn−1

R ,F ′)

// H1(PnR,F)×xn // H1(PnR,F) // H1(Pn−1

R ,F ′)

. . . // Hn−1(PnR,F)×xn // Hn−1(PnR,F) // Hn−1(Pn−1

R ,F ′)

// Hn(PnR,F)×xn // Hn(PnR,F) // 0

.

We will now show that this gives an isomorphism13

(53) ×xn : Hi(PnR,F)→∼ Hi(PnR,F)

for 0 < i < n. The inductive hypothesis gives us this except for i = 1 and i = n − 1, where

we have to pay a bit more attention. For the first, note that H0(PnR,F) // H0(Pn−1R ,F ′)

is surjective: this map corresponds to taking the set of all polynomials in x0, . . . , xn, and set-ting xn = 0. The last is slightly more subtle: Hn−1(Pn−1

R ,F ′) → Hn(PnR,F) is injective, andcorresponds to taking a Laurent polynomial in x0, . . . , xn−1 (where in each monomial, each xi

13boxedin

386

appears with degree at most −1) and multiplying by x−1n , which indeed describes the kernel of

Hn(PnR,F)×xn // Hn(PnR,F) . (This is a worthwhile calculation! See the exercise after the end

of this proof.) We have thus established (53) above.We will now show that the localization Hi(PnR,F)xn = 0. (Here’s what we mean by localiza-

tion. Notice Hi(PnR,F) is naturally a module over R[x0, . . . , xn] — we know how to multiply byelements of R, and by (53) we know how to multiply by xi. Then we localize this at xn to get anR[x0, . . . , xn]xn-module.) This means that each element α ∈ Hi(PnR,F) is killed by some powerof xi. But by (53), this means that α = 0, concluding the proof of the theorem.

Consider the Cech complex computing Hi(PnR,F). Localize it at xn. Localization and co-homology commute (basically because localization commutes with operations of taking quotients,images, etc.), so the cohomology of the new complex is H i(PnR,F)xn . But this complex computesthe cohomology of Fxn on the affine scheme Un, and the higher cohomology of any quasicoherentsheaf on an affine scheme vanishes (by Theorem 23.3.3 which we’ve just proved — in fact we usedthe same trick there), so Hi(PnR,F)xn = 0 as desired.

23.4.A. Exercise. Verify that Hn−1(Pn−1R ,F ′) → Hn(PnR,F) is injective (likely by verifying

that it is the map on Laurent monomials we claimed above). [Kirsten proved this. Scan iskirsten13.8.pdf

23.5 Application of cohomology: Hilbert polynomials andHilbert functions; degrees

We’ve already seen some powerful uses of this machinery, to prove things about spaces ofglobal sections, and to prove Serre vanishing. We’ll now see some classical constructions come outvery quickly and cheaply.

In this section, we will work over a field k. Define hi(X,F) := dimkHi(X,F).

Suppose F is a coherent sheaf on a projective k-scheme X. Define the Euler characteristicEuler characteristic

χ(X,F) =dimXX

i=0

(−1)ihi(X,F).

We will see repeatedly here and later that while Euler characteristics behave better than individualcohomology groups. As one sign, notice that for fixed n, and m ≥ 0,

h0(Pnk ,O(m)) =“n+m

m

”=

(m + 1)(m + 2) · · · (m+ n)

n!.

[Two other reasons: additive in exact sequences. Constant in “nice” families.]Notice that the expression on the right is a polynomial in m of degree n. (For later reference,

I want to point out that the leading term is mn/n!.) But it is not true that

h0(Pnk ,O(m)) =(m + 1)(m + 2) · · · (m+ n)

n!

for all m — it breaks down for m ≤ −n− 1. Still, you can check that

χ(Pnk ,O(m)) =(m + 1)(m + 2) · · · (m + n)

n!.

So one lesson is this: if one cohomology group (usual the top or bottom) behaves well in a certainrange, and then messes up, likely it is because (i) it is actually the Euler characteristic which isbehaving well always, and (ii) the other cohomology groups vanish in that range.

In fact, we will see that it is often hard to calculate cohomology groups (even h0), but it isoften easier calculating Euler characteristics. So one important way of getting a hold of cohomology

groups is by computing the Euler characteristics, and then showing that all the other cohomologygroups vanish. Hence the ubiquity and importance of vanishing theorems. (A vanishing theoremvanishing theoremsusually states that a certain cohomology group vanishes under certain conditions.)

The following exercise already shows that Euler characteristic behaves well.

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23.5.A. Exercise. Show that Euler characteristic is additive in exact sequences. In otherwords, if 0 → F → G → H → 0 is an exact sequence of coherent sheaves on X, then χ(X,G) =χ(X,F) + χ(X,H). (Hint: consider the long exact sequence in cohomology.) More generally, if

0→ F1 → · · · → Fn → 0

is an exact sequence of sheaves, show that

nX

i=1

(−1)iχ(X,Fi) = 0.

23.5.B. Exercise. Prove the Riemann-Roch theorem for line bundles on a nonsingular projectivecurve C over k: suppose L is an invertible sheaf on C. Show that χ(L) = degL + χ(C,OC).(Possible hint: Write L as the difference of two effective Cartier divisors, L ∼= O(Z − P ) (“zeros”minus “poles”). Describe two exact sequences 0→OC(−P )→OC →OP → 0 and 0→ L(−Z)→L→ OZ ⊗L → 0, where L(−Z) ∼= OC(P ).) 14

[Remark: can use this to define degree on a singular projective curve... more on this soon...]If F is a coherent sheaf on X, define the Hilbert function of F :

hF (n) := h0(X,F(n)).

The Hilbert function of X is the Hilbert function of the structure sheaf. The ancients were awarethat the Hilbert function is “eventually polynomial”, i.e. for large enough n, it agrees with somepolynomial, called the Hilbert polynomial (and denoted pF(n) or pX(n)). In modern language, we H function, polyexpect that this is because the Euler characteristic should be a polynomial, and that for n 0,the higher cohomology vanishes. This is indeed the case, as we now verify.

[At some point we need the fact that for large enough n, the Sn = H0(X,OX(n)).][Hilbert function of an ample invertible sheaf L. More generally, of a coherent sheaf F H function

“with respect to” L.]

23.5.1. Theorem. — If F is a coherent sheaf on a projective k-scheme X → Pnk , for m 0,

h0(X,F(m)) is a polynomial of degree equal to the dimension of the support of F . In particular,for m 0, h0(X,OX(m)) is polynomial with degree = dimX.

(Here OX(m) is the restriction or pullback of OPnk(1).) [The case F = 0 is special.]

I realize now that I will use the notion of associated primes of a module, not just of a ring. Ithink I only discussed associated primes of a ring last quarter, because I had hoped not to needthis slightly more general case. Now I don’t really need it, and if you want to ignore this issue, youcan just prove the second half of the theorem, which is all we will use anyway. But the argumentcarries through with no change, so please follow along if you can.

Proof. For m 0, hi(X,F(m)) = 0 by Serre vanishing 23.1.2, so instead we will prove that forall m, χ(X,F(m)) is a polynomial of degree equal to the dimension of the support of F . DefinepF(m) = χ(X,F(m)); we’ll show that pF(m) is a polynomial of the desired degree.

Our approach will be a little weird. We’ll have two steps, and they will be very similar. Ifyou can streamline, please let me know.

Step 1. We first show that for all n, if F is scheme-theoretically supported a linear subspaceof dimension k (i.e. F is the pushforward of a coherent sheaf on some linear subspace of dimensionk), then pF(m) is a polynomial of degree at most k. (In particular, for any coherent F , pF(m) isa polynomial of degree at most n.) The polynomiality, not the degree bound, is waht we need.

We prove this by induction on the dimension of the support. I’ll leave the base case k = 0(or better yet, k = −1) to you (exercise). Suppose now that X is supported in a linear space Λ ofdimension k, and we know the result for all k′ < k. Then let x = 0 be a hyperplane not containingΛ, so Λ′ = dim(x = 0) ∩ Λ = k − 1. Then we have an exact sequence15

(54) 0 // K // F ×x // F(1) // K′ // 0

where K (resp. K′) is the kernel (resp. cokernel) of the map ×x. Notice that K and K′ are bothsupported on Λ′. (This corresponds to an algebraic fact: over an affine open SpecA, the exact

14rr

15suppind

388

sequence is

0 // K // M×x // M // K′ // 0

and both K = ker(×x) = (0 : x) and K ′ ∼= M/xM are (A/x)-modules.) Twist (54) by O(m) andtake Euler-characteristics to obtain pF(m + 1) − pF (m) = pK′(m) − pK(m). By the inductivehypothesis, the right side of this equation is a polynomial of degree at most k − 1. Hence (by aneasy induction) p(m) is a polynomial of degree at most k. [Give them a quick explanation. Showthat there is a bijection between polynomials of degree ≤ k with p(0) = 0 and polynomials ofdegree ≤ k − 1 given by p(x)− p(x− 1) = q(x). Show that it is injective. Hence by dimensionalreasons it must be surjective.]

Step 2. We’ll now show that the degree of this polynomial is precisely dimSuppF . As Fis a coherent sheaf on a Noetherian scheme, it has a finite number of associated points, so wecan find a hypersurface H = (f = 0) not containing any of the associated points. (This is thatproblem from last quarter that we have been repeatedly using recently: problem 24(c) on set 5,which was exercise 1.19 in the class 11 notes.) In particular, dimH ∩ SuppF is strictly less thandimSuppF , and in fact one less by Krull’s Principal Ideal Theorem [REF!]. Let d = deg f . ThenI claim that ×f : F(−d) → F is an inclusion. Indeed, on any affine open set, the map is of the

form ×f : M → M (where f is the restriction of f to this open set), and the fact that f = 0contains no associated points means that this is an injection of modules. (Remember that those

ring elements annihilating elements of M are precisely the associated primes, and f is containedin none of them.) Then we have

0→ F(−d)→ F → K′ → 0.

Twisting by O(m) yields

0→ F(m − d)→ F(m)→ K′(m)→ 0.

Taking Euler characteristics gives pF(m) − pF (m − d) = pK′(m). Now by step 1, we know thatpF(m) is a polynomial. Also, by our inductive hypothesis, and Exercise 23.5.C below, the rightside is a polynomial of degree of precisely dim SuppF − 1. Hence p(m) is a polynomial of degreedimSuppF .

23.5.C. Exercise. Consider the short exact sequence of A-modules

0 //M×f // M // K′ // 0 .

Show that SuppK′ = Supp(M) ∩ Supp(A/f). 16

Notice that we needed the first part of the proof to ensure that pF (m) is in fact a polynomial;otherwise, the second part would just show that pF(m) is just a polynomial whenm is fixed modulod.

(For experts: here is a different way to avoid having two similar steps. If k is an infinite field,e.g. if it were algebraically closed, then we could find a hypersurface as in step 2 of degree 1, usingthat problem from last quarter mentioned in the proof. So what to do if k is not infinite? Notethat if you have a complex of k-vector spaces, and you take its cohomology, and then tensor withk, you get the same thing as if you tensor first, and then take the cohomology. By this trick, wecan assume that k is algebraically closed. In fancy language: we have taken a faithfully flat baseextension. I won’t define what this means here; it will turn up early in the third quarter.)

Example 1. [Caution, the subjects e.g. pPn(m) are not yet defined.] pPn(m) =`m+n

n

´, where

we interpret this as the polynomial (m + 1) · · · (m+ n)/n!.Example 2. Suppose H is a degree d hypersurface in Pn. Then from

0→OPn (−d)→OPn →OH → 0,

we have

pH(m) = pPn(m) − pPn(m− d) =“m+ n

n

”−“m+ n− d

n

”.

23.5.D. Exercise. Show that the twisted cubic (in P3) has Hilbert polynomial 3m + 1. [Referto Exercise 9.1.T, where the twisted cubic was defined.]

16addthis

389

23.5.E. Exercise. Find the Hilbert polynomial for the dth Veronese embedding of Pn (i.e. theclosed immersion of Pn in a bigger projective space by way of the line bundle O(d)). d-tuple embedding,

Veronese embeddingFrom the Hilbert polynomial, we can extract many invariants, of which two are particularlyimportant. The first is the degree. Classically, the degree of a complex projective variety of

degree of a projective

schemedimension n was defined as follows. We slice the variety with n generally chosen hyperplane.Then the intersection will be a finite number of points. The degree is this number of points. Ofcourse, this requires showing all sorts of things. Instead, we will define the degree of a projectivek-scheme of dimension n to be leading coefficient of the Hilbert polynomial (the coefficient of degree of a projective

schememn) times n!.For example, the degree of Pn in itself is 1. The degree of the twisted cubic is 3. [Refer to

Exercise 9.1.T, where the twisted cubic was defined.]

23.5.F. Exercise. Show that the degree is always an integer. Hint: show that any polynomialin m of degree k taking on only integral values must have coefficient of mk an integral multipleof 1/k!.

23.5.G. Exercise. Show that the degree of a degree d hypersurface is d (preventing a notationalcrisis).

23.5.H. Exercise. Suppose a curve C is embedded in projective space via an invertible sheaf ofdegree d. (In other words, this line bundle determines a closed immersion.) Show that the degreeof C under this embedding is d (preventing another notational crisis). (Hint: Riemann-Roch,Exercise 23.5.B.)

23.5.I. Exercise. Show that the degree of the dth Veronese embedding of Pn is dn. d-tuple embedding,

Veronese embedding23.5.J. Exercise (Bezout’s theorem). Suppose X is a projective scheme of dimension at least1, and H is a degree d hypersurface not containing any associated points of X. (For example, ifX is a projective variety, then we are just requiring H not to contain any irreducible componentsof X.) Show that degH ∩X = d degX.

This is a very handy theorem! For example: if two projective plane curves of degree m anddegree n share no irreducible components, then they intersect in mn points, counted with appro-priate multiplicity. The notion of multiplicity of intersection is just the degree of the intersectionas a k-scheme.

We trot out a useful example for a third time: let k = Q, and consider the parabola x = y2.We intersect it with the four usual suspects: x = 1, x = 0, x = −1, and x = 2, and see that weget 2 each time (counted with the same convention as with the last time we saw this example).

If we intersect it with y = 2, we only get one point — but that’s of course because this isn’ta projective curve, and we really should be doing this intersection on P2

k — and in this case, theconic meets the line in two points, one of which is “at ∞”.

[(More generally: if reduced and Cohen-Macaulay, we always get the right number of points.I could refer to the Cohen-Macaulay section 16.7 here.) ] 17 CM

23.5.K. Exercise. Determine the degree of the d-fold Veronese embedding of Pn in a differentway as follows. Let vd : Pn → PN be the Veronese embedding. To find the degree of the image, weintersect it with n hyperplanes in PN (scheme-theoretically), and find the number of intersectionpoints (counted with multiplicity). But the pullback of a hyperplane in PN to Pn is a degree dhypersurface. Perform this intersection in Pn, and use Bezout’s theorem. (If already you knowthe answer by the earlier exercise on the degree of the Veronese embedding, this will be easier.) d-tuple embedding,

Veronese embeddingThere is another nice bit of information residing in the Hilbert polynomial. Notice thatpX(0) = χ(X,OX), which is an intrinsic invariant of the scheme X, which does not depend onthe projective embedding.

Imagine how amazing this must have seemed to the ancients: they defined the Hilbert functionby counting how many “functions of various degrees” there are; then they noticed that when thedegree gets large, it agrees with a polynomial; and then when they plugged 0 into the polynomial— extrapolating backwards, to where the Hilbert function and Hilbert polynomials didn’t agree— they found a magic invariant!

17degreeofscheme

390

And now I can give you a nonsingular curve over an algebraically closed field that is not P1!Note that the Hilbert polynomial of P1 is (m + 1)/1 = m + 1, so χ(OP1 ) = 1. Suppose C is adegree d curve in P2. Then the Hilbert polynomial of C is

pP2(m) − pP2 (m− d) = (m+ 1)(m + 2)/2 − (m − d+ 1)(m − d+ 2)/2.

Plugging in m = 0 gives us −(d2 − 3d)/2. Thus when d > 2, we have a curve that cannot beisomorphic to P1! (I think I gave you an earlier exercise that there is a nonsingular degree dcurve. Note however that the calculation above didn’t use nonsingularity.) [Make sure to link toan exercise on this in FOAG!]

Now from 0→OP2 (−d)→ OP2 →OC → 0, using h1(OP2 (d)) = 0, we have that h0(C,OC) =1. As h0 − h1 = χ, we have

h1(C,OC) = (d − 1)(d − 2)/2.

Motivated by geometry, we define the arithmetic genus of a scheme X as 1 − χ(X,OX). This issometimes denoted pa(X). In the case of nonsingular complex curves, this corresponds to thearithmetic genus patopological genus. For irreducible reduced curves (or more generally, curves with h0(X,OX) ∼= k),pa(X) = h1(X,OX). (In higher dimension, this is a less natural notion.)

We thus now have examples of curves of genus 0, 1, 3, 6, 10, . . . (corresponding to degree 1or 2, 3, 4, 5, . . . ).

This begs some questions, such as: are there curves of other genera? (Yes.) Are there othergenus 1 curves? (Not if k is algebraically closed, but yes otherwise.) Do we have all the curves ofgenus 3? (Almost all, but not quite all.) Do we have all the curves of genus 6? (We’re missingmost of them.)

Caution: The Euler characteristic of the structure sheaf doesn’t distinguish between isomor-phism classes of nonsingular projective schemes over algebraically closed fields — for example,P1×P1 and P2 both have Euler characteristic 1, but are not isomorphic (as for example Pic P2 ∼= Zwhile Pic P1 × P1 ∼= Z⊕ Z).

Important Remark. We can restate the Riemann-Roch formula (Exercise 23.5.B) as:

h0(C,L)− h1(C,L) = degL− pa + 1.

This is the most common formulation of the Riemann-Roch formula.

23.5.2. Complete intersections. We define a complete intersection in Pn as follows. Pn is acomplete intersection in itself. A closed subscheme Xr → Pn of dimension r (with r < n) is acomplete intersection if there is a complete intersection Xr+1, and Xr is a Cartier divisor in classOXr+1

(d).

Exercise. Show that ifX is a complete intersection of dimension r in Pn, thenHi(X,OX(m)) =0 for all 0 < i < r and allm. Show that if r > 0, then H0(Pn,O(m))→ H0(X,O(m)) is surjective.

Now in my definition, Xr is the zero-divisor of a section of OXr+1(m) for some m. But

this section is the restriction of a section of O(m) on Pn. Hence Xr is the scheme-theoreticintersection of Xr+1 with a hypersurface. Thus inductively we can show that Xr is the scheme-theoretic intersection of n− r hypersurfaces. (By Bezout’s theorem, degXr is the product of thedegree of the defining hypersurfaces.)

Exercise. Show that complete intersections of positive dimension are connected. (Hint: showh0(X,OX) = 1.) [In fact geometrically connected.]

Exercise. Find the genus of the intersection of 2 quadrics in P3. (We get curves of moregenera by generalizing this!) [We need Bertini’s theorem!]

Exercise. Show that the rational normal curve of degree d in Pd is not a complete intersection

if d > 2.Exercise. Show that the union of 2 distinct planes in P4 is not a complete intersection. (This

is the first appearance of another universal counterexample!) Hint: it is connected, but you canslice with another plane and get something not connected.

This is another important scheme in algebraic geometry that is an example of many sorts ofbehavior. We will see more of it later!

Not done:

23.5.L. Extended exercise: cohomology of line bundles on P1 × P1.Assume that cohomology of structure sheaf is 1, 0, 0. (We’ll prove this later; you could prove

it yourself.) Figure out how to compute the cohomology of O(a, b) where a and b are positive.

391

Find the genus of class (a, b). Assume field is algebraically closed if you want. Show that there is anonsingular curve in class (2, g+1). Exhibit curves of every genus. (These are called hyperellipticcurves.)

23.6 Higher direct image sheaves

[Just as Euler characteristics are better, the analogue here is the Grothendieck group. Betterstill: the derived category.]

I’ll now introduce a notion generalizing these Cech cohomology groups. Cohomology groupswere defined for X → SpecA where the structure morphism is quasicompact and separated; forany quasicoherent F on X, we defined Hi(X,F).

We’ll now do something similar for quasicompact and separated morphisms π : X → Y : forany quasicoherent F on X, we’ll define Riπ∗F , a quasicoherent sheaf on Y .

We have many motivations for doing this. In no particular order:

(1) It “globalizes” what we were doing anywhere.(2) If 0→ F → G → H → 0 is a short exact sequence of quasicoherent sheaves on X, then

we know that 0 → π∗F → π∗G → π∗H is exact, and higher pushforwards will extendthis to a long exact sequence.

(3) We’ll later [where?!] see that this will show how cohomology groups vary in families,especially in “nice” situations. Intuitively, if we have a nice family of varieties, and afamily of sheaves on them, we could hope that the cohomology varies nicely in families,and in fact in “nice” situations, this is true. (As always, “nice” usually means “flat”,whatever that means.) higer direct image sheaf

There will be no extra work involved for us.Paragraph to place somewhere around here: In the notation Rjf∗F for higher pushforward

sheaves, the “R” stands for “right derived functor”, and “corresponds” to the fact that we get along exact sequence in cohomology extending to the right (from the 0th terms). More generally,next quarter we will see that in good circumstances, if we have a left-exact functor, there may bea long exact sequence going off to the right, in terms of right derived functors. Similarly, if wehave a right-exact functor (e.g. if M is an A-module, then ⊗AM is a right-exact functor from thecategory of A-modules to itself), there may be a long exact sequence going off to the left, in termsof left derived functors.

Suppose π : X → Y , and F is a quasicoherent sheaf on X. For each SpecA ⊂ Y , wehave A-modules Hi(π−1(SpecA),F). We will show that these patch together to form a qua-sicoherent sheaf. We need check only one fact: that this behaves well with respect to takingdistinguished open sets. In other words, we must check that for each f ∈ A, the natural mapHi(π−1(SpecA),F) → Hi(π−1(SpecA),F)f (induced by the map of spaces in the opposite di-

rection — Hi is contravariant in the space) is precisely the localization ⊗AAf . But this can be

verified easily: let Ui be an affine cover of π−1(SpecA). We can compute Hi(π−1(SpecA),F)using the Cech complex (51). But this induces a cover SpecAf in a natural way: If Ui = SpecAiis an affine open for SpecA, we define U ′i = Spec(Ai)f . The resulting Cech complex for SpecAfis the localization of the Cech complex for SpecA. As taking cohomology of a complex com-mutes with localization, we have defined a quasicoherent sheaf on Y by one of our definitions ofquasicoherent sheaves by Definition 2’ of a quasicoherent sheaf (Section 17.2).

23.6.1. (Something important happened in that last sentence — localization commuting withtaking cohomology. If you want practice with this notion, here is an exercise: suppose C0 →C1 → C2 is a complex in an abelian category, and F is an exact functor to another abeliancategory. Show that F applied to the cohomology of this complex is naturally isomorphic to thecohomology of F of this complex. Translation: taking cohomology commutes with exact functors.In the particular case of this construction, the exact functor in equation is the localization functor⊗AAf from A-modules to Af -modules. [I have an algebraic fact I need to do earlier.])

Define the ith higher direct image sheaf or the ith (higher) pushforward sheaf to bethis quasicoherent sheaf. 18 higher direct image

sheaf, higher pushfor-

ward sheaf18

H.C.III.8.6 says Rif∗F is quasicoherent if X is Noetherian.

392

23.6.2. Theorem. — 19

(a) R0π∗F is canonically isomorphic to π∗F .(b) Riπ∗ is a covariant functor from the category of quasicoherent sheaves on X to the

category of quasicoherent sheaves on Y , and a contravariant functor in Y -schemes X.(c) A short exact sequence 0 → F → G → H → 0 of sheaves on X induces a long exact

sequence

0 // R0π∗F // R0π∗G // R0π∗H //

R1π∗F // R1π∗G // R1π∗H // · · ·

of sheaves on Y . (This is often called the corresponding long exact sequence ofhigher pushforward sheaves.)long exact sequence

of higher pushforward

sheaves

(d) (projective pushforwards of coherent are coherent) If π is a projective morphism andOY is coherent on Y (this hypothesis is automatic for Y locally Noetherian), and Fis a coherent sheaf on X, then for all i, Riπ∗F is a coherent sheaf on Y . 20 21

Proof. Because it suffices to check each of these results on affine open sets, they all follow fromthe analogous statements in Cech cohomology (XXXX). [(this essentially follows from Corollary23.1.3)]

The following result is handy (and essentially immediate from our definition).

23.6.A. Exercise. Show that if π is affine, then for i > 0, Riπ∗F = 0. Moreover, if Y isquasicompact and quasiseparated, show that the natural morphism H i(X,F)→ Hi(Y, f∗F) (seefor example Exercise 23.7.F) is an isomorphism. (A special case of the first sentence is a specialcase we showed earlier, when π is a closed immersion. Hint: use any affine cover on Y , which willinduce an affine cover of X.) 22 23

(This is in fact a characterization. Serre’s criterion for affineness. Assume f quasicompactand separated. Then f is affine if and only if f∗ is exact on the category of quasicoherent sheaves.EGA II.5.2. I don’t think we’ll use this, so it could just be an exercise.)Serre’s criterion for

affineness

23.7 A useful categorical fact (shouldn’t be a full section)

Here is an algebra exercise that is very useful. [Decide where precisely to put this. Forexample, it was used earlier.]

23.7.A. Important algebra exercise. Suppose M1α // M2

β // M3 is a complex of

A-modules (i.e. β α = 0), and N is an A-module. (a) Describe a natural homomorphism of thecohomology of the complex, tensored with N , with the cohomology of the complex you get whenyou tensor with N , H(M∗)⊗A B → H(M∗ ⊗A N), i.e.

„ker β

imα

«⊗A N →

ker(β ⊗N)

im(α⊗N).

I always forget which way this map is supposed to go.(b) If N is flat, i.e. ⊗N is an exact functor, show that the morphism defined above is an isomor-phism. (Hint: This is actually a categorical question: if M∗ is an exact sequence in an abelian cat-egory, and F is a right-exact functor, then (a) there is a natural morphism FH(M∗)→ H(FM∗),and (b) if F is an exact functor, this morphism is an isomorphism.)flat

19bigpushforward

20t:projpushforwardcoherent2

21t:cohderpushforward

22H.E.III.8.2; I think I do more.

23affinehigherpushforward

393

[Answer for (a) — (b) is an easy variation: start with our complex · · · → Mp−1 → Mp →Mp+1 → · · · . Let Z denote the kernel, defined by the exactness of 0 → Zp → Mp → Mp+1.Apply F to get the complex (not necessarily exact) F (Zp) → FMp → FMp+1. But as 0 →(FZ)p → FMp → FMp+1 is exact, we obtain a morphism F (Zp) → (FZ)p. Let K denote theimage, defined by 0 → Zp → Mp → Kp → 0. Apply F to get the exact sequence F (Zp) →FMp → F (Kp) → 0. But we also have 0 → (FZ)p → FMp → (FK)p → 0, and our mapsF (Zp) → (FZ)p and FMp = FMp induces a map F (K)p → (FK)p. Finally, cohomology isdefined by 0→ Kp−1 → Zp → Hp → 0. Then apply F to get F (Kp−1)→ F (Zp)→ F (Hp)→ 0.By considering the two earlier maps of pieces to 0 → (FK)p−1 → (FZ)p → (FH)p → 0, we getthe desired answer.]

[Earlier note:

23.7.1. Stupid fact. Suppose I have a complex C• in an abelian category B, and a covariantfunctor F : A → B. Then there is a natural morphism FH• → H•F . Probably if F is exact, thenthis is an isomorphism. (State this in contravariant case?)] 24

Example: localization is exact, so S−1A is a flat A-algebra for all multiplicative sets S. Inparticular, Af is a flat A-algebra. We used (b) implicitly last day [hence move this back], whenI said that given a quasicompact, separated morphism π : X → Y , and an affine open subsetSpecA of Y , and a distinguished affine open SpecAf of that, the cohomology of any Cech complex

computing the cohomology π−1(SpecA), tensored with Af , would be naturally isomorphic to thecohomology of the complex you get when you tensor with Af .

Here is another example.

23.7.B. Exercise (higher pushforwards and base change). (a) Suppose f : Z → Y is anymorphism, and π : X → Y as usual is quasicompact and separated. Suppose F is a quasicoherentsheaf on X. Let

Wf ′ //

π′

X

π

Z

f // Yis a fiber diagram. [I wrote to myself during one of the reading courses: Do we need this to bea fiber diagram? I think not; we needed only the f = idcase. We may need non-fiber diagramcases.] Describe a natural morphism f∗(Riπ∗F) → Riπ′∗(f

′)∗F of sheaves on Z. 25 [This isMilne p. 117.] [We need the flat version of this!!! I feel like introducing flatness soon.](b) If f : Z → Y is an affine morphism, and for a cover SpecAi of Y , where f−1(SpecAi) =SpecBi, Bi is a flat A-algebra, show that the natural morphism of (a) is an isomorphism. (Youcan likely generalize this immediately, but this will lead us into the concept of flat morphisms,and we’ll hold off discussing this notion for a while.)

A useful special case if the following. If f is a closed immersion of a closed point in Y , theright side is the cohomology of the fiber, and the left side is the fiber of the cohomology. In otherwords, the fiber of the higher pushforward maps naturally to the cohomology of the fiber. We’ll

later see that in good situations this is an isomorphism, and thus the higher direct image sheafindeed “patches together” the cohomology on fibers.

More explicitly [fold in]:

23.7.C.Exercise. Show that if y ∈ Y , there is a natural morphismHi(Y, f∗F)y → Hi(f−1(y),F|f−1(y)).

(Hint: if you take a complex, and tensor it with a module, and take cohomology, there is a mapfrom that to what you would get if you take cohomology and tensor it with a module.) 26

[Older stuff...

23.7.D. Easy exercise. If π : X → Y [is such that pushforwards send quasicoherents toquasicoherents, e.g. qc+qs, or X Noetherian], show that π∗ is an exact functor from the categoryof quasicoherent sheaves on X to the category of quasicoherent sheaves on Y . If π is finite, show

24FHHF

25Rlowerstarfiber

26Rlowerstarfiber

394

that π∗ is an exact functor from the category of coherent sheaves on X to the category of coherentsheaves on Y .

(d) from Corollary 23.1.3 [what is this all about?]]

23.7.E. Exercise (projection formula). Suppose π : X → Y is quasicompact and separated,and E , F are quasicoherent sheaves on X and Y respectively. (a) Describe a natural morphism

(Riπ∗E)⊗ F → Riπ∗(E ⊗ π∗F).

(b) If F is locally free, show that this natural morphism is an isomorphism.27

make this part (a). This uses silly fact 23.7.1. [Do we really need finite rank? I don’t thinkso.] [Find out from Jarod which way the map goes.] [This natural map now comes up in theprojection formula“neat algebraic fact section”.]

In particular [fold in]:

23.7.F. Exercise. Suppose π : X → Y is a morphism, and F is a quasicoherent sheaf.[Assumption: π is quasicompact and quasiseparated, so that we have pushforwards; or at leastthat X Noetherian. See Theorem 20.2.1, which is unfortunately in the future.] Describe thenatural morphism Hi(X,F) → Hi(Y, f∗F). 28 [This is used in Exercise 23.6.A.] This is part ofthe story behind the Leray spectral sequence (Section 23.9).Leray

Here is another consequence.Exercise. Suppose that X is a quasicompact separated k-scheme, where k is a field. SupposeF is a quasicoherent sheaf on X. Let Xk = X ×Spec k Spec k, and f : Xk → X the projection.

Describe a natural isomorphism Hi(X,F) ⊗k k → Hi(Xk , f∗F). Recall that a k-scheme X is

geometrically integral if Xk is integral. Show that if X is geometrically integral and projective,

then H0(X,OX) ∼= k. (This is a clue that P1C is not a geometrically integral R-scheme.) [Perhapsgeometrically inte-

gral; make sure that

I actually defined this

before!

replace “projective” with “proper”: all we need is that pushforwards of coherents are coherent.][Kirsten’s argument: reduce to case k algebraically closed. H0(X,OX) is a finite-dimensionalintegral domain over k, necessarily a field, hence equal to k.]

23.8 Fun applications of the higher pushforward

Last day we proved that if π : X → Y is a projective morphism, and F is a coherent sheaf onX, then π∗F is coherent (under a technical assumption: if either Y and hence X are Noetherian;or more generally if OY is a coherent sheaf).

As a nice immediate consequence is the following. Finite morphisms are affine (from thedefinition) and projective (21.4); the converse also holds.

23.8.1. Corollary. — If π : X → Y is projective and affine and OY is coherent, then π isfinite. 29 [I wrote to myself: Fix OY . But why does it need fixing? Perhaps this isn’t necessary?Ditto in the next Theorem.]

In fact, more generally, if π is universally closed and affine, then π is finite, by [AM,

Exer. 5.35] [pointed out by Joe R, used in 23.17.1]. We won’t use this, so I won’t explain why.[Also: EGA III.4.4.2: if target is locally Noetherian, then finite = proper + quasifinite. SeeFulton, presumably Intersection theory, p. 429.]

Proof. By Theorem XXXX(d), π∗OX is coherent and hence finitely generated.

Here is another handy theorem.

27projectionformula, H.E.III.8.3

28cohpushforward

29projectiveaffinefinite, H.E.III.11.2

395

23.8.2. Theorem (relative dimensional vanishing). — If f : X → Y is a projectivemorphism and OY is coherent, then the higher pushforwards vanish in degree higher than themaximum dimension of the fibers. 30 31

[Presumably thi sis ture with proper replaced by projective?]This is false without the projective hypothesis. Here is an example of why.

Exercise. Consider the open immersion π : An − 0 → An. By direct calculation, show thatRn−1f∗OAn−0 6= 0. 32

Proof. Let m be the maximum dimension of all the fibers.The question is local on Y , so we’ll show that the result holds near a point p of Y . We may

assume that Y is affine, and hence that X → PnY .

Let k be the residue field at p. Then f−1(p) is a projective k-scheme of dimension at mostm. Thus we can find affine open sets D(f1), . . . , D(fm+1) that cover f−1(p). In other words, theintersection of V (fi) does not intersect f−1(p).

If Y = SpecA and p = [p] (so k = Ap/pAp), then arbitrarily lift each fi from an elementof k[x0, . . . , xn] to an element f ′i of Ap[x0, . . . , xn]. Let F be the product of the denominatorsof the f ′i ; note that F /∈ p, i.e. p = [p] ∈ D(F ). Then f ′i ∈ AF [x0, . . . , xn]. The intersection oftheir zero loci ∩V (f ′i) ⊂ PnAF

is a closed subscheme of PnAF. Intersect it with X to get another

closed subscheme of PnAF. Take its image under f ; as projective morphisms are closed, we get a

closed subset of D(F ) = SpecAF . But this closed subset does not include p; hence we can findan affine neighborhood SpecB of p in Y missing the image. But if f ′′i are the restrictions of f ′ito B[x0, . . . , xn], then D(f ′′i ) cover f−1(SpecB); in other words, over f−1(SpecB) is covered bym+1 affine open sets, so by the affine-cover vanishing theorem, its cohomology vanishes in degreeat least m + 1. But the higher-direct image sheaf is computed using these cohomology groups,hence the higher direct image sheaf Rif∗F vanishes on SpecB too.

23.8.A. Important exercise. Use a similar argument to prove semicontinuity of fiber dimensionof projective morphisms: suppose π : X → Y is a projective morphism where OY is coherent.Show that y ∈ Y | dim f−1(y) > k is a Zariski-closed subset. In other words, the dimensionof the fiber “jumps over Zariski-closed subsets”. (You can interpret the case k = −1 as the factthat projective morphisms are closed.) This exercise is rather important for having a sense of howprojective morphisms behave! Presumably the result is true more generally for proper morphisms.33

Here is another handy theorem, that is proved by a similar argument. We know that finitemorphisms are projective, and have finite fibers. [Add ref.] Here is the converse.

23.8.3. Theorem (projective + finite fibers = finite). — Suppose π : X → Y is such thatOY is coherent. Then π is projective and finite fibers if and only if it is finite. Equivalently, πis projective and quasifinite if and only it is finite. I have a note calling this H.E.III.11.2,but another note says that was Corollary 23.8.1. 34 proj+qf implies finite

(Recall that quasifinite = finite fibers + finite type. But projective includes finite type.)It is true more generally that proper + quasifinite = finite. (We’ll later see a generalization,

that proper and quasifinite implies finite, 23.17.D.)

Proof. We show it is finite near a point y ∈ Y . Fix an affine open neighborhood SpecA of y in Y .Pick a hypersurface H in PnA missing the preimage of y, so H ∩ X is closed. (You can take thisas a hint for Exercise 23.8.A!) Let H ′ = π∗(H ∩X), which is closed, and doesn’t contain y. LetU = SpecR−H′, which is an open set containing y. Then above U , π is projective and affine, sowe are done by the previous Corollary 23.8.1.

Here is one last potentially useful fact. (To be honest, I’m not sure if we’ll use it in thiscourse.)

30H.C.III.11.2; he has: of Noetherian schemes

31dimvanishing

32H.E.III.11.1

33semicont

34projqffinite

396

23.8.B. Exercise. Suppose f : X → Y is a projective morphism, with O(1) on X. Suppose Yis quasicompact and OY is coherent. Let F be coherent on X. Show that

(a) f∗f∗F(n) → F(n) is surjective for n 0. (First show that there is a natural mapfor any n! Hint: by adjointness of f∗ with f∗.) [Should I relate this to fact 23.7.1?]Translation: for n 0, F(n) is relatively generated by global sections.

(b) For i > 0 and n 0, Rif∗F(n) = 0.35

23.9 ? Leray spectral sequence

36

Joe is suspicious of this argument, so I should look carefully.Suppose

Xf //

h @@@

@@@@

@ Y

g~~~~

~~~~

~

Z,

with f and g (and hence h) quasicompact and separated. Suppose F is a quasicoherent sheaf onX. The Leray spectral sequence lets us find out about the higher pushforwards of h in terms ofthe higher pushforwards under g of the higher pushforwards under f .

23.9.1. Theorem (Leray spectral sequence). — There is a spectral sequence whose Ep,q2 -

term is Rjg∗(Rif∗F), abutting to Ri+jh∗F .An important special case is if Z = Spec k, or Z is some other base ring. Then this gives us

handle on the cohomology of F on X in terms of the cohomology of its higher pushforwards to Y .

Proof. We assume Z is an affine ring, say SpecA. Our construction will be “natural” and willhence glue. (At worst, we you can check that it behaves well under localization.)

Fix a finite affine cover of X, Ui. Fix a finite affine cover of Y , Vj . Create a double complex

Ea,b0 = ⊕|I|=a+1,|J|=b+1F(UI ∩ π−1VJ)

for a, b ≥ 0, with obvious Cech differential maps. By Exercise 12.1.M [forme: currentlycommented out] and Proposition 12.1.8, UI ∩ π−1VJ is affine (for all I, J).

Let’s choose the filtration that corresponds to first taking the arrow in the vertical (V )direction. For each I, we’ll get a Cech covering of UI . The Cech cohomology of an affine is trivialexcept for H0, so the E1 term will be 0 except when j = 0. There, we’ll get ⊕F(UI ). Then theE2 term will be Ep,q2 = Hp(X,F) = Γ(Z,Rph∗F) if q = 0 and 0 otherwise, and it will convergethere.

Let’s next choose the filtration that corresponds to first taking the arrow in the horizontal(U) direction. For each VJ , we will get a Cech covering of π−1VJ . The entries of E1 will thus be⊕JHi(f−1Vj ,F) = ⊕jΓ(Vj , Riπ∗F). Thus E2 will be as advertised in the statement of Leray.

Here are some useful examples.Consider hi(Pmk ×k Pnk ,OPm

k×kPn

k). We get 0 unless i = 0, in which case we get 1. (The same

argument shows that hi(PmA ×A PnA,OPmA×APn

A) ∼= A if i = 0, and 0 otherwise.) You should make

this precise:Exercise. Suppose Y is any scheme, and π : PnY → Y is the trivial projective bundle over Y .

Show that π∗OPnY∼= OY . More generally, show that Rjπ∗O(m) is a finite rank free sheaf on Y ,

and is 0 if j 6= 0, n. Find the rank otherwise. 37 [Possibly used in Ex. 29.9.D.]

35H.T.III.8.8abc, cut if we never use this. H requires Y Noetherian.

36s:leray

37e:Pnpush

397

More generally, let’s consider Hi(Pmk ×k Pnk ,O(a, b)). I claim that for each (a, b) at mostone cohomology group is non-trivial, and it will be i = 0 if a, b ≥ 0; i = m + n if a ≤ −m − 1,b ≤ −n− 1; i = m if a ≥ 0, b ≤ −n− 1, and i = n if a ≤ −m− 1, b = 0. I attempted to show thisto you in a special case, in the hope that you would see how the argument goes. I tried to showthat hi(P2

k ×k P1k,O(−4, 1)) is 6 if i = 2 and 0 otherwise. The following exercise will help you see

if you understood this.Exercise. Let A be any ring. Suppose a is a negative integer and b is a positive integer. Show

that Hi(PmA ×A PnA,O(a, b)) is 0 unless i = m, in which case it is a free A-module. Find the rankof this free A-module. (Hint: Use the previous exercise, and the projection formula, which wasExercise 1.3 of class 32, and exercise 17 of problem set 14.)

[EXPLAIN. f : P2 × P1 → P1. Show that Rjf∗O(−4, 1) is 0 unless it is j = 2. Thenyou get (Rjf∗O(−4)) ⊗ OP1 (2) (using the projection formula). So we can even interpret this as

x?0x

?1x

?2y

?1y

?2 of homogeneous degree (−4, 1), where the x’s powers are negative, and the powers on

the y’s are non-negative.]We can now find curves of any (non-negative) genus, over any algebraically closed field. An

integral projective nonsingular curve over k is hyperelliptic if admits a finite degree 2 morphism(or “cover”) of P1.

23.9.A. Exercise. (a) Find the genus of a curve in class (2, n) on P1k ×k P1

k. (A curve in class(2, n) is any effective Cartier divisor corresponding to invertible sheaf O(2, n). Equivalently, it isa curve whose ideal sheaf is isomorphic to O(−2,−n). Equivalently, it is a curve cut out by anon-zero form of bidegree (2, n).)(b) Suppose for convenience that k is algebraically closed of characteristic not 2. Show that thereexists an integral nonsingular curve in class (2, n) on P1

k × P1k for each n > 0. [Possibly go to

“geometrically integral”.] [Yu-jong’s example: yn0 x21 = x1x2yn1 + x2

2yn0 .]

23.9.B. Exercise. Suppose X and Y are projective k-schemes, and F and G are coherentsheaves on X and Y respectively. Recall that if π1 : X × Y → X and π2 : X × Y → Y are thetwo projections, then F G := π∗1F ⊗ π∗2G. Prove the following, adding additional hypotheses ifyou find them necessary.(a) Show that H0(X × Y,F G) = H0(X,F)⊗H0(Y,G).(b) Show that HdimX+dimY (X × Y,F G) = HdimX(X,F)⊗k HdimY (Y,G).(c) Show that χ(X × Y,F G) = χ(X,F)χ(Y,G).

I suspect that this Leray spectral sequence converges in this case at E2, so that hn(X ×Y,F G) =

Pi+j=n h

i(X,F)hj(Y,G). Or if this is false, I’d like to see a counterexample. It

might even be true that

Hn(X × Y,F G) = ⊕i+j=nHi(X,F)⊗Hj(Y,G).

[Kirsten gives an argument, assuming nothing. For (a) and (b), we only need quasicoherence— even less than I gave. See scan kirsten15.5.pdf.] [Ziyu’s argument: Consider

X × Y

π1

wwwwwwww

f

π2

##FFF

FFFF

FF

X

ψ1

##FFFFFFFF Y

ψ2xxxxx

xxxx

Spec k,

398

F and G coherent sheaves on X, Y respectively. We want to compute E2 term of the spectralsequence converging to Ri+jf∗(π∗1F ⊗ π∗2G) = Hi+j(X × Y, π∗1F ⊗ π∗2G). Now

E2i,j = Rjψ1∗(R

iπ∗1(π∗1F ⊗ π∗2G))Rjψ1∗(R

iπ∗1(π∗1F ⊗ π∗2G))Rjψ1∗(R

iπ∗1(π∗1F ⊗ π∗2G))= Rjψ1∗(F ⊗ Riπ∗1π∗2G) (proj. formula, if F loc. free)

= Rjψ1∗(F ⊗ ψ∗1Riψ2∗G) (flat base change)

= Rjψ1∗F ⊗Riψ2∗G (proj. formula, if Riψ2∗G loc. free)

= Hj(X,F)⊗k Hi(Y,G)]

23.10 ? Pushforward of coherent sheaves are coherent forproper morphisms

This is currently 21.6.3. The most important thing here is the statement.

23.11 Older notes (pre higher direct image sheaves)

Make sure to do maps between various cohomology groups. E.g. projection

formula. I think this is now done. 38

I think everything in this section (that I need) is done above now.Exercises you can do after this in Hartshorne: 4.1, 4.3, 4.7, 4.8, 4.9x, 5.1, 5.2, 5.3, 5.5, 5.6,

5.10, 7.3, 8.2, 8.3, 8.4, 11.2, 12.1, 12.2.A quick summary

• foundations: for any R, we have cohomology groups that are R-modules. H0. Functo-rial in F , and long exact sequence. If X in Y is a closed subscheme, then cohomologyof sheaf on X may be calculated on Y . Cohomology of projective space. Behaves wellwith respect to Rf . Cohomology of affine scheme.

• Cech cohomology with respect to a cover. We will soon see that it is independent ofcover. Once we have that, we have all of the above.

• Theorem: any coherent sheaf on a projective R-scheme where R is Noetherian, hi

is finitely generated R-module. Serre vanishing. Corollaries: Projective k-scheme hasfinite-dimensional space of global sections. [integral projective over k = k has 1 section;relax algebraic closure, and irreducibility and show what you get.] More generally,projective R-scheme where R Noetherian. Also, exact sequence, twisted sufficiently,become exact on global sections.

• Grothendieck (dimensional) vanishing for projective schemes; and for quasiprojectiveschemes, using affine minus Cartier, 9.1.V

• application: euler characteristic additive. (Remark: O(m), get polynomial.) Non-rational curve. Exercise: cohomology of O(m) on a complete intersection. Hilbert

function.• Riemann-Roch (Exercise 23.5.B) for smooth curves.• The following is done in Chapter ??: behaves well with respect to localization R→ Rf .

Hence we get higher direct image sheaves, and exact sequences of those. Restate theprevious item in these. They are coherent if the morphism is projective.Consequence: if f : X → Y is a projective morphism then the higher pushforwardsvanish in degree higher than the maximum dimension. Exercise: Note that this is falsewithout the projective hypothesis:projective + affine = finite. Proof: π∗OX coherent (Better: Joe R: affine + universally

38s:cohotake1

399

closed is finite.)projective + finite fibers = finiteaffine morphism: no higher cohomology. Also, Hi(X,F) = Hi(Y, f∗F).

23.12 Things you need to know

These modules will satisfy certain properties. H0(X,F) = Γ(X,F) , satisfying a long exact sequence.

If X → Y is a closed immersion [we don’t know that language yet], and F is a quasicoherent

sheaf on X, then Hi(X,F) = Hi(Y, f∗F) . H0(SpecR, M) = M from above. We’ll see that

Hi(SpecR, M) = 0 for i > 0. Also, Hi(PnR,OPnR

) = 0 for i > 0. (Contravariant in space,

covariant in sheaf.)

23.12.1. Proposition. — Hi(Pn, R,OPnR

(m)) = . . . . Show that they vanish when they are

supposed to, and are coherent otherwise. H0: polynomials in x0, . . . , xn of degree m. Hn:Laurent monomoials in x0, . . . , xn of degree m, each has degree at most −1.

In particular, it:

• vanishes above dimension if R = k; we’ll generalize this for SpecR, and even more,in 23.8.2

• always free R modules (don’t have this yet)• higher cohomology vanishes for m > −n− 1.• 1-dimensional for m = −n− 1 if field.• duality (don’t have yet; also must make precise later; just show identity of dimensions)

[Remark: proper reduced, get one-dimensional family of global sections. The only functionsare constants.]

Another a priori interesting consequence:

23.12.2. Corollary. — If 0 → F → G → H → 0 is an exact sequence of coherent sheaveswith F coherent, then for n 0, 0→ H0(X,F(n)) → H0(X,G(n)) → H0(X,H(n)) → 0 is alsoexact.

This result can also be shown directly, without the use of cohomology. [Give ref.]Place the following exercises. 1. Show that a reduced connected projective k-

scheme X has h0(X,OX) = 1. (Proof: any function is a constant. Any map givesa morphism X → A1. The image must be proper. The image scheme is reduced.)Follow-up: Show that a reduced (non-empty) projective k-scheme is connected iffh0(X,OX) = 1. (Easy.) Is it true that a connected projective k-scheme is reducedif h0(X,OX) = 1? Answer: No, e.g. x2 = 0 in P2. 2. Same thing with projectivereplaced by proper. 3. Show that a complete intersection of dimension at least 1 isconnected. Hint: Instead we show that if X is connected of dimension at least 2, andD is an effective ample Cartier divisor on X, then D is connected. Proof: We cantake X to be reduced (ample behaves well under taking reduction!) Choose nD suchthat h1(OX(−nD)) = 0 using Serre vanishing. Then h0(OnD) = 1 from the long exactsequence for nD.

23.13 Application of cohomology: Hilbert polynomials andHilbert functions; degrees

(Stuff not done before...)Hilbert’s syzygy theorem? Resolve in n steps. [See Auslander Buchsbaum theorem.] [Reflex- Hilbert syzygy theorem

ive sheaf: 2-syzygy sheaf: locally 0 → F → E1 → E2 where Ei are locally free. (From Pareschitalk in Georgia. 0-syzygy sheaf: any. 1-syzygy sheaf: torsion-free. ∞-syzygy sheaf: locally free.

400

Some theorem from the 70’s: any k-syzygy sheaf of rank less than k is locally free, or somethinglike that.)

Slicing with an effective Cartier divisor (defined in ??). Use this to prove:

23.13.1. Bezout’s theorem (divisor form). — [DONE EARLIER NOW] Slice a Cohen-Macaulay scheme with an effective Cartier divisor (see 7.5.4); degrees multiply. 39Bezout

[When will I do the general case? Caution: example where you don’t this doesn’t work:two planes meeting at a point. The first counterexample necessarily can’t arise before P4. Thisexample comes up all over the place.]2 planes meeting at

point Mention the Tor form.Complete intersections in projective space.

23.13.A. Exercise. (a) Define the twisted cubic. (We’ve seen this in some guise before.)Show that the twisted cubic 40 is not a complete intersection in P3. [Refer to Exercise 9.1.T,twisted cubicwhere the twisted cubic was defined.] (b) Degree and genus of complete intersections in Pn.

Perhaps do Ext1(F ,G), and Grothendieck’s theorem on vector bundles on P1? ExceptG’s theoremwe haven’t done Ext groups. Make sure to do this somewhere in these notes!!

23.14 Still to place

[DO [locally] constant sheaves? (Use connected affines.) Question: O∗X?]

23.14.1. Theorem(Grothendieck vanishing). — 41 Let X be a projective k-scheme dimen-sion n. Then for all i > n and all quasicoherent sheaves F on X, we have Hi(X,F) = 0.

Note that projective varieties satisfy Grothendieck vanishing. [change to dimensionalvanishing]dimensional vanishing

23.14.2. Remark. Quasicoherent sheaves on quasiprojective space satisfy Grothendieck vanish-

ing. We’ll show this later (27.3.6). But as a first case: quasiprojective that is projective minuseffective Cartier divisor is an example. (They don’t know what this is!) Here’s a sketch of how thegeneral case will follow, which will motivate some other constructions that will come up anyway.Say X = X′ −C, where X′ is projective, and C is some closed set, which can be given some sub-scheme structure (the induced reduced subscheme structure, 9.2.6). Then there is construction

called the blow-up, which gives you π : X → X′, which is an isomorphism above X, but so thatthe preimage of C is an effective Cartier divisor. Furthermore, X is projective (XXXX).

23.14.3. Remark. Grothendieck vanishing is true in general, see [H, Theorem III.2.7].42 Let Xbe a Noetherian topological space of dimension n [define]. Then for all i > N and all sheaves ofabelian groups F on X, we have Hi(X,F) = 0. [Apparently FAC no. 3, see my H notes p. 208.]

23.14.4. Theorem. — (Serre vanishing on projective schemes)43Serre vanishing1. F(n) is generated by global sections. (Do this somewhere.)2. Finite number by earlier exercise.3. Hence O(−n)⊕m → F → 0. [H, Cor. III.5.18]4. Use vanishing of H0(O(−n)).5. Use Grothendieck vanishing and induct down.(Do I want to generalize this to schemes with just an ample line bundle?)

23.14.5. Theorem (Serre). — Let X be a projective scheme over a noetherian ring A, O(1)a very ample invertible sheaf on X, and F a coherent sheaf on X. Then there is an integer n0

such that for all n ≥ n0, the sheaf F(n) can be generated by a finite number of global sections.

39Bezout

40twistedcubic

41Gvanishing

42H.T.III.2.7

43Svanishing

401

[[H, Thm. 5.17] gives a short hands-on proof. My H notes: This is taken to be the definition ofample in II.7, p. 153. Discussed futher in II.7.]

Proof. We use Noetherian induction ??. It suffices to show that we can do this at a single point.By Nakamaya’s lemma we need generate only M/mM . Look at use of Nakayama

0→ K→ O(−n)⊕m → F/mF → 0.

Wait... are we using Serre vanishing to prove this? And this to prove Serre vanishing?!

23.14.6. Theorem. — Suppose X is a projective scheme over a ring of finite type over a fieldk, and F is a coherent sheaf on X. Then hi(X,F) is a finitely-dimensional k-vector space. [H,Thm. III.5.2.1]. f.d. of coh

In particular, the space of global sections is finitely generated. For a hands-on cohomology-free proof of this fact, see [H, Thm. 5.19].

Proof.

0→ K→ O(−n)⊕m → F → 0.

Use the long exact sequence in cohomology and induct down.

23.14.7. Corollary. — In category of k-schemes, k algebraically closed. The only globalfunctions on a reduced scheme are the locally constants. The only global functions on a connectedreduced scheme are the constants.

This is the analogue of the following fact in complex analysis: the only holomorphic functionson a compact complex manifold are locally constant (because of the maximum principle).

Mild variation if k is not algebraically closed. Still constants. Spec C is an R-scheme...

Proof. In second case: the only finite-dimensional k-algebras that are domains is k itself. [Is thatthe Nullstellensatz?] In the first case: the only finite-dimensional k-algebras that are reduced areproducts of k’s. To see the connected components, take Spec of this (cf. hint to Exercise reffini-teunionofaffines).

Suppose F is a coherent sheaf on Pn. Can I show that χ(F(n)) is polynomial in n? Thiswould follow if I could resolve it by direct sums of line bundles. Some theorem of Hilbert.

23.15 Some more detail that I don’t want to tell them.

There are derived functors on the category of abelian sheaves on a topological space. [H, p.207].

Lemma [H] III.2.10. Let Y be a closed subset of X, let F be a sheaf of abelian groups onY , and let j : Y → X be the inclusion. Then Hi(Y,F) = Hi(X, j∗F), where j∗F is the extensionof F by zero outside Y . (That may need to be defined.)

23.15.1. Grothendieck vanishing theorem. — We’ve done the case of quasiprojectiveschemes (Theorem 27.3.6, see also Theorem 24.1.1). I have a remark about the general case atRemark 23.14.3. 44

Serre’s criterion for affineness. Let X be a (noetherian?) scheme. Then the followingconditions are equivalent. (i) X affine. (ii) all higher cohomology vanishies for all quasicoherentF . (iii) all higher cohomology vanishes for all coherent sheaves of ideals I.

The proof (iii) to (i) involves an appeal to an earlier Hartshorne exercise.

23.15.2. Cohomological vanishing. Affine covering number. Kodaira. Nakai-Moishezon.This is already discussed at 26.2.B. 45 Nak-Moi

44refer to H.T.III.2.7

45nakai2

402

23.15.3. To place. When can you extend a function over a point? Answer: normality. This“begs” local cohomology (see [W]).

Application: Hodge numbers of 2 birational smooth varieties are the same ([H] exercise).

23.15.A. Exercise. (Move somewhere) Find h0(Pn, TPn). Remark: infinitesimal automor-phisms. Dimension is dimension of automorphism group. Lie algebras remark.

23.15.4. Useless criterion for ampleness (exercise?). [Move somewhere.] Suppose L is aninvertible sheaf on L with sections si global sections of powers of L, with Xsi an affine cover of X.

Suppose also that hi(L⊗m) = 0 for i > 0, m 0. Then for any coherent sheaf F , hi(F ⊗Ln) = 0for all n 0, i > 0.

403

23.16 The categorical perspective

Functor of points. “Solutions to equations”. Covariant functor Rings to sets. Example: thefunctor x2 = y3. Example: An+1 − 0.

More complicated example: x2 = 0. Notice that we can get interesting solutions in rings thatare not reduced. This isn’t frightening at all.

Group functor (define), e.g. Gm.Projective space: R→ (r1, . . . , rn)/R∗.Max and Martin: other interpretations: An+1 − 0 modulo group action. projective rank 1

submodule of Rn+1; that doesn’t work; we need locally split, which turns out to be globally split,because some cohomology group vanishes on affine. (I don’t like this.)

“A scheme is a covariant functor rings to sets satisfying certain properties.” What properties?This is discussed a bit in [M] p. 160 (“Grothendieck’s existence problem”) but I won’t discussit.46

23.17 +The cup product in Cech cohomology

47 The cup product in Cech cohomology in general is suprisingly straightforward to define,and it is even more straightforward in this algebro-geometric setting. So it would be a shame notto define it here. However, unfortunately as of yet we will have no interesting examples of it, andthe interesting examples come later (for example, in our discussion of Serre duality). So I’ll makethis section a bonus “+” section.

Suppose F and G are sheaves with values in a tensor category. Our goal is to define a bilinearmap Hi(F) × Hj(G) → Hi+j(F ⊗ G) that is bilinear and associative, and symmetric if F = G.[Verify this last! There might be a sign.]

We’ll define this on the level of co-chains. Define Cp to be the group of Cech p-cochains, so⊕|I|=p+1F(UI ). If f ∈ Cp, then df is given by

(df)J = ⊕I⊂J (−1)place of J − I in JfI |Uj.

Suppose that f ∈ Ci(F) and g ∈ Cj(G). Define fg ∈ Ci+j (F ⊗ G) by

fg(K) := f(IK)|Uk⊗ g(JK)|UK

where IK and JK are defined as follows: IK consists of the first i+ 1 of the i+ j + 1 elements ofK, and JK consists of the last j + 1 elements of K. [Perhaps a picture would be nice.]

Notice that this is obviously associative and bilinear. The only problem is that it is a mapof cocycles, not classes!

Here is what we have to do now: (1) show that it maps cocycles × cocycles to cocycles, andcoboundary × cocycle to coboundary. (2) Show that it is independent of the order of the UI .(3) Show that it is independent of the choice of cover (i.e. you can add an element, and get thesame thing). We’ll show (2) and (3) at the same time, by showing that you can add a redundantelement to the beginning or the end. Then two different covers (with different orders) will givethe same thing: start with the first cover, add the second cover to the end, and then remove thefirst cover from the beginning.

23.17.A. Important exercise. Fix a covering indexed by an ordered set. Suppose thatf ∈ Ci(F ,U), g ∈ Cg . Verify that

d(fg) = (−1)?f dg + (−1)?g df

where the two mystery exponents are in terms of i and j. (Part of the exercise is to figure outthose exponents. The reason it is worth doing this exercise is that you will force yourself not tobe careless with signs.)

46Mp160a

47cupproduct

404

[Answer: d(fg) = (−1)if dg + g df . Solution: We use component language, so f ∈ Cp maybe written as f = fI ∈ F(UI ) | |I| = p+ 1. Recall that

(df)J =X

j∈J,|J|=p+2

(−1)(position of j in J)−1fJ−j ,

so for example (df)012 = f12 − f02 + f01. Then (f ⊗ g)K = fIgJ if k = i+ j+ 1, and I is the firstelements i + 1 of K, and J is the last j + 1 elements of K (so they “overlap by one”). Then for|K| = i+ j + 2,

(d(f ⊗ g))K =X

k∈K

(−1)(pos. of k in K)−1(f ⊗ g)K−k

We now calculate. Here |K| = i+ j + 2, I is the first i+ 1 elements of K, and J is the first j + 2

elements of K.

(f ⊗ (dg))K = fI(dg)K

= fIX

k∈J

(−1) (pos. of k in J)− 1gJ−k

= f1st i+ 1 in Kglast j + 1 in K

+X

k one of last j + 1 in K

f1st i+ 1 in K-kglast j + 1 in K − k(−1) (pos. of k in J)−1

= f1st i+ 1 in Kglast j + 1 in K

+X

k one of last j + 1 in K

f1st i+ 1 in K-kglast j + 1 in K − k(−1) (pos. of k in K)+i−1

Similarly, where now |I| = i+ 2 and |J | = j + 1,

((df) ⊗ g)K = (df)IgJ

=X

k∈I

(−1) (pos. of k in I)−1fI−kgJ

= (−1)i+2−1f1st i + 1 in Kglast j + 1 in K

+X

k one of first i+ 1 in K

f1st i+ 1 in K-kglast j + 1 in K − k(−1) (pos. of k in K)−1.

Thus

(−1)i(f ⊗ (dg))K + ((df) ⊗ g)K =X

k∈K

f1st i+ 1 in K − kglast j + 1 in K − k(−1) (pos. of k in K)−1

= d(f ⊗ g)

as desired. ]

23.17.B. Important exercise. Use the preceding exercise to show that ⊗ sends cocycles tococycles, and cocycle ⊗ coboundary to coboundary. Hence conclude that it descends to a mapHi(F) ×Hj(G)→ Hi+j (F ⊗ G) that is bilinear and associative.

We next show that we can add a redundant element to the cover on the end. Let I be theindex set of a cover, and I ′ the same as I with a redundant element added at the end. Then wehave an exact sequence of complex

0→ C•(I′)→ C•(I)→ C•(U, I)→ 0

where the last term is the Cech complex for U , which we have verified has vanishing cohomology.Our definition of f ⊗ g in the top row (using the bigger cover) maps immediately to f ⊗ g in themiddle row (using the smaller cover), and hence maps to the same term in cohomology (as the

cohomology of the final term is 0).The identical argument shows that we can add a redundant element to the beginning.Show it is independent of choice of cover as follows, and order of indexing of cover: show that

you can add an entry anywhere.

405

23.17.1. Facts about properness we’ll use. 48

We need fairly few facts about properness!1. Suppose X is geometrically reduced and geometrically over k. Then k → H0(X,OX) is

an isomorphism. Proof: Any global section gives a morphism X → A1k (we’ve shown earlier). The

first is reduced, and the latter is separated, so the morphism is determined by the map on points(proved earlier). [Some care is required here!]

Worse argument using Chow: X → Spec k implies X ′ → Spec k, but the only such areconstants, and then maps X → Spec k are determined by behaviour on dense open set.

2. Pushforward of coherent is coherent (with locally Noetherian hypotheses). Here’s theargument, using devissage, which is a variant of Noetherian induction. Proof: let π : X → Y devissagebe the morphism. Reduce to the case where Y is affine. Now, we define the notion of scheme-theoretic support of quasicoherent sheaf. Intuitively, it is the smallest closed subscheme on which scheme-th suppit lives. Over an affine open set, the ideal corresponds to those elements of the ring that kill theentire module. So for example, we can compute cohomology using this closed subscheme. Wenote that those coherent F for which the result holds are closed under exact sequences. (Littlecheck necessary.) We work by induction on the dimension of the scheme-theoretic support of X.The base case is the empty set (dimension −1). So we assume the result for coherent sheaves withsmaller support. [NOT NEEDED: We then work by induction on the order of vanishing of thenilpotents in the scheme-theoretic support (small argument necessary). So we may assume that

the scheme-theoretic support is reduced. We then work by induction on the number of irreduciblecomponents (we could have done this in either order). So we may assume that X is integral.]

So finally we are ready to use Chow’s lemma. Fix O(1) for f : X ′ → X. Then for anycoherent G on X ′, for d 0 Hq(X, f∗G(d)) → Hq(X′,G(d)) are isomorphisms. So apply this forG = f∗F . Now for d 0, there is a map F → f∗f∗F(d) (whoops, be careful about that d) thatis an isomorphism on a dense open subset. So its kernel and cokernel are coherent sheaves whosesupport are properly contained in X.

Misc. Valuative criterion [H, Thm. II.4.7]. I guess the hypotheses I’l use are finite typeand quasiseparated. An injective proper morphism of varieties is affine [Newstead, Lemma 2.5].Likely a monomorphism of proper varieties is a finite morphism? Is proper + affine = finite?

Show that a proper morphism of affine varieties over k is finite [H, II.4.6]? Yes: Proper andaffine implies finite, by the remark after 23.8.1. 49 Useful properties: [H, Cor. II.4.8], includingmagic diagrams (Exercise 3.3.L).

23.17.C. Exercise. Show that a proper morphism that separates points and tangent vectorsis a closed immersion. 50 [Repeated at 21.4.3.] (Closed points? For varieties?) Brian Ossermangave an example to show that properness is necessary: normalization of the node, minus one ofthe node-branches, see Figure 2.

23.17.D. Exercise. Proper and quasifinite implies finite. 51 (We’ve seen earlier that projectiveand quasifinite implies finite, 23.8.3.) Application: quasifinite morphisms from proper schemesare finite. Explicit example: P1 → P1 via |O(2)|. [This is hard! There is an easier argument forthat last explicit example, which is given earlier.]

23.18 ? Derived functors

Place this earlier in this part!To the reader: Read this when you are ready, at some point! We will use some ideas here

later, notably Ext-groups.Summary: when an abelian category has “enough injectives”, given a left-exact functor, we

can get derived functors. Example in the category of modules: Ext. Then abelian sheaves.In this section, Ri refers to derived functors.

48properfacts

49properaffinefinite, H.E.II.4.6, except H uses something hard

50separatepointstangentvectors2

51propqffinite

406

• Define injective object.• Define if a category has enough injectives.• Then get an injective resolution.• Unique up to homotopy.• Define derived functor.• Show that it is unique, i.e. independent of resolution.• Show that it satisfies long exact sequence• Exercise. define δ-functor. Show that this is a universal δ-functor. (This exercise isn’t

important for our exposition, but is useful.)• Example: Ext.

Perhaps a new section: abelian sheaves on a topological space have enough injectives. Hence:cohomology. More generally, higher pushforwards. Why do qcoh go to qcoh in good situations?Reason: next section.

Possible fact needed in Martin’s proof: that pushforwards are sheafification of the cohomologyof preimages.

Exercise: Grothendieck (composition of functor) spectral sequence.Application: Leray spectral sequence. Needed: pushforward of injective is injective.

23.19 Derived functor cohomology and Cech cohomology forquasicoherent sheaves

In this section only: one notation for derived functors, and the other is for Cech.Summary: they are the same. Ingredients: (i) Grothendieck (composition of functor) spectral

sequence. (ii) affine has no cohomology.Theorem: (ii).Proof: Martin Olsson. We show two facts by induction. (i) Up to i, cohomology of affine

vanishes. (ii) Up to i, cohomology of affine morphism vanishes.(i) implies (ii) is easy (for same i). We show (ii) for i implies (i) for i+ 1. Trick: Suppose α

is an element of a higher (i + 1) cohomology group. Then there is some affine cover on which αis trivial. U → X. Use the Leray spectral sequence.

Theorem: Cech = derived functor. Proof: Take a double complex, involving Cech andinjectives. It falls right out.

23.20 Grothendieck group

This might be a good time to discuss the Grothendieck group. (Related concepts for alter: Chernclasses, derived category.) Each sheaf is equal to its resolution. Tor depends only on this element.K∗ is a ring. Chern classes show two things are distinct in K-theory.

][primordial:

CHAPTER 24

? Derived functor cohomology

1

This was apparently introduced by Grothendieck in his Tohoku article [Gr] (as described at[H, p. 211], [KS, p. 138]).

[Barbara Fantechi suggests looking at the development of Kashiwara and Shapira [KS].] Fix acategory. I injective means Hom(·, I) is exact. (It is automatically left-exact.) (iff Ext1(A, I) = 0 injective objectfor all A, assuming A exists).

I’ve discussed this in the Tor section. We can generalize this immediately. If there are enoughinjectives/projectives, then there exists a derived functor. Properties: it is covariant, and satisfiesa long exact sequence. (This is called a δ-functor in Hartshorne p. 205.)

In fact, we need less for a given functor (p. 50 of [KS]). [Reason we care: proof of compositionof functor spectral sequence. We need that adjusted objects get sent to adjusted objects. Injectivesare sometimes not sent to injectives.]

Why is there only one derived functor? What is the universal property? [universal δ-functor.argument?]

OX -modules has enough injectives. Argument: (i) A-modules have enough injectives. (Why?[H] Prop. 2.1A refers to Godement, or Hilton-Stammbach.) 2

Then the category of OX -modules has enough injectives. [H.P.III.2.2]3 Slick proof: directsums of injectives is injective. Pushforwards of injectives are injective. Then consider the mapπ : X′ → X where X′ is X with the discrete topology. For each x ∈ X, Fx → Ix. The skyscrapersheaf with Ix is injective. Define I sheaf on X ′ that is ⊕Ix; it is injective. Note that π∗F → I.F → π∗π∗F → π∗I.

24.1 Why the derived category

For this discussion, I’ll assume that we have a covariant left-exact functor F : C → C ′ (e.g.global sections).

There is more information than just the cohomology groups. Why we care: (i) (bad reason)on principal, we don’t like losing information. (ii) (better reason) composition of functors; whatare composition of derived functors? We could use the Grothendieck spectral sequence.

We start with an object. We take a resolution. (That is a quasiisomorphic complex. Wecould define the resolution of a finite or bounded-below. Let K+(C) be the category of complexesbounded below.) We define the derived functor. But instead we could just stick with this resolu-tion. But of course it is not well-defined. But it is well-defined up to homotopy. So we define anew category Kom+(C) of complexes, with the same objects, but where morphisms are “glued”together. Then our construction gives a well-defined element of Kom(C) up to homotopy.

Then if the complex has enough injectives, or more generally a subcategory adapted to ourfunctor, every bounded-below complex is isomorphic (in Kom+(C)) to something with only injec-tive elements. We could define the right-derived functors as follows: hit any such with our functor.Instead, I’ll define the right-derived functor to be Kom+(C)→ Kom+(C′) that is obtained in such

1derivedfunctorcohomology

2H.P.III.2.1A

3H.P.III.2.2

407

408

a way. In fact, this is a bit sloppy; you get only an equivalence class; you get something on theright up to unique isomorphism.

If we invert quasiisomorphisms (we’ll have to define this, not just say that they are isomor-phisms!), we can then describe it as follows: take the isomorphic element that consists of allinjectives; hit it with F .

The reason we care: if we have 2 functors, and we compose them, then we get a doublecomplex. We then get a spectral sequence.

24.1.1. Nifty trick: Grothendieck vanishing. But I don’t seem to use it. 4

24.2 Other notes

In the proof of Theorem 24.2.3, I will need:

24.2.1. Proposition. — If I is injective on X, then I|U is injective on U . 5

I will also need: (1) higher Cech cohomology of injectives is 0. Caution here: the injectivesaren’t quasicoherent, so we haven’t defined Cech cohomology in general, so we need that higherCech cohomology of injectives with respect to a fixed affine cover is 0. (It would suffice to reduceto a sufficiently fine affine cover, I think.) (2) higher derived functor cohomology of affine schemes

vanishes (for any quasicoherent sheaf).

24.2.2. Derived functor cohomology and Cech cohomology agree.[In fact H0 and H1 are always the same, in any circumstance. H0 is clear, and H1 is sketched

in Milne’s online etale cohomology notes, “732”, Proposition 10.4.]We’ll now show that we are recovering (and extending?) our earlier definition of cohomology

of quasicoherent sheaves. Henc there is no problem with using the same notation for them both.Recall Cech cohomology was defined if X → Y is separated and quasicompact.

(It is possible that we don’t know yet that higher pushforwards via derived functors givequasicoherent sheaves; this result will show it.)

24.2.3. Theorem. — There is a natural isomorphism between Cech cohomology and derivedfunctor cohomology if X → Y is separated and quasicompact. 6

Proof. The proof below is for Y affine. The general case will then follow from Exercise 24.2.A.We are given a quasicoherent F . Fix an injective resolution

0→ F → I0 → I1 → · · · .Consider the double complex

......

0 // ⊕iI1(Ui)

OO

// ⊕i,jI1(Uij)

OO

// · · ·

0 // ⊕iI0(Ui)

OO

// ⊕i,jI0(Uij)

OO

// · · ·

0

OO

0

OO

(After finishing the proof, you should think over where this complex came from.)Let’s now apply the spectral sequence machinery to it (Section 3.8).

4Gvanishing, “H.T.III.2.7”

5IonopenisI

6cechDF, H.T.III.4.5; he says instead that X is Noetherian

409

First, let’s pick the filtration corresponding to choosing the rightward arrow first. As higherCech cohomology of injectives is 0, we get 0’s everywhere except in “column 0”, where we getIi(X) in row i. Then we take cohomology in the vertical direction, and we get derived functorcohomology of F on X.

Next, we pick the filtration corresponding to choosing the upward arrow first. By Proposi-tion 24.2.1, I|UJ

is injective on UJ , so we are computing the derived functor cohomology of F onUJ . Then the higher derived functor cohomology is 0, so all entries are 0 except possibly on row0. There, we have the complex computing the Cech cohomology of X.

24.2.A. Exercise. Finish the proof of the general case.7 [Idea: both are sheaves. We then needto just show that they are the same over the affine base. Caution: we don’t want to just showthat they are isomorphic over each affine open; we want to show they have a given isomorphism,which behaves well with respect to distinguished restriction.]

24.2.4. Cech cohomology. Cech cohomology is derived functor cohomology if X is aNoetherian separated scheme. [H, Theorem III.4.5] Let X be a noetherian separated scheme,U be an open affine cover of X, F a quasicoherent sheaf on X. Then for all p ≥ 0, the naturalmaps H∨p(U,F)→ Hp(X,F) are isomorphisms.

A related note: apparently that Cech cohomology with coefficients in abelian groups agreeswith derived functor cohomology for H0 and H1. Andy mentioned this in the etale cohomologyseminar. I’m not sure what that requires, and/or how this works. Also: For etale cohomology, weneed to let the cover grow before we have a long exact sequence.

Question: cup product in derived functor cohomology?

] [primordial:

7e:cechDF

CHAPTER 25

Running example: curves

1Where should this go?!Also, put the theory of dedekind domains in here.

25.1 Curves

Let’s now use our technology to study something explicit! For our discussion here, we willtemporarily define a curve to be an integral variety over k of dimension 1. (In particular, curvesare reduced, irreducible, separated, and finite type over k.)

I gave an incomplete proof to the following proposition. Because I don’t think I’ll use it,I haven’t tried to patch it. But if there is interest, I’ll include the proof with the hole, in caseone of you can figure out how to make it work. (We showed that each closed point gives adiscrete valuation, and we showed that each discrete valuation gives a morphism from the Speccorresponding discrete valuation ring to the curve, but we didn’t show that it was the local ringof the corresponding closed point. I would like to do this without invoking any algebra that wehaven’t yet proved.)

25.1.1. Proposition. — Suppose C is a projective nonsingular curve. Then each closed pointof C yields a discrete valuation ring, and hence a discrete valuation on FF (C). This gives abijection from closed points of C and discrete valuations on FF (C).

Thus a projective nonsingular curve is a convenient way of seeing all the discrete valuationsat once, in a nice geometric package.

There was something incomplete with this argument, so I cut it. Proof. This follows easily

from the valuative criterion of properness. But that is hard, and we can do this directly. SupposeC → Pn is a closed immersion. Suppose we have a discrete valuation on FF (C). Let (A,m)be the corresponding discrete valuation ring. We wish to show that there is a unique morphismSpecA→ C extending SpecFF (C)→ C. (This is precisely the valuative criterion of properness!)

We can see that there is at most one such morphism in two related ways. As C is separated(as it is projective), we can use the easy direction of the valuative criterion of separatedness.Alternatively, we can use the proposition from last day that a rational map has a unique extensionto a largest domain of definition.

So now we just need to show that there exists such a morphism. We show first that there is amorphism SpecA→ Pn. The rational map can be described as [a0; a1; · · · ;an] where ai ∈ A. Letm be the minimum valuation of the ai, and let t be a uniformizer of A (an element of valuation1). Then [t−ma0; t−ma1; . . . t−man] is another description of the morphism SpecFF (A)→ Pn,and each of the entries lie in A, and not all entries lie in m (as one of the entries has valuation0). This same expression gives a morphism SpecA→ Pn.

Finally, I claim that this morphism factors through the closed subscheme C. Indeed, on any

open subset of Pn, any function vanishing on C will pull back to the zero element on FF (A), andhence the zero element of A. (This is so tautological that it sounds confusing!)

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25.1.A. Exercise for those with arithmetic proclivities. Suppose A is the ring of integersin a number field (i.e. the integral closure of Z in a finite field extension K/Q — K = FF (A)).Show that there is a natural bijection between discrete valuations on K are in bijection with themaximal ideals of A.

I had wanted to ask you the following exercise (for those with arithmetic proclivities), butI won’t now: Suppose A is the ring of integers in a number field (i.e. the integral closure of Zin a finite field extension K/Q — K = FF (A)). Show that there is a natural bijection betweendiscrete valuations on K are in bijection with the maximal ideals of A.

25.1.2. Key Proposition. — Suppose C is a dimension 1 finite type k-scheme, and p is anonsingular point of it. Suppose Y is a projective k-scheme. Then any morphism C − p → Yextends to C → Y . [H, Prop. I.6.8] 2

Note: if such an extension exists, then it is unique: The non-reduced locus of C is a closedsubset (we checked this earlier for any Noetherian scheme), not including p, so by replacing C byan open neighborhood of p that is reduced, we can use our recently-proved theorem that mapsfrom reduced schemes to separated schemes are determined by their behavior on a dense open set(Proposition 13.1.1).

I’d like to give two proofs, which are enlightening in different ways.Proof 1. By restricting to an affine neighborhood of C, we can reduce to the case where C isaffine.

We next reduce to the case where Y = Pnk . [This argument was largely given in the proof tothe previous proposition.] Here is how. Choose a closed immersion Y → Pnk . If the result holdsfor Pn, and we have a morphism C → Pn with C − p mapping to Y , then C must map to Y aswell. Reason: we can reduce to the case where the source is an affine open subset, and the targetis Ank ⊂ Pnk (and hence affine). Then the functions vanishing on Y ∩ Ank pull back to functionsthat vanish at the generic point of C and hence vanish everywhere on C, i.e. C maps to Y .

Choose a uniformizer t ∈ m−m2 in the local ring. By discarding the points of the vanishingset V (t) aside from p, and taking an affine open subset of p in the remainder we reduce to thecase where t cuts out precisely m (i.e. m = (y)). Choose a dense open subset U of C − p wherethe pullback of O(1) is trivial. Take an affine open neighborhood SpecA of p in U ∪ p. Thenthe map SpecA− p→ Pn corresponds to n+ 1 functions, say f0, . . . , fn ∈ Am, not all zero. Letm be the smallest valuation of all the fi. Then [t−mf0; . . . ; t−mfn] has all entries in A, and notall in the maximal ideal, and thus is defined at p as well.

Proof 2. We extend the map SpecFF (C) → Y to SpecOC,p → Y as follows. Note that OC,p isa discrete valuation ring. We show first that there is a morphism SpecOC,p → Pn. The rationalmap can be described as [a0;a1; · · · ; an] where ai ∈ OC,p. Let m be the minimum valuation of the

ai, and let t be a uniformizer of OC,p (an element of valuation 1). Then [t−ma0; t−ma1; . . . t−man]is another description of the morphism SpecFF (OC,p)→ Pn, and each of the entries lie in OC,p,and not all entries lie in m (as one of the entries has valuation 0). This same expression gives amorphism SpecOC,p → Pn.

Our intuition now is that we want to glue the maps SpecOC,p → Y (which we picture as amap from a germ of a curve) and C − p → Y (which we picture as the rest of the curve). LetSpecR ⊂ Y be an affine open subset of Y containing the image of SpecOC,p. Let SpecA ⊂ Cbe an affine open of C containing p, and such that the image of SpecA − p in Y lies in SpecR,and such that p is cut out scheme-theoretically by a single equation (i.e. there is an element t ∈ Asuch that (t) is the maximal ideal corresponding to p. Then R and A are domains, and we havetwo maps R → A(t) (corresponding to SpecOC,p → SpecR) and R → At (corresponding to

SpecA− p→ SpecR) that agree “at the generic point”, i.e. that give the same map R→ FF (A).But At ∩ A(t) = A in FF (A) (e.g. by Hartogs’ theorem — elements of the fraction field of A

that don’t have any poles away from t, nor at t, must lie in A), so we indeed have a map R→ A

agreeing with both morphisms.

25.1.B. Exercise (Useful practice!). Suppose X is a Noetherian k-scheme, and Z is anirreducible codimension 1 subvariety whose generic point is a nonsingular point of X (so the localring OX,Z is a discrete valuation ring). Suppose X 99K Y is a rational map to a projective

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k-scheme. Show that the domain of definition of the rational map includes a dense open subsetof Z. In other words, rational maps from Noetherian k-schemes to projective k-schemes can beextended over nonsingular codimension 1 sets. We saw this principle in action with the Cremonatransformation, in Class 27 Exercise 4.6 [ref!]. (By the easy direction of the valuative criterion Cremonaof separatedness, or the theorem of uniqueness of extensions of maps from reduced schemes toseparated schemes — Theorem 3.3 of Class 27 [ref!] — this map is unique.)

25.1.3. Theorem. — If C is a nonsingular curve, then there is some projective nonsingularcurve C′ and an open immersion C → C ′. 3 [[H, Cor. I.6.10].]

This proof has a bit of a different flavor than proofs we’ve seen before. We’ll use makeparticular use of the fact that one-dimensional Noetherian schemes have a boring topology.

Proof. Given a nonsingular curve C, take a non-empty=dense affine open set, and take any non-constant function f on that affine open set to get a rational map C 99K P1 given by [1; f ]. As adense open set of a dimension 1 scheme consists of everything but a finite number of points, byProposition 25.1.2, this extends to a morphism C → P1.

We now take the normalization of P1 in the function field FF (C) of C (a finite extension ofFF (P1)), to obtain C ′ → P1. Now C′ is normal, hence nonsingular (as nonsingular = normal indimension 1). By the finiteness of integral closure, C ′ → P1 is a finite morphism. Moreover, finitemorphisms are projective, so by considering the composition of projective morphisms C ′ → P1 →Spec k, we see that C ′ is projective over k. Thus we have an isomorphism FF (C) → FF (C ′),hence a rational map C 99K C ′, which extends to a morphism C → C ′ by Key Proposition 25.1.2.

Finally, I claim that C → C ′ is an open immersion. If we can prove this, then we are done.I note first that this is an injection of sets:

• the generic point goes to the generic point• the closed points of C correspond to distinct valuations on FF (C) (as C is separated,

by the easy direction of the valuative criterion of separatedness)

Thus as sets, C is C ′ minus a finite number of points. As the topology on C and C ′ is the“cofinite topology” (i.e. the open sets include the empty set, plus everything minus a finite numberof closed points), the map C → C ′ of topological spaces expresses C as a homeomorphism of Conto its image im(C). Let f : C → im(C) be this morphism of schemes. Then the morphismOim(C) → f∗OC can be interpreted as Oim(C) → OC (where we are identifying C and im(C)

via the homeomorphism f). This morphism of sheaves is an isomorphism of stalks at all pointsp ∈ im(C) (it is the isomorphism the discrete valuation ring corresponding to p ∈ C ′), and ishence an isomorphism. Thus C → im(C) is an isomorphism of schemes, and thus C → C ′ is anopen immersion.

[(Exercise 13.4.K) is the exercise where normalization in the function field was done]We now come to the big theorem of today (although the Key Proposition 25.1.2 above was

also pretty big).

25.1.4. Theorem. — The following categories are equivalent.

(i) nonsingular projective curves, and surjective morphisms.(ii) nonsingular projective curves, and dominant morphisms.(iii) nonsingular projective curves, and dominant rational maps(iv) quasiprojective reduced curves, and dominant rational maps(v) function fields of dimension 1 over k, and k-homomorphisms.

[(Basically [H, Cor. 6.12]) I’ve defined curves to be nonsingular, so this is a little awkward.](All morphisms and maps are assumed to be k-morphisms and k-rational maps, i.e. they are

all over k. Remember that today we are working in the category of k-schemes.)This has a lot of implications. For example, each quasiprojective reduced curve is isomorphic

to precisely one projective nonsingular curve.This leads to a motivating question that I mentioned informally last day (and that isn’t in

the notes). Suppose k is algebraically closed (such as C). Is it true that all nonsingular projectivecurves are isomorphic to P1

k? Equivalently, are all quasiprojective reduced curves birational to

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A1k? Equivalently, are all transcendence degree 1 extensions of k generated (as a field) by a single

element? The answer (as most of you know) is no, but we can’t yet see that.

25.1.C. Exercise. Show that all nonsingular proper curves are projective. [Kirsten’s argument:if C is proper, then it has an open immersion into a projective. This morphism is proper, hencethe image is closed. An open immersion whose image is closed is the inclusion of a union ofconnected components of the target scheme, into the target scheme, and in partiuclar it is aclosed immersion.]

(We may eventually see that all reduced proper curves over k are projective, but I’m not sure;this will use the Riemann-Roch theorem, Exercise 23.5.B, and I may just prove it for projectivecurves.)

Before we get to the proof, I want to mention a sticky point that came up in class. If k = R,then we are allowing curves such as P1

C that “we don’t want”. One way of making this precise isnoting that they are not geometrically irreducible (as C(t)⊗R

C ∼= C(t)⊕C(t)). Another way is to

note that this function field K does not satisfy k ∩ K = k in K. If this bothers you, then addit to each of the 5 categories. (For example, in (i)–(iii), we consider only nonsingular projective

curves whose function field K satisfies k ∩K = k in K.) If this doesn’t bother you, please ignorethis paragraph!

Proof. Any surjective morphism is a dominant morphism, and any dominant morphism is adominant rational map, and each nonsingular projective curve is a quasiprojective curve, so we’veshown (i) → (ii) → (iii) → (iv). To get from (iv) to (i), we first note that the nonsingular pointson a quasiprojective reduced curve are dense. (One method, suggested by Joe: we know thatnormalization is an isomorphism away from a closed subset.) Given a dominant rational mapbetween quasiprojective reduced curves C → C ′, we get a dominant rational map between theirnormalizations, which in turn gives a dominant rational map between their projective modelsD 99K D′. The the dominant rational map is necessarily a morphism by Proposition 25.1.2, andthen this morphism is necessarily projective and hence closed, and hence surjective (as the imagecontains the generic point of D′, and hence its closure). Thus we have established (iv) → (i).

It remains to connect (i). Each dominant rational map of quasiprojective reduced curvesindeed yields a map of function fields of dimension 1 (their fraction fields). Each function fieldof dimension 1 yields a reduced affine (hence quasiprojective) curve over k, and each map of twosuch yields a dominant rational map of the curves.

[Unimportant but fun exercise. 0, 1,∞ ∈ P1. Permute them. Write down six automorphismscorresponding to permuting them. Action on k(t). Field of invariants. Get the j-invaraint ofelliptic curves.]

25.1.5. Degree of a morphism between projective nonsingular curves.[To place: the well-behavedness of degree is intuitively quite reasonable from complex anal-

ysis. Consider z 7→ zm. Have a picture.]We conclude with a useful fact: Any non-constant morphism from one projective nonsingular

curve to another has a well-behaved degree, in a sense that we will now make precise. We willalso show that any non-constant finite morphism from one nonsingular curve to another has awell-behaved degree in the same sense.

Suppose f : C → C ′ is a surjective (or equivalently, dominant) map of nonsingular projectivecurves.

It is a finite morphism. Here is why. (If we had already proved that quasifinite projectiveor proper morphisms with finite fibers were finite, we would know this. Once we do know this,the contents of this section would extend to the case where C is not necessarily nonsingular.)Let C′′ be the normalization of C ′ in the function field of C. Then we have an isomorphism

FF (C) ∼= FF (C′′) which leads to birational maps C oo //___ C′′ which extend to morphisms as

both C and C′′ are nonsingular and projective. Thus this yields an isomorphism of C and C ′′.But C′′ → C is a finite morphism by the finiteness of integral closure.

We can then use the following proposition, which applies in more general situations.

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25.1.6. Proposition. — Suppose that π : C → C ′ is a surjective finite morphism, where Cis an integral curve, and C ′ is an integral nonsingular curve. Then π∗OC is locally free of finiterank.

As π is finite, π∗OC is a finite type sheaf on O′C . In case you care, the hypothesis “integral”on C′ is redundant.

Before proving the proposition. I want to remind you what this means. Suppose d is therank of this allegedly locally free sheaf. Then the fiber over any point of C with residue field K isthe Spec of an algebra of dimension d over K. This means that the number of points in the fiber,counted with appropriate multiplicity, is always d.

As a motivating example, consider the map Q[y]→ Q[x] given by x 7→ y2. (We’ve seen thisexample before.) I picture this as the projection of the parabola x = y2 to the x-axis. (i) Thefiber over x = 1 is Q[y]/(y2 − 1), so we get 2 points. (ii) The fiber over x = 0 is Q[y]/(y2) —we get one point, with multiplicity 2, arising because of the nonreducedness. (iii) The fiber overx = −1 is Q[y]/(y2 + 1) ∼= Q[i] — we get one point, with multiplicity 2, arising because of thefield extension. (iv) Finally, the fiber over the generic point Spec Q(x) is Spec Q(y), which is onepoint, with multiplicity 2, arising again because of the field extension (as Q(y)/Q(x) is a degree2 extension). We thus see three sorts of behaviors (as (iii) and (iv) are the same behavior). Notethat even if you only work with algebraically closed fields, you will still be forced to this third typeof behavior, because residue fields at generic points tend not to be algebraically closed (witness

case (iv) above).Note that we need C ′ to be nonsingular for this to be true. (I gave a picture of the normal-

ization of a nodal curve as an example. A picture would help here.)Let’s prove it.All we will really need is that C is reduced of pure dimension 1.Let’s discuss again what this means. [This is mostly a repeat.] Suppose d is the rank of this

allegedly locally free sheaf. Then the fiber over any point of C with residue field K is the Specof an algebra of dimension d over K. This means that the number of points in the fiber, countedwith appropriate multiplicity, is always d.

Proof. (For experts: we will later see that what matters here is that the morphism is finite andflat. But we don’t yet know what flat is.)

The question is local on the target, so we may assume that C ′ is affine. Note that π∗OC istorsion-free (as Γ(C,OC) is an integral domain). Our plan is as follows: by an important exercisefrom last quarter (Exercise 5.2 of class 15; problem 10 on problem set 7), if the rank of the coherentsheaf π∗OC is constant, then (as C ′ is reduced) π∗OC is locally free. We’ll show this by showingthe rank at any closed point of C ′ is the same as the rank at the generic point. [This is anotherexample of the usefulness of the generic point.]

The notion of “rank at a point” behaves well under base change, so we base change tothe discrete valuation ring OC′,p, where p is some closed point of C ′. Then π∗OC is a finitelygenerated module over a discrete valuation ring which is torsion-free. By the classification offinitely generated modules over a principal ideal domain, any finitely generate module over aprincipal ideal domain A is a direct sum of modules of the form A/(d) for various d ∈ A. But if Ais a discrete valuation ring, and A/(d) is torsion-free, then A/(d) is necessarily A (as for exampleall ideals of A are of the form 0 or a power of the maximal ideal). Thus we are done.

Remark. If we are working with complex curves, this notion of degree is the same as thenotion of the topological degree. [See for example the following exercise, which should be placedsomewhere when we do ramification. Suppose C → C ′ is a map of C-curves. Then above mostpoints of C′, we have the right number of preimages.]

25.2 Degree of invertible sheaves on curves

Suppose C is a projective curve, and L is an invertible sheaf. We will define degL.Let s be a non-zero rational section of L. For any p ∈ C, recall the valuation of s at p

(vp(s) ∈ Z). (Pick any local section t of L not vanishing at p. Then s/t ∈ FF (C). vp(s) := vp(s/t).We can show that this is well-defined.)

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Define deg(L, s) (where s is a non-zero rational section of L) to be the number of zerosminus the number of poles, counted with appropriate multiplicity. (In other words, each pointcontributes the valuation at that point times the degree of the field extension.) We’ll show thatthis is independent of s. (Note that we need the projective hypothesis: the sections x and 1 ofthe structure sheaf on A1 have different degrees.)

[Do they know that the transition functions of the dual line bundle are the inverse of thetransition functions of the original? Make sure they also know what happens in higher rank.]

Notice that deg(L, s) is additive under products: deg(L, s) + deg(M, t) = deg(L⊗M, s⊗ t).Thus to show that deg(L, s) = deg(L, t), we need to show that deg(OC , s/t) = 0. Hence it sufficesto show that deg(OC , u) = 0 for a non-zero rational function u on C. Then u gives a rational mapC 99K P1. By our recent work (Proposition 25.1.2 above), this can be extended to a morphismC → P1. The preimage of 0 is the number of 0’s, and the preimage of ∞ is the number of ∞’s.But these are the same by our previous discussion of degree of a morphism! Finally, supposep 7→ 0. I claim that the valuation of u at p times the degree of the field extension is precisely thecontribution of p to u−1(0). (A similar computation for ∞ will complete the proof of the desiredresult.) This is because the contribution of p to u−1(0) is precisely

dimkOC,p/(u) = dimk OC,p/mvp(u) = vp(u) dimk OC,p/m.

We can define the degree of an invertible sheaf L on an integral singular projective curveC as follows: if ν : C → C be the normalization, let degC L := degC ν

∗L. Notice that if s is ameromorphic section that has neither zeros nor poles at the singular points of C, then degC Lis still the number of zeros minus the number of poles (suitably counted), because the zeros andpoles of ν∗L are just the same as those of L.

25.2.A. Exercise. Suppose f : C → C ′ is a degree d morphism of integral projective nonsingularcurves, and L is an invertible sheaf on C ′. Show that degC f

∗L = d degC′ L. [Proof by Kirstenin scan kirsten13.2.pdf.]

[Remark: finite flat morphisms, degree is well-defined, and constant if the target is constant.We’ll see later that if (R,m) is a local ring [hypotheses], then flat [and coherent] equals free.In a special case, if (R,m) is a principal ideal domain, then flat equals free equals torsion-free.We have: free implies flat clearly; and flat implies torsion-free by an easy algebraic argument.Torsion-free to free must require a little more. Or wait ... maybe Nakayama is all that is needed!Take generators.... In any case, I show later that degree is well-defined for maps of projectivecurve to projective curves, in the previous paragraph.]

25.2.1. Degree of a Cartier divisor on a curve.[Remark: this is designed so that the number of poles equals number of zeroes for a rational

function. Motivating examples: (x−1)(x−2)/(x+1)(x+2). (x−1)x/(x+1). (x2−2)/(x+1)(x−1)over Q.]

Suppose D is an effective Cartier divisor on a projective curve, or a Cartier divisor on aprojective nonsingular curve (over a field k). (I should really say: suppose D is a Cartier divisoron a projective curve, but I don’t think I defined Cartier divisors in that generality.) Then definethe degree of D (denoted degD) to be the degree of the corresponding invertible sheaf.Exercise. If D is an effective Cartier divisor on a projective nonsingular curve, say D =

Pnipi,

prove that degD =Pni deg pi, where deg pi is the degree of the field extension of the residue

field at pi over k.

25.3 Older notes

[Earlier notes: We’ll need nonsingularity before we get here. We’ll also need to define “curve”!

State DVR facts here. Finish showing the equivalence of the 2 categories: (curves with morphisms)and (curves with rational maps). Link it to a third: transcendence degree 1 extensions of k.]

A curve here is an irreducible variety over k of dimension 1 (reduced + irreducible = integral;separated; finite type).

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Nonsingular: regular local rings are all DVR’s.Every nonsingular curve is isomorphic to an open subset of a projective curve [H, Cor. I.6.10].

Every curve is birationally equivalent to a nonsingular projective curve [H, Cor. I.6.12]. Thefollowing 3 categories are equivalent.(i) nonsingular projective curves, and surjective morphisms.(ii) nonsingular projective curves, and dominant morphisms.(iii) nonsingular projective curves, and dominant rational maps(iv) quasiprojective curves, and dominant rational maps(v) fields of transcendence degree 1 over k, and k-homomorphisms (Basically [H, Cor. 6.12])

The key fact will be the following

25.3.1. Theorem. — If C is a nonsingular curve, then there is some projective nonsingularcurve C′ and an open immersion C → C ′. 4

We’ll prove this later.

25.3.2. Vignette: maps of curves to projective schemes.If we new the big curve theorem 25.3.1, then we would know that there is one projective curve

in each equivalence class. Reason: if there were two, then we’d have two morphisms between them,whose composition is a birational map. But this extends to a morphism by Proposition 25.1.2,and this morphism is an isomorphism, using Proposition 13.1.1.

So we’re reduced to proving the big curve theorem 25.3.1.Given a nonsingular curve C, get a rational map to P1. This extends to a morphism by

Proposition 25.1.2 (for all missing points, of which there are a finite number XXXX).I want to take the normalization in the function field (Exercise 13.4.K), so we obtain C ′ → P1

finite where C′ is normal, hence smooth by DVR facts [put good ref in Section A.2.7]. Finiteimplies projective. Composition of projectives is projective, so C ′ is projective. We have C → C ′

by Exercise 13.3.J. How do we know that this is an open immersion?

25.4 The Riemann-Roch formula for nonsingular projectivecurves, and applications

[This is now done earlier in Exercise 23.5.B.][This is surely later!] I’m ready to prove this! 5 Riemann-RochProof: fix an invertible sheaf L. We define its degree as follows. We take a rational section,

with zeros Z and poles P , and we take degL = Z − P . Then do:

0→OC(−P )→ L→ OZ ⊗ L → 0.

Also

0→OC(−P )→OC →OP → 0.

Apply euler characteristic, and a computation of χ(OP ) = h0(OP ).

Use Riemann-Roch to give a second proof that all curves are projective. [Problem: our proofof Riemann-Roch kind of assumes this. We need that h0(OC(−P )) <∞.]

All Riemann surfaces are algebraic. Method: prove Riemann-Roch in the analytic category[give ref]. Stick it in projective space. Use Chow’s Theorem, stated earlier in the GAGA section,33.

25.5 Fun with Riemann-Roch

Needed: criterion for basepoint freeness (easy) and closed immersion (hard). [base-point-freedom]

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I seem to remember that we need that any negative degree invertible sheaf has no sections;and a degree 0 sheaf has a section if and only if it is trivial.

We might need Serre duality.

25.5.1. Genus 0. All genus 0 curves with a k-point are P1k.

Caution: x2 + y2 + z2 = 0. (Also: one that is not geometrically irreducible?)Over any field: always a plane conic.

25.5.2. Genus 1. 6

This section is very useful, for counterexamples too!We assume the field is algebraically closed. (Exercise: what works in general?)Pic1 E ∼= E.Group law on an elliptic curve!In particular: it is not a finitely generated group.We have paid back an IOU: an example of a scheme such that any open set has nontrivial

class group.degree 2: double cover of P1. Riemann-Hurwitz [ref] implies branched at 4 points if in

characteristic 0.Hence: one-dimensional family.Degree 3: plane cubic: explicit description.

25.5.3. Genus 2. K gives map to P1. Double cover of B. Hence: have classified them all.3-dimensional family. Not dependent on base field!

25.5.4. Genus 3. Hyperelliptic or quartic plane curve. Hence: 2 families of them. Remark:one is limit of the other.

25.5.A. Exercise. Genus g: hyperelliptic, or embedded in Pg−1. In case of genus 4: completeintersection of degree 2 and degree 3. Genus 5: complete intersection of three quadrics. Highergenus: not complete intersection. I need Bezout for this.Bezout

25.6 More fun with curves

We already know enough to study curves in a great deal of detail, so this seems like a goodway to end this quarter. We get much more mileage if we have a few facts involving differentials,so I’ll introduce these facts, and take them as a black box. The actual black boxes we’ll need arequite small, but I want to tell you some of the background behind them.

For this topic, we will assume that all curves are projective geometrically integral nonsingularcurves over a field k. We will sometimes add the hypothesis that k is algebraically closed.

Most people are happy with working over algebraically closed fields, and all of you should ig-nore the adverb “geometrically” in the previous paragraph. For those interested in non-algebraicallyclosed fields, an example of a curve that is integral but not geometrically integral is P1

C over R.Upon base change to the algebraic closure C of R, this curve has two components.

25.6.1. Differentials on curves. There is a sheaf of differentials on a curve C, denoted

ΩC , which is an invertible sheaf. (In general, nonsingular k-varieties of dimension d will have asheaf of differentials over k that will be locally free of rank k. And differentials will be defined invastly more generality.) We will soon see that this invertible sheaf has degree equal to twice the

genus minus 2: deg ΩC = 2gc − 2 . For example, if C = P1, then ΩC ∼= O(−2).

Differentials pull back: any surjective morphism of curves f : C → C ′ induces a natural mapf∗ΩC′ → ΩC .

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25.6.2. The Riemann-Hurwitz formula. Whenever we invoke this formula (in this sec-tion), we will assume that k is algebraically closed and characteristic 0. These conditions aren’tnecessary, but save us some extra hypotheses. Suppose f : C → C ′ is a dominant morphism. Thenit turns out f∗ΩC′ → ΩC is an inclusion of invertible sheaves. (This is a case when inclusions ofinvertible sheaves does not mean what people normally mean by inclusion of line bundles, whichare always isomorphisms.) Its cokernel is supported in dimension 0:

0→ f∗ΩC′ → ΩC → [dimension 0]→ 0.

The divisor R corresponding to those points (with multiplicity), is called the ramification divisor.ramification divisor

We can study this in local coordinates. We don’t have the technology to describe this preciselyyet, but you might still find this believable. If the map at q ∈ C ′ looks like u 7→ un = t, thendt 7→ d(un) = nun−1du, so dt when pulled back vanishes to order n− 1. Thus branching of thissort u 7→ un contributes n− 1 to the ramification divisor. (More correctly, we should look at themap of Spec’s of discrete valuation rings, and then u is a uniformizer for the stalk at q, and t isa uniformizer for the stalk at f(q), and t is actually a unit times un. But the same argumentworks.)

Now in a recent exercise on pullbacks of invertible sheaves under maps of curves, we knowthat a degree of the pullback of an invertible sheaf is the degree of the map times the degree ofthe original invertible sheaf. Thus if d is the degree of the cover, deg ΩC = d deg ΩC′ + degR.Conclusion: if C → C ′ is a degree d cover of curves, then Riemann-Hurwitz for-

mula2gc − 2 = d(2gC′ − 2) + degR

Here are some applications.Example. When I drew a sample branched cover of one complex curve by another, I showed

a genus 2 curve covering a genus 3 curve. Show that this is impossible. (Hint: degR ≥ 0.)Example: Hyperelliptic curves. Hyperelliptic curves are curves that are double covers of P1

k. hyperelliptic curvesIf they are genus g, then they are branched over 2g + 2 points, as each ramification can happento order only 1. (Caution: we are in characteristic 0!) You may already have heard about genus1 complex curves double covering P1, branched over 4 points.

Application 1. First of all, the degree of R is even: any cover of a curve must be branchedover an even number of points (counted with multiplicity).

Application 2. The only connected unbranched cover of P1k is the isomorphism. Reason: if

degR = 0, then we have 2− 2gC = 2d with d ≥ 1 and gc ≥ 0, from which d = 1 and gC = 0.Application 3: Luroth’s theorem. Suppose g(C) = 0. Then from the Riemann-Hurwitz Luroth’s theorem

formula, g(C′) = 0. (Otherwise, if gC′ were at least 1, then the right side of the Riemann-Hurwitzformula would be non-negative, and thus couldn’t be −2, which is the left side. This has a non-obvious algebraic consequence, by our identification of covers of curves with field extensions (class28 Theorem 1.5 [fix ref]). Hence all subfields of k(x) containing k are of the form k(y) wherey = f(x). (Here we have the hypothesis where k is algebraically closed. We’ll patch that later.)Kirsten said that an algebraic proof was given in Math 210. 7

25.7 Onwards

[Better transition!]

25.7.1. Serre duality. (We are not requiring k to be algebraically closed.) In general,nonsingular varieties will have a special invertible sheaf KX which is the determinant of ΩX . Thisinvertible sheaf is called the canonical bundle, and will later be defined in much greater generality[ref]. In our case, X = C is a curve, so KC = ΩC , and from here on in, we’ll use KC instead ofΩC . The reason it is called the dualizing sheaf is because it arises in Serre duality. Serre duality

states that H1(C,K) ∼= k, or more precisely that there is a trace morphism H1(C,K)→ k that

is an isomorphism. (Example: if C = P1, then we indeed have h1(P1,O(−2)) = 1.)

7H.E.IV.2.5.5

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Further, for any coherent sheaf F , the natural map

H0(C,F)⊗k H1(C,K⊗ F∨)→ H1(C,K)

is a perfect pairing. Thus in particular, h0(C,F) = h1(C,K ⊗ F∨). Recall we defined thearithmetic genus of a curve to be h1(C,OC). Then h0(C,K) = g as well.

Recall that Riemann-Roch for a invertible sheaf L states that

h0(C,L)− h1(C,L) = degL− g + 1.

Applying this to L = K, we get

degK = h0(C,K)− h1(C,K) + g − 1 = h1(C,O)− h0(C,O) + g − 1 = g − 1 + g − 1 = 2g − 2

as promised earlier.

25.7.2. A criterion for when a morphism is a closed immersion.We’ll also need a criterion for when something is a closed immersion. To help set it up, let’s

observe some facts about closed immersions. Suppose f : X → Y is a closed immersion. Thenf is projective, and it is injective on points. This is not enough to ensure that it is a closedimmersion, as the example of the normalization of the cusp shows (Figure 1). Another example isthe Frobenius morphism from A1 to A1, given by k[t]→ k[u], u→ tp, where k has characteristicp.

Figure 1. Projective morphisms that are injective on points neednot be closed immersions [class34fig2]

The additional information you need is that the tangent map is an isomorphism at all closedpoints. (Exercise: show this is false in those two examples.)

25.7.3. Theorem. — Suppose k is an algebraically closed field, and f : X → Y is a projectivemorphism of finite-type k-schemes that is injective on closed points and injective on tangentvectors of closed points. Then f is a closed immersion. 8

8sepptstan

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[Remark: this is called “unramified”.]The example of Spec C → Spec R shows that we need the hypothesis that k is algebraically

closed.We need the hypothesis of projective morphism, as shown by the following example (which

was described at the blackboard, see Figure 2). We map A1 to the plane, so that its image is acurve with one node. We then consider the morphism we get by discarding one of the preimagesof the node. Then this morphism is an injection on points, and is also injective on tangentvectors, but it is not a closed immersion. (In the world of differential geometry, this fails to be anembedding because the map doesn’t give a homeomorphism onto its image.)

Figure 2. We need the projective hypothesis in Theorem 25.7.3 [class34fig]

Suppose f(p) = q, where p and q are closed points. We will use the hypothesis that X andY are k-schemes where k is algebraically closed at only one point of the argument: that the mapinduces an isomorphism of residue fields at p and q.

(For those of you who are allergic to algebraically closed fields: still pay attention, as we’lluse this to prove things about curves over k where k is not necessarily algebraically closed.)

This is the hardest result of today. We will kill the problem in old-school French style: deathby a thousand cuts.

Proof. [Cut, as this was a hypothesis: “Note that f is a projective morphism. (Any morphismfrom a projective k-scheme to a separated k-scheme is necessarily projective, by our fun-but-useful“property P result”, Class 25 Proposition 1.25. [ref])”]

We may assume that Y is affine, say SpecB.I next claim that f has finite fibers, not just finite fibers above closed points: the fiber

dimension for projective morphisms is upper-semicontinuous (Class 32 Exercise 2.3 [ref!]), so thelocus where the fiber dimension is at least 1 is a closed subset, so if it is non-empty, it must containa closed point of Y . [Perhaps a little more: finitely presented dimension 0 schemes over a field arefinite sets — but it would be nice to have something cleaner.] Thus the fiber over any point is adimension 0 finite type scheme over that point, hence a finite set.

Hence f is a projective morphism with finite fibers, thus affine, and even finite (Class 32Corollary 2.4 [ref!]).

[So far everything is reasonable. Things get weird starting here.]Thus X is affine too, say SpecA, and f corresponds to a ring morphism B → A. We wish

to show that this is a surjection of rings, or (equivalently) of B-modules. We will show that forany maximal ideal n of B, Bn → An is a surjection of Bn-modules. (This will show that B → Ais a surjection. Here is why: if K is the cokernel, so B → A→ K → 0, then we wish to show thatK = 0. Now A is a finitely generated B-module, so K is as well, being a homomorphic image ofA. Thus SuppK is a closed set. If K 6= 0, then SuppK is non-empty, and hence contains a closed

point [n]. Then Kn 6= 0, so from the exact sequence Bn → An → Kn → 0, Bn → An is not asurjection.)

If An = 0, then clearly Bn surjects onto An, so assume otherwise. I claim that An = A⊗BBn

is a local ring. Proof: SpecAn → SpecBn is a finite morphism (as it is obtained by base change

422

from SpecA → SpecB), so we can use the going-up theorem. An 6= 0, so An has a prime ideal.Any point p of SpecAn maps to some point of SpecBn, which has [n] in its closure. Thus there isa point q in the closure of p that maps to [n]. But there is only one point of SpecAn mapping to[n], which we denote [m]. Thus we have shown that m contains all other prime ideals of SpecAn,so An is a local ring.

Injectivity of tangent vectors means surjectivity of cotangent vectors, i.e. n/n2 → m/m2 is asurjection, i.e. n → m/m2 is a surjection. Claim: nAn = mAn. Reason: By Nakayama’s lemmafor the local ring An and the An-module mAn, we conclude that nAn = mAn.

Next apply Nakayama’s Lemma to the Bn-module An. The element 1 ∈ An gives a generatorfor An/nAn = An/mAn, which equals Bn/nBn (as both equal k), so we conclude that 1 alsogenerates An as a Bn-module as desired.

25.7.A. Exercise. Use this to show that the dth Veronese morphism from Pnk , corresponding tothe complete linear series (see Class 22 [ref!]) |OPn

k(d)|, is a closed immersion. Do the same for

the Segre morphism from Pmk ×Spec k Pnk . (This is just for practice for using this criterion. Thisis a weaker result than we had before; we’ve earlier checked this over an arbitrary base ring [ref],and we are now checking it only over algebraically closed fields.)d-tuple embedding,

Veronese embedding

25.8 A series of useful remarks

Suppose now that L is an invertible sheaf on a curve C (which as always in this discus-sion is projective, geometrically integral and nonsingular, over a field k which is not necessarilyalgebraically closed). I’ll give a series of small useful remarks that we will soon use to great effect.

25.8.1. h0(C,L) = 0 if degL < 0. Reason: if there is a non-zero section, then the degree of Lcan be interpreted as the number of zeros minus the number of poles. But there are no poles, sothis would have to be non-negative. A slight refinement gives:

25.8.2. h0(C,L) = 0 or 1 if degL = 0. This is because if there is a section, then the degree of L isthe number of zeros minus the number of poles. Then as there are no poles, there can be no zeros.Thus the section (call it s) vanishes nowhere, and gives a trivialization for the invertible sheaf.(Recall how this works: we have a natural bijection for any open set Γ(U,L)↔ Γ(U,OU ), wherethe map from left to right is s′ 7→ s′/s, and the map from right to left is f 7→ sf .) Thus if thereis a section, L ∼= O. But we’ve already checked that for a geometrically integral and nonsingularcurve C, h0(C,L) = 1.

25.8.3. Suppose p is any closed point of degree 1. (In other words, the residue field of p is k.)Then h0(C,L)−h0(C,L(−p)) = 0 or 1. Reason: consider 0→OC(−p)→OC →Op → 0, tensorwith L (this is exact as L is locally free) to get

0→ L(−p)→ L→ L|p → 0.

Then h0(C,L|p) = 1, so as the long exact sequence of cohomology starts off

0→ H0(C,L(−p))→ H0(C,L)→ H0(C,L|p),we are done.9 [Careful here: I’d earlier used Lp, but that looks like stalk language. L|p is what Ishould use.]

25.8.4. Suppose for this remark that k is algebraically closed. (In particular, all closed pointshave degree 1 over k.) Then if h0(C,L) − h0(C,L(−p)) = 1 for all closed points p, then L isbase-point-free, and hence induces a morphism from C to projective space. (Note that L has afinite-dimensional vector space of sections: all cohomology groups of all coherent sheaves on aprojective k-scheme are finite-dimensional.) Reason: given any p, our equality shows that thereexists a section of L that does not vanish at p.I like base-point-free,

and make sure I use it9Lminusp

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25.8.5. Next, suppose p and q are distinct points of degree 1. Then h0(C,L)−h0(C,L(−p−q)) =0, 1, or 2 (by repeating the argument of 25.8.3 twice). If h0(C,L)− h0(C,L(−p− q)) = 2, thennecessarily10

(55) h0(C,L) = h0(C,L(−p)) + 1 = h0(C,L(−q)) + 1 = h0(C,L(−p− q)) + 2.

I claim that the linear system L separates points p and q, by which I mean that the correspondingmap f to projective space satisfies f(p) 6= f(q). Reason: there is a hyperplane of projective spacepassing through p but not passing through q, or equivalently, there is a section of L vanishing atp but not vanishing at q. This is because of the last equality in (55).

25.8.6. By the same argument as above, if p is a point of degree 1, then h0(C,L)−h0(C,L(−2p)) =0, 1, or 2. I claim that if this is 2, then map corresponds to L (which is already seen to be base-point-free from the above) separates the tangent vectors at p. To show this, I need to show thatthe cotangent map is surjective. To show surjectivity onto a one-dimensional vector space, I justneed to show that the map is non-zero. So I need to give a function on the target vanishing at theimage of p that pulls back to a function that vanishes at p to order 1 but not 2. In other words, Iwant a section of L vanishing at p to order 1 but not 2. But that is the content of the statementh0(C,L(−p))− h0(C,L(−2p)) = 1.

25.8.7. Combining some of our previous comments: suppose C is a curve over an algebraicallyclosed field k, and L is an invertible sheaf such that for all closed points p and q, not necessarilydistinct, h0(C,L)− h0(C,L(−p− q)) = 2, then L gives a closed immersion into projective space,as it separates points and tangent vectors, by Theorem 25.7.3.

25.8.8. We now bring in Serre duality. I claim that degL > 2g − 2 implies

h0(C,L) = degL− g − 1.

This is important — remember this! Reason: h1(C,L) = h0(C,K⊗L∨); but K⊗L∨ has negativedegree (as K has degree 2g−2), and thus this invertible sheaf has no sections. Thus Riemann-Rochgives us the desired result.Exercise. Suppose L is a degree 2g − 2 invertible sheaf. Show that it has g − 1 or g sections, andit has g sections if and only if L ∼= K.

25.8.9. We now come to our most important conclusion. Thus if k is algebraically closed,then degL ≥ 2g implies that L is basepoint free (and hence determines a morphism to projectivespace). Also, degL ≥ 2g + 1 implies that this is in fact a closed immersion. Remember this!

25.8.10. [Make sure to refer to this entire subsubsection later on.]11 I now claim (for the peoplewho like fields that are not algebraically closed) that the previous remark holds true even if

k is not algebraically closed. Here is why: suppose C is our curve, and Ck := C ⊗k k is thebase change to the algebraic closure (which we are assuming is connected and nonsingular), withπ : Ck → C (which is an affine morphism, as it is obtained by base change from the affine

morphism Spec k→ Spec k). Then H0(C,L)⊗k k ∼= H0(Ck, π∗L) for reasons I explained last day

(see the first exercise on the class 33 notes, and also on problem set 15).

Ck

π // C

Spec k // Spec k

Let s0, . . . , sn be a basis for the k-vector space H0(C,L); they give a basis for the k-vectorspace H0(Ck , π

∗L). If L has degree at least 2g, then these sections have no common zeros onCk; but this means that they have no common zeros on C. If L has degree at least 2g + 1, thenthese sections give a closed immersion Ck → Pn

k. Then I claim that f : C → Pnk (given by the

same sections) is also a closed immersion. Reason: we can check this on each affine open subsetU = SpecA ⊂ Pnk . Now f has finite fibers, and is projective, hence is a finite morphism (and in

10septan

11inowclaim

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particular affine). Let SpecB = f−1(U). We wonder if A → B is a surjection of rings. But we

know that this is true upon base changing by k: A⊗k k→ B ⊗k k is surjective. So we are done.We’re now ready to take these facts and go to the races.

25.9 Genus 0

25.9.1. Claim. — Suppose C is genus 0, and C has a k-valued point. Then C ∼= P1k.

Of course C automatically has a k-point if k is algebraically closed. Thus we see that allgenus 0 (integral, nonsingular) curves over an algebraically closed field are isomorphic to P1.

If k is not algebraically closed, then C needn’t have a k-valued point: witness x2+y2 +z2 = 0in P2

R. We have already observed that this curve is not isomorphic to P1R, because it doesn’t have

an R-valued point.

Proof. Let p be the point, and consider L = O(p). Then degL = 1, so we can apply what we knowabove: first of all, h0(C,L) = 2, and second of all, these two sections give a closed immersion into P1

k. But the only closed immersion of a curve into P1k is the isomorphism!

As a fun bonus, we see that the weird real curve x2 + y2 + z2 = 0 in P2R has no divisors

of degree 1 over R; otherwise, we could just apply the above argument to the corresponding linebundle.

Our weird curve shows us that over a non-algebraically closed field, there can be genus 0curves that are not isomorphic to P1

k. The next result lets us get our hands on them as well.

25.9.2. Claim. — All genus 0 curves can be described as conics in P2k.

[Error! k = k is required for that criterion. So add a comment to that effect. Just refer to25.8.10.]

Proof. Any genus 0 curve has a degree −2 line bundle — the canonical bundle K. Thus anygenus 0 curve has a degree 2 line bundle: L = K∨. We apply our machinery to this bundle:h0(C,L) = 3 ≥ 2g + 1, so this line bundle gives a closed immersion into P2.

25.9.A. Exercise. Suppose C is a genus 0 curve (projective, geometrically integral and nonsin-gular). Show that C has a point of degree at most 2.

We will use the following result later.

25.9.3. Claim. — Suppose C is not isomorphic to P1k (with no restrictions on the genus of

C), and L is an invertible sheaf of degree 1. Then h0(C,L) < 2.12

Proof. Otherwise, let s1 and s2 be two (independent) sections. As the divisor of zeros of si isthe degree of L, each vanishes at a single point pi (to order 1). But p1 6= p2 (or else s1/s2 hasno poles or zeros, i.e. is a constant function, i.e. s1 and s2 are dependent). Thus we get a mapC → P1 which is basepoint free. This is a finite degree 1 map of nonsingular curves, which inducesa degree 1 extension of function fields, i.e. an isomorphism of function fields, which means thatthe curves are isomorphic. But we assumed that C is not isomorphic to P1

k.

Older notes: All genus 0 curves with a point are isomorphic to P1. (Caution:they need not have a point. x2 + y2 + z2 = 0. This is an example of a scheme thatbecomes isomorphic to P1 after base change.)

All genus 0 curves are plane conics. (Hint: degree 2 invertible sheaf.) Any isisomorphic to P1 after a degree 2 extension.

12deg1

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25.10 Genus ≥ 2

It might make most sense to jump to genus 1 at this point, but the theory of elliptic curvesis especially rich and beautiful, so I’ll leave it for the end.

In general, the curves have quite different behaviors (topologically, arithmetically, geometri-cally) depending on whether g = 0, g = 1, or g > 2. This trichotomy extends to varieties of higherdimension. I gave a very brief discussion of this trichotomy for curves. For example, arithmeti-cally, genus 0 curves can have lots and lots of points, genus 1 curves can have lots of points, andby Faltings’ Theorem (Mordell’s Conjecture) any curve of genus at least 2 has at most finitely Faltings, Mordellmany points. (Thus we knew before Wiles that xn + yn = zn in P2 has at most finitely many

solutions for n ≥ 4, as such curves have genus`n−1

2

´> 1.) Geometrically, Riemann surfaces of

genus 0 are positively curved, Riemann surfaces of genus 1 are flat, and Riemann surfaces of genus1 are negatively curved. We will soon see that curves of genus at least 2 have finite automor-phism groups, while curves of genus 1 have some automorphisms (a one-dimensional family), and(we’ve seen earlier) curves of genus 1 (over an algebraically closed field) have a three-dimensionalautomorphism group.

Older notes:

25.10.A. Exercise. Classification of genus 3 nonhyperelliptics. Then hyperelliptics.We showed what happens to hyperelliptic curves under the canonical embedding inExercise 22.6.D. Basically, we did it for hyperelliptic curves described in a certainway. But you can show that any double cover of P1 can be described in that way;this should probably be an exercise somewhere.

(Define hyperelliptic curve, and trigonal curve.) [Hyperelliptic curve was defined hyperelliptic and trigo-

nal curvesearlier in 22.6.C.]

Exercises: Curves of genus 4 and 5. Canonical curve in general. 13 In the canonical curvehyperelliptic case.

Promised example. A motivating example for schemes: genus 3 curve with noautomorphisms. No 2 points have isomorphic open subsets.

Exercise still to file. Normalization exact sequence. (Need pushforward in coho-mology.) Then similarly genus of a nodal curve. genus of nodal curve,

normalization exact se-

quence25.10.1. Genus 2. [In the third sentence, I have a note to myself: “Here k = k. So reduce

to the k = k case earlier.”] Fix a curve C of genus 2. Then K is degree 2, and has 2 sections. Iclaim that K is base-point-free. Otherwise, if p is a base point, then K(−p) is a degree 1 invertiblesheaf with 2 sections, and we just showed (Claim 25.9.3) that this is impossible. Thus we have adouble cover of P1. Conversely, any double cover C → P1 arises from a degree 2 invertible sheafwith at least 2 sections, so by one of our useful facts, if g(C) = 2, this invertible sheaf must be thecanonical bundle (as the only degree 2 invertible sheaf on a genus 2 curve with at least 2 sectionsis KC). Hence we have a natural bijection between genus 2 curves and genus 2 double covers ofP1.

We now specialize to the case where k = k, and the characteristic of k is 0. (All we will need,once we actually prove the Riemann-Hurwitz formula, is that the characteristic be distinct from2.) Then the Riemann-Hurwitz formula shows that the cover is branched over 6 points. We willsee next day that a double cover is determined by its branch points. Hence genus 2 curves are inbijection with unordered sextuples of points on P1. There is thus a 3-dimensional family of genus2 curves — we have found them all!

(This is still a little imprecise; we would like to say that the moduli space of genus 2 curvesis of dimension 3, but we haven’t defined what we mean by moduli space!)

More generally, we may see next week (admittedly informally) that if g > 1, the curves ofgenus g “form a family” of dimension 3g − 3. (If we knew the meaning of “moduli space”, wewould say that the dimension of the moduli space of genus g curves Mg is 3g − 3.) What goeswrong in genus 0 and 1? The following table (as yet unproved by us!) might help.

13canonicalcurves

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genus dimension of family of curves dimension of automorphism group of curve

0 0 31 1 12 3 03 6 04 9 05 12 0...

......

You can probably see the pattern. This is a little like the behavior of the Hilbert function: thedimension of the moduli space is “eventually polynomial”, so there is something that is better-behaved that is an alternating sum, and once the genus is sufficiently high, the “error term”becomes zero. The interesting question then becomes: why is the “right” notion the secondcolumn of the table minus the third? (In fact the second column is h1(C, TC), where TC is thetangent bundle — not yet defined — and the third column is h0(C, TC). All other cohomologygroups of the tangent bundle vanish by dimensional vanishing.)

25.11 Hyperelliptic curves

[At some point earlier, when discussing RH, make sure to point out what kind of branchingcan happen if the degree is 2.] As usual, we begin by working over an arbitrary field k, andspecializing only when we need to. A curve C of genus at least 2 is hyperelliptic if it admits adegree 2 cover of P1. This map is often called the hyperelliptic map.

Equivalently, C is hyperelliptic if it admits a degree 2 invertible sheaf L with h0(C,L) = 2.hyperelliptic curve

25.11.A. Exercise. Verify that these notions are the same. Possibly in the course of doingthis, verify that if C is a curve, and L has a degree 2 invertible sheaf with at least 2 (linearlyindependent) sections, then L has precisely two sections, and that this L is base-point free andgives a hyperelliptic map.

The degree 2 map C → P1 gives a degree 2 extension of function fields FF (C) over FF (P1) ∼=k(t). If the characteristic is not 2, this extension is necessarily Galois, and the induced involutionon C is called the hyperelliptic involution.hyperelliptic involution

25.11.1. Proposition. — If L corresponds to a hyperelliptic cover C → P1, then L⊗(g−1) ∼=KC .

Proof. Compose the hyperelliptic map with the (g − 1)th Veronese map:d-tuple embedding,

Veronese embedding

CL // P1

OP1

(g−1)// Pg−1.

The composition corresponds to L⊗(g−1). This invertible sheaf has degree 2g − 2, and the imageis nondegenerate in Pg−1, and hence has at least g sections. But one of our useful facts (andindeed an exercise) was that the only invertible sheaf of degree 2g− 2 with (at least) g sections isthe canonical sheaf.

25.11.2. Proposition. — If a curve (of genus at least 2) is hyperelliptic, then it is hyperellipticin “only one way”. In other words, it admits only one double cover of P1.

Proof. If C is hyperelliptic, then we can recover the hyperelliptic map by considering the canonical

map: it is a double cover of a degree g − 1 rational normal curve (by the previous Proposition),and this double cover is the hyperelliptic cover (also by the proof of the previous Proposition).

Next, we invoke the Riemann-Hurwitz formula. We assume the char k = 0, and k = k, sowe can invoke this black box. However, when we actually discuss differentials, and prove theRiemann-Hurwitz formula, we will see that we can just require char k 6= 2 (and k = k).

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The Riemann-Hurwitz formula implies that hyperelliptic covers have precisely 2g + 2 (dis-tinct) branch points. We will see in a moment that the branch points determine the curve(Claim 25.11.3).

Assuming this, we see that hyperelliptic curves of genus g correspond to precisely 2g + 2points on P1 modulo S2g+2, and modulo automorphisms of P1. Thus “the space of hyperellipticcurves” has dimension

2g + 2− dim Aut P1 = 2g − 1.

(As usual, this is not a well-defined statement, because as yet we don’t know what we mean by“the space of hyperelliptic curves”. For now, take it as a plausibility statement.) If we believethat the curves of genus g form a family of dimension 3g − 3, we have shown that “most curvesare not hyperelliptic” if g > 2 (or on a milder note, there exists a hyperelliptic curve of each genusg > 2).

25.11.3. Claim. — Assume char k 6= 2 and k = k. Given n distinct points on P1, there isprecisely one cover branched at precisely these points if n is even, and none if n is odd. 14

In particular, the branch points determine the hyperelliptic curve. (We also used this factwhen discussing genus 2 curves last day.)

Proof. Suppose we have a double cover of A1, C → A1, where x is the coordinate on A1. Thisinduces a quadratic field extension K over k(x). As char k 6= 2, this extension is Galois. Let σ bethe hyperelliptic involution. Let y be an element of K such that σ(y) = −y, so 1 and y form abasis for K over the field k(x) (and are eigenvectors of σ). Now y2 ∈ k(x), so we can replace yby an appropriate k(x)-multiple so that y2 is a polynomial, with no repeated factors, and monic.(This is where we use the hypothesis that k is algebraically closed, to get leading coefficient 1.)Thus y2 = xn+an−1xn−1 + · · ·+a0. The branch points correspond to those values of x for whichthere is exactly one value of y, i.e. the roots of the polynomial. As we have no double roots, thecurve is nonsingular. Let this cover be C ′ → A1. Both C and C′ are normalizations of A1 in thisfield extension, and are thus isomorphic. Thus every double cover can be written in this way, andin particular, if the branch points are r1, . . . , rn, the cover is y2 = (x− r1) · · · (x− rn).

[Rewrite this paragraph.] We now consider the situation over P1. A double cover can’t bebranched over an odd number of points by the Riemann-Hurwitz formula. Given an even numberof points r1, . . . , rn in P1, choose an open subset A1 containing all n points. Construct the doublecover of A1 as explained in the previous paragraph: y2 = (x − r1) · · · (x − rn). Then take thenormalization of P1 in this field extension. Over the open A1, we recover this cover. We just needto make sure we haven’t accidentally acquired a branch point at the missing point ∞ = P1 − A1.But the total number of branch points is even, and we already have an even number of points, sothere is no branching at ∞.

Remark. If k is not algebraically closed (but of characteristic not 2), the above argument showsthat if we have a double cover of A1, then it is of the form y2 = af(x), where f is monic,

and a ∈ k∗/(k∗)2 . So (assuming the field doesn’t contain all squares) a double cover does notdetermine the same curve. Moreover, see that this failure is classified by k∗/(k∗)2. Thus we have

lots of curves that are not isomorphic over k, but become isomorphic over k. These are oftencalled twists of each other.

(In particular, even though haven’t talked about elliptic curves yet, we definitely have twoelliptic curves over Q with the same j-invariant, that are not isomorphic.)

[Somewhere around here, mention the fact taht we get a basis for differentials. Perhaps atthe end of this section.]

25.12 Curves of genus 3

Suppose C is a curve of genus 3. Then K has degree 2g − 2 = 4, and has g = 3 sections.

25.12.1. Claim. — K is base-point-free, and hence gives a map to P2.

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Proof. We check base-point-freeness by working over the algebraic closure k. For any point p, byRiemann-Roch,

h0(C,K(−p)) − h0(C,O(p)) = deg(K(−p)) − g + 1 = 3− 3 + 1 = 1.

But h0(C,O(p)) = 0 by one of our useful facts, so

h0(C,K(−p)) = 2 = h0(C,K)− 1.

Thus p is not a base-point of K, so K is base-point-free.

The next natural question is: Is this a closed immersion? Again, we can check over algebraicclosure. We use our “closed immersion test” (again, see our useful facts). If it isn’t a closedimmersion, then we can find two points p and q (possibly identical) such that

h0(C,K)− h0(C,K(−p− q)) = 2,

i.e. h0(C,K(−p − q)) = 2. But by Serre duality, this means that h0(C,O(p + q)) = 2. We havefound a degree 2 divisor with 2 sections, so C is hyperelliptic. (Indeed, I could have skippedthat sentence, and made this observation about K(−p − q), but I’ve done it this way in order togeneralize to higher genus.) Conversely, if C is hyperelliptic, then we already know that K givesa double cover of a nonsingular conic in P2 (also known as a rational normal curve of degree 2).

Thus we conclude that if C is not hyperelliptic, then the canonical map describes C as adegree 4 curve in P2.

Conversely, any quartic plane curve is canonically embedded. Reason: the curve has genus 3(we can compute this — see our discussion of Hilbert functions), and is mapped by an invertiblesheaf of degree 4 with 3 sections. Once again, we use the useful fact saying that the only invertiblesheaf of degree 2g − 2 with g sections is K.Exercise. Show that the nonhyperelliptic curves of genus 3 form a family of dimension 6. (Hint:Count the dimension of the family of nonsingular quartics, and quotient by Aut P2 = PGL(3).)

The genus 3 curves thus seem to come in two families: the hyperelliptic curves (a family ofdimension 5), and the nonhyperelliptic curves (a family of dimension 6). This is misleading —they actually come in a single family of dimension 6.

In fact, hyperelliptic curves are naturally limits of nonhyperelliptic curves. We can write downan explicit family. (This next paragraph will necessarily require some hand-waving, as it involvestopics we haven’t seen yet.) Suppose we have a hyperelliptic curve branched over 2g+2 = 8 pointsof P1. Choose an isomorphism of P1 with a conic in P2. There is a nonsingular quartic meetingthe conic at precisely those 8 points. (This requires Bertini’s theorem, so I’ll skip that argument.)Then if f is the equation of the conic, and g is the equation of the quartic, then f 2+t2g is a family

of quartics that are nonsingular for most t (nonsingular is an open condition as we will see). Thet = 0 case is a double conic. Then it is a fact that if you normalize the family, the central fiber(above t = 0) turns into our hyperelliptic curve. Thus we have expressed our hyperelliptic curveas a limit of nonhyperelliptic curves.

25.13 Genus at least 3

We begin with two exercises in general genus, and then go back to genus 4.Exercise. Suppose C is a genus g curve. Show that if C is not hyperelliptic, then the canonicalbundle gives a closed immersion C → Pg−1. (In the hyperelliptic case, we have already seen thatthe canonical bundle gives us a double cover of a rational normal curve.) Hint: follow the genus3 case. Such a curve is called a canonical curve. [A little more effort is required in the case whencanonical curvethe base field is not algebraically closed. See kirsten16.3.pdf]Exercise. Suppose C is a curve of genus g > 1, over a field k that is not algebraically closed. Showthat C has a closed point of degree at most 2g − 2 over the base field. (For comparison: if g = 1,there is no such bound!)

We next consider nonhyperelliptic curves C of genus 4. Note that degK = 6 and h0(C,K) = 4,so the canonical map expresses C as a sextic curve in P3. We shall see that all such C are completeintersections of quadric surfaces and cubic surfaces, and vice versa.

By Riemann-Roch, K⊗2 has degK⊗2 − g+ 1 = 12− 4 + 1 = 9 sections. That’s one less than

dimSym2 Γ(C,K) =`4+1

2

´. Thus there is at least one quadric in P3 that vanishes on our curve C.

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Translation: C lies on at least on quadric Q. Now quadrics are either double planes, or the unionof two planes, or cones, or nonsingular quadrics. (They corresponds to quadric forms of rank 1,2, 3, and 4 respectively.) Note that C can’t lie in a plane, so Q must be a cone or nonsingular. Inparticular, Q is irreducible.

Now C can’t lie on two (distinct) such quadrics, say Q and Q′. Otherwise, as Q and Q′

have no common components (they are irreducible and not the same!), Q ∩ Q′ is a curve (notnecessarily reduced or irreducible). By Bezout’s theorem, it is a curve of degree 4. Thus our curveC, being of degree 6, cannot be contained in Q ∩Q′.

We next consider cubics. By Riemann-Roch, K⊗3 has degK⊗3 − g + 1 = 18 − 4 + 1 = 15

sections. Now dimSym3 Γ(C,K) has dimension`4+2

3

´= 20. Thus C lies on at least a 5-dimensional

vector space of cubics. Admittedly 4 of them come from multiplying the quadric Q by a linearform (?w+?x+?y+?z). But hence there is still one cubic K whose underlying form is not divisibleby the quadric form Q (i.e. K doesn’t contain Q.) Then K and Q share no component, so K ∩Qis a complete intersection. By Bezout’s theorem, we obtain a curve of degree 6. Our curve C hasdegree 6. This suggests that C = K ∩Q. In fact, K ∩Q and C have the same Hilbert polynomial,and C ⊂ K ∩Q. Hence C = K ∩Q by the following exercise.

Exercise. Suppose X ⊂ Y ⊂ Pn are a sequence of closed subschemes, where X and Y havethe same Hilbert polynomial. Show that X = Y . Hint: consider the exact sequence

0→ IX/Y →OY →OX → 0.

Show that if the Hilbert polynomial of IX/Y is 0, then IX/Y must be the 0 sheaf. [Handy trick:

For m 0, IX/Y (m) is generated by global sections and is also 0. This of course applies with Ireplaced by any coherent sheaf.]

We now consider the converse, and who that any nonsingular complete intersection C of aquadric surface with a cubic surface is a canonically embedded genus 4 curve. It is not hardto check that it has genus 3 (again, using our exercises involving Hilbert functions). Exercise.Show that OC(1) has 4 sections. (Translation: C doesn’t lie in a hyperplane.) Hint: long exactsequences! Again, the only degree 2g− 2 invertible sheaf with g sections is the canonical sheaf, soOC(1) ∼= KC , and C is indeed canonically embedded.

Exercise. Conclude that nonhyperelliptic curves of genus 4 “form a family of dimension9 = 3g − 3”. (Again, this isn’t a mathematically well-formed question. So just give a plausibilityargument.)

On to genus 5!Exercise. Suppose C is a nonhyperelliptic genus 5 curve. The canonical curve is degree 8 in

P4. Show that it lies on a three-dimensional vector space of quadrics (i.e. it lies on 3 independentquadrics). Show that a nonsingular complete intersection of 3 quadrics is a canonical genus 5curve.

In fact a canonical genus 5 is always a complete intersection of 3 quadrics.Exercise. Show that the complete intersections of 3 quadrics in P4 form a family of dimension

12 = 3g − 3.This suggests that the nonhyperelliptic curves of genus 5 form a dimension 12 family.So we’ve managed to understand curves of genus up to 5 (starting with 3) by thinking of

canonical curves as complete intersections. Sadly our luck has run out.Exercise. Show that if C ⊂ Pg−1 is a canonical curve of genus g ≥ 6, then C is not a complete

intersection. (Hint: Bezout.)

25.14 Genus 1

Finally, we come to the very rich case of curves of genus 1.Note that K is an invertible sheaf of degree 2g − 2 = 0 with g = 1 section. But the only

degree 0 invertible sheaf with a section is the trivial sheaf, so we conclude that K ∼= O.Next, note that if degL > 0, then Riemann-Roch and Serre duality gives

h0(C,L) = h0(C,L)− h0(C,K⊗L∨) = h0(C,L)− h0(C,L∨) = degL

as an invertible sheaf L∨ of negative degree necessarily has no sections.

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An elliptic curve is a genus 1 curve E with a choice of k-valued point p. (Note: it is not the

same as a genus 1 curve — some genus 1 curves have no k-valued points. However, if k = k, thenany closed point is k-valued; but still, the choice of a closed point should always be consideredpart of the definition of an elliptic curve.) [Example?]

Note that OE(2p) has 2 sections, so the argument given in the hyperelliptic section showsthat E admits a double cover of P1. One of the branch points is 2p: one of the sections of OE(2p)vanishes to p of order 2, so there is a point of P1 consists of p (with multiplicity 2). Assume now

that k = k, so we can use the Riemann-Hurwitz formula. Then the Riemann-Hurwitz formulashows that E has 4 branch points (p and three others). Conversely, given 4 points in P1, we get amap (y2 = · · · ). This determines C (as shown in the hyperelliptic section [ref; it is just above]).Thus elliptic curves correspond to 4 points in P1, where one is marked p, up to automorphismsof P1. (Equivalently, by placing p at ∞, elliptic curves correspond to 3 points in A1, up to affinemaps x 7→ ax+ b.)

If the three other points are temporarily labeled q1, q2, q3, there is a unique automorphismof P1 taking p, q1, q2 to (∞, 0, 1) respectively (as Aut P1 is three-transitive).[Make sure to havethis an exercise earlier!] Suppose that q3 is taken to some number λ under this map. Notice thatλ 6= 0, 1,∞.

• If we had instead sent p, q2, q1 to (∞, 0, 1), then q3 would have been sent to 1− λ.• If we had instead sent p, q1, q3 to (∞, 0, 1), then q2 would have been sent to 1/λ.• If we had instead sent p, q3, q1 to (∞, 0, 1), then q2 would have been sent to 1− 1/λ =

(λ− 1)/λ.• If we had instead sent p, q2, q3 to (∞, 0, 1), then q2 would have been sent to 1/(1− λ).• If we had instead sent p, q3, q2 to (∞, 0, 1), then q2 would have been sent to 1− 1/(1−λ) = λ/(λ − 1).

Thus these six values (in bijection with S3) yield the same elliptic curve, and this ellipticcurve will (upon choosing an ordering of the other 3 branch points) yield one of these six values.

Thus the elliptic curves over k corresponds to k-valued points of P1 − 0, 1, λ, modulo theaction of S3 on λ given above. Consider the subfield of k(λ) fixed by S3. By Luroth’s theorem, itmust be of the form k(j) for some j ∈ k(λ). Note that λ should satisfy a sextic polynomial over

k(λ), as for each j-invariant, there are six values of λ in general.At this point I should just give you j:

j = 28 (λ2 − λ+ 1)3

λ2(λ− 1)2.

But this begs the question: where did this formula come from? How did someone think ofit?

Far better is to guess what j is. We want to come up with some j(λ) such that j(λ) =j(1/λ) = · · · . Hence we want some expression in λ that is invariant under this S3-action. A sillychoice would be the product of the six numbers λ(1/λ) · · · as this is 1.

A better idea is to add them all together. Unfortunately, if you do this, you’ll get 3. (Hereis one reason to realize this can’t work: if you look at the sum, you’ll realize that you’ll getsomething of the form “degree at most 3” divided by “degree at most 2” (before cancellation).Then if j′ = p(λ)/q(λ), then λ satisfies (at most) a cubic over j. But we said that λ should satisfya sextic over j′. The only way we avoid a contradiction is if j′ ∈ k.

Our next attempt is to add up the six squares. When you do this by hand (it isn’t hard),you get

j′′ =2λ6 − 6λ5 + 9λ4 − 8λ3 + 9λ2 − 6λ+ 2

λ2(λ− 1)2.

This works just fine: k(j) ∼= k(j′′). If you really want to make sure that I’m not deceiving you,you can check (again by hand) that

2j/28 =2λ6 − 6λ5 + 12λ4 − 14λ3 + 12λ2 − 6λ+ 2

λ2(λ − 1)2.

The difference is 3.

Place this earlier! From 216 notes: Let me first give you an exercise I shouldhave given you last day.

Exercise. (a) Suppose C is a projective curve. Show that C − p is affine. (Hint: show thatn 0, O(np) gives an embedding of C into some projective space Pm, and that there is some

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hyperplane H meeting C precisely at p. Then C − p is a closed subscheme of Pn −H.)(b) If C is a geometrically integral nonsingular curve over a field k (i.e. all of our standingassumptions, minus projectivity), show that it is projective or affine.15

***

25.14.1. Degree 3. Consider the degree 3 invertible sheaf OE(3p). We consult our useful facts.By Riemann-Roch, h0(E,OE(3p)) = deg(3p) − g + 1 = 3. As degE > 2g, this gives a closedimmersion. Thus we have a closed immersion E → P2

k as a cubic curve. Moreover, there is a line

in P2k meeting E at point p with multiplicity 3. (Remark: a line in the plane meeting a smooth

curve with multiplicity at least 2 is said to be a tangent line. A line in the plane meeting a smoothcurve with multiplicity at least 3 is said to be a flex line.) tangent line, flex line

We can choose projective coordinates on P2k so that p maps to [0; 1; 0], and the flex line is

the line at infinity z = 0. Then the cubic is of the following form:

?x3 + 0x2y + 0xy2 + 0y3

+ ?x2z + ?xyz + ?y2z

+ ?xz2 + ?yz2

+ ?z3 = 0

The co-efficient of x is not 0 (or else this cubic is divisible by z). We can scale x so that thecoefficient of x3 is 1. The coefficient of y2z is not 0 either (or else this cubic is singular atx = z = 0). As k is algebraically closed, we can scale y so that the coefficient of y2z is 1. (Moreprecisely, we are changing variables, say y′ = ay for some a ∈ k.) If the characteristic of k is not2, then we can then replace y by y+?x+?z so that the coefficients of xyz and yz2 are 0, and if thecharacteristic of k is not 3, we can replace x by x+?z so that the coefficient of x2z is also 0. Inconclusion, if k is algebraically closed of characteristic not 2 or 3, we can write our elliptic curvein the form

y2z = x3 + ax2z + bz3.

This is called Weierstrass normal form. (If only some of the “bonus hypotheses” k = k, char k 6= Weierstrass normal form2, 3 is true, then we can perform only some of the reductions of course.)

Notice that we see the hyperelliptic description of the curve (by setting z = 1, or moreprecisely, by working in the distinguished open set z 6= 0 and using inhomogeneous coordinates).In particular, we can compute the j-invariant.

Here is the geometric explanation of why the double cover description is visible in the cubicdescription.

I drew a picture of the projective plane, showing the cubic, and where it met the z-axis (theline at infinity) — where the z-axis and x-axis meet — it has a flex there. I drew the lines throughthat point — vertical lines. Equivalently, you’re just taking 2 of the 3 sections: x and z. Theseare two sections of O(3p), but they have a common zero — a base point at p. So you really gettwo sections of O(2p).

Exercise. Show that O(4p) embeds E in P3 as the complete intersection of two quadrics.[Reason: at least 2. None of them are reducible. Hence they share no components, so by Bezoutwe win.]

25.14.2. The group law.

25.14.3. Theorem. — The closed points of E are in natural bijection with Pic0(E), viax↔ x− p. In particular, as Pic0(E) is a group, we have endowed the closed points of E with agroup structure.

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432

For those of you familiar with the complex analytic picture, this isn’t surprising: E is iso-morphic to the complex numbers modulo a lattice: E ∼= C/Λ.

This is currently just a bijection of sets. Given that E has a much richer structure (it has ageneric point, and the structure of a variety), this is a sign that there should be a way of definingsome scheme Pic0(E), and that this should be an isomorphism of schemes.

Proof. For injectivity: O(x− p) ∼= O(y − p) implies O(x− y) ∼= O. But as E is not genus 0, thisis possible only if x = y.

For surjectivity: any degree 1 invertible sheaf has a section, so if L is any degree 0 invertiblesheaf, then O(L(p)) ∼= O(x) for some x.

Note that more naturally, Pic1(E) is in bijection with the points of E (without any choice ofpoint p).

From now on, we will conflate closed points of E with degree 0 invertible sheaves on E.Remark. The 2-torsion points in the group are the branch points in the double cover! Reason:q is a 2-torsion point if and only if 2q ∼ 2p if and only if there is a section of O(2p) vanishingat q to order 2. (This is characteristic-independent.) Now assume that the characteristic is 0.(In fact, we’ll only be using the fact that the characteristic is not 2.) By the Riemann-Hurwitzformula, there are 3 non-trivial torsion points. (Again, given the complex picture E ∼= C/Λ, thisisn’t surprising.)Follow-up remark. An elliptic curve with full level n-structure is an elliptic curve with an iso-morphism of its n-torsion points with (Z/n)2 . (This notion will have problems if n is divisible bychar k.) Thus an elliptic curve with full level 2 structure is the same thing as an elliptic curve

with an ordering of the three other branch points in its degree 2 cover description. Thus (if k = k)these objects are parametrized by the λ-line (see the discussion last day).Follow-up to the follow-up. There is a notion of moduli spaces of elliptic curves with full level nstructure. Such moduli spaces are smooth curves (where this is interpreted appropriately), andhave smooth compactifications. A weight k level n modular form is a section of K⊗k where K is

the canonical sheaf of this “modular curve”.But let’s get back down to earth.

25.14.4. Proposition. — There is a morphism of varieties E → E sending a (degree 1) pointto its inverse.

In other words, the “inverse map” in the group law actually arises from a morphism ofschemes — it isn’t just a set map. This is another clue that Pic0(E) really wants to be a scheme.

Proof. It is the hyperelliptic involution y 7→ −y! Here is why: if q and r are “hyperellipticconjugates”, then q + r ∼ 2p = 0.

We can describe addition in the group law using the cubic description. (Here a picture isabsolutely essential, and at some later date, I hope to add it.) To find the sum of q and r on thecubic, we draw the line through q and r, and call the third point it meets s. Then we draw the line

between p and s, and call the third point it meets t. Then q+r = t. Here’s why: q+r+s = p+s+tgives (q − p) + (r − p) = (s− p).

(When the group law is often defined on the cubic, this is how it is done. Then you have toshow that this is indeed a group law, and in particular that it is associative. We don’t need to dothis — Pic0 E is a group, so it is automatically associative.)

Note that this description works in all characteristics; we haven’t required the cubic to be inWeierstrass normal form.

25.14.5. Proposition. — There is a morphism of varieties E×E → E that on degree 1 pointssends (q, r) to q + r.

“Proof”. We just have to write down formulas for the construction on the cubic. This isno fun, so I just want to convince you that it can be done, rather than writing down anythingexplicit. They key idea is to define another map E × E → E, where if the input is (a, b), theoutput is the third point where the cubic meets the line, with the natural extension if the linedoesn’t meet the curve at three distinct points. Then we can use this to construct addition onthe cubic.

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25.14.6. Aside: Discussion on group varieties and group schemes. 16 group schemes[forme: This should be moved back ages ago to ??.]A group variety X over k is something that can be defined as follows: We are given an

element e ∈ X(k) (a k-valued point of X), and maps i : X → X, m : X ×X → X. They satisfythe hypotheses you’d expect from the definition of a group.

(i) associativity:

X ×X ×X(m,id) //

(id,m)

X ×X

m

X ×X m // X

commutes.

(ii) Xe,id // X ×X m // X and X

id,e // X ×X m // X are both the identity.

(iii) Xi,id // X ×X m // X and X

id,i // X ×X m // X are both e.

More generally, a group scheme over a base B is a scheme X → B, with a section e : B → X,and B-morphisms i : X → X, m : X ×B X → X, satisfying the three axioms above.

More generally still, a group object in a category C is the above data (in a category C),satisfying the same axioms. The e map is from the final object in the category to the groupobject.

You can check that a group object in the category of sets is in fact the same thing as a group.(This is symptomatic of how you take some notion and make it categorical. You write down itsaxioms in a categorical way, and if all goes well, if you specialize to the category of sets, you getyour original notion. You can apply this to the notion of “rings” in an exercise below.)

25.14.7. The functorial description. It is often cleaner to describe this in a functorial way.Notice that if X is a group object in a category C, then for any other element of the category,the set Hom(Y,X) is a group. Moreover, given any Y1 → Y2, the induced map Hom(Y2,X) →Hom(Y1, X) is group homomorphism.

We can instead define a group object in a category to be an object X, along with morphismsm : X × X → X, i : X → X, and e : final object → X, such that these induce a naturalgroup structure on Hom(Y,X) for each Y in the category, such that the forgetful maps are grouphomomorphisms. This is much cleaner!

Exercise. Verify that the axiomatic definition and the functorial definition are the same.[Anssi did this explicitly. Jarod did it using Yoneda. To get the multiplicative map: Hom(S,G)×Hom(S,G) → Hom(S,G) for all S (naturally) by Yoneda gives G × G → G.] [Kirsten’s proofkirsten16.13.pdf]

Exercise. Show that (E, p) is a group scheme. (Caution! we’ve stated that only the closedpoints form a group — the group Pic0. So there is something to show here. The main idea is thatwith varieties, lots of things can be checked on closed points. First assume that k = k, so theclosed points are dimension 1 points. Then the associativity diagram is commutative on closedpoints; argue that it is hence commutative. Ditto for the other categorical requirements. Finally,deal with the case where k is not algebraically closed, by working over the algebraic closure.)

[Perhaps discuss it over Z[a1, . . . , an].]We’ve seen examples of group schemes before. For example, A1

k is a group scheme under

addition. Gm = Spec k[t, t−1] is a group scheme.Easy exercise. Show that A1

k is a group scheme under addition, and Gm is a group schemeunder multiplication. You’ll see that the functorial description trumps the axiomatic descrip-tion here! (Recall that Hom(X,A1

k) is canonically Γ(X,OX), and Hom(X,Gm) is canonicallyΓ(X,OX)∗.)

Exercise. Define the group scheme GL(n) over the integers.Exercise. Define µn to be the kernel of the map of group schemes Gm → Gm that is “taking

nth powers”. In the case where n is a prime p, which is also char k, describe µp. (I.e. how manypoints? How “big” = degree over k?)

[Possible remark: Lie groups should be group objects in the category of manifolds. Presum-ably the morphisms we allow are any differentiable morphisms?]

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Exercise. Define a ring scheme. Show that A1k is a ring scheme.ring scheme

25.14.8. Hopf algebras. Here is a notion that we’ll certainly not use, but it is easy enoughto define now. Suppose G = SpecA is an affine group scheme, i.e. a group scheme that is anaffine scheme. The categorical definition of group scheme can be restated in terms of the ringA. Then these axioms define a Hopf algebra. For example, we have a “comultiplication map”A→ A⊗ A. Exercise. As A1

k is a group scheme, k[t] has a Hopf algebra structure. Describe thecomultiplication map k[t]→ k[t]⊗k k[t]. [Answer: f 7→ 1⊗ f + f ⊗ 1 in Hom(k[t], k[t]⊗ k[t].]

25.14.9. Some additional points to place later.Andy on principal homogeneous spaces. Much of this may come from Milne.Consider group schemes flat and (locally?) of finite type over a base scheme X.Exercise: define action of a group scheme on another X-scheme. (One possible answer:

S ×X G→ S that induces action of HomX(T,G) on HomX(T, S) for any X-scheme T .Theorem: for a n X-scheme S with G-action, TFAE: (i) S faithfully flat and lft over X and

S ×X G→ S ×X S given by (s, g) 7→ (s, sg) is an isomorphism. (ii) there is a flat over Ui → Xso that for all i, S ×X Ui → G×X Ui with its G×X Ui action.

Such a scheme is called a G-torsor or principal homogeneous space. A trivial torsor is iso-morphic to G (with its G-action). (Fact: PHS(G/X) = Hom(π1(X), G).)

Of course G-torsors inherit nice properties of G, e.g. smooth, etale, proper.We need a G-action on S early on — the action on T -valued points. A sheaf S is a G-torsor

if there exists Ui → X so that HomUi(·, G) is isomorphic to S|Ui

.Every schemy G-torsor gives a sheafy G-torsor. This is an injection, basically by Yoneda.So we have a map PHS → H1(X,G).Theorem: A sheafy G-torsor is representable if one of the following holds. 1) G→ X affine.

2) G smooth and separated over X and dimX ≤ 1. 3) G is smooth and proper over X withgeometrically connected fibers and regular. etc.

Older notes:

25.14.A. Exercise. Group law on nodal cubic. (Refer backward to find solutions ofy2 = x2 + x3, in section on rational maps.) What group do you get? Then cuspidalcubic.

Pappus theorem etc. Variety without finite generated global sections. Affine inaffine not distinguished. Can’t glue a variety to itself. Poncelet’s porism?

Exercise: get lots of field-valued points on x3 + y3 = z3.

25.15 Fun counterexamples using elliptic curves

We have a morphism (×n) : E → E that is “multiplication by n”, which sends p to np. Ifn = 0, this has degree 0. If n = 1, it has degree 1. Given the complex picture of a torus, youmight not be surprised that the degree of ×n is n2. If n = 2, we have almost shown that it hasdegree 4, as we have checked that there are precisely 4 points q such that 2p = 2q. All that reallyshows is that the degree is at least 4.

25.15.1. Proposition. — For each n > 0, the “multiplication by n” map has positive degree.In other words, there are only a finite number of n torsion points.

Proof. We prove the result by induction; it is true for n = 1 and n = 2.If n is odd, then assume otherwise that nq = 0 for all closed points q. Let r be a non-trivial

2-torsion point, so 2r = 0. But nr = 0 as well, so r = (n− 2[n/2])r = 0, contradicting r 6= 0.If n is even, then [×n] = [×2] [×(n/2)], and by our inductive hypothesis both [×2] and

[×(n/2)] have positive degree.

In particular, the total number of torsion points on E is countable, so if k is an uncountablefield, then E has an uncountable number of closed points (consider an open subset of the curve as

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y2 = x3 + ax+ b; there are uncountably many choices for x, and each of them has 1 or 2 choicesfor y).

Thus almost all points on E are non-torsion. I’ll use this to show you some pathologies.An example of an affine open set that is not distinguished. I can give you an affine scheme X andan affine open subset Y that is not distinguished in X. Let X = E − p, which is affine (easy, orsee Exercise 25.14).

Let q be another point on E so that q − p is non-torsion. Then E − p − q is affine (Exer-cise 25.14). Assume that it is distinguished. Then there is a function f on E − p that vanisheson q (to some positive order d). Thus f is a rational function on E that vanishes at q to orderd, and (as the total number of zeros minus poles of f is 0) has a pole at p of order d. But thend(p− q) = 0 in Pic0 E, contradicting our assumption that p− q is non-torsion.

[older notes: 17 This is referred to in 9.1.V and 7.3.1.]An example of a scheme that is locally factorial near a point p, but such that no affine open

neighborhood of p has ring that is a Unique Factorization Domain.Consider p ∈ E. Then an open neighborhood of E is of the form E − q1 − · · · − qn. I claim

that its Picard group is nontrivial. Recall the exact sequence:

Zn(a1,...,an)7→a1q1+···+anqn // PicE // Pic(E − q1 − · · · − qn) // 0 .

But the group on the left is countable, and the group in the middle is uncountable, so the group

on the right is non-zero.Example of variety with non-finitely-generated space of global sections.[Older notes: Mentioned in 8.4.E. Mumford says on [M, p. 123] that this has a stronger

relation to Hilbert’s 14th problem than I thought. 18]This is related to Hilbert’s fourteenth problem, although I won’t say how.Before we begin we have a preliminary exercise.Exercise. Suppose X is a scheme, and L is the total space of a line bundle corresponding to

invertible sheaf L, so L = Spec⊕n≥0(L∨)⊗n. Show that H0(L,OL) = ⊕H0(X, (L∨)⊗n).Let E be an elliptic curve over some ground field k, N a degree 0 non-torsion invertible sheaf

on E, and P a positive-degree invertible sheaf on E. Then H0(E,Nm ⊗ Pn) is nonzero if andonly if either (i) n > 0, or (ii) m = n = 0 (in which case the sections are elements of k). Thus thering R = ⊕m,n≥0H

0(E,Nm ⊗ Pn) is not finitely generated.Now let X be the total space of the vector bundle N ⊕ P over E. Then the ring of global

sections of X is R.

25.16 +More serious stuff: “the Picard group has dimensiong,” “the moduli space of genus g curves has dimension 3g− 3”

[Perhaps discuss moduli in here? Or refer them to a first definition of moduli, things currentlyin the appendix?]

I’ll conclude the quarter by showing the following.

• If C has genus g, then “Pic0(C) has dimension g”.• “The moduli space of curves of genus g “is dimension 3g − 3.”

We’ll work over an algebraically closed field k of characteristic 0. We haven’t yet made the abovenotions precise, so what follows are just plausibility arguments. (It is worth trying to think of away of making these notions precise! There are several ways of doing this usefully.)

25.16.1. The Picard group has dimension g: “dim Pic0 C = g”. There are quotesaround this equation because so far, Pic0 C is simply a set, so this will just be a plausibilityargument. Let p be any (closed, necessarily degree 1) point of C. Then twisting by p gives an

isomorphism of Picd C and Picd+1 C, via L ↔ L(p). Thus we’ll consider Picd C, where d 0 (in

fact d > degK = 2g − 2 will suffice). Say dim Picd C = h. We ask: how many degree d effectivedivisors are there (i.e. what is the dimension of this family)? The answer is clearly d, and Cd

surjects onto this set (and is usually d!-to-1).

17undistinguishedopen

18hugeglobal

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But we can count effective divisors in a different way. There is an h-dimensional family ofline bundles by hypothesis, and each one of these has a (d− g+1)-dimensional family of non-zerosections, each of which gives a divisor of zeros. But two sections yield the same divisor if one is amultiple of the other. Hence we get: h+ (d − g + 1) − 1 = h+ d− g.

Thus d = h+ d− g, from which h = g as desired.Note that we get a bit more: if we believe that Picd has an algebraic structure, we have a

fibration (Cd)/Sd → Picd, where the fibers are isomorphic to Pd−g . In particular, Picd is reduced,and irreducible.

25.16.2. The moduli space of genus g curves has dimension 3g − 3. Let Mg be the setof nonsingular genus g curves, and pretend that we can give it a variety structure. Say Mg hasdimension p. By our useful Riemann-Roch facts, if d 0, and D is a divisor of degree d, thenh0(C,O(D)) = d − g + 1. If we take two general sections s, t of the line bundle O(D), we geta map to P1, and this map is degree d. Conversely, any degree d cover f : C → P1 arises fromtwo linearly independent sections of a degree d line bundle. Recall that (s, t) gives the same mapto P1 as (s′, t′) if and only (s, t) is a scalar multiple of (s′, t′). Hence the number of maps to P1

arising from a fixed curve C and a fixed line bundle L correspond to the choices of two sections(2(d−g+1)), minus 1 to forget the scalar multiple, for a total of 2d−2g+1. If we let the the line

bundle vary, the number of maps from a fixed curve is 2d−2g+1+dim Picd(C) = 2d−g+1. If we

let the curve also vary, we see that the number of degree d genus g covers of P1 is p+ 2d− g + 1 .

But we can also count this number using the Riemann-Hurwitz formula. I’ll need one be-lievable fact: there are a finite number of degree d covers with a given set of branch points. (Inthe complex case, this is believable for the following reason. If C → P1 is a branched cover of P1,branched over p1, . . . , pr, then by discarding the branch points and their preimages, we have anunbranched cover C ′ → P1 − p1, . . . , pr. Then you can check that (i) the original map C → P1

is determined by this map (because C is the normalization of P1 in this function field extensionFF (C′)/FF (P1)), and (ii) there are a finite number of such covers (corresponding to the mon-odromy data around these r points; we have r elements of Sd once we take branch cuts). Thislast step is where the characteristic 0 hypothesis is necessary.)

By the Riemann-Hurwitz formula, for a fixed g and d, the total amount of branching is2g + 2d − 2 (including multiplicity). Thus if the branching happens at no more than 2g + 2d− 2points, and if we have the simplest possible branching at 2g+ 2d− 2 points, the covering curve isgenus g. Thus

p+ 2d− g + 1 = 2g + 2d− 2,

from which p = 3g − 3 .

Thus there is a 3g − 3-dimensional family of genus g curves! (By showing that the spaceof branched covers is reduced and irreducible, we could again “show” that the moduli space isreduced and irreducible.)

25.16.3. Older notes: The first glimmerings of moduli. Two facts in this section:there is a g-dimensional space of divisors on a genus g curve. There are 3g−3 “moduli” of a genusg curves. These arguments are quite informal. Mention Riemann. Someone (Luca Migliorini?)mentioned that Riemann’s argument was more differential geometric, involving cutting up a curve.19 [See my CIME/Cetraro notes.]

Show g-dimensional space of divisors on genus g curve, informally. Method: (1) this isPicard varietyindependent of degree d (so long as we are over degree d curve). (2) by Serre duality, we knowSerre duality, make sure

we already have itthat h1(D) = 0 for degD 0. How many divisors of degree d 0? Answer 1: d. Answer 2: pchoices for divisor. Then d− g+1 choice of section, −1 to projectivize. Hence d− g+ p = d, fromwhich g = p.

25.16.A. Exercise. What happens if we are working over a curve over a field k that is notnecessarily algebraically closed? (Answer: there is a “period”.)

If k is algebraically closed, then there is a parameter space of dimension 3g − 3. Method:moduli space of curveswe need the Riemann-Hurwitz formula. Count degree d branched covers of P1 in two ways. GetRiemann-Hurwitzh+ g + 2(d− g + 1) − 1 = 2d+ 2g − 2 from which h = 3g − 3.

To FOAG:

19glimmeringsofmoduli

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• Show that E satisfies the universal property of Pic0 (Theorem IV.4.11)• Given a genus g curve. Show that there exists a ≤ (g + 1)-to-1 cover of P1.

(Hint: degree 2g − 2 in Pg−1. We get degree g cover of P1.)• Given a plane curve, how many tangents through a given point?• A genus g curve has at most 84(g− 1) automorphisms, following Ex. IV.2.5,

in characteristic 0. This relies on us knowing that a genus g curve has a finitenumber of automorphisms. One argument: no infinitesimal automorphisms= sections of tangent bundle, and also the automorphism group scheme isfinite type over k.

25.16.4. The ring ⊕nΓ(C,O(np)). Handy to consider the ring ⊕∞n=0H0(C,O(np)).

H0(O)∼ // H0(O(p)) // H0(O(2p)) // · · ·

By RR, for n > 0, H0(O(np)) = n.We write down bases. H0(O): 1.n = 1: z. We get multiplication by z. Order of vanishing at p is 1.n = 2. z2 plus a new one: y. Order of vanishing is 2 and 0 respectively. (Interpret

in terms of double cover.)n = 3. z3, yz, x, degp x = 0.

n = 4. z4, yz2, xz, y2. That’s it.n = 5. z5, yz3, xz, y2z, xy. We have a basis.

n = 6. Multiply them by z. Add x2 and y3. Both vanish to degree 0. Hence afterscaling x or y, we get x2 = y3 + · · · .

25.17 Place later

Have them prove Riemann-Roch for vector bundles on a smooth projective curve. Define thedegree of a vector bundle on a curve as the degree of the determinant of the bundle. This behaves degree of a vector bun-

dle on a smooth projec-

tive curve

well in exact sequences by Exercise 17.6.H. The right side of Riemann-Roch behaves well in exactsequence, as does the left. Show the result for line bundles, and show that any vector bundle canbe filtered.

Do Riemann-Roch for coherent sheaves on a projective curve as follows. Use the direct sumdecomposition into torsion plus locally free (done in notes). Define the degree of a coherentsheaf. This will require classification of torsion sheaves on DVR’s, which should also be doneearlier. Consequence: as euler characteristics behave well in exact sequences, it follows thatdegree behaves well in exact sequence. At least mention the fact that det makes sense, and thatthis behaves well in exact sequence, and the degree is just a shadow of that. Also I’d like tomention this fact for arbitrary schemes, at least with the finite resolution property (any coherentsheaf has a finite locally free resolution), which holds for nonsingular quasiprojective things overa field.

Fun random consequence of Riemann-Roch for vector bundles: if π : C ′ → C is an evendegree map of genus 0 curves over a field k, then C is rational over k. Method: π∗OC′ is then anodd degree vector bundle, from which we can find a degree 1 vector bundle.

][primordial:

CHAPTER 26

Base-point-free, ampleness, very ampleness

[This will move earlier in the notes! EGA.II p. 78 is very ample. p. 83 is ample. In short:

very ample: relative notion. if projective morphism, and pullback of O(1). EGA.II.4.4.

ample: global notion; cohomological; algebraic; Serre vanishing; “positive” [“negative”] polarization.

EGA II.4.5 (“Polarization”: in Kleiman, it is the choice of an ample divisor. §is on polarization.)

relatively ample: ample over every affine. f -ample. EGA II.4.6] [immersion = isomorphism to an open

subscheme of a closed subscheme.]

My goal is to discuss properties of invertible sheaves on schemes (an “absolute” notion), andproperties of invertible sheaves on a scheme with a morphism to another scheme (a “relative”notion, meaning that it makes sense in families). The notions fit into this table:

absolute relative

base-point-free relatively base-point-freeample relatively amplevery ample over a ring very ample

This is admittedly horrible terminology. Warning: my definitions may have some additionalhypotheses not used in EGA. The additional hypotheses exclude some nasty behavior which tendsnot to come up in nature; indeed, I have only seen these notions used in the circumstances inwhich I will describe them. In particular, for proper/projective things in my exposition[but maybe will say more]. Also, quasicompact quasiseparated morphisms. There arevery few facts to know, and there is fairly little to prove.

26.0.1. Definition of base-point-free and relative base-point-free (review [ref ]). Recall that if Fis a quasicoherent sheaf on a scheme X, then F is generated by global sections if for any x ∈ X,the global sections generate the stalk Fx. Equivalently: F is the quotient of a free sheaf. Latermake sure this equivalence is given. If F is a finite type quasicoherent sheaf, then we justneed to check that for any x, the global sections generate the fiber of F , by Nakayama’s lemma.If furthermore F is invertible, we need only check that for any x there is a global section notvanishing there. In the case where F is invertible, we give “generated by global sections” a specialname: base-point-free. bpf, generated

26.0.A. Exercise (globally generated ⊗ globally generated = globally generated forfinite type sheaves). Suppose F and G are finite type sheaves on a scheme X that are generatedby global sections. Show that F⊗G is also generated by global sections. In particular, if L andMare invertible sheaves on a scheme X, and both L and M are base-point-free, then so is L ⊗M.(This is often summarized as “base-point-free + base-point-free = base-point-free”. The symbols+ is used rather than ⊗, because Pic is an abelian group.) [Is this true for quasicoherentsheaves in general? In fact we don’t need finite type.] [Jason says: (F ⊗ G)x = Fx ⊗ Gxshould give it to you without finite type assumption, with quasicoherent of course.]

If π : X → Y is a morphism of schemes that is quasicompact and quasiseparated (so pushfor-wards of quasicoherent sheaves are quasicoherent sheaves [ref ]), and F is a quasicoherent sheaf onX, we say that F is relatively generated [caution: used to be called “relatively generated relatively genby global sections”] if π∗π∗F → F is a surjection of sheaves. As this is a morphism of quasi-coherent sheaves, this can be checked over any affine open subset of the target, and correspondsto “generated by global sections” above each affine. In particular, this notion is affine-local onthe target. [Possibly we can drop those hypotheses. They may not be in EGA.] If F islocally free [or invertible?], this notion is called relatively base-point-free. relatively base-point-

free[I should check that if the target is affine, then relatively base-point-free is the same asbase-point-free, and also for generated. Reason: it means surjectivity of p∗p∗F → F .]

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26.0.2. Definition of very ample. Suppose X → Y is a projective morphism. Then X = ProjS•for some graded algebra, locally generated in degree 1; given this description, X comes with O(1).Then any invertible sheaf [of this sort] on X is said to be very ample (for the morphism π). Thenotion of very ample is local on the base. (This is “better” than the notion of projective, whichisn’t local on the base, as we’ve seen [ref ]. Recall why: a morphism is projective if there exists anO(1). Thus a morphism X → Y ∪ Y ′ could be projective over Y and over Y ′, but not projectiveover Y ∪ Y ′, as the “O(1)” above Y need not be the same as the “O(1)” above Y ′. On the otherhand, the notion is “very ample” is precisely the data of “an O(1)”.) You’ll recall Exercise 21.3.Bthat given such an invertible sheaf, then X = Projπ∗L⊗n, where the algebra on the right has the

desired form. (It isn’t necessarily the same graded algebra as you originally used to construct X.)Notational remark: If Y is implicit, it is often omitted from the terminology. For example,

if X is a complex projective scheme, the phrase “L is very ample on X” often means that “L isvery ample for the structure morphism X → Spec C”.

[[[Correction: According to both EGA.4.4.2 (and, roughly, Hartshorne), it should be that X should

be an open immersion into a Proj over Y . In particular, this implies finite type and separated. I have only

ever seen this as useful when the open immersion is the identity. [Example: affine finite type morphisms

over affines: the structure sheaf is very ample.] Remark: if π is quasicompact, then very ample is

equivalent to (i) π∗L quasicoherent, (ii) surjectivity of the canonical homomorphism π∗π∗L → L, and

the induced X → Proj Sym∗(π∗L) is a locally closed immersion (EGA Prop. II.4.4.4). I’m not keen on

this, because this Proj could be not of finite type. This really messes me up. But at least then if π is

quasicompact, then this notion is local on the base (EGA Cor. II.4.4.5).]]]

26.0.B. Exercise (very ample + very ample = very ample). If L and M are invertiblesheaves on a scheme X, and both L and M are base-point-free, then so is L ⊗M. Hint: Segre.SegreIn particular, tensor powers of a very ample invertible sheaf are very ample. 1

26.0.C. Tricky exercise+ (very ample + relatively generated = very ample). 2 SupposeL is very ample, andM is relatively generated, both on X → Y . Show that L⊗M is very ample.(Hint: Reduce to the case where the target is affine. L induces a map to PnA, and this correspondsto n + 1 sections s0, . . . , sn of L. We also have a finite number m of sections t1, . . . , tm of Mwhich generate the stalks. Consider the (n + 1)m sections of L ⊗M given by sitj . Show that

these sections are base-point-free, and hence induce a morphism to P(n+1)m−1. Show that it is aclosed immersion.) Say all this better. [Look away from where tj vanishes. We get amap to a smaller projective space, and we get a closed immersion there. Then usethe Cancellation Theorem 12.1.19 to show that we get a closed immersion from thefirst part.] [Or: separates points and tangent vectors.]

26.0.3. Definition of ample and relatively ample. Suppose X is a quasicompact scheme. Later(rel) amplesee why we need this. We say an invertible sheaf L on X is ample if for all finite type sheavesF , F ⊗Ln is generated by global sections for n 0. (“After finite twist, it is generated by globalsections.”) This is an absolute notion, not depending on a morphism. 3 [Hartshorne p. 153.EGA ample: X quasicompact, Xf where f ∈ Γ(L⊗n) form a base for the topology;equivalent by EGA II.4.5.5.] Exercise (which might be useful ACTUALLY NOT). Consider L on

X. Show that TFAE. (a) L ample. (b) As f runs through sections of Γ(X,L⊗n) (as n varies), the Xf

form a base for the topology. (c) There is a cover by some Xf that are affine.

26.0.4. Example. (a) If X is an affine scheme, and L is any invertible sheaf on X, then L isample. 4

(b) If X → SpecB is a projective morphism and L is a very ample invertible sheaf on X, then Lis ample (by Serre vanishing 23.1.2). [We may need B Noetherian here.] 5

We now give the relative version of this notion. Suppose π : X → Y is a morphism, and Lis an invertible sheaf on X. Suppose that for every affine open subset SpecB of Y there is an n0

such that F ⊗L⊗n restricted to the preimage of SpecB is relatively generated by global sections

1H.E.II.5.12a, EGA Prop. II.4.4.9(ii). EGA.III.4.4.8.

2H.E.II.7.5d

3d:ample

4affineprojample

5evaa

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for n ≥ n0. (In particular, π is quasicompact and quasiseparated — that was a hypothesisfor relatively generated.) Then we say that L is relatively ample (with respect to π; although relatively amplethe reference to the morphism is often suppressed when it is clear from the context). It is alsosometimes called π-ample. Warning: the n0 depends on the affine open; we may not be able totake a single n0 for all affine open sets. We can, however, if Y is quasicompact, and hence we’llsee this quasicompactness hypothesis on Y often.

[Caution: EGA has a more expansive definition, that applies when π is notquasiseparated.] relatively ample, π-

ample, ample, very

ample

[Kirsten seemed to suggest that I had two definitions in the notes. But her two definitionsseemed the same. See kirsten20.5.pdf.]

Example. The examples of 26.0.4 naturally generalize.(a) If X → Y is an affine morphism, and L is any invertible sheaf on X, then L is relatively ample.(b) If X → Y is a projective morphism and L is a very ample invertible sheaf on X, then L isrelatively ample. [We may need Y locally Noetherian here.]

26.0.5. Easy Lemma. — Fix a positive integer n.6

(a) If L is an invertible sheaf on a scheme X, then L is ample if and only if L⊗n is ample.(b) If π : X → Y is a morphism, and L is an invertible sheaf on X, then L is relatively

ample if and only if L⊗n is relatively ample.

In general, statements about ample sheaves (such as (a) above) will have immediate analoguesfor statements about relatively ample sheaves where the target is quasicompact (such as (b)above), and I won’t spell them out in the future. [What did I mean about (b)?! There isno quasicompact target hypothesis there!]

Proof. We prove (a); (b) is then immediate.Suppose L is ample. Then for any finite type sheaf F on X, there is some m0 such that

for m ≥ m0, F ⊗ L⊗m0 is generated by global sections. Thus for m′ ≥ m0/n, F ⊗ (L⊗n)m′

isgenerated by global sections, so L⊗n is ample.

Suppose next that L⊗n is ample, and let F by any finite type sheaf. Then there is some m0

such that (F)⊗ (Ln)m, (F ⊗L)⊗ (Ln)m, (F ⊗L⊗2)⊗ (Ln)m, . . . (F ⊗L⊗(m−1))⊗ (Ln)m, are all

generated by global sections for m ≥ m0. In other words, for m′ ≥ nm0, F ⊗ L⊗m′is generated

by global sections. Hence L is ample.

Example: any positive degree invertible sheaf on a curve is ample. Reason: a high tensorpower (such that the degree is at least 2g + 1) is very ample give ref.

26.0.6. Proposition. — In each of the following, X is a scheme, L is an ample invertiblesheaf (hence X is quasicompact), and M is an invertible sheaf.

(a) (ample + generated = ample) If M is generated by global sections, then L ⊗M isample. 7

(b) (ample + ample = ample) If M is ample, then L⊗M is ample. 8

Similar statements hold for quasicompact and quasiseparated morphisms and relatively ample andrelatively generated.

Proof. (a) Suppose F is any finite type sheaf. Then by ampleness of L, there is an n0 such thatfor n ≥ n0, F ⊗ L⊗n is generated by global sections. Hence F ⊗ L⊗n ⊗M⊗n is generated byglobal sections. Thus there is an n0 such that for n ≥ n0, F ⊗ (L⊗M)⊗n is generated by globalsections. Hence L⊗M is ample.

(b) As M is ample, M⊗n is base-point-free for some n > 0. But L⊗n is ample, so by (a)(L⊗M)⊗n is ample, so by Lemma 26.0.5, L⊗M is ample.

6multipleample; H.P.II.7.5; (a) is EGA.II.4.5.6(i)

7H.E.II.7.5a

8H.E.II.7.5c, EGA II.4.5.7.

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26.1 Every ample on a proper has a tensor power that isvery ample

We’ll spend the rest of our discussion of ampleness considering consequences of the followingvery useful result.

26.1.1. Theorem. — Suppose π : X → Y is proper and Y = SpecB is affine. If L is ample,then some tensor power of L is very ample. 9 EGA.II.4.5.6. [Weaker conclusion withoutproper hypotheses: ample + generated EGA II.4.5.8.] [H.T.II.7.6. jazzed up.]

The converse follows from our earlier discussion, that very ample implies ample, Exam-ple 26.0.4(b).

Proof. I hope to type in a short proof at some point. For now, I’ll content myself with referringto Hartshorne Theorem II.7.6. (He has more hypotheses, but his argument essentially applies inthis more general situation.)

Proof. Replace L by a suitable tensor power that is base-point-free. Then we havesections s1, . . . , sn such that the loci where each one doesn’t vanish (call them Xs1 ,. . . , Xsn) cover X.

Quick reduction: We have n sections of L so that Xsi ’s cover X. Each is finitelygenerated B-algebra. Choose generators bij of them. Then hit by large enoughpowers of si so that they extend to sections cij of some power of L. Rescale L again,so that these are all sections of L. Use si’s and these cij ’s to map to projective space.Then where xi 6= 0, where sni 6= 0, we get that Xsi is a closed immersion into that affinespace. Thus we have just described X as a closed immersion into an open immersion.But its image is closed, hence we get a closed immersion into everything.

26.1.A. Exercise. Suppose π : X → Y is proper and Y is quasicompact. Show that if L isrelatively ample on X, then some tensor power of L is very ample. [Give an example to show thatthe result is not true without the quasicompact hypothesis.]

Serre vanishing ?? holds for any relatively ample invertible sheaf for a proper morphism to aNoetherian base. More precisely:

26.1.2. Corollary (Serre vanishing, take two). — Suppose π : X → Y is a propermorphism, Y is quasicompact, and L is a π-ample invertible sheaf on X. Then for any coherentsheaf F on X, for m 0, Riπ∗F ⊗ L⊗m = 0 for all i > 0. 10Serre vanishing

Proof. By Theorem 26.1.1, L⊗n very ample for some n, so π is projective. Apply Serre vanishingto F ⊗L⊗i for 0 ≤ i < n.

The converse holds, i.e. this in fact characterizes ampleness. For convenience, we state it forthe case of an affine target. [Do we need proper for this?]

26.1.3. Theorem (Serre’s criterion for ampleness). — Suppose that π : X → Y = SpecBis a proper morphism, and L is an invertible sheaf on X such that for any finite type sheaf F onX, F ⊗L⊗n is generated by global sections for n 0. Then L is ample. 11 [Serre’s criterion foramplitude]

Essentially the same statement holds for relatively ample and quasicompact target. Exercise.Give and prove the statement. WHOOPS! Serre’s criterion for ampleness is my definition

of ampleness, see Definition 26.0.3. Thus this exercise is bogus, and I should figureout what Hartshorne is actually proving in Proposition III.5.3.

My cliff’s notes, I think on Hartshorne Prop. III.5.3: we’ll show that for eachcoherent sheaf F on X, F ⊗ L⊗n is generated by global sections for all n.

9multample

10Serrevanishingample

11H.P.III.5.3

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First, let me do the case F = OX . Fix a point p. For n ≥ n0, we can find sectionof L⊗n not vanishing at x [explain]. Then they don’t vanish in a neighborhood. Findsections for n = n0, . . . , n = 2n0 − 1. Take intersection of all these open sets. Wehave an open set where for n > n0 this works. Now do this repeatedly, and usequasicompactness.

Next, let’s do the general case, with the same trick. For n ≥ n1, we can find asection of F ⊗L⊗n not vanishing at p. Go from n1 to n1 + n0. Then take intersectionof those open sets.

26.1.B. Exercise. Use Serre’s criterion for ampleness to prove that the pullback of ample sheafon a projective scheme by a finite morphism is ample. Hence if a base-point-free invertible sheafon a proper scheme induces a morphism to projective space that is finite onto its image, then itis is ample. [The same argument will show that the pullback of a relatively ample invertible sheafon a projective scheme under a finite map is also relatively ample.]

26.1.4. Theorem. — (a) π : X → SpecB, L ample, M invertible. Then L⊗n ⊗ M is ample and

generated for n 0. (b) If moreover π is proper, then L⊗n⊗M is very ample for n 0. [H.E.II.7.5b,

jazzed up. for projective: ample = all sufficient large multiples are very ample — H.E.II.7.5e] Ditto for

relatively ample over a quasicompact base. Proof of (a) L⊗n′⊗ M generated. for n′ 0. L⊗n′′

is

generated and ample for n′′ 0. Tensor these two together.

26.1.5. Key Corollary. — Suppose π : X → SpecB is proper, and L and M are invertiblesheaves on X with L ample. Then L⊗n ⊗M is very ample for n 0. 12 [H.E.II.7.5b,jazzed up. for projective: ample = all sufficient large multiples are very ample —H.E.II.7.5e]

[Proof/hint: va + bpf = va.]

26.1.C. Exercise. Give and prove the corresponding statement for a relatively ample invertiblesheaf over a quasicompact base.

Proof. The theorem says that L⊗n is very ample for n 0. By the definition of ampleness,L⊗n ⊗M is generated for n 0. Tensor these together, using the above.

A key implication of the key corollary is:

26.1.6. Corollary. — Any invertible sheaf on a projective X → SpecB is a difference of twovery ample invertible sheaves. 13

Proof. If M is any invertible sheaf, choose L very ample. Corollary 26.1.5 states that M⊗L⊗nis very ample. As L⊗n is very ample (Exercise 26.0.B), we can write M as the difference of twovery ample sheaves: M∼= (M⊗L⊗n)⊗ (L⊗n)∨.

As always, we get a similar statement for relatively ample sheaves over a quasicompact base.Here are two interesting consequences of Corollary 26.1.6.

26.1.D. Exercise. Suppose X a projective A-scheme, where A is a Noetherian ring. Show thatevery invertible sheaf is the difference of two effective Cartier divisors. Thus the groupificationof the semigroup of effective Cartier divisors is the Picard group. Hence if you want to provesomething about Cartier divisors on such a thing, you can study effective Cartier divisors. [Ican’t prove this. I discussed this with Kirsten and Joe. Here is the upshot. I think we can dothis when X is irreducible, and A is general. More relevant to us: we can do the case whereA is a field k. If k is infinite, we’re okay. If k is finite, of order q, we give a different proof.Say we have n associated points p1, . . . , pn, in PN . For each I ⊂ 1, . . . , n, suppose the k-vector space of degree d polynomials vanishing at pi for i ∈ I is fdI . By inclusion-exclusion, the

number of degree d polynomials vanishing on none of the pi is IE =PI (−1)|I|qf

dI . We wish

to show that this is positive for d 0. Now for d 0, if XI is the scheme-theoretic closure

12ava

13ivava

444

of the pi, fdI is a`N+dN

´− gI(d) where gI(d) = h0(XI ,O(d)) is a positive degree polynomial in

d. Then IE = q

“

N+dN

”PI(−1)|I|q−gI (d). Let’s first discuss the case where each pi has positive

dimensional Zariski closure, so each gI(d) has positive degree. Then all but the first term go tozero. Next, suppose the first m points have zero-dimensional Zariski closure, and the rest havepositive dimensional Zariski closure. Suppose degk pi = βi for 1 ≤ i ≤ m. Then the terms ofPI⊂1,...,m(−1)|I|q−gI(d). For such I, gI(d) =

Pi∈I βi. Thus these terms of q

−“

N+dN

”

IE that

don’t go to zero can be rewritten asQmi=1(1−q−βi) > 0. This latter fraction is very believable: the

chance that a polynomial vanishes at pi is precisely q−βi , and these are independent (as d→∞).This is a very neat proof! It treats finite fields differently, and the dimension 0 very differently.

My original argument should work for Noetherian A, except the source must be integral,or possibly it can have two associated points, and there is a little combinatorial issue to see ifthere can be more — for example, you can’t find two associated points such at every hyperplanecontains at least one of them. What about 3? 4? Some of this depends on characteristic. Hmm!

I point out that we need this problem to prove Riemann-Roch for curves over a field!] [SeeKirsten’s write-up kirsten20.9.pdf.]

(This is false if projective is replaced by proper — ask Sam Payne for an example.)I stated this for quasiprojective, but my argument doesn’t quite work. I think I

need the notion of very ample for quasiprojective things! It might be patchable, ifwe know we can extend line bundles on X to coherent sheaves on X.

26.1.E. Important exercise. Suppose C is a generically reduced projective k-curve. Then wecan define degree of an invertible sheaf M as follows. Show that M has a meromorphic sectionthat is regular at every singular point of C. Thus our old definition (number of zeros minusnumber of poles, using facts about discrete valuation rings) applies. Prove the Riemann-Rochtheorem for generically reduced projective curves. (Hint: our original proof [ref!] essentially willcarry through without change.) [Make sure to say this is over a field!]

Exercise: L ample, n 0, L⊗ separates k-jets.[Suppose we knew that for n 0, then L⊗n has no higher cohomology. Then Serre vanishing

applies; the old argument works without change.]

26.2 ? Other criteria for ampleness / other adjectives fordivisors / other vanishing theorems / Properties of line

bundles / Cartier divisors

[This section needs to be cleaned up a lot.] [I’m not talking about vanishingtheorems. The reason we want them is because of the geometric significance of theglobal sections of a sheaf, and the computability of the Euler characteristic.]

In this section, X is projective over an algebraically closed field.We’ve described Serre’s criterion for ampleness. I’ll now discuss Kleiman’s criterion for

ampleness (as described in his paper “Toward a numerical theory of ampleness”, which appearedin Annals). You certainly know enough at this point to read the paper easily.

We can make sense of the multiplicity of an invertible sheaf along a projective curves. (Tech-nically speaking, we’ve only dealt with the case of generically reduced curves, and that’s all weneed in what follows. But it extends to the nonreduced case.)

We will consider an equivalence relation on invertible sheaves (hence on divisors). Two invert-ible sheaves are numerically equivalent if they meet each integral curve with the same multiplicity.Then the group of invertible sheaves modulo numerical equivalence is a finite free Z-module (i.e.isomorphic to Zn for some n). [This is proved in Kleiman Chapter 5, by reduction tothe case of a nonsingular surface, which is known by Lang-Neron (he calls this thetheorem of the base, or the Neron-Severi theorem).] This group is called the Neron-Severi

group of X, and is denoted NS(X). 14 The rank n of NS(X) is called the Picard number of V .Neron-Severi

Picard number We define the group of curves on X as a formal sum of integral curves, by which we mean asum

PnCC as C runs over all integral curves in X, and nC = 0 ∈ Z, with nC = 0 for all but

14neronseveri

445

finitely many C. We say that such a formal sum is effective if nC ≥ 0 for all C. If each nC is0 or 1, there is no harm in interpreting this as the union of those C such that nC = 1. We saythat two such sums are numerically equivalent if they meet every Cartier divisor with the samemultiplicity.

We thus have a pairing from Cartier divisors on X and formal sums of curves on X tothe integers. This pairing is additive in both variables: (L, C + C ′) = (L, C) + (L, C′) and(L ⊗ L′, C) = (L, C) = (L′, C). This obviously descends to a pairing of the Neron-Severi groupwith formal sums of curves modulo numerical equivalence. (I think of the invertible sheaves asalgebraic codimension 1 cohomology classes. Notice that they pull back under morphismsX ′ → X.I think of the formal sums of curves as algebraic dimension 1 homology classes. Notice that theypush forward under morphisms X → X ′′.)

Kleiman proves [Proposition 1 (p. 31)] that this intersection pairing is perfect.The ample divisors in NS(X) form a cone, called (unsurprisingly) the ample cone. Here

roughly is why: ample ⊗ ample is ample. Moreover, it is a full-dimensional cone: fix an amplesheaf L. Then given any invertible sheaf, if you twist by enough copies of L, you will end up withan ample bundle. Moreover, this cone contains no lines: we cannot have both L and L∗ be ample,or else there is some multiple of L such that both it and its dual are very ample; but then theyboth have sections, and hence (by an earlier result [ref ]) they are both trivial. But the only waythe trivial bundle could be very ample on an irreducible X is if X is a point. (Note: it is not yet

obvious from what I’ve told you that if L and L′ are numerically equivalent, and L is ample, thenL′ is ample too! But it is true.) ample cone

A Cartier divisor is numerically effective, or nef if it meets every curve with non-negativedegree. Then nef divisors clearly form a cone: the nef cone. Note also that an ample divisor isnef: it is true for very ample divisors, and if it is true for L⊗n, then it is true for L. Thus theample cone is contained in the nef cone. Kleiman’s criterion for ampleness for ampleness statesthat the ample cone is the interior of the nef cone. This is easy to remember, and makes it easyto remember that (as stated earlier) ampleness depends only on numerical criteria; the tensorproduct of two ample invertible sheaves is ample; and the pullback of an ample invertible sheafunder a finite morphism is ample. Also, (i) if L is an ample invertible sheaf X if and only if“Lred” (the restriction of L to Xred) is ample on Xred,[Who cares? But use Serre. Better:devissage. Mention that very ample doesn’t work this way] 15 and (ii) “ample + nef =ample”: if L is ample and M is nef, then L ⊗M is ample. nef

Kl criterion26.2.A. Exercise. Verify Kleiman’s criterion in the case where X is a curve.

26.2.B. Exercise. Show that there is no numerical criterion for very ampleness. (Hint: find acounterexample for curves.)

[Remark: Example 2 on p. 35 of Kleiman gives an example of Mumford, wherethere is a non-ample divisor that meets every curve positively. V is a general ruledsurface of even type (?) with unisecant C (?) of high genus over a field of charac-

teristic 0. E is a ruling. D = C − 12(C2)E meets every curve X positively (why?), but

D2 = 0.]SKIP THIS. An easier numerical criterion is Nakai’s criterion, which is also

discussed in Kleiman’s paper. 16 [This is also discussed at 23.15.2.] Ample iff Nakai critdeg Y > 0 for every s-dimensional closed integral subscheme. (also in Kleiman’spaper). Note: implies pullback of ample under finite is finite., and red fact. Explainvaguely how to define L⊗nY : moving lemma, or sneakier, using Euler-characteristic.[History: Nakai did the surface case. Grothendieck, Kleiman, Moishezon, Mumford,Nakai, Zariski.]

Special case, for surfaces: Naikai-Moishezon criterion. H.T.V.1.10. Finite type.D2 > 0 and D · C > 0.

By popular demand, we may as well discuss other adjectives for divisors, whichyou may see in talks.

An invertible sheaf is effective if it has a non-zero section, or equivalently (as we are assumingX is integral) if it corresponds to an effective Cartier divisor. Effective divisors also obviouslyform a cone (as the tensor product of two effective invertible sheaves is also effective). We say an eff, big div

15H.E.III.5.7b

16nakai

446

invertible sheaf L (or Cartier divisor) is big if h0(X,L⊗n) grows like as polynomial of dimensionn as n→∞. For example, one can show that ample divisors are big. Big divisors are importantfor two reasons. First, a sufficiently large multiple of them will give a rational map to projectivespace that will map X birationally onto its image. Second, big and nef invertible sheaves are auseful generalization of ample invertible sheaves, and for example satisfy vanishing theorems. Itis a fact that the big divisors are the interior of the effective cone.eff., big

If X is normal, then recall that the Cartier divisors are a subgroup of the Weil divisors. Thenthe above notions, which are a priori about invertible sheaves, and hence about Cartier divisors,can be lifted to notions about Weil divisors. We say that a Weil divisor is Q-Cartier if somemultiple is Cartier. For example, it makes sense to define when a Weil divisor is ample. (In thisQ-Cartiercase, it is not true that a sufficiently large multiple is very ample! Only a sufficiently large Cartiermultiple.)Q-divisor

Remark: Since we can intersect Cartier divisors and curves, we can also intersect Q-Cartierdivisors with curves. Warning: something weird could happen: the result could be nonintegral.Exercise. Use this notion to compute the intersection number of two rulings on a conic. Showthat their intersection number is 1/2.

26.2.1. Things to possibly mention, that I won’t in class..Snapper’s Theorem (from p. 295 of Kleiman’s criterion paper). Let F be a coherent sheaf onSnapper’s Th

V , and s = dimSuppF . Let L1, . . . , Lr be r invertible sheaves on V . then the Euler characteristicχ(F ⊗L⊗n1 ⊗ · · · L⊗nr ) is a polynomial in n1, . . . , nt of total degree s.

Exercise. Prove it as follows. By “devissage”, reduce to the case where V is a variety, andany torsion sheaf on V satisfies this fact. Then work by induction on t.devissage

Then if SuppF ≤ t, define (L1 · · · Lt · F) to be the coefficient of the monomial n1 · · ·nt inthis Euler characteristic. Show that this is an integer. Show that it is additive in L. Show thatit is “additive” in F .

26.3 +Quasiaffine and quasiprojective morphisms

Two new classes of morphisms that we’re not going to use, and aren’t so central. Reasonqa and qproj morphismwe care: many facts we prove about quasiprojective things because we are using projective as acrutch.

26.3.1. Quasiaffine morphisms. EGA II.5.1. A scheme is quasiaffine if there exists aquasicompact open immersion. Note that the quasicompact condition is automatic if you are inthe category of Noetherian schemes. Note that this also implies that X is quasicompact.

Exercise: this is equivalent to OX being ample. [Think this through.] [Here we see that weneed quasicompactness.]

We say that X → Y is a quasiaffine morphism if OX is ample. This (we know) can bechecked on an affine cover of the base. Remark: it doesn’t automatically mean to me that X is aquasicompact open immersion inside an affine.

Note that quasiaffine implies separated.I should show that it is closed under base change. [Do!] It is also closed under composition

[here we really use quasicompactness] by Serre’s criterion for ampleness and the Leray spectralsequence. Then we have a Cancellation Theorem 12.1.19-type result [edit]: If g is quasiseparatedand h is quasiaffine then so is f

Xf //

h @@@

@@@@

Y

g~~

~~~~

~

Z

26.3.2. Quasiprojective morphisms. We say that π : X → Y is quasiprojective, or Xquasiprojective over Y , or for π, if locally on the base X is a quasicompact open immersion intosomething projective.

Hence implies quasicompact, separated, finite type.

447

Examples: projective morphism, affine finite type morphism, more generally quasiaffine finitetype morphism.

Remark: it doesn’t necessarily mean globally as far as I can tell!We can redefine very ample in this setting. Note that this notion is local on the base!Everything works [check!!!] — there are three things to check.

• very ample+va = va.• va + rel gen = va• ample + ft means that some multiple is very ample. (Hence in particular ample+ft

implies separated. Presumably ample implies separated.)

I should check that (i) it is closed under composition (here quasicompactness comes into it),and I may need further hypotheses. Then there is a cancellation theorem fact 12.1.19, which isthe same as above.

]

Part VII

Useful properties andconstructions (not currently

ordered)

[primordial:

CHAPTER 27

? Blowing up a scheme along a closed subscheme

1 2

[Note: we can only blow up a “coherent” subscheme, one whose sheaf of idealsis coherent. If your scheme is Noetherian, there is no issue.] [Make sure they know:If p is an associated point then the scheme theoretic closure of X − p in X is not

X.] [Explain how to make a model. Perhaps use β or βXY for the blow-up mor-phism?] [Make sure that birational and desingularization are both defined at somepoint. Mention resolution of singularities, weak factorization, structure theorem forbirational projective morphisms [H, Thm. 7.17], 27 and possibly alterations.]

Also, add the following exercise. Suppose p1, . . . , . . . , pn on a quasiprojective

k-variety (indeed A-variety). Show that there is an affine open set containing all ofthem. (This observation is useful in the invariant theory of finite groups. I coulddescribe it.)

We’ll next discuss an important construction in algebraic geometry (and especially the geo-metric side of the subject), the blow-up of a scheme along a closed subscheme (cut out by a finitetype ideal sheaf). We’ll start with a motivational example that will give you a picture of theconstruction in a particularly important case (and the historically earliest case), in Section 27.1.I’ll then give a formal definition, in terms of universal property, Section 27.2. This definition won’timmediately have a clear connection to the motivational example! We’ll deduce some consequencesof the definition (assuming that the blow-up actually exists). We’ll prove that the blow-up alwaysexists, by describing it quite explicitly, in Section 27.3. As a consequence, the blow-up morphismis projective, and we’ll deduce more consequences from this. In Section 27.4, we’ll do a numberof explicit computations, and see that in practice, it is possible to compute many things by hand.I’ll then mention a couple of useful facts: (i) the blow-up a nonsingular variety in a nonsingularvariety is still nonsingular, something we’ll have observed in our explicit examples, and (ii) Castel-nuovo’s criterion, that on a smooth surface, “(−1)-curves” (P1’s with normal bundle O(−1)) canbe “blown down”.

27.1 Motivational example

3 We’re going to generalize the following notion, which will correspond to “blowing up” the originof A2

k (over an algebraically closed field k). Because this is just motivation, I’ll be informal.

Consider the subset of A2 × P1 corresponding to the following. We interpret P1 as the linesthrough the origin. Consider the subset (p ∈ A2, [`] ∈ P1) | p ∈ `). [Refer to model.]

I’ll now try to convince you that this is nonsingular (informally). Now P1 is smooth, andfor each point [`] in P1, we have a smooth choice of points on the line `. Thus we are verifyingsmoothness by way of the fibration over P1.

Let’s make this more algebraic. Let x and y be coordinates on A2, and X and Y be projectivecoordinates on P1 (“corresponding” to x and y); we’ll consider the subset Bl(0,0) A2 of A2 × P1

corresponding to xY − yX = 0. We could then verify that this is nonsingular (by looking at twocovering patches).

1blowupnotes

2c:blowup

3blmot

451

452

Notice that the preimage of (0, 0) is a curve and hence a divisor (an effective Cartier divisor,as the blown-up surface is nonsingular). Also, note that if we have some curve singular at theorigin, this could be partially desingularized. (A desingularization or a resolution of singularities

of a variety X is a proper birational morphism X → X from a nonsingular scheme. We areinterested in desingularizations for many reasons. For example, we understand nonsingular curvesquite well [ref ], and we could hope to understand other curves through their desingularizations.This philosophy holds true in higher dimension as well.) For example, the curve y2 = x3 + x2,desing, res of singwhich is nonsingular except for a node at the origin, then we can take the preimage of the curveminus the origin, and take the closure of this locus in the blow-up, and we’ll obtain a nonsingular

curve; the two branches of the node downstairs are separated upstairs. (This will later be anexercise, once we’ve defined things properly. [ref ] The result will be called the proper transformof the curve.)proper transform

Let’s generalize this. First, we can blow up An at the origin (or more informally, “blow upthe origin”), getting a subvariety of An × Pn−1. More algebraically, If x1, . . . , xn are coordinateson An, and X1, . . . , Xn are projective coordinates on Pn−1, then the blow-up Bl~0 An is given by

the equations xiXj − xjXi = 0. Once again, this is smooth: Pn−1 is smooth, and for each point

[`] ∈ Pn−1, we have a smooth choice of p ∈ `.We can extend this further, by blowing up An+m along a coordinate m-plane An by adding

m more variables xn+1, . . . , xn+m to the previous example; we get a subset of An+m × Pn−1.Then intuitively, we could extend this to blowing up a nonsingular subvariety of a nonsingular

variety. We’ll make this more precise. In the course of doing so, we will accidentally generalizethis notion greatly, defining the blow-up of any finite type sheaf of ideals in a scheme. In general,blowing up may not have such an intuitive description as in the case of blowing up somethingnonsingular inside something nonsingular — it does great violence to the scheme — but even then,it is very useful (as we’ll see when we use it to prove dimensional vanishing for quasiprojectiveschemes Theorem 27.3.6). [Caution — this argument is flawed.] The result will be verypowerful, and will touch on many other useful notions in algebra (such as the Rees algebra) thatwe won’t discuss here.Rees algebra

Our description will depend only the closed subscheme being blown up, and not on coordi-nates. That remedies a defect was already present in the first baby example, blowing up the planeat the origin. It is not obvious that if we picked different coordinates for the plane (preservingthe origin as a closed subscheme) that we wouldn’t have two different resulting blow-ups.

As is often the case, there are two ways of understanding this notion, and each is usefulin different circumstances. The first is by universal property, which lets you show some thingswithout any work. The second is an explicit construction, which lets you get your hands dirtyand compute things (and implies for example that the blow-up morphism is projective).

27.2 Blowing up, by universal property

4

I’ll start by defining the blow-up using the universal property. The disadvantage of startinghere is that this definition won’t obviously be the same as the examples I just gave. It won’t evenlook related!

Suppose X → Y is a closed subscheme corresponding to a finite type sheaf of ideals. (If Yis locally Noetherian, the “finite type” hypothesis is automatic, so Noetherian readers can ignoreit.)

The blow-up X → Y is a fiber diagramblow up

EXY //

BlX Y

β

X

// Y

4blowupuniversalproperty

453

such that EXY is an effective Cartier divisor on BlX Y (and is the scheme-theoretical pullbackof X on Y ), such any other such fiber diagram5

(56) D //

W

X

// Y,

where D is an effective Cartier divisor on W , factors uniquely through it:

D //

W

EXY

//

BlX Y

X

// Y.(Recall that an effective Cartier divisor is locally cut out by one equation that is not a zero-divisor;equivalently, it is locally cut out by one equation, and contains no associated points. This latterdescription will prove crucial.) BlX Y is called the blow-up (of Y along X, or of Y with centerX). EXY is called the exceptional divisor. (Bl and β stand for “blow-up”, and E stands for“exceptional”.)

By a universal property argument, if the blow-up exists, it is unique up to unique isomor-phism. (We can even recast this more explicitly in the language of Yoneda’s lemma: consider thecategory of diagrams of the form (56), where morphisms are of the form

D //

W

D′

//

W ′

X

// Y.Then the blow-up is a final object in this category, if one exists.)

If Z → Y is any closed subscheme of Y , then the (scheme-theoretic) pullback β−1Z is calledthe total transform of Z. We will soon see that β is an isomorphism away from X (Observa-

tion 27.2.2). β−1(Z −X) is called the proper transform or strict transform of Z. (We will use thefirst terminology. We will also define it in a more general situation.) We’ll soon see that the propertransform is naturally isomorphic to BlZ∩X Z, where by Z ∩ X we mean the scheme-theoreticintersection (the blow-up closure lemma 27.2.4). 6 blowing up

exceptional divisor

universal property

total, proper=strict

trans

We will soon show that the blow-up always exists, and describe it explicitly. But first, wemake a series of observations, assuming that the blow up exists.

27.2.1. Observation. If X is the empty set, then BlX Y = Y .7 More generally, if X is aCartier divisor, then the blow-up is an isomorphism. (Reason: idY : Y → Y satisfies the universalproperty.)

27.2.A. Exercise. If U is an open subset of Y , then BlU∩X U ∼= β−1(U), where β : BlX Y → Y

is the blow-up. (Hint: show β−1(U) satisfies the universal property!) 8

Thus “we can compute the blow-up locally.”

27.2.B. Exercise. Show that if Yα is an open cover of Y (as α runs over some index set), andthe blow-up of Yα along X ∩ Yα exists, then the blow-up of Y along X exists.

5blcategory

6ptdef1

7blempty

8bllocal

454

27.2.2. Observation. Combining Observation 27.2.1 and Exercise 27.2.A, we see that theblow-up is an isomorphism away from the locus you are blowing up:

β|BlX Y−EXY : BlX Y −EXY → Y −Xis an isomorphism. 9

27.2.3. Observation. If X = Y , then the blow-up is the empty set: the only map W → Ysuch that the pullback of X is a Cartier divisor is ∅ → Y . In this case we have “blown Y out ofexistence”!

27.2.C. Exercise (blow-up preserves irreducibility and reducedness). Show that if Y isirreducible, and X doesn’t contain the generic point of Y , then BlX Y is irreducible. Show thatif Y is reduced, then BlX Y is reduced.10 [Give this after the blow-up closure lemma? Kirstenproved it directly.]

The following blow-up closure lemma is useful in several ways. At first, it is confusing tolook at, but once you look closely you’ll realize that it is not so unreasonable.

Suppose we have a fibered diagram

W cl. imm.//

Z

X

cl. imm.// Ywhere the bottom closed immersion corresponds to a finite type ideal sheaf (and hence the upperclosed immersion does too). The first time you read this, it may be helpful to consider the specialcase where Z → Y is a closed immersion.

Then take the fibered product of this square by the blow-up β : BlX Y → Y , to obtain

Z ×Y EXY //

Z ×Y BlX Y

EXY

Cartier // BlX Y.

The bottom closed immersion is locally cut out by one equation, and thus the same is true of thetop closed immersion as well. However, it need not be a non-zero-divisor, and thus the top closedimmersion is not necessarily an effective Cartier divisor.

Let Z be the scheme-theoretic closure of Z ×Y BlX Y −W ×Y BlX Y in Z ×Y BlX Y . Notethat in the special case where Z → Y is a closed immersion, Z is the proper transform, as definedin §27.2. For this reason, it is reasonable to call Z the proper transform of Z even if Z isn’t aclosed immersion. Similarly, it is reasonable to call Z×Z BlX Y the total transform even if Z isn’ta closed immersion.

Define EZ → Z as the pullback of EXY to Z, i.e. by the fibered diagram

EZ //

_

cl. imm.

Z _

cl. imm.

proper transform

Z ×Y EXY //

Z ×Y BlX Y

total transform

EXY Cartier // BlX Y.

Note that EZ is Cartier on Z (as it is locally the zero-scheme of a single function that does not

vanish on any associated points of Z).

27.2.4. Blow-up closure lemma. — (BlZW,EZW ) is canonically isomorphic to (Z,EZ).11

This is very handy.

9P.H.II.7.13b

10blred

11blclosure

455

The first three comments apply to the special case where Z →W is a closed immersion, andthe fourth basically tells us we shouldn’t have concentrated on this special case.

(1) First, note that if Z → Y is a closed immersion, then this states that the proper transform(as defined in §27.2) is the blow-up of Z along the scheme-theoretic intersection W = X ∩ Z.

(2) In particular, it lets you actually compute blow-ups, and we’ll do lots of examples soon.For example, suppose C is a plane curve, singular at a point p, and we want to blow up C atp. Then we could instead blow up the plane at p (which we have already described how to do,even if we haven’t yet proved that it satisfies the universal property of blowing up), and then takethe scheme-theoretic closure of C − p in the blow-up. [I believe this is discussed in Eisenbud andHarris.]

(3) More generally, ifW is some nasty subscheme of Z that we wanted to blow-up, and Z werea finite type k-scheme, then the same trick would work. We could work locally (Exercise 27.2.A),so we may assume that Z is affine. If W is cut out by r equations f1, . . . , fr ∈ Γ(OZ), thencomplete the f ’s to a generating set f1, . . . , fn of Γ(OZ). This gives a closed immersion Y → An

such that W is the scheme-theoretic intersection of Y with a coordinate linear space Ar .

27.2.5. (4) Most generally still, this reduces the existence of the blow-up to a specific special case.(If you prefer to work over a fixed field k, feel free to replace Z by k in this discussion.) Supposethat for each n, Bl(x1,...,xn) Spec Z[x1, . . . , xn] exists. Then I claim that the blow-up always exists.

Here’s why. We may assume that Y is affine, say SpecB, and X = SpecB/(f1 , . . . , fn). Then wehave a morphism Y → AnZ given by xi 7→ fi, such that X is the scheme-theoretic pullback of the

origin. Hence by the blow-up closure lemma, BlX Y exists.12

27.2.D. Tricky Exercise+. Prove the blow-up closure lemma. Hint: obviously, construct mapsin both directions, using the universal property. The following diagram may or may not help.

EZ

zzvvvvvvvvvvvvvvvvvvvvvv

Cartier //55

uu

_

cl. imm.

Z55

uu

zzvvvvvvvvvvvvvvvvvvvvvvv _

cl. imm.

EWZ Cartier//

BlW Z

W

//

$$III

IIIII

IIZ

$$IIIIIIIIII EXY

Cartier//

zzuuuuuuuuuBlX Y

zzuuuuuuuuu

X // Y

Cut in class notes. [By the universal property of blow-up of BlW Z, there is a unique map

(Z,EZ)→ (BlW Z,EWZ). By the universal property of blow-up of BlX Y , there is a unique mor-phism (BlW Z,EWZ)→ (BlX Y,EXY ), and hence (BlW Z,EWZ)→ (Z×Y BlX Y,W ×X EXY ),

and hence to (BlW Z,EWZ) → (Z,EZ). The compositions in both directions are isomorphismson dense open subsets not containing any associated points. Hence both compositions are isomor-phisms.] This diagram is from blowup.tex. HARD, but details not discussed

EWZ

Cartier

44

444

444

44

444

44EZ

Cartier

cl. imm.// W ×X EXY

cl. imm.

// EXY

Cartier

BlW Z

555

5555

5555

5555

Zcl. imm.// Z ×Y BlX Y

// BlX Y

W

cl. imm.

// X

cl. imm.

Z // Y

12blAn

456

[If you blow up no generic points, the blow-up is surjective. Who cares?]

27.2.E. Exercise. If Y and Z are closed subschemes of a given scheme X, show that BlY Y ∪Z ∼=BlY ∩Z Z. (In particular, if you blow up a scheme along an irreducible component, the irreduciblecomponent is blown out of existence.) [Ziyu: We use the blow-up closure lemma twice. Denoteβ : BlY Y ∪ Z → Y ∪ Z.

Y

cl. imm.// Y ∪ Z

Y

cl. imm.// Y ∪ Zso BlY Y ∪ Z ∼= β−1(Y ∪ Z − Y ). Similarly,

Y ∩ Z

cl. imm. // Z

Y

cl. imm.// Y ∪ Z

so BlY ∩Z Z ∼= β−1(Z − Y ∩ Z). But Y ∪ Z − Y = Z − Y ∩ Z.]

27.3 The blow-up exists, and is projective

13It is now time to show that the blow up always exists. I’ll give two arguments, becauseI find them enlightening in two different ways. Both will imply that the blow-up morphism isprojective. Hence the blow-up morphism is projective, hence quasicompact, proper, finite type,separated. In particular, if Y → Z is projective (resp. quasiprojective, quasicompact, proper,finite type, separated), so is BlX Y → Z. The blow-up of a k-variety is a k-variety (using the factthat irreducibility, reducedness are preserved, Exercise 27.2.C). 14

Approach 1. As explained above (§27.2.5), it suffices to show that Bl(x1,...,xn) Spec Z[x1, . . . , xn]

exists. But we know what it is supposed to be: the locus in Spec Z[x1, . . . , xn]×ProjZ[X1, . . . ,Xn]such that xiXj − xjXi = 0. We’ll show this soon.Approach 2. We can describe the blow-up all at once as a Proj.

27.3.1. Theorem (Proj description of the blow-up). — Suppose X → Y is a closedsubscheme cut out by a finite type sheaf of ideals I →OY . Then

Proj`OY ⊕ I ⊕ I2 ⊕ I3 ⊕ · · ·

´→ Y

satisfies the universal property of blowing up. [Make sure that at some point we check thatI2 makes sense.] 15

We’ll prove this soon (Section 27.3.2), after seeing what this gives us. (The reason we hada finite type requirement is that I wanted this Proj to exist; we needed the sheaf of algebras to

satisfy the conditions stated earlier.)But first, we should make sure that the preimage of X is indeed an effective Cartier divisor.

We can work affine-locally (Exercise 27.2.A), so I’ll assume that Y = SpecB, and X is cut out bythe finitely generated ideal I. Then

BlX Y = Proj`B ⊕ I ⊕ I2 ⊕ · · ·

´.

(We are slightly abusing notation by using the notation BlX Y , as we haven’t yet shown that thissatisfies the universal property. But I hope that by now you trust me.)

The preimage ofX isn’t just any effective Cartier divisor; it corresponds to the invertible sheafO(1) on this Proj. Indeed, O(1) corresponds to taking our graded ring, chopping off the bottom

13blexists

14H.P.II.7.16

15blun

457

piece, and sliding all the graded pieces to the left by 1; it is the invertible sheaf corresponding tothe graded module

I ⊕ I2 ⊕ I3 ⊕ · · ·(where that first summand I has grading 0). But this can be interpreted as the scheme-theoreticpullback of X, which corresponds to the ideal I of B:

I`B ⊕ I ⊕ I2 ⊕ · · ·

´→ B ⊕ I ⊕ I2 ⊕ · · · .

Thus the scheme-theoretic pullback of X → Y to ProjOY ⊕I⊕I2⊕· · · , the invertible sheaf

corresponding to I ⊕ I2 ⊕ I3 ⊕ · · · , is an effective Cartier divisor in class O(1). Once we haveverified that this construction is indeed the blow-up, this divisor will be our exceptional divisorEXY . 16

Moreover, we see that the exceptional divisor can be described beautifully as a Proj over X:

(57) EXY = ProjXB/I ⊕ I/I2 ⊕ I2/I3 ⊕ · · · .

We’ll later see that in good circumstances (if X is a local complete intersection in somethingnonsingular, or more generally a local complete intersection in a Cohen-Macaulay scheme) this isa projective bundle (the “projectivized normal bundle”).

27.3.2. Proof of the universal property, Theorem 27.3.1. Let’s prove that this Proj

construction satisfies the universal property. 17 Then approach 1 will also follow, as a specialcase of approach 2. You may ask why I bothered with approach 1. I have two reasons: one isthat you may find it more comfortable to work with this one nice ring, and the picture may begeometrically clearer to you (in the same way that thinking about the blow-up closure lemma inthe case where Z → Y is a closed immersion is more intuitive). The second reason is that, asyou’ll find in the exercises, you’ll see some facts more easily in this explicit example, and you canthen pull them back to more general examples.

Proof. Reduce to the case of affine target R with ideal I. Reduce to the case of affine source, withprincipal effective Cartier divisor t. (A principal effective Cartier divisor is cut out by a single non-zero-divisor.) Thus we have reduced to the case Spec S → SpecR, corresponding to f : R → S.Say (x1, . . . , xn) = I, with (f(x1), . . . , f(xn)) = (t). We’ll describe one map Spec S → ProjR[I]that will extend the map on the open set Spec St → SpecR. It is then unique: a map to a separatedR-scheme is determined by its behavior away from the associated points (proved earlier [ref ]). Wemap R[I] to S as follows: the degree one part is f : R→ S, and f(Xi) (where Xi corresponds toxi, except it is in degree 1) goes to f(xi)/t. Hence an element X of degree d goes to X/(td). On theopen setD+(X1), we get the map R[X2/X1, . . . , Xn/X1]/(x2−X2/X1x1, . . . , xiXj−xjXi, . . . )→S (where there may be many relations) which agrees with f away from D(t). Thus this map doesextend away from V (I).

Here are some applications and observations arising from this construction of the blow-up.

27.3.3. Observation. We can verify that our initial motivational examples are indeed blow-ups. For example, blowing up A2 (with co-ordinates x and y) at the origin yields: B = k[x, y],I = (x, y), and ProjB⊕ I⊕ I2 = ProjB[X, Y ] where the elements of B have degree 0, and X andY are degree 1 and correspond to x and y.

27.3.4. Observation. Note that the normal bundle to a Cartier divisor D is the invertible sheafO(D)|D, the invertible sheaf corresponding to the D on the total space, then restricted to D.[This was discussed earlier in the section on differents, §22.2.E.] (Reason: if D corresponds to theideal sheaf I, then recall that I = O(D)∨, and that the conormal sheaf was I/I2 = I|D.) Theideal sheaf corresponding to the exceptional divisor is O(1), so the invertible sheaf correspondingto the exceptional divisor is O(−1). (I prefer to think of this in light of approach 1, but thereis no real difference.) Thus for example in the case of the blow-up of a point in the plane, theexceptional divisor has normal bundle O(−1). [Possibly refer forward to Castelnuovo’s criterion.]In the case of the blow-up of a nonsingular subvariety of a nonsingular variety, the blow up turns

16P.H.II.7.13a

17pfblun

458

out to be nonsingular (a fact discussed soon in §27.5.1), and the exceptional divisor is a projectivebundle over X, and the normal bundle to the exceptional divisor restricts to O(−1).

27.3.5. More serious application: dimensional vanishing for quasicoherent sheaveson quasiprojective schemes. Here is something promised long ago. I want to point outsomething interesting here: in proof I give below, we will need to potentially blow up arbitraryclosed schemes. We won’t need to understand precisely what happens when we do so; all we needis the fact that the exceptional divisor is indeed a (Cartier) divisor.

CUT THIS IN CLASS:

27.3.6. Theorem (dimensional vanishing for quasicoherent sheaves on quasiprojectiveschemes). — Suppose π : X → Y is a quasiprojective morphism all of whose fibers are ofdimension at most d, where Y (and hence X) is locally Noetherian. Then for any quasicoherentsheaf F on X, Riπ∗F = 0 for i > d. 18 Technically I haven’t defined quasiprojectivemorphisms for the class. Say: over any affine open we get something quasiprojective.This isn’t quite right. In FOAG, say the more general setting, and then specialize.

Proof. As higher pushforwards are computed locally on the target, we may assume that Y is affine.It suffices to show that X may be covered by d+1 affine open sets, as we can use those affine setsto compute cohomology, and the Cech complex for that cover vanishes after the dth step. As X isquasiprojective, there exists a open immersion X → X ′ into a closed subscheme of PnY . Blow upX′ along X′ −X, where we give X ′ −X any scheme structure, for example the reduced inducedsubscheme structure. (We use the Noetherian hypothesis here: we need X ′ − X to be cut outby a finite type sheaf of ideals.) Let BlX ′ → X′ be this blow-up. Then the blow-up morphismBlX′ → X′ is projective, X → BlX ′ is an open immersion, and there is a Cartier divisor E (theexceptional divisor) that is supported precisely on the complement of X in BlX ′. Now BlX′ isthe scheme-theoretic closure of X in BlX ′ (as the complement supports a Cartier divisor, whichcontains no associated points). Hence BlX ′ → Y has fiber dimension over Y bounded by d.Wait, this may be false! Thus by an earlier result, we can cover BlX ′ by d + 1 affine opensets. But then each affine open set U of BlX ′, minus the Cartier divisor U ∩ E, is again affine(we’ve shown that affine open sets minus Cartier divisors are always affine [ref ]). Hence we havecovered X with a finite number of affine open sets as desired.

Another case: intersection theory. We can intersect Cartier divisors with everything. To intersect two things generally, blow up one. Intersect upstairs. Pushforward. Possibly refer forward to intersection theory starred chapter.

Do this for proper morphisms eventually, using Chow.

27.4 Explicit computations

In this section you will do a number of explicit of examples, to get a sense of how blow-upsbehave, how they are useful, and how one can work with them explicitly.19 For convenience, allof the following are over an algebraically closed field k of characteristic 0.

27.4.1. Example: Blowing up the plane along the origin. Let’s first blow up the plane A2k

along the origin, and see that the result agrees with our discussion in §27.1. Let x and y be thecoordinates on A2

k. The the blow-up is Projk[x, y,X, Y ] where xY − yX = 0. This is naturally a

closed subscheme of A2k × P1

k, cut out (in terms of the projective coordinates X and Y on P1k) by

xY − yX = 0. We consider the two usual patches on P1k: [X;Y ] = [s; 1] and [1; t]. The first patch

yields Spec k[x, y, s]/(sy − x), and the second gives Spec k[x, y, t]/(y − xt). Notice that both arenonsingular: the first is naturally Spec k[y, s] ∼= A2

k, the second is Spec k[x, t] ∼= A2k.

Let’s describe the exceptional divisor. We first consider the first (s) patch. The ideal isgenerated by (x, y), which in our ys-coordinates is (ys, y) = (y), which is indeed principal. Thus

18genGrothvan

19blexplicit

459

on this patch the exceptional divisor is generated by y. Similarly, in the second patch, theexceptional divisor is cut out by x. (This can be a little confusing, but there is no contradiction!)

27.4.2. The proper transform of a nodal curve. Consider next the curve y2 = x3 +x2 inside theplane A2

k. Let’s blow up the origin, and compute the total and proper transform of the curve. (Bythe blow-up closure lemma, the latter is the blow-up of the nodal curve at the origin.) In the firstpatch, we get y2−s2y2−s3y3 = 0. This factors: we get the exceptional divisor y with multiplicitytwo, and the curve 1− s2−y3 = 0. Easy exercise: check that the proper transform is nonsingular.Also, notice where the proper transform meets the exceptional divisor: at two points, s = ±1.This corresponds to the two tangent directions at the origin. (Notice that s = y/x.)20

27.4.A. Exercise. Describe both the total and proper transform of the curve C given byy = x2 − x in Bl(0,0) A2. Verify that the proper transform of C is isomorphic to C. Interpret theintersection of the proper transform of C with the exceptional divisor E as the slope of C at theorigin.

27.4.B. Exercise: blowing up a cuspidal plane curve. Describe the proper transform of cuspthe cuspidal curve C ′ given by y2 = x3 in the plane A2

k. Show that it is nonsingular. Show thatthe proper transform of C meets the exceptional divisor E at one point, and is tangent to E there.[It would be good to have a picture here of a slinky]. 21

27.4.C. Exercise. (a) Desingularize the tacnode y2 = x4 by blowing up the plane at the origin(and taking the proper transform), and then blowing up the resulting surface once more.(b) Desingularize y8 − x5 = 0 in the same way. How many blow-ups do you need?(c) Do (a) instead in one step by blowing up (y, x2).

27.4.D. Exercise. Blowing up a nonreduced subscheme of a nonsingular scheme can give yousomething singular, as shown in this example. Describe the blow up of the ideal (x, y2) in A2

k.

What singularity do you get? (Hint: it appears in a nearby exercise, at least if√−1 is in your field

k.) Start: A2 × P1. In the first patch we get k[x, y, s]/(sy2 − x) ∼= k[y, s]. In the secondpatch we get k[x, y, t]/(y2 − xt). This is singular at the origin — no linear terms.)

27.4.E. Exercise. Blow up the cone point z2 = x2 + y2 at the origin. Show that the resultingsurface is nonsingular. Show that the exceptional divisor is isomorphic to P1.22

27.4.F. Harder but enlightening exercise. If X → Pn is a projective scheme, show that theexceptional divisor of the blow up the affine cone over X at the origin is isomorphic to X, andthat its normal bundle is OX(−1). (I prefer approach 1 here, but both work.)

In the case X = P1, we recover the blow-up of the plane at a point. In particular, we againrecover the important fact that the normal bundle to the exceptional divisor is O(−1).

27.4.G. Exercise. Show that the multiplicity of the exceptional divisor in the total transformof a subscheme of An when you blow up the origin is the lowest degree that appears in a definingequation of the subscheme. (For example, in the case of the nodal and cuspidal curves above,Example 27.4.2 and Exercise 27.4.B respectively, the exceptional divisor appears with multiplicity2.) This is called the multiplicity of the singularity. [Perhaps show it is independent of multiplicity of a singembedding?]

27.4.H. Exercise. Suppose Y is the cone x2 + y2 = z2, and X is the ruling of the cone x = 0,y = z. Show that BlX Y is nonsingular. (In this case we are blowing up a codimension 1 locusthat is not a Cartier divisor [ref ]. Note that it is Cartier away from the cone point, so you shouldexpect your answer to be an isomorphism away from the cone point.)

27.4.I. Harder but useful exercise (blow-ups resolve base loci of rational maps toprojective space). (I find this easier via method 1.) Suppose we have a scheme Y , an invertiblesheaf L, and a number of sections s0, . . . , sn of L. Then away from the closed subscheme X

20blnode

21blcusp

22blconic

460

cut out by s0 = · · · = sn = 0, these sections give a morphism to Pn. Show that this morphismextends to a morphism BlX Y → Pn, where this morphism corresponds to the invertible sheaf(π∗L)(−EXY ), where π : BlX Y → Y is the blow-up morphism. In other words, “blowing upthe base scheme resolves this rational map”. (Hint: it suffices to consider an affine open subsetof Y where L is trivial.) 23 24 Fix the reference, and also at HHHH! I should do threeexamples. (1) Pencil of elliptic curves in P2. The base locus is nine points. Blowingthem up gives an elliptic fibration. (2) the cubic surface in P3. (3) the Cremonatransformation P2 99K P2. Remind them what it is. Show that it is the resolutionof the rational map. Avoid painful computation as possible. (4) Should I describeCremona in higher dimension? Possibly starred. Then give the fact that there is onerational normal curve through an appropriate number of points.

Exercise: the quadric surface projected to P2 from a point on the surface. It involves blowingup a point, then blowing down two lines. The reverse rational map P2 99K P1 × P1 correspondsto taking the slopes of the lines to two fixed points.

27.5 Two stray facts

There are two stray facts I want to mention.

27.5.1. Blowing up a nonsingular in a nonsingular. 25HARD. Better title. The firstis that if you blow up a nonsingular subscheme of a nonsingular locally Noetherian scheme, theresult is nonsingular. I didn’t have the time to prove this, but I discussed some of the mathematics

behind it. (This is harder than our previous discussion. Also, it uses a flavor of argumentthat in general I haven’t gotten to, about local complete intersections and Cohen-Macaulayness.)Moreover, for a local complete intersection X → Y cut out by ideal sheaf I, I/I2 is locally free(Theorem 22.4.15). Then it is a fact (unproved here) that for a local complete intersection, thenatural map Symn I/I2 → In/In+1 is an isomorphism. Of course it suffices to prove this foraffine open sets. More generally, if A is Cohen-Macaulay (recall that I’ve stated that nonsingularschemes are Cohen-Macaulay), and x1, . . . , xr ∈ m is a regular sequence, with I = (x1, . . . , xr),then the natural map is an isomorphism. You can read about this at p. 110 of Matsumura’sCommutative Algebra. from H. This is stated in 16.7.4.

Assuming this fact, we conclude that if X → Y is a complete intersection in a nonsingularscheme (or more generally cut out by a regular sequence in a Cohen-Macaulay scheme), theexceptional divisor is the projectivized normal bundle (by (57)). (Exercise: Blow up (xy, z) in A3,and verify that the exceptional divisor is indeed the projectivized normal bundle.)

In particular, in the case where we blow up a nonsingular subvariety in a nonsingular variety,the exceptional divisor is nonsingular. We can then show that the blow-up is nonsingular asfollows. The blow-up BlX Y remains nonsingular away from EXY , as it is here isomorphic to thenonsingular space Y −X. Thus we need check only the exceptional divisor. Fix any point of theexceptional divisor p. Then the dimension of EXY at p is precisely the dimension of the Zariskitangent space (by nonsingularity). Moreover, the dimension of BlX Y at p is one more than thatof EXY (by Krull’s Principal Ideal Theorem 15.6.3, as the latter is an effective Cartier divisor),and the dimension of the Zariski tangent space of BlX Y at p is at most one more than that ofEXY . But the first of these is at most as big as the second, so we must have equality, whichmeans that BlX Y is nonsingular at p.

27.5.A. Exercise. Suppose X is an irreducible nonsingular subvariety of a nonsingular varietyY , of codimension at least 2. Describe a natural isomorphism Pic BlX Y ∼= PicY ⊕ Z. (Hint:compare divisors on BlX Y and Y . Show that the exceptional divisor EXY gives a non-torsionelement of Pic(BlX Y ) by describing a P1 on BlX Y which has intersection number −1 with EXY .)

(If I had more time, I would have used this to give Hironaka’s example of anonprojective proper nonsingular threefold. If you are curious and have ten minutes,

23H.E.II.7.13.3

24H.Example.7.17.3

25blnsns

461

please ask me! It includes our nonprojective proper surface as a closed subscheme,and indeed that is how we can show nonprojectivity.)

Application: Another example of a nonprojective proper smooth threefold. Hi-ronaka’s example. It includes our nonprojective proper surface as a closed sub- Hironakascheme, and indeed that is how we can show nonprojectivity. 26 This is a smoothversion of 29.3.7.

Use Hironaka’s example to show that a quotient by Z/2 need not exist. We need to have asmooth space, because we want the complement of our affine open subset to support an effectiveCartier divisor. (My earlier argument is false for this reason.)

Remark: blowing up something singular and ci in something nonsingular donebelow, give 2 refs: (x, y2) and (xy, z).

27.5.2. Castelnuovo’s criterion.Perhaps this should go in a section on intersection theory on a surface. A curve

in a nonsingular surface that is isomorphic to P1 with normal bundle O(−1) is called a (−1)-curve. We’ve shown that if we blow up a nonsingular point of a surface at a (reduced) point, (−1)-curve, (−2)the exceptional divisor is a (−1)-curve. Castelnuovo’s criterion is the converse: if we have aquasiprojective surface containing a (−1)-curve, that surface is obtained by blowing up anothersurface at a reduced nonsingular point. (We say that we can “blow down” the (−1)-curve.)In the case where X is a plane conic z2 = x2 + y2 (cf. 27.4.E), we see that the...(−2)-curve.

What happens to the normal bundle of a curve when you blow up? Answer: ittwists by that point.

Example of use: P1-bundles over a curve. Get different surfaces.Blowing up a cubic surface.A less important, long exercise about the Cremona transformation. Describe it Cremona

explicitly. Ask them to verify that it is the graph of the map [x; y; z]→ [1/x; 1/y; 1/z]. Itsends curves of degree d with multiplicity e1, e2, e3 to what? [M] has some discussionof this, but I can’t remember where.

27.6 Stray ideas (not used)

Do regular sequence a1, . . . , an (defined in 16.7.6). Rees algebra ⊕In ∼= R[x1, . . . , xn] modulo regular sequence2× 2 minors of »

a1 · · · anx1 · · · xn

–

27.6.1. Deformation to the normal cone. Deformation to the normal cone: suppose X → Yis a closed immersion. BlX×0Y ×P1. Show that the exceptional divisor is the normal cone plus itsprojectivization. It is flat (no associated points) hence equidimensional. Show that the multiplicityof the singularity is the degree of the normal cone. Hence define the Hilbert-Samuel multiplicityof a singularity. If you had some nonreduced structure (associated point) at the point you do this,there is no difference. Define a node to be something whose Zariski-tangent space is 2-dimensional,and the tangent cone consists of two geometric points (split or non-split).

] [primordial:[primordial:

26Hironakabl

CHAPTER 28

? Varieties over non-algebraically closed fields

This isn’t a real chapter. I’m just collecting facts about varieties over non-algebraically closedfields that might later be sprinkled elsewhere, or else collected in an appendix.

Throughout, suppose X is a k-variety, and let Xk = X ×Speck Spec k. We’d like to verify

facts about Xk by verifying them over k.

28.1 Nonsingularity

[Nonsingularity is not yet defined.] Our goal is to show that if Xk is nonsingular, then sois X, and that this is a characterization if k is separably closed (e.g. characteristic 0). We notethat it may be false if k is not separably closed, as shown by the example of (the spectrum of) aninseparable extension.

We begin with some facts about fields. Suppose k → k′ is finite degree d Then k′ ⊗k k is ak-vector space of dimension d. What is it as an algebra? If k′/k is separaable, then it is k×· · ·×k(d times).

Proof: k′ = k[x]/f(x), where f(x) has distinct roots in k. (Needed: single generation of finiteseparable extensions — theorem of the principle element.)

Remark: If k′/k is not separable, then this is not true. Proof: x ∈ k′, xp = a ∈ k, x /∈ k.k[x]/xp − a. Then (y − x) ∈ k[x]/(xp − a) satisfies (y − x)p = 0, but y 6= x.

Suppose now that X is a pure-dimensional k-variety, p a closed point (so by the Nullstellen-satz, it has residue field k′/k that is finite). Assume this is a separable extension. We have

0→ m→ A→ k′ → 0.

Define m := m⊗k k, A = A⊗k k. Then by flatness of field extensions, we have

0→ m→ A→ k′ ⊗k k→ 0.

Then

d points //

p

Spec k′ ⊗k k

// SpecA

Spec k′ // Spec k.

By applying ⊗kk to0→ m2 → m→ m/m2 → 0,

we obtain m/m2 ∼= (m/m2)⊗kk. If p is nonsingular, then each of the d points above are nonsingular,and vice versa.

Remark. If one of the d points is nonsingular, then p is nonsingular. (We won’t need this

fact.) Reason: base change to the Galois closure of k. Then they are all exchanged by the Galoisgroup action.

463

464

28.2 Other useful facts (possibly earlier)

28.2.1. If R is pure-dimension d, then R is pure dimension d.Proof:

SpecR //

Spec k

SpecR // Spec k

Use the going-up theorem. (This should be true for more general field extensions!)

28.2.2. Normality behaves well under base change. (This was done in the text somewhere...)That means that it may be checked on the algebraic closure.

28.2.3. The degree of invertible sheaf behaves well under ⊗kk. that was done in the textsomewhere. For example, we discussed it when discussing differentials of curves.

28.2.4. hi(F) behaves well under ⊗kk. Reason: it is an exact functor! This is a special case ofcohomology and flat base change.

28.3 Geometric points

][primordial:

CHAPTER 29

Flatness

Here is the outline of flatness I came to at the end of teaching it in 216.

• SECTION 1. Definition. Basic properties (localization; stalk/prime-local; base change;transitivity)

• SECTION 2. dimension behaves well for flat morphisms (for ft k-schemes). exercise:associated to associated (for ft k-schemes). exercise. flat morphisms are (usually) openmaps (flat+lft of lN).

• SECTION 3. Tor: want long exact sequence. Construct using frees resolution. Exactsequence, all but first are flat, then first is flat.

• Symmetry (spectral sequence). Tensor exact sequence of flats with anything, remains

exact.• SECTION 3b. Ideal-theoretic criterion Tor1(M,A/I) = 0 for all I. Flatness over PID’s

(DVR’s, hence over nonsingular curves; then flat limit), dual numbers.• coherent modules over Noetherian local rings: flat=free.• local criterion.• SECTION 4: CBC. [Restate as “higher direct image commutes with flat pullback”?]

cohomology and flat base change.• flat implies constant χ. Important consequence: degree of line bundles on flat proj.

family of curves. Hence: nonprojective surface. Impossibility of gluing scheme to itself.• semicontinuity. (lb are trivial in Z-closed locus. Glimpse of relative picard)• Grauert. Cohomology and base change.• Important special case: OY → π∗OX iso. (Implied by flat with geometrically integral

fibers). (Rigidity lemma; didn’t use flatness. Soon: Zariski connectedness lemma,implies Stein factorization and ZMT)

Other facts not described in class.

• open locus where something is flat, using the local criterion?• CM fact• flat descent• completion is flat (required?)• flatness over Artinian ring

[height=codimension is precisely when we’ll use the criterion of essentially of finite type overfield. State this when we prove that fact. Also, we have a recurring theme: stalks vs. fiber.Example: when is it locally free? Ample?]

[Place later: My goal is twofold. (1) To tell you facts about flat morphisms and sheaves.These won’t be intuitively obvious, so you should use these facts to build new intuition. (2) Tohelp you build new intuition, I’m going to try to draw pictures, so you’ll see examples of flatmorphisms, and get a sense of when a morphism is flat.] flat

We come next to the important concept of flatness. This topic is also not a hard topic, and wecould have dealt with it as soon as we had discussed quasicoherent sheaves and morphisms. Butit is an intuitively unexpected one, and the algebra and geometry are not obviously connected,so we’ve left it for relatively late. It is answer to many of your geometric prayers, but you justhaven’t realized it yet.

The notion of flatness apparently was first defined in Serre’s landmark “GAGA” paper.

[attributed to Serre’s GAGA paper [S] in [EGA] 0.6.0, p. 54; there he provedGAGA, see 33.]

Here are some of the reasons it is a good concept. We would like to make sense of the notion of“fibration” in the algebraic category (i.e. in algebraic geometry, as opposed to differential geometry

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466

[try to avoid this phrase, or to define it properly the first time I use it]), and it turnsout that flatness is essential to this definition. It turns out that flat is the right algebraic versionof a “nice” or “continuous” family, and this notion is more general than you might think. Forexample, the double cover A1 → A1 over an algebraically closed field given by y 7→ x2 is a flatfamily, which we interpret as two points coming together to a fat point. The fact that the degreeof this map always is 2 is a symptom of how this family is well-behaved. Another key exampleis that of a family of smooth curves degenerating to a nodal curve [make figure]. One can provethings about smooth curves by first proving them about the nodal curve, and then showing thatthe result behaves well in flat families. In general, we’ll see that certain things behave well in nicefamilies, such as cohomology groups (and even better Euler characteristics) of fibers.

There is a second flavor of prayer that is answered. It would be very nice if tensor product (ofquasicoherent sheaves, or of modules over a ring) were an exact functor, and certain statementsof results and proofs we have seen would be cleaner if this were true. Those modules for whichtensoring is always exact are flat (this will be the definition!), and hence for flat modules (orquasicoherent sheaves, or soon, morphisms) we’ll be able to get some very useful statements. Aflip side of that is that exact sequences of flat modules remain exact when tensored with any othermodule.

In this chapter, we’ll discuss flat morphisms. When introducing a new notion, I prefer tostart with a number of geometric examples, and figure out the algebra on the fly. In this case,

because there is enough algebra, I’ll instead discuss the algebra at some length and then laterexplain why you care geometrically. This will require more patience than usual on your part.

29.1 Algebraic definition and easy facts

Many facts about flatness are easy or immediate, and a few are tricky. I’m going to try tomake clear which is which, to help you remember the easy facts and the idea of proof for theharder facts.

The definition of a flat A-module is very simple. Recall that if 1flat A-module

(58) 0→ N ′ → N → N ′′ → 0

is a short exact sequence of A-modules, and M is another A-module, then

M ⊗A N ′ →M ⊗A N →M ⊗A N ′′ → 0

is exact. In other words, M⊗A is a right-exact functor. We say that M is a flat A-module ifM⊗A is an exact functor, i.e. if for all exact sequences (58),

0→M ⊗A N ′ →M ⊗A N →M ⊗A N ′′ → 0

is exact as well.flat A-moduleExercise. If N ′ → N → N ′′ is exact and M is a flat A-module, show that M ⊗A N ′ →

M ⊗A N → M ⊗A N ′′ is exact. Hence any exact sequence of A-modules remains exact upontensoring with M . (We’ve seen things like this before, so this should be fairly straightforward.)

We say that a ring homomorphism B → A is flat if A is flat as a B-module. (We don’t careabout the algebra structure of A.)flat ring homomorphism

Here are two key examples of flat ring homomorphisms:

(i) free modules A-modules are clearly flat.(ii) Localizations are flat: Suppose S is a multiplicative subset of B. Then B → S−1B is

a flat ring morphism.

Exercise. Verify (ii). We have used this before: localization is an exact functor.Here is a useful way of recognizing when a module is not flat. Flat modules are torsion-free.

More precisely, if x is a non-zero-divisor of A, and M is a flat A-module, then M×x // M is

injective. Reason: apply the exact functor M⊗A to the exact sequence 0 // A×x // A .

We make some quick but important observations:

1short

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29.1.1. Proposition (flatness is a stalk/prime-local property). — An A-module M isflat if and only if Mp is a flat Ap-module for all primes p. 23

Proof. Suppose first that M is flat. Given any exact sequence of Ap-modules (58),

0→M ⊗A N ′ →M ⊗A N →M ⊗A N ′′ → 0

is exact too. But M ⊗A N is canonically isomorphic to M ⊗ApNp (exercise: show this!), so Mp

is a flat Ap-module.Suppose next that M is not flat. Then there is some short exact sequence (58) that upon

tensoring with M becomes

(59) 0→ K →M ⊗A N ′ →M ⊗A N →M ⊗A N ′′ → 0

where K 6= 0 is the kernel of M ⊗A N ′ →M ⊗A N . (You should recall that M ⊗ · is exact [ref].)Then as K 6= 0, K has non-empty support, so there is some prime p such that Kp 6= 0. Then4

(60) 0→ N ′p → Np → N ′′p → 0

is a short exact sequence of Ap-modules (recall that localization is exact — see (ii) before the

statement of the Proposition), but is no longer exact upon tensoring (over Ap) with Mp (as5

(61) 0→ Kp →Mp ⊗ApN ′p →Mp ⊗Ap

Np →Mp ⊗ApN ′′p → 0

is exact). (Here we use that localization commutes with tensor product.) [ref; more generally,tensor product commutes with direct limits.]

29.1.2. Proposition (flatness is preserved by change of base ring). — If M flat A-module, A→ B is a homomorphism, then M ⊗A B is a flat B-module. 6

Proof. Exercise (Hint: consider the natural isomorphism (M ⊗A B) ⊗B · ∼= M ⊗B (B ⊗A ·).)(M ⊗A B) ⊗B · = M ⊗B (B ⊗A ·) is a composition of exact functors and hence is exact.Wrong reason, right identity. [Rob says: this hint doesn’t apply here, but applies in the nextproblem.]

29.1.3. Proposition (transitivity of flatness). — If B is a flat A-algebra, and M is B-flat,then it is also A-flat. 7

Proof. Exercise. (The same hint as in the previous proposition applies.) As with the previousproposition, the result holds because the composition of two exact functors is exact:M ⊗A · = M ⊗B (B ⊗A ·).

The extension of this notion to schemes is straightforward.

29.1.4. Definition: flat quasicoherent sheaves, and more generally, flat OX -modules. We saythat a quasicoherent sheaf F on a scheme X is flat (over X) if for all x ∈ X, Fx is a flat OX,x-module. In light of Proposition 29.1.1, we can check this notion on affine open cover of X. 8

(Remark: applies to OX -modules as well.)

29.1.5. Definition: flat morphism. Similarly, we say that a morphism of schemes π : X → Y isflat if for all x ∈ X, OX,x is a flat OY,π(x)-module. Again, we can check locally, on maps of affine

schemes. 9

We can combine these two definitions into a single definition.

2H.P.III.9.1Ad

3flatstalk

4shortk2

5shortk3

6H.P.III.9.1Ab

7H.P.III.9.1Ac

8d:flatqcs

9d:flatmorphism

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29.1.6. Definition: flat quasicoherent sheaf over some base. Suppose π : X → Y is a morphismof schemes, and F is a quasicoherent sheaf on X. We say that F is flat over Y if for all x ∈ X,Fx is a flat OY,π(x)-module.flat qcs; flat morphism;

flat qcs over a base Definitions 29.1.4 and 29.1.5 correspond to the cases X = Y and F = OX respectively.This definition can be extended without change to the category of ringed spaces, but we

won’t need this.All of the Propositions above carry over naturally. For example, flatness is preserved by base

change. (More explicitly: suppose π : X → Y is a morphism, and F is a quasicoherent sheaf onX, flat over Y . If Y ′ → Y is any morphism, and p : X ×Y Y ′ → X is the projection, then p∗Fis flat over Y ′.) 10 Also, flatness is transitive. (More explicitly: suppose π : X → Y and F is aquasicoherent sheaf on X, flat over Y . Suppose also that ψ : Y → Z is a flat morphism. Then Fis flat over Z.) 11

We also have other statements easily. For example: open immersions are flat.12

29.1.A. Exercise. If X is a scheme, and η is the generic point for an irreducible component,show that the natural morphism SpecOX,η → X is flat. (Hint: localization is flat.)13

We earlier proved the following important fact, although we did not have the language offlatness at the time.

29.1.7. Theorem (cohomology commutes with flat base change). — Suppose

X′g′ //

f ′

X

f

Y ′

g // Y

is a fiber diagram, and f (and thus f ′) is quasicompact and separated (so higher pushforwardsexist). Suppose also that g is flat, and F is a quasicoherent sheaf on X. Then the naturalmorphisms g∗Rif∗F → Rif ′∗(g

′∗F) are isomorphisms.14

A special case that is often useful is the case where Y ′ is the generic point of a componentof Y . In other words, in light of Exercise 29.1.A, the stalk of the higher pushforward of F at thegeneric point is the cohomology of F on the fiber over the generic point. This is a first example ofsomething important: understanding cohomology of (quasicoherent sheaves on) fibers in terms ofhigher pushforwards. (We would certainly hope that higher pushforwards would tell us somethingabout higher cohomology of fibers, but this is certainly not a priori clear!)

(I might dig up the lecture reference later, but I’ll tell you now where proved it: where wedescribed this natural morphism, I had a comment that if we had exactness of tensor product,them morphisms would be an isomorphism.)

We will spend the rest of this chapter as follows. First, we will ask ourselves: what are theflat modules over particularly nice rings? More generally, how can you check for flatness? Andhow should you picture it geometrically? We will then prove additional facts about flatness, andusing flatness, answering the essential question: “why do we care?”

29.1.8. Faithful flatness. The notion of faithful flatness is handy, although we probably won’tuse it. We say that an extension of rings B → A is faithfully flat if for every A-module M , M isA-flat if and only if M ⊗A B is B-flat. We say that a morphism of schemes X → Y is faithfullyflat if it is flat and surjective. These notions are the “same”, as shown by the following exercise.faithfully flat

Exercise. Show that B → A is faithfully flat if and only if SpecA→ SpecB is faithfully flat.Note that faithful flatness is preserved by base change, as surjectivity is, and flatness is too.

[forme: Wait, is surjectivity preserved by base change? Milne states that faithfulflatness is preserved by base change on p. 50 of [MilneLET].]

10H.P.III.9.2b

11H.P.III.9.2c

12H.P.III.9.2a

13flatgeneric

14H.P.III.9.3, except with better hypotheses

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29.2 The “Tor” functors, and a “cohomological” criterion forflatness

In order to prove more facts about flatness, it is handy to have the notion of Tor. (Tor is shortfor “torsion”. The reason for this name is that the 0th and/or 1st Tor-group measures commontorsion in abelian groups (aka Z-modules).) If you have never seen this notion before, you maywant to just remember its properties, which are natural. But I’d like to prove everything anyway— it is surprisingly easy.

The idea behind Tor is as follows. Whenever we see a right-exact functor, we always hopethat it is the end of a long-exact sequence. Informally, given a short exact sequence (58), we arehoping to see a long exact sequence 15

(62) · · · // TorAi (M,N ′) // TorAi (M,N) // TorAi (M,N ′′) // · · ·

// TorA1 (M,N ′) // TorA1 (M,N) // TorA1 (M,N ′′)

// M ⊗A N ′ // M ⊗A N // M ⊗A N ′′ // 0.

More precisely, we are hoping for covariant functors TorAi (·,N) from A-modules to A-modules

(giving 2/3 of the morphisms in that long exact sequence), with TorA0 (M,N) ≡ M ⊗A N , and

natural δ morphisms TorAi+1(M,N ′′) → TorAi (M,N ′) for every short exact sequence (58) giving

the long exact sequence. (In case you care, “natural” means: given a morphism of short exactsequences, the natural square you would write down involving the δ-morphism must commute.I’m not going to state this explicitly.)

It turns out to be not too hard to make this work, and this will later motivate derivedfunctors. I’ll now define TorAi (M,N). Take any resolution R of N by free modules:

· · · // A⊕n2 // A⊕n1 // A⊕n0 // N → 0.

More precisely, build this resolution from right to left. Start by choosing generators of N as anA-module, giving us A⊕n0 → N → 0. Then choose generators of the kernel, and so on. Notethat we are not requiring the ni to be finite, although if N is a finitely-generated module and Ais Noetherian (or more generally if N is coherent and A is coherent over itself), we can choosethe ni to be finite. Truncate the resolution, by stripping off the last term. Then tensor with M(which may lose exactness!). Let ToriA(M,N)R be the homology of this complex at the ith stage(i ≥ 0). The subscript R reminds us that our construction depends on the resolution, althoughwe will soon see that it is independent of the resolution.

We make some quick observations.• TorA0 (M,N)R ∼= M ⊗A N (and this isomorphism is canonical). Reason: as tensoring is rightexact, and A⊕n1 → A⊕n0 → N → 0 is exact, we have that M⊕n1 → M⊕n0 → M ⊗A N → 0 isexact, and hence that the homology of the truncated complex M⊕n1 →M⊕n0 → 0 is M ⊗A N .• If M is flat, then TorAi (M,N)R = 0 for all i.

Now given two modules N and N ′ and resolutions R and R′ of N and N ′, we can “lift” anymorphism N → N ′ to a morphism of the two resolutions:

· · · //

A⊕ni //

· · · //

A⊕n1 //

A⊕n0 //

N

// 0

· · · // A⊕n′i

// · · · // A⊕n′1

// A⊕n′0

// N ′ // 0

Denote the choice of lifts by R → R′. Now truncate both complexes and tensor with M . Maps ofcomplexes induce maps of homology, so we have described maps (a priori depending on R → R′)

TorAi (M,N)R → TorAi (M,N ′)R′ .

We say two maps of complexes f, g : C∗ → C′∗ are homotopic if there is a sequence of maps homotopic maps of com-

plexes15

TorLES

470

w : Ci → C′i+1 such that f−g = dw+wd. Two homotopic maps give the same map on homology.

(Exercise: verify this if you haven’t seen this before.)Crucial Exercise: Show that any two lifts R → R′ are homotopic.We now pull these observations together.

(1) We get a covariant functor from TorAi (M,N)R → TorAi (M,N ′)R′ (independent of thelift R → R′).

(2) Hence for any two resolutionsR andR′ we get a canonical isomorphism TorAi (M,N)R ∼=Tor1i (M,N)R′ . Here’s why. Choose lifts R → R′ and R′ → R. The compositionR → R′ → R is homotopic to the identity (as it is a lift of the identity map N → N).Thus if fR→R′ : TorAi (M,N)R → Tor1i (M,N)R′ is the map induced by R → R′, andsimilarly fR′→R is the map induced by R → R′, then fR′→R fR→R′ is the identity,and similarly fR→R′ fR′→R is the identity.

(3) Hence the covariant functor doesn’t depend on the resolutions!

Finally:(4) [Careful here!] For any short exact sequence (58) we get a long exact sequence of Tor’s (62).Here’s why: given a short exact sequence (58), choose resolutions of N ′ and N ′′. Then use these to

get a resolution for N in the obvious way (see below; the map A⊕(n′0→n

′′0 ) → N is the composition

A⊕n′0 → N ′ → N along with any lift of An

′′0 → N ′′ to N) so that we have a short exact sequence

of resolutions

0

0

0

· · · //

A⊕n′1

//A⊕n

′0

// N ′

// 0

· · · //A⊕(n′

1+n′′1 )

//A⊕(n′

0+n′′0 )

// N

// 0

· · · //A⊕n

′′1

//A⊕n

′′0

// N ′′

// 0

0 0 0

Then truncate (removing the column (58)), tensor with M (obtaining a short exact sequence ofcomplexes) and take cohomology, yielding a long exact sequence.

We have thus established the foundations of Tor!Note that if N is a free module, then TorAi (M,N) = 0 for all M and all i > 0, as N has itself

as a resolution.

29.2.1. Remark: Projective resolutions. We used very little about free modules in the aboveconstruction; in fact we used only that free modules are projective, i.e. those modules M such thatfor any surjection M ′ → M ′′, it is possible to lift any morphism M → M ′′ to M → M ′. This issummarized in the following diagram.

M

exists

||M ′ // //M ′′

Equivalently, Hom(M, ·) is an exact functor (Hom(N, ·) is always left-exact for any N). (Moregenerally, we can define the notion of a projective object in any abelian category.) Hence (i) weproj. module; proj. obj.

in ab. cat. can computeTorAi (M,N) by taking any projective resolution of N , and (ii) TorAi (M,N) = 0 forany projective A-module N .

29.2.2. Remark. Flat resolutions work too. Prove this (unimportant exercise). Hint: use aspectral sequence. (Give it to them.) Philosophy: you can resolve by acyclics.

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29.2.3. Remark: Generalizing this construction. The above description was low-tech, butimmediately generalizes drastically. All we are using is that M⊗A is a right-exact functor. Ingeneral, if F is any right-exact covariant functor from the category of A-modules to any abeliancategory, this construction will define a sequence of functors LiF (called left-derived functors of F )such that L0F = F and the Li’s give a long-exact sequence. We can make this more general still.We say that an abelian category has enough projectives if for any object N there is a surjectiononto it from a projective object. Then if F is any right-exact functor from an abelian categorywith enough projectives to any abelian category, then F has left-derived functors.

29.2.A. Exercise. The notion of an injective object in an abelian category is dual to the notionof a projective object. Define derived functors for (i) covariant left-exact functors (these arecalled right-derived functors), (ii) contravariant left-exact functors (also right-derived functors),and (iii) contravariant right-exact functors (these are called left-derived functors), making explicitthe necessary assumptions of the category having enough injectives or projectives. Define derived functor

derived functors; inj. ob-

jects; enough injectives;

enough projectives

Ext.[According to [MilneLET] p. 2, they were used systematically in Cartan and Eilenberg’s

Homological Algebra. They were used on abelian categories first by Buchsbaum.]Here are two quick practice exercises, giving useful properties of Tor.Important exercise. If B is A-flat, then we get isomorphism B ⊗ TorAi (M,N) ∼= TorBi (B ⊗

M,B⊗N). (This is tricky rather than hard; it has a clever one-line answer. Here is a fancier factthat experts may want to try: if B is not A-flat, we don’t get an isomorphism; instead we get aspectral sequence.)

29.2.B. Exercise- (good practice if you haven’t played with Tor before). If x is not a

0-divisor, show that TorAi (A/x,M) is 0 for i > 1, and for i = 0, get M/xM , and for i = 1, get(M : x) (those things sent to 0 upon multiplication by x). [Put this in a brace.] (We’ll use this inExercise 29.2.D.) [Also used in the local slicing criterion, 29.13.2?] [Also used when showing thatflatness over pid is the same as torsion-free, Cor. 29.2.12.]16

29.2.4. “Symmetry” of Tor. The natural isomorphism M ⊗ N → N ⊗M extends to thefollowing.

29.2.5. Theorem. — There is a natural isomorphism Tori(M,N) ∼= Tor(N,M). [Kirsten also Eisenbud Ex. 6.1 p. 171refers to McCleary “A User’s Guide to Spectral Sequences” pg. 52 Prop. 2.17 Chap. 2.]

Proof. Take two resolutions of M and N :

· · · → A⊕m1 → A⊕m0 →M → 0

and· · · → A⊕n1 → A⊕n0 → N → 0.

Consider the double complex obtained by tensoring their truncations.

...

...

...

· · · // A⊕(m2+n2) //

A⊕(m1+n2) //

A⊕(m0+n2) //

0

· · · // A⊕(m2+n1) //

A⊕(m1+n1) //

A⊕(m0+n1) //

0

· · · // A⊕(m2+n0) //

A⊕(m1+n0) //

A⊕(m0+n0) //

0

0 0 0.

16TorAmodprincipal

472

Apply our spectral sequence machinery. We compute the homology of this complex in twoways.

We start by using the vertical arrows. Notice that the ith column is precisely the truncatedresolution of N , tensored with A⊕mi . Thus the homology in the vertical direction in the ithcolumn is 0 except in the bottom element of the column, where it is N⊕mi . We next takehomology in the horizontal direction. In the only non-zero row (the bottom row), we see preciselythe complex computing Tori(N,M). After using these second arrows, the spectral sequence hasconverged. Thus the ith homology of the double complex is (naturally isomorphic to) Tori(N,M).

Similarly, if we began with the arrows in the horizontal direction, we would conclude thatthe ith homology of the double complex is Tori(M,N). Reverse the order for a cleanerdescription!

This gives us a quick but very useful result. Recall that if 0→ N ′ → N → N ′′ → 0 is exact,then so is the complex obtained by tensoring with M if M is flat. (Indeed that is the definitionof flatness!) But in general we have an exact sequence

TorA1 (M,N ′′)→M ⊗A N ′ →M ⊗A N →M ⊗A N ′′ → 0

Hence we conclude:

29.2.6. Proposition. — If 0 → N ′ → N → N ′′ → 0 is exact, and N ′′ is flat, then 0 →M ⊗A N ′ →M ⊗A N →M ⊗A N ′′ → 0 is exact. 17

Note that we would have cared about this result long before learning about Tor. This givessome motivation for learning about Tor. Presumably one can also show this directly by some sortof diagram chase. (Is there an easy proof?)

One important consequence is the following. Suppose we have a short exact sequence ofsheaves on Y , and the rightmost element is flat (e.g. locally free). Then if we pull this exactsequence back to X, it remains exact. (I think we may have used this.)

29.2.7. Flatness in exact sequences. [CLUMP OF IDEAS][PLACE THIS APPROPRI-ATELY; BEFORE SYMMETRY OF TOR?]

Suppose 0→M ′ →M →M ′′ → 0 is an exact sequence of A-modules.

29.2.8. Proposition. — If M and M ′′ are both flat, then so is M ′. If M ′ and M ′′ are bothflat, then so is M .

Proof. We use the characterization of flatness that N is flat if and only if Tori(N,N′) = 0 for all

i > 0, N ′. The result follows immediately from the long exact sequence for Tor.

Important remark: Finite rank vector bundles. In particular, over a smooth curve [moregenerality?], this result specializes to the case where the same statements hold, with “flat” replacedby “finite rank locally free.” This will be useful.

29.2.9. Unimportant remark. This begs the question: if M ′ and M are both flat, is M ′′ flat?(The argument above breaks down.) The answer is no: over k[t], consider 0 → tk[t] → k[t] →k[t]/t → 0 (geometrically: the closed subscheme exact sequence for a point on A1). The moduleon the right has torsion, and hence is not flat. The other two modules are free, hence flat.

29.2.C. Easy exercise. (We will use this shortly.) If 0 → M0 → M1 → · · · → Mn → 0 is anexact sequence, and Mi is flat for i > 0, show that M0 is flat too. (Hint: break the exact sequenceinto short exact sequences.) 18

We now come to the next result about flatness that will cause us to think hard.[BIGGER CONSEQUENCE 1]

29.2.10. An ideal-theoretic criterion for flatness. We come now to a useful fact. Observethat Tor1(M,N) = 0 for all N implies that M is flat; this in turn implies that Tori(M,N) = 0for all i > 0.

The following is a very useful variant on this.

17flat2

18firstflat

473

29.2.11. Key theorem. — M is flat if and only if TorA1 (M,A/I) = 0 for all ideals I.19

(The interested reader can tweak the proof below a little to show that it suffices to considerfinitely generated ideals I, but we won’t use this fact.)

Proof. [The M ’s and N ’s are screwed up here, up to the first line of paragraph two as well.] We

have already observed that if N is flat, then TorA1 (M,R/I) = 0 for all I. So we assume that

TorA1 (M,A/I) = 0, and hope to prove that TorA1 (M,N) = 0 for all A-modules N , and hence thatM is flat.

By induction on the number of generators of N , we can prove that TorA1 (M,N) = 0 for allfinitely generated modules N . (The base case is our assumption, and the inductive step is asfollows: if N is generated by a1, . . . , an, then let N ′ be the submodule generated by a1, . . . ,an−1, so 0→ N ′ → N → A/I → 0 is exact, where I is some ideal. Then the long exact sequence

for Tor gives us 0 = TorA1 (M,N ′)→ TorA1 (M,N)→ TorA1 (M,A/I) = 0.)We conclude by noting that N is the union (i.e. direct limit) of its finitely generated submod-

ules. As ⊗ commutes with direct limits, Tor1 commutes with direct limits as well. (This requiressome argument!) [Give it!]

Ben gives the following discussion. Hom(M ⊗N,S) ∼= Hom(M,Hom(N,S)), so ⊗N isa left-adjoint, and hence commutes with colimits. Jonathan and Ben explained howthis goes, at least if the index catogry is linear. With the well-ordering principlewe are okay in general. This uses: any right exact functor from A-modules thatcommutes with colimits has derived functors that commute with colimits. We usethe fact taht hjomology commutes with colimits. Colimits commtue with homologyhence colimit is an exact functor. More generall, universal δ-functors even if not aderived functor.

Here is a sketch of an alternate conclusion. We wish to show that for any exact 0→ N ′ → N ,0 → M ⊗ N ′ → M ⊗ N is also exact. Suppose

Pmi ⊗ n′i 7→ 0 in M ⊗ N . Then that equality

involves only finitely many elements of N . Work instead in the submodule generated by theseelements of N . Within these submodules, we see that

Pmi ⊗ n′i = 0. Thus this equality holds

inside M ⊗N ′ as well.(I may try to write up a cleaner argument. Joe pointed out that the cleanest thing to do

is to show that injectivity commutes with direct limits. [Comes from homology commuting withdirect limits.])

This has some cheap but important consequences.Recall (or reprove) that flatness over a domain implies torsion-free.

29.2.12. Corollary to Theorem 29.2.11. — Flatness over principal ideal domain is thesame as torsion-free. 20

This follows directly from the proposition. [Did I prove Tor1 = torsion for a PID? Yes, seeexercise 29.2.B.]

29.2.D. Important exercise (flatness over the dual numbers). This fact is important in

deformation theory and elsewhere. Show that M is flat over k[t]/t2 if and only if the natural map Eisenbud Cor 6.2M/tM → tM is an isomorphism. [Use: k[t]/t2 has only three ideals.] 21 [Apparently, an earlierexercise computes the ideals of k[t]/tp, Z/25. Why did I note this?]

[BIG CONSEQUENCE]

29.2.13. Important Theorem (for coherent modules over Noetherian local rings, flatequals free). — Suppose (A,m) is a local ring, and M is a coherent A-module (e.g. if A is

Noetherian, then M is finitely generated). Then M is flat if and only if it is free. 22

[flat = projective = free]

29.2.14. (It is true more generally, although we won’t use those facts: apparently we can replacecoherent with finitely presented, which only non-Noetherian people care about; or we can give up

19Torideal

20flatpid

21flatdual

22flatfree

474

coherent completely if A is Artinian, although I haven’t defined this notion. Reference: Mumfordp. 296. I may try to clean the proof up to work in these cases. Perhaps look at Nakayama,Artinian version 15.3.C.) 23

Proof. Clearly we are going to be using Nakayama’s lemma. Now M/mM is a finite-dimensionalvector space over the field A/m. Choose a basis, and lift it to elements m1, . . . , mn ∈ M . Thenconsider An → M given by ei 7→ mi. We’ll show this is an isomorphism. This is surjective byNakayama’s lemma: the image is all of M modulo the maximal ideal, hence is everything. Let Kbe the kernel, which is finitely generated by coherence:

0→ K → An →M → 0.

Tensor this with A/m. As M is flat, the result is still exact (Proposition 29.2.6):

0→ K/mK → (A/m)n →M/mM → 0.

But (A/m)n →M/mM is an isomorphism, so K/mK = 0. As K is finitely generated, K = 0.

Here is an immediate corollary (or really just a geometric interpretation).

29.2.15. Corollary. — Suppose F is coherent over a locally Noetherian scheme X. Then Fis flat over X if and only if it is locally free. 24

(Reason: we have shown that local-freeness can be checked at the stalks.) [give ref!] Checkhypotheses, and get rid of Noetherian if I can. Recall these facts: coherent plus freestalks means locally free. coherent plus flat implies free for local rings. Noetherian-ness seems unnecessary.

Make sure to give an very useful exercise: Show that if 0 → F ′ → F → F ′′ → 0is exact, over locally Noetherian scheme presumably, and the last two (or the firstand third) is coherent and locally free, so is the third. [Perhaps this can be proveddirectly? Not clear to me!]

This is a useful fact. Here’s a consequence that we prove earlier by other means: if C → C ′

is a surjective map of nonsingular irreducible projective curves, then π∗OC is locally free. ref?In general, this gives us a useful criterion for flatness: Suppose X → Y finite, and Y integral.

Then f is flat if and only if dimFF (y) f∗(OX )y ⊗ FF (y) is constant. So the normalization of aMumford p. 301

node is not flat (I drew a picture here).[PLACE EARLIER. Also, add normalizatioon of curve example to show how nonsingularity

is necessary.]

29.2.16. A useful special case: flatness over nonsingular curves. When are morphismsto nonsingular curves flat? Local rings of nonsingular curves are discrete valuation rings, whichare principal ideal domains, so for them flat = torsion-free (Prop. 29.2.12). Thus, any map froma scheme to a nonsingular curve where all associated points go to a generic point is flat. (I drewseveral pictures of this.)

Here’s a version we’ve seen before: a map from an irreducible curve to a nonsingular curve.Here is another important consequence, which we can informally state as: we can take flat

limits over one-parameter families. More precisely: suppose A is a discrete valuation ring, and let0 be the closed point of SpecA and η the generic point. Suppose X is a scheme over A, and Yis a scheme over X|η . Let Y ′ be the scheme-theoretic closure of Y in X. Then Y ′ is flat over A.Then Y ′|0 is often called the flat limit of Y .flat limit

(Suppose A is a discrete valuation ring, and let η be the generic point of SpecA. SupposeX is proper over A, and Y is a closed subscheme of Xη . Exercise: Show that there is only oneclosed subscheme Y ′ of X, proper over A, such that Y ′|η = Y , and Y ′ is flat over A. Aside

for experts: For those of you who know what the Hilbert scheme is, by taking the case of X asprojective space, this shows that the Hilbert scheme is proper, using the valuative criterion forproperness.) [We don’t need properness here. So perhaps state it without properness, and thenadd in properness, and ask them to show that the limit is also proper. This may require thevaluative criterion.] [Jared’s argument. Let A be a DVR with maximal ideal m and generic pointη. Claim. If X → SpecA is any morphism, and Y → Xη = X ×A FF (η) is a closed subscheme,

23NakArtuse

24finiteflat, used in Serre duality

475

there exists a unique closed subscheme Y ′ → X flat over A and Y ′η = Y ′ ×A FF (η) = Y . Proof.

Let Y ′ be the scheme-theoretic image of Y under g : Xη → X. To show that Y ′ ×A FF (η) ∼= Y ,it suffices to assume X = SpecB is affine. We have a corresponding diagram of rings.

Bφ(t)/J At ⊗A Boooo Bψ

oo

FF (η) = At

OO

Aoo

φ

OO

Since ψ is localization, (Jc)e = J (where Jc = ψ−1(J) and (Jc)e = ψ(ψ−1(J))Bφ(t)). The

scheme-theoretic image of Y under g is the closed subscheme defined by the ideal J ′ = ker( Bψ // Bφ(t) // Bφ(t)/J ) =

ψ−1(J) = Jc. Then B/J ′ ⊗A At = (B/J ′)φ(t) = Bφ(t)/(J′)e = Bφ(t)/J . This is exactly the

statement Y ′×AFF (η) ∼= Y . Since SpecFF (η) → SpecA is an open immersion, Y → Y ′ is open,and we claim that the associated primes of Y ′ are contained in Y . Again, it suffices to assumethat X = SpecB is affine. Let p = Ann(b) ∈ Ass(B/J ′) = p ∈ AssB | p ⊃ J ′ with b ∈ B. Thenpe ⊃ (J ′)e = J and pe = Ann(ψ(b)). Since (pe)c = p, pe is an associated prime ofY which mapsto p in Y ′. Since SpecA is a nonsingular curve, we can use the criterion for flatness dealing withthat. Rest is skipped.]

29.2.E. Exercise (an interesting explicit example of a flat limit). (Here the base is A1,not a discrete valuation ring. You can either restrict to the discrete valuation ring that is thestalk near 0, or generalize the above discussion appropriately.) Let X = A3 × A1 → Y = A1 overa field k, where the coordinates on A3 are x, y, and z, and the coordinates on A1 are t. Define Xaway from t = 0 as the union of the two lines y = z = 0 (the x-axis) and x = z− t = 0 (the y-axistranslated by t). Find the flat limit at t = 0. (Hint: it is not the union of the two axes, althoughit includes it. The flat limit is non-reduced.)

29.2.17. Stray but important remark: flat morphisms are (usually) open. I’mdiscussing this here because I have no idea otherwise where to put it.

29.2.F. Exercise. Prove that flat and locally finite type morphisms of locally Noetherianschemes are open. (Hint: reduce to the affine case. Use Chevalley’s theorem (Exercise 15.5.1) toshow that the image is constructable. Reduce to a target that is the spectrum of a local ring.Show that the generic point is hit.) I think Joe mentioned that all we should need islocally of finite presentation. Older note: use “going down theorem” for flat ringmaps. See [AM, Ex. 7.25]25 [One more version of open: if π is a morphism of finite type flat lft morphisms are

openof Noetherian schemes, then the locus on the source where π is flat is open, H.E.III.9.4, EGAIV3.11.1.1.]

26[Kirten’s argument is kirsten18.24.pdf.][Some more tor fun: Suppose (R,m) is a local ring. I’m not sure if we need Noetherian.

Proposition: If Tori(M,N) = 0 for all N , i > 0, then M is free, and vice versa. (All we need isTor1(M,k) = 0 and Nakayama.) We need that M is finite type.

One direction is trivial. For the other, choose

0→ K → R? →M → 0

so that the map on the right is an isomorphism modulo m. Then apply ⊗Rk to get K = mK,from which we are done.

Corollary: M has a finite free resolution iff it has finite Tor dimension. Method: Construct0 → K → R? → M → 0. K will have lower Tor-dimension than M (assuming M doesn’t haveTor dimension 0).

(I believe Auslander-Buchsbaum says something like: (R,m) is regular iff all M have thisproperty.)

29.3 Flat implies constant Euler characteristic

25flatlftopen, H.E.III.9.1

26H.E.III.9.4

476

I should state the correct converse to this important theorem. If the pushforwardof sufficiently high twists of F is locally free, then we get flatness. Hartshorne’sstatement is a very special case of this. I heard this from Barbara.

We come to an important consequence of flatness. We’ll see that this result implies manyanswers and examples to questions that we would have asked before we even knew about flatness.

29.3.1. Important Theorem. — Suppose f : X → Y is a projective morphism, and F isa coherent sheaf on X, flat over Y . Then

P(−1)ihi(XyF|y) is a locally constant function in

y ∈ Y . In other words, the Euler characteristic of F is constant in the fibers.27

[The proof requires Y locally Noetherian, so likely add it to the hypotheses. Also, the proofis short, so give it here.]

This is first sign that cohomology behaves well in families. (We’ll soon see a second: theSemicontinuity Theorem 29.6.4.) Before getting to the proof, I’ll show you some of its manyconsequences. (A second proof will be given after the semicontinuity discussion.)

The theorem also gives a necessary condition for flatness. It also sufficient if target is integraland locally Noetherian, although we won’t use this. (Reference: You can translate HartshorneTheorem III.9.9 into this.) I seem to recall that both the necessary and sufficient conditions aredue to Serre, but I’m not sure of a reference. [Very possibly [FAC]] It is possible that integralityis not necessary, and that reducedness suffices, but I haven’t checked.

29.3.2. Corollary. — Assume the same hypotheses and notation as in Theorem 29.3.1. Thenthe Hilbert polynomial of F is locally constant as a function of y ∈ Y .

Thus for example a flat family of varieties in projective space will all have the same degree andgenus (and the same dimension!). Another consequence of the corollary is something remarkablyuseful.

29.3.3. Corollary. — An invertible sheaf on a flat projective family of connected nonsingularcurves has locally constant degree on the fibers.

[Prove this in a better way?! Well, it looks like a bizarre way to define degree.But it is actually reasonable; consider for example the Snapper’s Theorem way todefine the intersection of a bunch of Cartier divisors.]

Proof. An invertible sheaf L on a flat family of curves is always flat (as locally it is isomorphicto the structure sheaf). Hence χ(Ly) is constant. From the Riemann-Roch formula χ(Ly) =deg(Ly)− g(Xy) + 1, using the local constancy of χ(Ly), the result follows.

Riemann-Roch holds in more general circumstances, and hence the corollary does too. Tech-nically, in the example I’m about to give, we need Riemann-Roch for the union of two P1’s, whichI haven’t shown. This can be shown in three ways. (i) I’ll prove that Riemann-Roch holds forprojective generically reduced curves later. (ii) You can prove it by hand, as an exercise. (iii) Youcan consider this curve C inside P1 × P1 as the union of a “vertical fiber” and “horizontal fiber”.Any invertible sheaf on C is the restriction of some O(a, b) on P1 × P1. Use additivity of Euler

characteristics on 0 → OP1×P1 (a − 1, b − 1) → OP1×P1 (a, b) → OC(a, b) → 0, and note that wehave earlier computed the χ(OP1×P1 (c, d)).

This result has a lot of interesting consequences.

29.3.4. Example of a proper non-projective surface. We can use it to show that a certain propersurface is not projective. Here is how. 28proper non-projective

surface Fix any field with more than two elements. We begin with a flat projective family of curveswhose X → P1, such that the fiber X0 over 0 is isomorphic to P1, and the fiber X∞ over ∞ isisomorphic to two P1’s meeting at a point, X∞ = Y∞ ∪ Z∞. For example, consider the family ofconics in P2 (with projective coordinates x, y, z) parameterized by P1 (with projective coordinatesλ and µ given by

λxy + µz(x+ y + z) = 0.

This family unfortunately is singular for [λ;µ] = [0; 1] (as well as [1; 0] and one other point), sochange coordinates on P1 so that we obtain a family of the desired form.

27flateuler

28nonprojectivesurface

477

We now take a break from this example to discuss an occasionally useful construction.

29.3.5. Gluing two schemes together along isomorphic closed subschemes. Suppose X ′ andX′′ are two schemes, with closed subschemes W ′ → X′ and W ′′ → X′′, and an isomorphismW ′ →W ′′. Then we can glue together X ′ and X′′ along W ′ ∼= W ′′. We define this more formallyas the coproduct: gluing along closed sub-

schemesW ′ ∼= W ′′ //

X′

X′′ // ?.

Exercise. Prove that this coproduct exists. Hint: work by analogy with our product con-struction. If the coproduct exists, it is unique up to unique isomorphism. Start with judiciouslychosen affine open subsets, and glue. [Possibly better: describe the topological space, then thesheaf. Show that this is a scheme. Worry later (maybe) that this is the coproduct.

Warning: You might hope that if you have a scheme X with two disjoint closed subschemesW ′ and W ′′, and an isomorphism W ′ → W ′′, then you should be able to glue X to itself alongW ′ → W ′′. This is not always possible! I’ll give an example shortly. [ref ] You can still makesense of the quotient as an algebraic space, which I will not define here. [ref forward to algspace]

29.3.6. Back to the non-projective surface. Now take two copies of the X we defined above; callthem X′ and X′′. Glue X′ to X′′ by identifying X′0 with Y ′′∞ (in any way you want) and Y ′∞ withX′′0 . (Somewhat more explicitly: we are choosing an isomorphism X ′0 ∪ Y ′∞ with X′′0 ∪ Y ′′∞ that“interchanges the components”.) I claim that the resulting surface X is proper and not projectiveover the base field k. The first is an exercise.

Exercise. Show that X is proper over k. (Hint: show that the union of two proper schemesis also proper.)

Suppose now that X is projective, and is embedded in projective space by an invertible sheaf(line bundle) L. Then the degree of L on each curve of X is non-negative [ref ]. For any curveC ⊂ X, let degC be the degree of L on C (or equivalently, the degree of C under this projectiveembedding). Pull L back to X ′. Then this is a line bundle on a flat projective family, so thedegree is constant in fibers. Thus [check that I say what I mean here.]

degX′0 = deg(Y ′∞ ∪ Z′∞) = deg Y ′∞ + degZ′∞ > deg Y ′∞.

(Technically, we have not shown that the middle equality holds, so you should think about whythat is clear.) Similarly degX ′′0 > deg Y ′′∞. But after gluing, X ′0 = Y ′′∞ and X′′0 = Y ′∞, so we havea contradiction.

29.3.7. Remark. This is a stripped down version of Hironaka’s example in dimension 3 (§27.5.A).[Hartshorne, but actually I should give a better version.] Hironaka’s example has theadvantage of being nonsingular. I’ll present that example (and show how this one comes fromHironaka) when we discuss blow-ups. (I think it is a fact that nonsingular proper surfaces over afield are always projective.) 29 Hironaka’s example

29.3.8. Unimportant remark. You can do more fun things with this example. For example, weknow that projective surfaces can be covered by three affine open sets. This can be used to givean example of (for any N) a proper surface that requires at least N affine open subsets to coverit (see my paper with Mike Roth on my preprints page, Example 4.9).

29.3.9. Problematic nature of the notion of “projective morphism”. This example shows thatthe notion of being projective isn’t a great notion. There are four possible definitions that mightgo with this notion. (1) We are following Grothendieck’s definition. This notion is not local onthe base. For example, by following the gluing above for the morphisms X ′ → P1 and X′′ → P1,we obtain a morphism π : X → P1 ∪ P1, where the union on the right is obtained by gluing the 0of the first P1 to the ∞ of the second, and vice versa. Then away from each node of the target, πis projective. (You could even give some explicit equations if you wanted.) Give some explicit

29Hironaka

478

equations. However, we know that π is not projective, as ρ : P1 ∪ P1 → Spec k is projective, butwe have already shown that ρ π : X → Spec k is not projective.

(2) Hartshorne’s definition is designed for finite type k-schemes, and is definitely the wrongone for schemes in general.

(3) You could make our notion “local on the base” by also requiring more information: e.g. thenotion of a projective morphism could be a morphism of schemes X → Y along with an invertiblesheaf L on X that serves as an O(1). This is a little unpleasant; when someone says “considera projective surface”, they usually wouldn’t want to have any particular projective embeddingpreferred.

(4) Another possible notion is that of locally projective: π : X → Y is locally projective ifthere is an open cover of Y by Ui such that over each Ui, π is projective (in our original sense(1)). The disadvantage is that this isn’t closed under composition, as is shown by our exampleX → P1 ∪ P1 → Spec k.

29.3.10. Example: You can’t always glue a scheme to itself along isomorphic disjoint subschemes.I’ll use a variation of the above example. We’ll see that you can’t glue X to itself along anisomorphism X0

∼= Y∞. (To make this a precise statement: there is no morphism π : X → Wsuch that there is a curve C →W such that π−1(W −U) = X −X0 − Y∞, and π maps both X0

and Y∞ isomorphically to W .) A picture here is essential!If there were such a scheme W , consider the point π(Y∞ ∩ Z∞) ∈ W . It has an affine

neighborhood U ; let K be its complement. Consider π−1(K). This is a closed subset of X,missing Y∞ ∩ Z∞. Note that it meets Y∞ (as the affine open U can contain no P1’s) and Z∞.Discard all components of π−1(K) that are dimension 0, and that contain components of fibers;call what’s left K ′. Caution: I need to make sure that I don’t end up discarding the points onY∞ and Z∞. I could show that π−1(K) has pure codimension 1, but I’d like to avoid doing that.For now, assume that is the case; I may patch this later. Then K ′ is an effective Cartier divisor,

inducing an invertible sheaf on the surface X, which in turn is a flat projective family over P1.Thus the degree of K ′ is constant on fibers. Then we get the same sort of contradiction:

degK′ Y∞ = degK′ X0 = degK′ Y∞ + degK′ Z∞ > degK′ Y∞.

This led to a more wide-ranging discussion. A surprisingly easy theorem (which you can findin Mumford’s Abelian Varieties [MAV] for example) states that if X is a projective k-schemewith an action by a finite group G, then the quotient X/G exists, and is also a projective scheme.(One first has to define what one means by X/G!) Refer to this in a section on quotients inalgebraic geometry. If you are a little careful in choosing the isomorphisms used to build ournonprojective surface (picking X ′0 → Y ′′∞ and X′′0 → Y ′∞ to be the “same” isomorphisms), then

there is a Z/2-action on X (“swapping the P1’s”), we have shown that the quotient W does notexist as a scheme, hence giving another proof (modulo things we haven’t shown) that X is not

projective. This is currently bogus, so this should be moved to the blow-up section.

29.4 Proof of Important Theorem on constancy of Eulercharacteristic in flat families

Now you’ve seen a number of interesting results that seem to have nothing to do with flatness.I find this a good motivation for this motivation: using the concept, we can prove things that

were interested in beforehand. It is time to finally prove Theorem 29.3.1.

Proof. The question is local on the base, so we may reduce to case Y is affine, say Y = SpecB,so X → PnB for some n. We may reduce to the case X = PnB (as we can consider F as a sheaf onPnB). We may reduce to showing that Hilbert polynomial F(m) is locally constant for all m 0(as by Serre vanishing for m 0, the Hilbert polynomial agrees with the Euler characteristic).Now consider the Cech complex C∗ for F . Note that all the terms in the Cech complex are flat.

Twist by O(m) for m 0, so that all the higher pushforwards vanish. Hence Γ(C∗(m)) is exactexcept at the first term, where the cohomology is Γ(π∗F(m)). We tack on this module to the frontof the complex, so it is once again exact. Thus by exercise 29.2.C, as we have an exact sequencein which all but the first terms are known to be flat, the first term is flat as well. As it is finitely

479

generated, it is also free by an earlier fact (flat and finitely generated over a Noetherian local ringequals free [Ben says: flat + locally of finite presentationn over arbitrary local ring. Except wemay need Noetherian hypotheses to get from the local ring fact to the general fact.]), and thushas constant rank.

We’re interested in the cohomology of the fibers. To obtain that, we tensor the Cech resolutionwith k(y) (as y runs over Y ) and take cohomology. Now the extended Cech resolution (withΓ(π∗F(m)) tacked on the front) is an exact sequence of flat modules, and hence remains exactupon tensoring with k(y) (or indeed anything else). (Useful translation: cohomology commuteswith base change.) [We need flatness here! Not cohomology and flat base change, but the factthat cohomology of flat commutes with base change. Mention earlier thes paired results, e.g. C•

is exact implies C•⊗A is exact if A is flat or C• is flat. Thus Γ(π∗F(m))⊗k(y) ∼= Γ(π∗F(m)|y).Thus the dimension of the Hilbert function is the rank of the locally free sheaf at that point, whichis locally constant. Say that better!

29.5 Start of next day’s class

At this point, you’ve already seen a large number of facts about flatness. Don’t be over-whelmed by them; keep in mind that you care about this concept because we have answeredquestions we cared about even before knowing about flatness. Here are three examples. (i) Ifyou have a short exact sequence where the last is locally free, then you can tensor with anythingand the exact sequence will remain exact. (ii) We described a morphism that is proper but notprojective. (iii) We showed that you can’t always glue a scheme to itself.

Here is a summary of what we know, highlighting the hard things.

• definition; basic properties (pullback and localization). flat base change commutes withhigher pushforwards

• Tor: definition and symmetry. (Hence tensor exact sequences of flats with anythingand keep exactness.)

• ideal-theoretic criterion: Tor1(M,A/I) = 0 for all I. (flatness over PID = torsion-free;over dual numbers) (important special case: DVR)

• for coherent modules over Noetherian local rings, flat=locally free• flatness is open in good circumstances (flat + lft of lN is open; we should need only

weaker hypotheses)• euler characteristics behave well in projective flat families. In particular, the degree of

an invertible sheaf on a flat projective family of curves is locally constant.

29.6 Cohomology and base change theorems

Here is the type of question we are considering. We’d like to see how higher pushforwardsbehave with respect to base change. For example, we’ve seen that higher pushforward commuteswith flat base change. A special case of base change is the inclusion of a point, so this question

specializes to the question: can you tell the cohomology of the fiber from the higher pushforward?The next group of theorems I’ll discussed deal with this issue. I’ll prove things for projectivemorphisms. The statements are true for proper morphisms of Noetherian schemes too; the onefact you’ll see that I need is the following: that the higher direct image sheaves of coherent sheavesunder proper morphisms are also coherent. (I’m largely following Mumford’s Abelian Varieties[MAV]. The geometrically interesting theorems all flow from the following neat but unmotivatedresult.

[Geometry into linear algebra:]

29.6.1. Key theorem. — Suppose π : X → SpecB is a projective morphism of Noetherian[needed?] schemes, and F is a coherent sheaf on X, flat over SpecB. Then there is a finite

480

complex

0→ K0 → K1 → · · · → Kn → 0

of finitely generated projective B-modules and an isomorphism of functors 30

(63) Hp(X ×B A,F ⊗B A) ∼= Hp(K∗ ⊗B A)

for all p in the category of B-algebras A. 31

In fact, Ki will be free for i > 0. For i = 0, it is projective hence flat hence locally free on Y(Theorem 29.2.13).

Translation/idea: Given π : X → SpecB, we will have a complex of vector bundles onthe target that computes cohomology (higher-pushforwards), “universally” (even after any basechange). The idea is as follows: take the Cech complex, produce a “quasiisomorphic” complex (acomplex with the same cohomology) of free modules. We’ll first construct the complex so that(63) holds for B = A, and then show the result for general A later. Let’s put this into practice.[Cut-and-pasted; reword later]

29.6.2. Lemma. — Let C∗ be a complex of B-modules such that Hi(C∗) are finitely generatedB-modules, and that Cp 6= 0 only if 0 ≤ p ≤ n. Then there exists a complex K∗ of finitelygenerated B-modules such that Kp 6= 0 only if 0 ≤ p ≤ n and Kp is free for p ≥ 1, and ahomomorphism of complexes φ : K∗ → C∗ such that φ induces isomorphisms Hi(K∗)→ Hi(C∗)for all i.

Note that Ki is B-flat for i > 0. Moreover, if Cp are B-flat, then K0 is B-flat too.For all of our purposes except for a side remark, I’d prefer a cleaner statement, where C∗

is a complex of B-modules, with Cp 6= 0 only if p ≤ n (in other words, there could be infinitelymany non-zero Cp’s). The proof is then about half as long [We only need K0 for the eulercharacteristic statement.] [Rob and Joe: this is the mapping cone, not the mappingcylinder.] 32 [Refer to other mentions of mapping cone, ??.]

Proof.Step 1. We’ll build this complex inductively, and worry about K0 when we get there.

Km

φm

δm // Km+1δm+1 //

φm+1

Km+2 //

φm+2

· · ·

· · · // Cm−1 // Cmδm

// Cm+1

δm+1// Cm+2 // · · · .

We assume we’ve defined (Kp, φp, δp) for p ≥ m+1 such that these squares commute, and the toprow is a complex, and φp defines an isomorphism of cohomology Hq(K∗)→ Hq(C∗) for q ≥ m+2and a surjection ker δm+1 → Hm+1(C∗), and the Kp are finitely generated B-modules.

We’ll adjust the complex to make φm+1 an isomorphism of cohomology, and then again tomake φm a surjection on cohomology. Let Bm+1 = ker(δm+1 : Hm+1(K∗)→ Hm+1(C∗)). Thenwe choose generators, and make these Km

1 . We have a new complex. We get the 0-maps oncohomology at level m. We then add more in to surject on cohomology on level m.

Now what happens when we get to m = 0? We have maps of complexes, where everythingin the top row is free, and we have an isomorphism of cohomology everywhere except for K0,where we have a surjection of cohomology. Replace K0 by K0/ ker δ0 ∩kerφ0. Then this gives anisomorphism of cohomology [check]. [You can get around this check.]

Step 2. We need to check that K0 is B-flat. Note that everything else in this quasiisomor-phism is B-flat. Here is a clever trick: construct the mapping cylinder (call it M∗):

0→ K0 → C0 ⊕K1 → C1 ⊕K2 → · · · → Cn−1 ⊕Kn → Cn → 0.

[Describe the maps, because they aren’t obvious to the reader. Verify the sign issue. Check thisby hand.] Then we have a short exact sequence of complexes

0→ C∗ →M∗ → K∗[1]→ 0

30keymum

31presemi

32mappingcone3

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(where K∗[1] is just the same complex as K∗, except slid over by one) yielding isomorphismsof cohomology H∗(K∗) → H∗(C∗), from which H∗(M∗) = 0. (This was an earlier exercise:given a map of complexes induces an isomorphism on cohomology, the mapping cylinder is exact.[Ben wants a better statement: given K∗ → C∗, the boundary maps in LES for 0 → C∗ →M∗ → K∗[1] → 0 are precisely the maps on cohomology. Special case: quasiisomorphism iffH∗(M) = 0.]) Now look back at the mapping cylinder M∗, which we now realize is an exactsequence. All terms in it are flat except possibly K0. Hence K0 is flat too (also by an earlierexercise)! [This is the mapping cone, not the mapping cylinder.] 33 [Refer to other mentions

of mapping cone, ??.]

29.6.3. Lemma. — Suppose K∗ → C∗ is a morphism of finite complexes of flat B-modulesinducing isomorphisms of cohomology (a “quasiisomorphism”). Then for every B-algebra A, themaps Hp(C∗ ⊗B A)→ Hp(K∗ ⊗B A) are isomorphisms.

Proof. Consider the mapping cylinder M∗, which we know is exact. Then M∗ ⊗B A is stillexact! (The reason was our earlier exercise that any exact sequence of flat modules tensored withanything remains flat.) But M∗ ⊗B A is the mapping cylinder of K∗ ⊗B A → C∗ ⊗B A, so thisis a quasiisomorphism too.

Now let’s prove the theorem!Proof of theorem 29.6.1. Choose a finite covering (e.g. the standard covering). Take the Cechcomplex C∗ for F . Apply the first lemma to get the nicer version K∗ of the same complex C∗.Apply the second lemma to see that if you tensor with B and take cohomology, you get the sameanswer whether you use K∗ or C∗.

We are now ready to put this into use. We will use it to discuss a trio of facts: theSemicontinuity Theorem, Grauert’s Theorem, and the Cohomology and Base Change Theorem.(We’ll prove the first two.) The theorem of constancy of euler characteristic in flat families alsofits in this family.

These theorems involve the following situation. Suppose F is a coherent sheaf on X, π : X →Y projective, Y (hence X) Noetherian, and F flat over Y .

Here are two related questions. Is Rpπ∗F locally free? Is φp : Rpπ∗F ⊗ k(y)→ Hp(Xy ,Fy)an isomorphism?

We have shown Key theorem 29.6.1, that if Y is affine, say Y = SpecB, then we can computethe pushforwards of F by a complex of locally free modules

0→M0 →M1 → · · · →Mn → 0

where in fact Mp is free for p > 1. Moreover, this computes pushforwards “universally”: after abase change, this remains true.

We have already shown constancy of Euler characteristic. [Mumford [MAV] p. 50Cor. 1(b); EGA III.2 Thm 7.9.4. in the proper case.] This also follows from the keyresult.

Now the dimension of the left is uppersemicontinuous by uppersemicontinuity of fiber di-mension of coherent sheaves [ref ]. The semicontinuity theorem states that the dimension of theright is also uppersemicontinuous. More formally: [Mumford p. 50, Cor. 1(a). HartshorneTheorem III.12.8]. EGA III.2, theorem 7.7.5, I.

29.6.4. Semicontinuity theorem. — Suppose X → Y is a projective morphisms of Noetherianschemes, and F is a coherent sheaf on X flat over Y . Then for each p ≥ 0, the function Y → Zgiven by y 7→ dimk(y)H

p(Xy ,Fy) is upper semicontinuous on Y . 34

So “cohomology groups jump in projective flat families”. Again, we can replace projectiveby proper once we’ve shown finite-dimensionality of higher pushforwards (which we haven’t). Forpedants: can the Noetherian hypotheses be excised?

Here is an example of jumping in action. Let C be a positive genus nonsingular projectiveirreducible curve, and consider the projection π : E × E → E. Let L be the invertible sheaf

33mappingcone4

34semicontinuity, H.T.III.12.8

482

(line bundle) corresponding to the divisor that is the diagonal, minus the section p0 ∈ E. thenLp0 is trivial, but Lp is non-trivial for any p 6= p0 (as we’ve shown earlier [ref? in “fun withcurves”]). Thus h0(E,Lp) is 0 in general, but jumps to 1 for p = p0. Heard from Joe Harris.

[Example where things if we give up projectivity is easy: open immersion. Ex-ample where it fails if we give up flatness?] Remark. Deligne showed that in the smoothcase, at least over C, there is no jumping of cohomology of the structure sheaf. Hartshornecredits this to Deligne, and gives a reference (Example III.12.9.3). 35 Add the ref.

Proof. The result is local on Y , so we may assume Y is affine. Let K∗ be a complex as in thekey theorem 29.6.1. By localizing further, we can assume K∗ is locally free. So we are computingcohomology on any fiber using a complex of vector bundles.

Then for y ∈ Ydimk(y)H

p(Xy ,Fy) = dimk(y) ker(dp ⊗A k(y)) − dimk(y) im(dp−1 ⊗A k(y))= dimk(y)(K

p ⊗ k(y)) − dimk(y) im(dp ⊗A k(y)) − dimk(y) im(dp−1 ⊗A k(y))(Side point: by taking alternating sums of these terms, we get a second proof of Theo-

rem 29.3.1 that χ(Xy ,Fy) =P

(−1)ihi(Xy ,Fy) is a constant function of y. I mention this becauseif extended the fact that higher cohomology of coherents is coherent under proper pushforwards,we’d also have Theorem 29.3.1 in this case.)

Now dimk(y) im(dp ⊗A k(y)) is a lower semicontinuous function on Y . Reason: the locuswhere the dimension is less than some number q is obtained by setting all q × q minors of thematrix Kp → Kp+1 to 0. So we’re done!

29.7 Line bundles are trivial in a Zariski-closed locus[horrible title!], and glimpses of the relative Picard scheme

[Miscellaneous facts. Over an algebraically closed field: If V is projective then the torsiongroup that numerically equivalent to 0 modulo algebraically equivalent to 0 is a finite group,Kleiman’s numerical criterion chapter (2?), Cor. 1, §2, a result due to Matsusaka. Grothendieckproves it for non-projective V , as stated in Keliman. Moreover, Pic modulo those algebraicallyequivalent to 0 is a free group of finite rank ρ(V ), provided that we know this to the case for anon-singular surface (from the last chapter of Kleiman, Remark 1, §1).]

29.7.1. Proposition. — Suppose L is an invertible sheaf on an integral projective scheme Xsuch that both L and L∨ have non-zero sections. Then L is the trivial sheaf.

As usual, “projective” may be replaced by “proper”. The only fact we need (which we haven’tproved) is that the only global functions on proper schemes are constants. (We haven’t provedthat. It follows easily from the valuative criterion of properness — but we haven’t proved thateither!)

Proof. Suppose s and t are the non-zero sections of L and L∨. Then they are both non-zero at thegeneric point (or more precisely, in the stalk at the generic point). (Otherwise, they would be thezero-section — this is where we are using the integrality of X.) Under the map L⊗L∨ →O, s⊗ tmaps to st, which is also non-zero. But the only global functions (global sections of OX) are theconstants, so st is a non-zero constant. But then s is nowhere 0 (or else st would be somewherezero), so L has a nowhere vanishing section, and hence is trivial (isomorphic to OX) [ref?].

Now suppose X → Y is a flat projective morphism with integral fibers. (It is a “flat family ofgeometrically integral schemes”.) Suppose that L is an invertible sheaf. Then the locus of y ∈ Ywhere Ly is trivial on Xy is a closed set. Reason: the locus where h0(Xy ,Ly) ≥ 1 is closed bythe Semicontinuity Theorem 29.6.4, and the same holes for the locus where h0(Xy ,L∨y ) ≥ 1.

(Similarly, if L′ and L′′ are two invertible sheaves on the family X, the locus of points ywhere L′y ∼= L′′ is a closed subset: just apply the previous paragraph to L := L′ ⊗ (L′′)∨.)

35H.Example.III.12.9.3

483

In fact, we can jazz this up: for any L, there is in a natural sense a closed subscheme whereL is trivial. More precisely, we have the following theorem.

29.7.2. Seesaw Theorem. — Suppose π : X → Y is a projective flat morphism to a Noetherianscheme, all of whose fibers are geometrically integral schemes, and L is an invertible sheaf on X.Then there is a unique closed subscheme Y ′ → Y such that for any fiber diagram

X ×Y Zg //

ρ

X

π

Z

f // Ysuch that g∗L ∼= ρ∗M for some invertible sheaf M on Z, then f factors (uniquely) throughY ′ → Y . [MAV, p. 89], generalized appropriately. I’m a little concerned that I’massuming reducedness. [I’ll mention this when discussing closed subfunctors.] [Thiscomes up with the moduli space of curves. Also the existence of the Picard variety.]36

I want to make three comments before possibly proving this.• I have no idea why it is called the seesaw theorem.• As a special case, there is a “largest closed subscheme” on which the invertible sheaf is the

pullback of a trivial invertible sheaf.• Also, this is precisely the statement that the functor is representable Y ′ → Y , and that

this morphism is a closed immersion.I’m not going to use this, so I won’t prove it. But a slightly stripped down version of this

appears in Mumford (p. 89), and you should be able to edit his proof so that it works in thisgenerality. Picard scheme

There is a lesson I want to take away from this: this gives evidence for existence of a veryimportant moduli space: the Picard scheme. The Picard scheme PicX/Y → Y is a scheme over

Y which represents the following functor: Given any T → Y , we have the set of invertible sheaveson X×Y T , modulo those invertible sheaves pulled back from T . In other words, there is a naturalbijection between diagrams of the form

L

X ×T Y //

X

T // Y

and diagrams of the form

PicX/Y

T

;;xxxxxxxxx // Y.It is a hard theorem (due to Grothendieck) that (at least if Y is reasonable, e.g. locally Noetherian— I haven’t consulted the appropriate references) PicX/Y → Y exists, i.e. that this functor isrepresentable. In fact PicX/Y is of finite type.

We’ve seen special cases before when talking about curves: if C is a geometrically integralcurve over a field k, of genus g, PicC = PicC/k is a dimension g projective nonsingular variety.

Given its existence, it is easy to check that PicX/Y is a group scheme over Y , using ourfunctorial definition of group schemes.

29.7.A. Exercise. Do this!The group scheme has a zero-section 0 : Y → PicX/Y . This turns out to be a closed

immersion. The closed subscheme produced by the Seesaw theorem is precisely the pullback of

36seesaw

484

the 0-section. I suspect that you can use the Seesaw theorem to show that the zero-section is aclosed immersion.

29.7.B. Exercise. Show that the Picard scheme forX → Y (with our hypotheses: the morphismis flat and projective, and the fibers are geometrically integral) is separated over Y by showingthat it satisfies the valuative criterion of separatedness.

29.8 Cohomology and base change theorems

[Earlier things we will use: cohomology and flat base change. Remark: if allhigher pushforwards are flat then cohomology and base change always commute.Special case of EGA III.2, 6.9.9.2.]

We’re in the midst of discussing a family of theorems involving the following situation. Sup-pose F is a coherent sheaf on X, π : X → Y projective, Y (hence X) Noetherian, and F flat overY .

Here are two related questions. Is Rpπ∗F locally free? Is φp : Rpπ∗F ⊗ k(y)→ Hp(Xy ,Fy)an isomorphism?

[repeated par] We have shown Key theorem 29.6.1, that if Y is affine, say Y = SpecB,then we can compute the pushforwards of F by a complex of locally free modules

0→M0 →M1 → · · · →Mn → 0

where in fact Mp is free for p > 1. Moreover, this computes pushforwards “universally”: after abase change, this remains true.

We have already shown the constancy of Euler characteristic, and the semicontinuity theorem.I’m now going to discuss two big theorems, Grauert’s theorem and the Cohomology and basechange theorem, that are in some sense the scariest in Hartshorne, coming at the end of ChapterIII (along with the semicontinuity theorem). I hope you agree that semicontinuity isn’t that scary(given the key fact). I’d like to discuss applications of these two theorems to show you why youcare; then given time I’ll give proofs. I’ve found the statements worth remembering, even thoughthey are a little confusing.

Note that if Rpπ∗F is locally free and φp is an isomorphism, then the right side is locallyconstant. The following is a partial converse.

29.8.1. Grauert’s Theorem. — If Y is reduced, then hp locally constant implies Rpπ∗F islocally free and φp is an isomorphism. [Mumford p. 50 Cor. 2, most of it; the rest is oncbc. Hartshorne Cor. III.12.9, except he does only the integral case.] 37

[Perhaps a better conclusion: replace φp with any base change. “Pushforward commuteswith any base change.”]

29.8.2. Cohomology and base change theorem. — Assume φp is surjective. Then thefollowing hold.

(a) φp is an isomorphism, and the same is true nearby. [Rk: The hypothesis is triviallysatisfied in the common case Hp = 0. If Hp = 0 at a point, then it is true nearby bysemicontinuity.]

(b) φp−1 is surjective (=isomorphic) if and only if Rpπ∗F is locally free. [This in turnimplies that hp is locally constant.]

DIRECT PROOF?!! [This is Hartshorne Theorem 12.11. Part (b) generalizesMumford p. 50 Cor. 2.] Brian O: (a) implies (b) and (c); and when base is reduced,(b) implies (a). (a) coho and base change commute for F in degrees i and i+ 1; (b)the function y 7→ hi+1 is locally constant; (c) Ri+1f∗F locally free. From EGA III.2§7. Ref from another source: EGA III.?.7.7. 38

Again, perhaps that conclusion about φp−1 being surjective should be replacedwith: φp is surjective at all points implies that the (p − 1)st cohomology commuteswith any base change.

37C.H.III.12.9, Grauert

38cbc, H.T.III.12.11

485

Notice that (a) is about just what happens over the reduced scheme, but (b) has a neattwist: you can check things over the reduced scheme, and it has implications over the scheme asa whole! [Mention that you should feel free to use this when p = 0, and p− 1 gives you somethingautomatic.]

Here are a couple of consequences.

29.8.A. Exercise. Suppose Hp(Xy ,Fy) = 0 for all y ∈ Y . Show that φp−1 is an isomorphism

for all y ∈ Y . (Hint: cohomology and base change (b).) [[MAV, p. 53] Cor. 3] 39

29.8.B. Exercise. Suppose Rpπ∗F = 0 for p ≥ p0. Show that Hp(Xy ,Fy) = 0 for all y ∈ Y ,k ≥ k0. (Same hint. You can also do this directly from the key theorem above.) [[MAV, p. 53]Cor. 4.] 40

29.9 When the pushforward of the functions on X are thefunctions on Y

Many fun applications happen when a certain hypothesis holds, which I’ll now describe.We say that π satisfies (*) if it is projective, and the natural morphism OY → π∗OX is an

isomorphism. Here are two statements that will give you a feel for this notion. [Perhaps addNoetherian hypotheses?] First:

29.9.A. Important exercise. Suppose π is a projective flat family, each of whose fibers are(nonempty) integral schemes, or more generally whose fibers satisfy h0(Xy) = 1. Then (*) holds.(Hint: consider

OY ⊗ k(y) // (π∗OX)⊗ k(y) φ0// H0(Xy ,OXy ) ∼= k(y) .

The composition is surjective, hence φ0 is surjective, hence it is an isomorphism (by the Coho-mology and base change theorem 29.8.2 (a)). Then thanks to the Cohomology and base changetheorem 29.8.2 (b), π∗OX is locally free, thus of rank 1. If I have a map of invertible sheavesOY → π∗OX that is an isomorphism on closed points, it is an isomorphism (everywhere) byNakayama.) 41

Note in the previous exercise: we are obtaining things not just about closed points!Second: we will later prove a surprisingly hard result, that given any projective (proper)

morphism of Noetherian schemes satisfying (*) (without any flatness hypotheses!), the fibers areall connected (“Zariski’s connectedness lemma” 29.14.1). [Also: rigidity lemma, which doesn’tuse flatness.]

29.9.B. Exercise (the Hodge bundle; important in Gromov-Witten theory). Supposeπ : X → Y is a projective flat family, all of whose geometric fibers are connected reduced curvesof arithmetic genus g. Show that R1π∗OX is a locally free sheaf of rank g. This is called theHodge bundle. [Hint: use cohomology and base change (b) twice, once with p = 2, and once withp = 1.] Hodge bundle

Here is the question we’ll address in this section. Given an invertible sheaf L on X, wewonder when it is the pullback of an invertible sheafM on Y . Certainly it is necessary for it to betrivial on the fibers. We’ll see that (*) holds, then this basically suffices. Here is the idea: givenL, how can we recover M? Thanks to the next exercise, it must be π∗L. [(Picture of this linebundle: we get one section at each spot.)]

29.9.C. Exercise. Suppose π : X → Y satisfies (*). Show that if M is any invertible sheaf onY , then the natural morphismM→ π∗π∗M is an isomorphism. In particular, we can recoverMfrom π∗M by pushing forward. (Hint: projection formula.)

39mp53c3

40mp53c4

41geointstar

486

29.9.1. Proposition. — Suppose π : X → Y is a morphism of locally Noetherian integralschemes with geometrically integral fibers (hence by Exercise 29.9.A satisfying (*)). Suppose alsothat Y is reduced, and L is an invertible sheaf on X that is trivial on the fibers of π (i.e. Ly is atrivial invertible sheaf on Xy). Then π∗L is an invertible sheaf on Y (call itM), and L = π∗M.42 I think I need geometrically integral fibers here, in order to use Grauert.

Proof. To show that there exists such an invertible sheaf M on Y with π∗M ∼= L, it suffices toshow that π∗L is an invertible sheaf (call it M) and the natural homomorphism π∗M→ L is anisomorphism.

Now by Grauert’s theorem 29.8.1, π∗L is locally free of rank 1 (again, call it M), andM⊗OY

k(y) → H0(Xy ,Ly) is an isomorphism. We have a natural map of invertible sheavesπ∗M = π∗π∗L → L. To show that it is an isomorphism, we need only show that it is surjective,i.e. show that it is surjective on the fibers, which is done.

Here are some consequences.A first trivial consequence: if you have two invertible sheaves on X that agree on the fibers

of π, then they differ by a pullback of an invertible sheaf on Y .

29.9.D. Exercise. Suppose X is an integral Noetherian scheme. Show that Pic(X × P1) ∼=PicX×Z. (Side remark: If X is non-reduced, this is still true, see Hartshorne Exercise III.12.6(b).It need only be connected of finite type over k. Presumably locally Noetherian suffices.) Extendthis to X × Pn. Extend this to any Pn-bundle over X.43 [Exercise 23.9.1 may help.] 44

29.9.E. Exercise. Suppose X → Y is the projectivization of a vector bundle F over a reducedlocally Noetherian scheme (i.e. X = Proj Sym∗ F). Then I think we’ve already shown in anexercise that it is also the projectivization of F ⊗L. If Y is reduced and locally Noetherian, showthat these are the only ways in which it is the projectivization of a vector bundle. (Hint: notethat you can recover F by pushing forward O(1).)

29.9.F. Exercise. Suppose π : X → Y is a projective flat morphism over a Noetherian integralscheme, all of whose geometric fibers are isomorphic to Pn (over the appropriate field). Show thatthis is a projective bundle if and only if there is an invertible sheaf on X that restricts to O(1) onall the fibers. (One direction is clear: if it is a projective bundle, then it has a projective O(1).In the other direction, the candidate vector bundle is π∗O(1). Show that it is indeed a locallyfree sheaf of the desired rank. Show that its projectivization is indeed π : X → Y .) [I haven’tchecked the latter.] [Perhaps mention the universal conic to show that this doesn’talways hold. Mention Tsen’s theorem to show that it always holds over a curve.]

29.9.G. Exercise (an example of a Picard scheme). Show that the Picard scheme of P1k over

k is isomorphic to Z.

29.9.2. Harder (but worthwhile) exercise (another example of a Picard scheme). Show that if Eis an elliptic curve over k (a geometrically integral and nonsingular genus 1 curve with a markedk-point), then PicE is isomorphic to E×Z. Hint: Choose a marked point p. (You’ll note that thisisn’t canonical.) Describe the candidate universal invertible sheaf on E × Z. Given an invertiblesheaf on E ×X, where X is an arbitrary Noetherian scheme, describe the morphism X → E× Z.Is there something in Hartshorne IV on this?

29.10 The rigidity lemma

The rigidity lemma is another useful fact about morphisms π : X → Y such that π∗OX(condition (*) of the previous section). It is quite powerful, and quite cheap to prove, so we mayas well do it now.

42generalization of E.H.III.12.4; no finite type conditions on Y .

43H.E.III.12.5, extended.

44e:Pnpushuse

487

29.10.1. Rigidity lemma (first version). — Suppose we have a commutative diagram

X

e closed, e∗OX = OY AAA

AAAA

f // Z

g quasi–proj.~~

~~~~

~

Y

where Y is locally Noetherian, where f takes Xy for some y ∈ Y . Then there is a neighborhoodU ⊂ Y of y on which this is true. Better: over U , f factors through the projection to Y , i.e. thefollowing diagram commutes for some choice of h:

X|Uf //

e

!!CCC

CCCC

CZ|U

U

h

==||||||||

Proof. This proof is very reminiscent of an earlier result, when we showed that a projectivemorphisms with finite fibers is a finite morphism.

We can take g to be projective. We can take Y to be an affine neighborhood of y. ThenZ → PnY for some n. Choose a hyperplane of Pny missing f(Xy), and extend it to a hyperplane

H of PnY . (If Y = SpecB, and y = [n], then we are extending a linear equation with coefficientsin B/n to an equation with coefficients in B.) Pull back this hyperplane to X; the preimage is aclosed subset. The image of this closed subset in Y is also a closed set K ⊂ Y , as e is a closed map.But y /∈ K, so let U = Y −K. Over U , f(Xy) misses our hyperplane H. Thus the map Xy → PnUfactors through Xy → AnU . Thus the map is given by n functions on X|U . But e∗OX ∼= OY , sothese are precisely the pullbacks of functions on U , so we are done.

29.10.2. Rigidity lemma (second version). — Same thing, with the condition on g changedfrom “projective” to simply “finite type”.

Proof. Shrink Y so that it is affine. Choose an open affine subset Z ′ of Z containing the f(Xy).Then the complement the pullback of K = Z − Z ′ to X is a closed subset of X whose image inY is thus closed (as again e is a closed map), and misses y. We shrink Y further such that f(X)lies in Z′. But Z′ → Y is quasiprojective, so we can apply the previous version.

Here is another mild strengthening.

29.10.3. Rigidity lemma (third version). — If X is reduced and g is separated, and Y isconnected, and there is a section Y → X, then we can take U = Y .

Proof. We have two morphisms X → Z: f and f s e which agree on the open set U . But we’veshown earlier that any two morphisms from a reduced scheme to a separated scheme agreeing ona dense open set are the same.ref

Here are some nifty consequences.

29.10.4. Corollary (abelian varieties are abelian). — Suppose A is a projective integralgroup variety (an abelian variety) over a field k. Then the multiplication map m : A×A→ A iscommutative. Make the hypotheses better.

Proof. Consider the commutator map c : A×A→ A that corresponds to (x, y) 7→ xyx−1y−1. Wewish to show that this map sends A×A to the identity in A. Consider A×A as a family over thefirst factor. Then over x = e, c maps the fiber to e. Thus by the rigidity lemma (third version),the map c is a function only of the first factor. But then c(x, y) = c(x, e) = e.

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29.10.A. Exercise. By a similar argument show that any map f : A → A′ from one abelianvariety to another is a group homomorphism followed by a translation. (Hint: reduce quickly tothe case where f sends the identity to the identity. Then show that “f(x+y)−f(x)−f(y) = e”.)

29.11 Proof of Grauert’s theorem

I’ll prove Grauert, but not Cohomology and Base Change. It would be wonderful if Coho-mology and Base Change followed by just mucking around with maps of free modules over a ring.

29.11.A. Exercise++. Find such an argument.We’ll need a preliminary result.

29.11.1. Lemma. — Suppose Y = SpecB is a reduced Noetherian scheme, and f : M → Nis a homomorphism of coherent free (hence projective, flat) B-modules. If dimk(y) im(f ⊗ k(y))is locally constant, then there are splittings M = M1 ⊕M2 and N = N1 ⊕N2 with f killing M1,and sending M2 isomorphically to N1.

Proof. [Here ⊗k means A/m.] Note that f(M) ⊗ k ∼= f(M ⊗ k) from that surjection. Saybetter. From 0→ f(M)→ N → N/f(M)→ 0 we have

f(M) ⊗ k //

N ⊗ k //

N/f(M) ⊗ k

// 0

f(M ⊗ k) // N ⊗ k // N ⊗ k/f(M ⊗ k) // 0

from which (N/f(M)) ⊗ k ∼= (N ⊗ k)/f(M ⊗ k). Now the one on the right has locally constantrank, so the one on the left does too, hence is locally free, and flat, and projective. Hence0→ f(M)→ N → N/f(M)→ 0 splits, so let N2 = N/f(M), N1 = f(M). Also, N and N/f(M)

are flat and coherent, hence so is f(M).We now play the same game with

0→ ker f →M → f(M)→ 0.

f(M) is projective, hence this splits. Let ker f = M1.

Now let’s prove Grauert’s theorem 29.8.1. We can use this lemma to rewrite

Mp−1dp−1 // Mp dp // Mp+1

as Zp−1 ⊕Kp−1 // Bp ⊕Hp ⊕Kp // Bp+1 ⊕Kp+1 where dp−1 sends Kp−1 isomor-

phically onto Bp (and is otherwise 0), and dp sends Kp isomorphically onto Bp+1. Here Hp isa projective module, so we have local freeness. Thus when we tensor with some other ring, thisstructure is preserved as well; hence we have isomorphism.

29.12 Dimensions behave well for flat morphisms; flatness ofgiven relative dimension

There are a few easier statements about flatness that I could have said much earlier.Here’s a basic statement about how dimensions behave in flat families.

29.12.1. Proposition. — Suppose f : X → Y is a flat morphism of schemes all of whosestalks are localizations of finite type k-algebras, with f(x) = y. (For example, X and Y could

489

be finite type k-schemes.) Then the dimension of Xy at x plus the dimension of Y and y is the

dimension X at x. 45

In other words, there can’t be any components contained in a fibers; and you can’t have anydimension-jumping.

In class, I first incorrectly stated this with the weaker hypotheses that X and Y are justlocally Noetherian. Kirsten pointed out that I used the fact that height = codimension, which isnot true for local Noetherian rings in general. However, we have shown it for local rings of finitetype k-schemes. Joe suggested that one could work around this problem.

Proof. This is a question about local rings, so we can consider SpecOX,x → SpecOY,y . We mayassume that Y is reduced. We prove the result by induction on dimY . If dimY = 0, the result isimmediate, as Xy = X and dimy Y = 0.

Now for dimY > 0, I claim there is an element t ∈ m that is not a zero-divisor, i.e. is notcontained in any associated prime, i.e. (as Y is reduced) is not contained in any minimal prime.Let p1, . . . , pn be the (finite number of) minimal primes. If m ⊂ p1 ∪ · · · ∪ pn, then in the firstquarter we showed (in an exercise) that m ⊂ pi for some i. But as m is maximal, and pi is minimal,we must have m = pi, and dimY = 0.

Now by flatness t is not a zero-divisor of OX,x. (Recall that non-zero-divisors pull back tonon-zero-divisors.) dimOY,y/t = dimOY,y − 1 by Krull’s principal ideal theorem 15.6.3 (here weuse the fact that codimension = height), and dimOX,x/t = dimOX,x − 1 similarly. .

29.12.2. Corollary. — Suppose f : X → Y is a flat finite-type morphism of locally Noetherianschemes, and Y is irreducible. Then the following are equivalent.

• Every irreducible component of X has dimension dimY + n.• For any point y ∈ Y (not necessarily closed!), every irreducible component of the fiberXy has dimension n.

((ii) follows H’s wording.) 46

[I suspect that we can replace the hypothesis that Y is irreducible with the hypothesis thatY is equidimensional, and this mild generalization is useful.]

29.12.A. Exercise. Prove this.Important definition: If these conditions hold, we say that π is flat of relative dimension n.

This definition will come up when we define smooth of relative dimension n [ref ].

29.12.B. Exercise.(a) Suppose π : X → Y is a finite-type morphism of locally Noetherian schemes, and Y isirreducible. Show that the locus where π is flat of relative dimension n is an open condition.(b) Suppose π : X → Y is a flat finite-type morphism of locally Noetherian schemes, and Y isirreducible. Show that X can be written as the disjoint union of schemes X0 ∪ X1 ∪ · · · whereπ|Xn : Xn → Y is flat of relative dimension n.

29.12.C. Important exercise. Use a variant of the proof of Proposition 29.12.1 to show thatif f : X → Y is a flat morphism of finite type k-schemes (or localizations thereof), then anyassociated point of X must map to an associated point of Y . (I find this an important pointwhen visualizing flatness!) [Kirsten’s argument is kirsten19.18.pdf.] [Ziyu: We want to prove: ifmx is an associated prime of 0 in the local ring OX,x, then my is an associated prime of 0 in the

local ring OY,y. If not, there exists t ∈ my which is not a zero-divisor, by flatness f#t is not azero-divisor, which contradicts that all elements of mx are zero-divisors. Rob responds: Is theresomething missing here?]

29.13 The local criterion for flatness

45H.P.III.9.5

46H.C.III.9.6

490

I’ll end our discussion of flatness with the statement of two results which can be quite useful.(Translation: I’ve seen them used.) They are both called the local criterion for flatness.

In both situations, assume that (B, n)→ (A,m) is a local morphism of local Noetherian rings(i.e. a ring homomorphism with nA ⊂ m), and that M is a finitely generated A-module. Of coursewe picture this in terms of geometry:

M

Spec(A,m)

Spec(B, n).

The local criteria for flatness are criteria for when M is flat over A. In practice, these are used intwo circumstances: to check when a morphism to a locally Noetherian scheme is flat, or when acoherent sheaf on a locally Noetherian scheme is flat.

We’ve shown that to check if M is flat, we need check if TorB1 (B/I,M) = 0 for all ideals I.The (first) local criterion says we need only deal with the maximal ideal.

29.13.1. Theorem (local criterion for flatness). — M is B-flat if and only if TorB1 (B/n,M) =0. I used this in my Murphy’s Law paper. Also stated in Eisenbud. The proof usesthe Artin-Rees lemma and the Krull intersection theorem.

(You can see a proof in [E, p. 168].)An even more useful variant is the following. Suppose t is a non-zero-divisor of B in m (ge-

ometrically: a Cartier divisor on the target passing through the generic point). If M is flat over B,

then t is not a zero-divisor ofM (we’ve checked this before: tensor 0 // B×t // B // B/(t) → 0

with M). Also, M/tM is a flat B/tB-module (flatness commutes with base change ref). The nextresult says that this is a characterization.

29.13.2. Theorem (local slicing criterion for flatness). — M is B-flat if and only ifM/tM is flat over B/(t). 47

[Proof from spring ’07 reading class. → is easy, so we do the ←. The first criterion implies

TorB/(t)1 (k,M/tM) = 0. But then (*) Tor

B/(t)1 (k,M/tM) = TorB1 (k,M) yields the result. Proof

of (*): F2 → F1 → F0 → M → 0. Apply B/(t)⊗. Now TorBi (B/(t), M) = M/tM (if i = 0) or 0(if i > 0), from class 41 (just before 3.4). Then B/(t)⊗B F is a free resolution of M/tM , and theresult follows.]

This is also sometimes called the local criterion for flatness. The proof is short (given thefirst local criterion). You can read it in [E, p. 169]. Give a proof later.

29.13.A. Important exercise (for those who know what a Cohen-Macaulay scheme is):The Cohen-Macaulay/Flatness theorem. Suppose π : X → Y is a map of locally Noetherianschemes, where both X and Y are equidimensional, and Y is nonsingular. Show that if any twoof the following hold, then the third does as well: 48

• π is flat.• X is Cohen-Macaulay.• Every fiber Xy is Cohen-Macaulay of the expected dimension.

I concluded the section on flatness by reviewing everything we have learned about flatness,in a good order.

The following is bogus. Proof. There is one direction to show. We wish to show that TorBi (M, N) =

0 for all finitely generated B-modules N . Consider 0 → M → M → M/tM → 0. Apply ⊗BN , and

consider the long exact sequence. I claim that TorBi (M/tM, N) = 0. If so, then ×t : TorB

i (M, N) →

TorBi (M, N) is an isomorphism. Also, TorB

i (M, N) is a coherent A-module. But then by considering tas an element of m, Nakayama shows that this group is 0. So we’re done once we establish the followinglemma.

47H.L.III.10.3A; H refers to Bourbaki [1,III, §5] or Altman and Kleiman [1, V, §3]

48cmf

491

Goal: Suppose M/tM is a flat B/(t)-module. Then for any B-module, TorBi (M/tM, N) = 0.

29.13.3. Lemma. — If t ∈ B is a nonzero divisor which doesn’t kill any elements of M, then for

any B/(t)-module N, TorB/(t)i (N, M/tM) = TorB

i (N, M) for all i ≥ 0.

Proof. Take a resolution R of M . (MAKE XY-MATRIX!) Consider 0 // R×t // R // R/tR // 0 .

The top two rows are exact, hence the bottom row is exact (by LES). Truncate. Tensor with N , which

may not be left exact in the columns. But the top vertical map is 0.

492

29.14 Condition (*) without flatness: the connectednesslemma and applications

[Move this later, into formal functions. Also presumably move rigidity lemma in here. Pointout here that they can read the rigidity lemma immediately.]

29.14.1. Zariski’s connectedness lemma. — Suppose π : X → Y is a projective morphismof Noetherian schemes which satisfies (*). Then π−1(y) is connected for every y ∈ Y . 49

[Hartshorne Cor III.11.3][forme: Is this the Enriques-Zariski connectedness principal?]We need to prove this. (Remark: we need the Noetherian hypotheses for this, I think!) [Vague

idea: assume Y = SpecA, Γ(X,OX) = A, y = [m]. Then assume Y is a complete Noetherianlocal ring.

Given this, we get Stein factorization and ZMT:

29.14.2. Stein Factorization Theorem. — Any projective morphism π : X → Y of Noe-therian schemes can be factored into g f , where f : X → Y ′ is a projective morphism satisfying(*), hence with connected fibers, and g : Y ′ → Y is a finite morphism.50 [Make into diagram. Apicture is necessary here. We can replace projective by proper by Remark 32.2.2.]Stein factorization

Proof. As π∗OX is a coherent sheaf on Y , Y ′ := Specf∗OXg // Y is a finite morphism. By

our fun property P facts, as g is separated (as finite), f is projective. Moreover, f∗OX = OY ′

[more explanation?].

29.14.A. Exercise. [Daniel Allcock has some good exmaples where it is useful. One example:In Castelnuovo’s criterion, ZMT is necessary to ensure that the contraction is unique.] [FromJames McKernan:] Suppose π : X → Y and π′ : X → Y ′ are two morphisms contracting thesame connected sets of X, and Y and Y ′ are normal, show that π and π′ are the same (or moreprecisely, that there is an isomorphism i : Y → Y ′ such that π′ = i π). Informal translation: IfX → Y has connected fibers, then Y is determined by just knowing the locus contracted on X.[Reason: We can identify Y and Y ′ as topological spaces. Use Stein factorization to show thatOY = f∗OX .]

29.14.3. Zariski’s Main Theorem. — If π is furthermore a birational morphism of integralschemes, and Y is normal, then (*) holds, and hence (by the connectedness lemma) for everyy ∈ Y , π−1(y) is connected. 51 Slightly stronger version (mixing in Mumford V below): X → YZariski’s Main Theorembirational proper morphism of Noetherian integral schemes, y a normal point of Y . Then f−1(y)is connected.

Proof. The question is local, so we assume Y = SpecB. Let A = Γ(Y, π∗OX), so A is a coherentB-algebra, hence a finitely generated A-module. But then both are integral domains with thesame quotient field, and A→ B is an isomorphism.

[quasifinite and separated implies open subset of finite. (Use: a quasifinite sep-arated map from an algebraic space to a scheme means that the algebraic space is ascheme.)]

29.14.4. Brian Conrad comments November 2007 ZMT. 52 ZMT says that if f : U → Sis quasi-finite and separated and S is qcqs scheme then f factors as (necessarily quasi-compact)open immersion into a finite S-scheme X.

In such generality (without f assumed to be finitely presented, just finite type...) it is rathermore difficult and the proof is in section 18.12 of EGA IV4. You don’t want to delve into that.The case of f finitely presented is in section 8.10 or so of EGA IV3, where it uses passage to finite

49connectednesslemma

50stein, H.C.III.11.5

51ZMT, H.T.III.11.4, M.III.9 ZMT V

52ZMTbc

493

type over Z case and then excellence considerations (to allow working with normalizations) andGrothendieck’s brilliant idea to induct on dimension by deleting the closed point from a localscheme. It is again too difficult. The noetherian case is proved in EGA III1, in section 4.4 I think(and as such is basically an exercise in Hartshorne). You could do that case. But personally Ithink it is far better to say nothing about the proof (leave it as an exercise with hints) and insteadshow various impressive ways it is *used* (such as the local structure theorem for etale maps).That might inspire them to read EGA on their own to learn more about it. I actually read EGAIV by starting with ZMT (in the finitely presented form) and then reading backwards to keepmotivated.

29.15 216 notes end here

Coming up: local criterion (hard). Slicing theorem. Picture. Fiber dimension,including: no jumping: But 2-plane example. See H Prop. 9.5.

Flatness over nonreduced schemes.Exercise: first-order deformation. p. 177 (E?) Hard Exercise. Show that the

first-order deformations of a smooth affine scheme are trivial. Show that the first-order deformations of a smooth scheme are given by H1(T ). (Walk them throughit!)

+CM flat.Why is flat = nice? 3) It turns out to be what we need, which is less than we think we need,

in several circumstances. E.g. smooth pullback of classes; flat suffices. etale descent; flat suffices.And: completion of Noetherian local ring is flat.

29.15.1. Equational criterion for flatness. [See also Eisenbud Lemma 6.4 p. 163, 6.5.]Mumford gives an interesting variant of flatness (his wording): M is R-flat if for all m1, . . . ,mn ∈M and a1, . . . , an ∈ R such that

Paimi = 0, there are equations mi =

Pkj=1 bijm

′j for some

m′j ∈ M , bij ∈ R such thatPni=1 bijai = 0 for all j. Good for exposition: clearly free’s are

flat, and any direct limit of flats is flat. Example of flat but not free: Q is a flat Z-module (atworst, because it is the direct limit of free Z-modules). The basic facts he mentions [M, p. 296]:A preserved by base change, B can check at all prime ideals, B’ M an S-module, S anR-algebra,then M is R-flat iff for all prime ideals P ⊂ S, MP is Ri−1(P )-flat. C. If R is local ring, and eiher

R is artinian, or M is finitely presented, then M is flat implies M is free. D. R domain, thenflat implies torsion-free. E. M B-module, B is an A-algebra. f ∈ B has the property that for allmaximal ideals m ⊂ A, multiplication by f is injective in M/mM . Then M A-flat implies M/fM

flat over A. F. O Noetherian local ring, then O is flat O-module. He refers to Bourbaki Ch. ICommutative algebra.53

29.15.2. Joe asked for a proof of the following: the fibral flatness theorem. Suppose

X

flat, lfpr @

@@@@

@@// Y

flat, lfpr

~~~~

~~~

S

Then X → Y is flat iff Xs → Ys is flat. This seems to come up in the proof of Serre duality infamilies. Some further hypotheses seem to be necessary. fibral flatness theorem

[Joe explained why a fancier fibral flatness theorem gives you the following useful fact. Sup-pose C → B is a family of curves, and s is a section hitting the smooth locus. Then s is a Cartierdivisor. Proof. There is a fibral flatness theorem of the following form. X → Y over S, and F isa sheaf on X, and we want to check if it is flat over Y . The conclusion is that it is, if it is fiberby fiber over s. (geometric points? all points?) The hypotheses are something like F , X, Y allflat over S. Here’s how we apply it in this case. X = Y = E, F = I(s). (the section is a closedsubscheme). From 0 → I → OE → Os → 0, we get that I is flat over S. Now by fibral flatness,Is is flat. It is also coherent, hence locally, hence locally free of rank 1.]

53Mflat, p. 295–7

494

From p. 132 of Fundamental Algebraic Geometry, we have the following statement. Let Sbe a Noetherian scheme, and let f : X → S and g : Y → S be finite type flat morphisms. Letπ : Y → X be any morphism over S. Let y ∈ Y , x = π(y), s = g(y) = f(x). If the restrictedmorphism πs : Ys → Xs between fibers over s is flat at ∈ Ys, then π is flat at y ∈ Y . He refersto Altman and Kleiman chapter V, and describes this as a consequence of what he calls the localcriterion of flatness. Altman-Kleiman, Introduction to Grothendieck duality theory, Lecture Notesin Math., Vol. 146, Springer-Verlag, Berlin, 1970.

29.15.3. ++Flattening stratifications.Given a projective (proper?) morphism π : X → Y of (locally?) Noetherian schemes, and a

coherent sheaf F on X, there is a unique stratification S =‘Si of S by locally closed subschemes

Si such that a given morphism g : T → S inducing g∗F on X ×S T is flat over T iff g factorsthrough g : T →‘

Si → S. For the proof, we may assume that S and T are affine.Aaron has a proof at http://www.math.utah.edu/ bertram/courses/hilbert/ps/flat.ps (uses

cohomology and base change).][primordial:

CHAPTER 30

Smooth, etale, unramified

[From Fulton’s ’07 colloquium talk: EGA defines unramified by: f : X → Y unramified if:consider f × f : X ×X → Y × Y . The image includes the diagonal. If the diagonal is open, thenf is unramified.]

1[The definition of etale in [MAV] p. 65-66 for algebraic varieties over an algebraically closedfield k is: f : X → Y is etale if (i) f flat and (ii) for all x ∈ X, y = f(x) ∈ Y , if mx and my are themaximal ideals in Ox and Oy, then f∗(my)Ox = my. He gives an equivalent definition for formalthings, and says cf. Mumford’s “Introduction to algebraic geometry” p. 353, which is likely thered book.]

[Place the following fact, which is done in Jack’s stalk notes: closed immersion plus flat =open immersion, or something like that.]

[I had been wondering about how to prove the left-exactness of some long exact sequence fordifferentials with a smooth hypothesis. Barbara had a suggestion at Oberwolfach in 2006. Here

are my only notes for that. f : X → Y smooth at x ∈ X, f(x) = y. f∗ : OY,y → OX,x ∼=ohY,y [[t1, . . . , tn]]. Cn × Y → Y → Z. A→ B → B[x1, . . . , xn].

[Tell them that we now talk about smooth k-varieties. This is strictly stronger than nonsin-gular!]

[I need a section on n-forms on a nonsingular variety on an algebraically closed field. Showthat it satisfies adjunction. Should I call this K? I’m not sure if the canonical bundle shouldbe defined as this, or as the relative dualizing sheaf. We’ll use this in the Serre duality section.I should refer forward to this from the section on differentials on varieties over an algebraicallyclosed field.]

[Have I shown that the composition of flat of relative dimension m and flat ofrelative dimension n is flat of relative dimension m+n?] [Possibly much earlier, whendiscussing nonsingularity, do “smooth over a field”?] [Hartshorne suggests we look atAltman-Kleiman, Matsumura, and Grothendieck SGA 1 exp. I–III.] [Describe thesenotions at a point?] [Formally unramified is discussed in Neron Models p. 37.]

Unramified morphisms. A local homomorphism f : A → B is unramified ifB/f(mA)B is a finite separable extension of A/mA, or equivalently, if (a) f(mA)B = mB ,and (b) the field B/mB is finite and separable over A/mA. This agrees with the defi-nition in algebraic number theory where one considers only discrete valuation rings.I should at least refer the reader to algebraic number theory.

30.1 Definitions and easier consequences

We will next describe analogues of some important notions in differential geometry — thefollowing particular types of maps of manifolds. They naturally form a family of three.

• Submersions are maps that induce surjections of tangent spaces everywhere. They areuseful in the notion of a fibration.

• Covering spaces are maps that induce isomorphisms of tangent spaces, or equivalently,are local isomorphisms.

• Immersions are maps that induce injections of tangent spaces.

1c:seu

495

496

Warning repeated from earlier: “immersion” is often used in algebraic geometry with a differentmeaning. We won’t use this word in an algebro-geometric context (without an adjective suchas “open” or “closed”) in order to avoid confusion. [Likely add picture.] (A fourth notion isrelated to these three: a map of manifolds is an embedding if it is an immersion that is an inclusionof sets, where the source has the subspace topology. This is analogous to locally closed immersionin algebraic geometry.) [Is it?] [Should “immersion” be “local embedding”?]submersion, cov space,

immersion, embedding

smooth and etale and

unramified morphisms

We will define algebraic analogues of these three notions: smooth, etale, and unramified. Inthe case of nonsingular varieties over an algebraically closed field, we could take the differential geo-metric definition. We would like to define these notions more generally. Indeed, one of the pointsof algebraic geometry is to generalize “smooth” notions to singular situations. Also, we’ll want tomake arguments by “working over” the generic point, and also over nonreduced subschemes. Wemay even want to do things over nonalgebraically closed fields, or over the integers.

Our definitions will be combinations of notions we’ve already seen, and thus we’ll see that theyhave many good properties. We’ll see (§30.2.1) that in the category of nonsingular varieties overalgebraically closed fields, we recover the differential geometric definition. Our three definitionswon’t be so obviously a natural triplet, but I’ll mention the definition given in EGA (§30.4.1), andin this context once again the definitions are very similar.

Let’s first consider some examples of things we want to be analogues of “covering space” and“submersion”, and see if they help us make good definitions.

We’ll start with something we would want to be a covering space. Consider the parabolax = y2 projecting to the x-axis, over the complex numbers. (This example has come up again andagain!) We might reasonably want this to be a covering space away from the origin. We mightalso want the notion of covering space to be an open condition: the locus where a morphism is acovering space should be open on the source. This is true for the differential geometric definition.(More generally, we might want this notion to be preserved by base change.) But then this shouldbe a “covering space” over the generic point, and here we get a non-trivial residue field extension(C(y)/C(y2)), not an isomorphism. Thus we are forced to consider (the Spec’s of) certain finiteextensions of fields to be covering spaces. (We’ll see soon that we just want separable extensions.)

Note also in this example there are no (non-empty) Zariski-open subsets U ⊂ X and V ⊂ Vwhere the map sends U into V isomorphically. This will later lead to the notion of the etaletopology, which is a bizarre sort of topology (not even a topology in the usual sense, but a“Grothendieck topology”, §37.0.5).et and G topology

30.1.1. 2Here is an issue with smoothness: we would certainly want the fibers to be smooth, soreasonably we would want the fibers to be nonsingular. But we know that nonsingularity overa field does not behave well over a base change (consider Spec k(t)[u]/(up − t) → Spec k(t) andbase change by Spec k(t)[v]/(vp − t) → Spec k(t), where char k = p). [Refer back to wherethis example appeared before.] We can patch that by noting that nonsingularity behaveswell over algebraically closed fields, and hence we could require that all the geometric fibers arenonsingular. But that isn’t quite enough. For example, a horrible map from a scheme X to acurve Y that maps a different nonsingular variety to a each point Y (X is an infinite disjointunion of these) should not be considered a submersion in any reasonable sense. Also, we mightreasonably not want to consider Spec k → Spec k[ε]/(ε2) to be a submersion (for example, thisisn’t surjective on tangent spaces, and more generally the picture “doesn’t look like a fibration”).[Add pictures of these two pathologies.] Both problems are failures of π : X → Y to be anice, ”continuous” family. Whenever we are looking for some vague notion of “niceness” we knowthat “flatness” will be in the definition. (This is the reason we waited so long before introducingthe notion of smoothness — we needed to develop flatness first!)

One last issue: we will require the geometric fibers to be varieties, so we can think of them as“smooth” in the old-fashioned intuitive sense. We could impose this by requiring our morphismsto be locally of finite type, or (a stronger condition) locally of finite presentation. (I shouldhave defined “locally of finite presentation” back when we defined “locally of finitetype” and the many other notions satisfying the affine covering lemma. It isn’t anyharder. A morphism of affine schemes SpecA→ SpecB is locally of finite presentation ifit corresponds to B → B[x1, . . . , xn]/(f1, . . . , fr) — A should be finitely generated overB, and also have a finite number of relations. This notion satisfies the hypotheses

2nogensm

497

of the affine covering lemma. A morphism of schemes π : X → Y is locally of finite

presentation if every map of affine open sets SpecA→ SpecB induced by π is locally offinite presentation. If you work only with locally Noetherian schemes, then these twonotions are the same.) I haven’t thought through why Grothendieck went with the strictercondition.

Finally, we define our three notions!

30.1.2. Definition. A morphism π : X → Y is smooth of relative dimension n provided thatit is locally of finite presentation and flat of relative dimension n, and ΩX/Y is locally free of

rank n. [EGA definition: lfpr + flat, and for all y ∈ Y , the fiber is regular, Cor. smoothIV.17.5.2. FOAG-smooth is EGA-smooth by Cor. IV.17.5.2. EGA-smooth impliesFOAG-smooth by Prop. IV.17.2.3(i).]

A morphism π : X → Y is etale provided that it is locally of finite presentation and flat, andΩX/Y = 0. [Equivalence with EGA definition: Corollary IV.17.6.2,using a) and c). I etale

believe Milne in [MilneLET] assumes etale morphisms are quasicompact. [MilneLET,p. 13]: A homomorphism of rings f : A → B is etale if the map on Spec’s is etale.Equivalently, it is etale if (a) B is a finitely generated A-algtebra; B is A-flat; and(c) for all maximal ideals n of B, Bn/F (p)Bn is a finite separable field extension ofAp/pAp, where p = f−1(n). Except Milne seems to have incorporated a finite typeassumption.] [Say: etale is the intuition of “local isomorphism”.]

A morphism π : X → Y is unramified provided that it is locally of finite presentation,and ΩX/Y = 0. [Equivalence with EGA definition: Theorem IV.17.4.1, or better unramified

Corollary IV.17.4.2. Milne [MilneLET, p. 13] gives the following definition for rings.A local homomorphism f : A → B of local rings is unramified if B/f(mA)B is a finiteseparable field extension of A/mA, or, equivalently, if (a) f(mA)B = mB, and (b) thefield B/mB is finite and separable over A/mA. This agrees with the definition inalgebraic number theory where one only considers DVRs. A morhpism φ : Y → Xof schemes is unramified if it is of finite type and the maps of stalks are unramifiedfor all points of the source. It suffices to check the condition for the closed pointsyofY . He sayss: this agrees with our definition once you have the assumption thatthe morphism of of finite type, and gives a reference [MilneEC, I.3.5].]

30.1.3. Examples.

• AnY → Y , PnY → Y are smooth morphisms of relative dimension n. 3

• Locally finitely presented open immersions are etale. 4

• Unramified. Locally finitely presented locally closed immersions are unramified.

30.1.4. Quick observations and comments.

30.1.5. All three notions are local on the target, and local on the source, and are preserved bybase change. 5 That’s because all of the terms arising in the definition have these properties.Exercise. Show that all three notions are open conditions. State this rigorously and prove it.(Hint: Given π : X → Y , then there is a largest open subset of X where π is smooth of relativedimension n, etc.)

30.1.6. Note that π is etale if and only if π is smooth and unramified, if and only if π is flat andunramified.6 [EGA IV.17.6.2]

30.1.7. Jacobian criterion. The smooth and etale definitions are perfectly set up to use aJacobian criterion. Exercise. Show that SpecB[x1, . . . , xn]/(f1, . . . , fr) → SpecB is smooth ofrelative dimension n (resp. etale) if it is flat of relative dimension n (resp. flat) and the corank ofJacobian matrix is n (resp. the Jacobian matrix is full rank). Jac matrix

3H.III.10.0.1

4H.P.III.10.1a

5H.P.III.10.1b

6H.E.III.10.3c

498

30.1.A. Exercise: smoothness etc. over an algebraically closed field. Show that if kis an algebraically closed field, X → Spec k is smooth of relative dimension n if and only if Xis a disjoint union of nonsingular k-varieties of dimension n. (Hint: use the Jacobian criterion.)Show that X → Spec k is etale if and only if it is unramified if and only if X is a union of pointsisomorphic to Spec k. More generally, if k is a field (not necessarily algebraically closed), showthat X → Spec k is etale if and only if it is unramified if and only if X is the disjoint union ofSpec’s of finite separable extensions of k. [Make sure to do etale = unramified+flat earlier]7

30.1.8. A morphism π : X → Y is smooth if it is locally of finite presentation and flat, and in anopen neighborhood of every point x ∈ X in which π is of constant relative dimension, ΩX/Y is

locally free of that relative dimension. [Make sure that I’ve shown that flat morphisms ofgiven relative dimension are open.] (Exercise. Show that π is smooth if X can be written asa disjoint union X =

‘n≥0 Xn where π|Xn is smooth of relative dimension n.) This notion isn’t

really as “clean” as “smooth of relative dimension n”, but people often use the naked adjective“smooth” for simplicity.smooth

Exercise. Show that etale is the same as smooth of relative dimension 0. In other words,show that etale implies relative dimension 0. (Hint: if there is a point x ∈ X where π has positiverelative dimension, show that ΩX/Y is not 0 at x. You may want to base change, to consider just

the fiber above π(x).)

30.1.9. Note that unramified doesn’t have a flatness hypothesis, and indeed we didn’t expect it, aswe would want the inclusion of the origin into A1 to be unramified. Thus seemingly pathologicalthings of the sort we excluded from the notion of “smooth” and “unramified” morphisms areunramified. For example, if X =

‘z∈C Spec C, then the morphism X → A1

C sending the point

corresponding to z to the point z ∈ A1C is unramified. Such is life.

Exercise. Suppose Xf //

h=gf

@@@

@@@@

Y

g~~

~~~~

~

Z

are locally finitely presented morphisms.

(a) Show that if h is unramified, then so is f . [True without lfpr hypotheses, withunr replaced by funr.] (Hint: Cancellation Theorem 12.1.19 for properties of mor-phisms.)

(b) Suppose g is etale. Show that f is smooth (resp. etale, unramified) if and only if his. (Hint: Observe that ΩX/Y → ΩX/Y is an isomorphism from the relative cotangent

sequence [ref ].)8

Regularity vs. smoothness. Suppose char k = p, and consider the morphism Spec k(u) →Spec k(up). Then the source is nonsingular, but the morphism is not etale (or smooth, or unram-

ified).In fact, if k is not algebraically closed, “nonsingular” isn’t a great notion, as we saw [ref ]

when we had to work hard to develop the theory of nonsingularity. Instead, “smooth (of somedimension)” over a field is much better. You should almost go back in your notes and throwout our discussion of nonsingularity. But don’t — there were a couple of key concepts thathave been useful: discrete valuation rings (nonsingularity in codimension 1) and nonsingularityat closed points of a variety (nonsingularity in top codimension). This is partially repeatedfrom earlier in this chapter.

30.2 Harder facts

I want to segregate three facts which require more effort, to emphasize that the earlier factsare automatic given what we know.

7fiber

8Petale

499

30.2.1. Connection to differential-geometric notion of smoothness.The following exercise makes the connection to the differential-geometric notion of smooth-

ness. Unfortunately, we will need this fact in the next section on generic smoothness.9

30.2.A. Trickier Exercise. Suppose π : X → Y is a morphism of smooth (pure-dimensional)varieties over a field k. Let n = dimX − dimY . Suppose that for each closed point x ∈ X, theinduced map on the Zariski tangent space Tf : Tx → Ty is surjective. Show that f is smooth of

relative dimension n. 10 (Hint: The trickiest thing is to show flatness. Use the (second) localcriterion for flatness.) 11

[Better: use Cohen-Macaulay/flatness theorem.]I think this is the easiest of the three “harder” facts, and it isn’t so bad.For pedants: I think the same argument works over a more arbitrary base. In other words,

suppose in the following diagram of pure-dimensional Noetherian schemes, Y is reduced.

X //

smooth @

@@@@

@@Y

smooth~~

~~~~

~

Z

Let n = dimX−dimY . Suppose that for each closed point x ∈ X, the induced map on the Zariskitangent space Tf : Tx → Ty is surjective. Show that f is smooth of relative dimension n. I thinkthe same argument works, with a twist at the end using Exercise 30.1.9(b). Please correct me ifI’m wrong!

30.2.2. The relative cotangent sequence is left-exact in good circumstances.

[If I don’t need full smoothness for this, e.g. if I only need Ω locally free, possiblyalso with lfpr, move Exercise 30.2.B back into the differential section, and the restof the discussion out of the “harder” section.]

Recall the relative cotangent sequence 22.2.15. Suppose Xf // Y

g // Z be morphismsof schemes. Then there is an exact sequence of quasicoherent sheaves on X

f∗ΩY/Z → ΩX/Z → ΩX/Y → 0.

We have been always keeping in mind that if you see a right-exact sequence, you shouldexpect that this is the tail end of a long exact sequence. In this case, you should expect that thenext term to the left (the “H1 term”) should depend just on X/Y , and not on Z, because the lastterm on the right does. Indeed this is the case: these “homology” groups are called Andre-Quillenhomology groups. You might also hope then that in some mysteriously “good” circumstances, thisfirst “H1” on the left should vanish, and hence the relative cotangent sequence should be exacton the left. Indeed that is the case, as is hinted by the following exercise.

30.2.B. Exercise on differentials. [How to do it? EGA 17.2.3(ii), relying on 0.20.5.7,which is in EGA IV vol. 1.]. IfX → Y is a smooth morphism, show that the relative cotangentsequence is exact on the left as well. This should be moved back into the differential bit,and should also replace some stuff in the nonsingular variety section. The analogousfact is true for the conormal exact sequence. Matsumura p. 193 proves this usingformal smoothness. 12

30.2.C. Unimportant exercise. Predict a circumstance in which the relative conormal sequenceis left-exact. [Add ref. Also, give a literature reference for this fact.]

30.2.3. Corollary. Suppose f is etale. Then the pullback of differentials f∗ΩY/Z → ΩX/Z is

an isomorphism. (This should be very believable to you from the picture you should have in yourhead!) [EGA IV.17.2.4]

9diffcon

10H.P.III.10.4, except he does it over an algebraically closed field

11yucky

12weirddiff

500

30.2.D. Exercise. Show that all three notions are preserved by composition. (More precisely,in the smooth case, smooth of relative dimension m composed with smooth of relative dimensionn is smooth of relative dimension n + m.) You’ll need Exercise 30.2.B in the smooth case. 13

[They’ll also need that an extension of a locally free by a locally free is locally free.]

30.2.E. Easy exercise. Show that all three notions are closed under products. (More precisely,in the case of smoothness: If X,Y → Z are smooth of relative dimension m and n respectively,then X ×Z Y → Z is smooth of relative dimension m + n.) 14 (Hint: This is a consequence ofbase change and composition, as we have discussed earlier [ref ]. Consider X ×Z Y → Y → Z.)

30.2.F. Exercise: smoothness implies surjection of tangent sheaves. Continuing theterminology of the above, Suppose X → Y is a smooth morphism of Z-schemes. Show that0 → TX/Y → TX/Z → f∗TY/Z → 0 is an exact sequence of sheaves, and in particular,

TX/Z → f∗TY/Z is surjective, paralleling the notion of submersion in differential geometry. (Re-

call TX/Y = Hom(ΩX/Y ,OX) and similarly for TX/Z , TY/Z .) Did I indeed define these?

[Hence conclude that smooth morphisms yield surjections of Zariski-tangent spaces?Answer: Apply Hom(·,OX) to 0 → f∗ΩY/Z → ΩX/Z → ΩX/Y → 0. We obtain 0 →TX/Y → TX/Z → (f∗ΩY/Z)∨ → 0; right-exactness comes because ΩX/Y is locally free.

We need a Hom fact that I should have proved! ]15 I’d like to connect this to the first“harder” fact, and to say something about Zariski tangent spaces, but I haven’t yetmanaged to.

30.2.4. Characterization of smooth and etale in terms of fibers.By Exercise 30.1.A, we know what the fibers look like for etale and unramified morphisms;

and what the geometric fibers look like for smooth morphisms. 16 [The converse is true tomany of these, see EGA comment.] There is a good characterization of these notions ingeometrically nonsingu-

lar fibers terms of the geometric fibers, and this is a convenient way of thinking about the three definitions.

30.2.G. Exercise: characterization of etale and unramified morphisms in terms offibers. Suppose π : X → Y is a morphism locally of finite presentation. Prove that π isetale if and only if it is flat, and the geometric fibers (above Spec k → Y , say) are unions of

Spec’s of fields (with discrete topology), each a finite separable extension of the field k. [EGAIV(4).17.6.2] [Cor: if π is a morphism of varieties over an algebraically closed field,that is etale. Then f is dominating, and k(X) is a finite separable extension of k(Y ).]

Prove that π is unramified if and only if the geometric fibers (above Spec k→ Y , say) are unions

of Spec’s of fields (with discrete topology), each a finite separable extension of the field k. [Also,Hartshorne’s definition of unramified: A morphism f : X → Y is unramified if for eachunramifiedx 7→ y we have myOx = mx, and k(x) is a separable algebraic extension of k(y). That’sHartshorne’s definition in the finite type over k case, Exercise III.10.3, so I shouldbe careful! 17] (Hint: a finite type sheaf that is 0 at all points must be the 0-sheaf.)

There is an analogous statement for smooth morphisms, that is harder. (That’s why thisdiscussion is in the “harder” section.)

30.2.H. Harder exercise. [HARD. Mumford p. 304, big theorem. EGA definition:lfpr + flat, and for all y ∈ Y , the fiber is regular, Cor. IV.17.5.2.] Suppose π : X → Yis locally of finite presentation. Show that π is smooth of relative dimension n if and only if π isflat, and the geometric fibers are disjoint unions of n-dimensional nonsingular varieties (over theappropriate field). 18 19]]]

[[[In the case of varieties, I should be able to check only over closed points.]]]

13H.P.III.10.1c

14H.P.III.10.1d

15smoothT

16H.III.10.0.3

17H.E.III.10.3u

18M.III.10.T.3 and T.3’

19H.T.III.10.2 is where H shows this is equivalent to his earlier definition. He does this in a special case,

for schemes ft/k.

501

30.3 Generic smoothness in characteristic 0

We will next see a number of important results that fall under the rubric of “generic smooth-ness”. All will require working over a field of characteristic 0 in an essential way. So far, we havehad to add a few caveats here and there for people encountering positive characteristic. This isprobably the first case where positive characteristic people should just skip this section.

Our first result is an algebraic analog of Sard’s theorem. Sard’s theorem

30.3.1. Proposition (generic smoothness in the source). — Let k be a field of character-istic 0, and let π : X → Y be a dominant morphism of integral finite-type k-schemes. Then thereis a non-empty (=dense) open set U ⊂ X such that π|U is smooth.20

We’ve basically seen this argument before, when we showed that a variety has an open subsetthat is nonsingular.

Proof. Define n = dimX − dimY (the “relative dimension”). Now FF (X)/FF (Y ) is a finitelygenerated field extension of transcendence degree n. It is separably generated by n elements (aswe are in characteristic 0, Theorem A.2.1). Thus Ω has rank n at the generic point. Its rank isat least n everywhere. By uppersemicontinuity of fiber rank of a coherent sheaf [ref ], it is rankn for every point in a dense open set. Recall that on a reduced scheme, constant rank implieslocally free of that rank (Exercise 17.10.B); hence Ω is locally free of rank n on that set. Also, byopenness of flatness [ref ], it is flat on a dense open set. Let U be the intersection of these twoopen sets.

For pedants: In class, I retreated to this statement above. However, I think the followingholds. Suppose π : X → Y is a dominant finite type morphism of integral schemes, wherecharFF (Y ) = 0 (and hence charFF (X) = 0 from FF (Y ) → FF (X)). Then there is a non-empty open set U ⊂ X such that π|U is smooth.

The proof above needs the following tweak. Define n = dimX − dimY . Let η be the genericpoint of Y , and let Xη be fiber of π above η; it is non-empty by the dominant hypothesis. Then Xηis a finite type scheme over FF (Y ). I claim dimXη = n. Indeed, π is flat near Xη (everythingis flat over a field, and flatness is an open condition), and we’ve shown for a flat morphism thedimension of the fiber is the dimension of the source minus the dimension of the target. Thenproceed as above.

Please let me know if I’ve made a mistake![Possibly give this as an exercise: Tweak the above argument to prove the fol-

lowing more general statement: ...]

30.3.2. In §30.1.1, we saw an example where this result fails in positive characteristic, involvingan inseparable extension of fields. Here is another example, over an algebraically closed field ofcharacteristic p: A1

k = Spec k[t]→ Spec k[u] = A1k, given by u 7→ tp. The earlier example (§30.1.1)

is what is going on at the generic point. [Put this in index.] 21

I AM MYSTIFIED AS TO WHY THIS IS SO HARD! WHERE DID I GET THIS FROM? Proof.

[of the first one] Reduce to the case where Y is irreducible and affine. Reduce to the case where X is

irreducible and affine. Reduce to the case where f is dominant. Using the key argument in the proof to

Chevalley’s theorem, reduce to showing the result for integral extensions SpecA → SpecB. Reduce to

showing it for SpecB[t]/(tn + · · · ) → SpecB. Everything up to here is basically trivial. Now use the

Jacobian criterion to show that the singular locus is where g(t) = tn + · · · = 0, and g′(t) = 0. Those t

for which g(t) = g′(t) = 0 are given by the resultant. The resultant vanishes on a closed set. It doesn’t

vanish at the generic point, because g is irreducible (it is the minimal polynomial for t).

If the source of π is smooth over a field, the situation is even nicer.

20genericsmoothness

21badp

502

30.3.3. Theorem (generic smoothness in the target). — Suppose f : X → Y is amorphism of k-varieties, where char k = 0, and X is smooth over k. Then there is a dense opensubset of Y such that f |f−1(U) is a smooth morphism. 22[Put this in index]

(Note: f−1(U) may be empty! Indeed, if f is not dominant, we will have to take such a U .)For pedants: I think the following generalization holds, assuming that my earlier notes to

pedants aren’t bogus. Generalize the above hypotheses to the following morphisms of Q-schemes.(Requiring a scheme to be defined over Q is precisely the same as requiring it to “live in charac-teristic 0”, i.e. the morphism to Spec Z has image precisely [(0)].)

locally Noetherian, irr. X

smooth

##GGGGG

GGGGfinite type // Y

dominant, smoothwww

wwwwww

locally Noetherian, integral

Z

integral

To prove this, we’ll use a neat trick.

30.3.4. Lemma. — Suppose π : X → Y is a morphism of schemes that are finite type over k,where char k = 0. Define

Xr = closed points x ∈ X — rank Tπ,x ≤ r .

Then dim f(Xr) ≤ r. (Note that Xr is a closed subset; it is cut out by determinantal equations.Hence by Chevalley’s theorem [ref ], its image is constructable, and we can take its dimension.)23

For pedants: I think the only hypotheses we need are that π is a finite type morphism oflocally Noetherian schemes over Q. The proof seems to work as is, after an initial reduction toverifying it on an arbitrary affine open subset of Y .

Here is an example of the lemma, to help you find it believable. Suppose X is a nonsingularsurface, and Y is a nonsingular curve. Then for each x ∈ x, the tangent map Tπ,x : Tx → Tπ(x) isa map from a two-dimensional vector space to a one-dimensional vector space, and thus has rank1 or 0. If π is dominant, then we have a picture like this [ADD IT]. The tangent map has rank 0at this one point. The image is indeed rank 0. The tangent map has rank at most 1 everywhere.The image indeed has rank 1.

Now imagine that π contracted X to a point. Then the tangent map has rank 0 everywhere,and indeed the image has dimension 0.Proof of lemma. We can replace by X by an irreducible component of Xr, and Y by the closureof that component’s image of X in Y . (The resulting map will have all of X contained in Xr .This boils down to the following linear algebra observation: if a linear map ρ : V1 → V2 has rankat most r, and V ′i is a subspace of Vi, with ρ sending V ′1 to V ′2 , then the restriction of ρ to V ′1has rank at most that of ρ itself.) Thus we have a dominant morphism f : X → Y , and we wishto show that dimY ≤ r. By generic smoothness on the source (Proposition 30.3.1), there is anonempty open subset U ⊂ X such that f : U → Y is smooth. But then for any x ∈ X, thetangent map Tx,X → Tπ(x),Y is surjective (by smoothness , Exercise 30.2.F, except that this

doesn’t quite work...), and has rank at most r, so dimY = dimπ(x) Y ≤ dimTπ(x),Y ≤ r.

There’s not much left to prove the theorem.Proof of Theorem 30.3.3. Reduce to the case Y smooth over k (by restricting to a smaller open

set, using generic smoothness of Y , Proposition 30.3.1). Say n = dimY . dim f(Xn−1) ≤ n− 1 bythe lemma, so remove this as well. Then the rank of Tf is at least r for each closed point of X.But as Y is nonsingular of dimension r, we have that Tf is surjective for every closed point of X,hence surjective. Thus f is smooth by Hard Exercise 30.2.A.

22genericsmoothness2

23H.P.III.10.6

503

30.3.5. The Kleiman-Bertini theorem. The Kleiman-Bertini theorem is elementary toprove, and extremely useful, for example in enumerative geometry.

Throughout this discussion, we’ll work in the category of k-varieties, where k is an alge-braically closed field of characteristic 0. The definitions and results generalize easily to the non-algebraically closed case, and I’ll discuss this parenthetically.

30.3.6. Suppose G is a group variety. Then I claim that G is smooth over k. Reason: It isgenerically smooth (so it has a dense open set U that is smooth), and G acts transitively on itself(so we can cover G with translates of U).24

We can generalize this. We say that a G-action a : G×X → X on a variety X is transitiveif it is transitive on closed points. (If k is not algebraically closed, we replace this by saying that

it is transitive on k-valued points. In other words, we base change to the algebraic closure, andask if the resulting action is transitive. Note that in characteristic 0, reduced = geometricallyreduced, so G and X both remain reduced upon base change to k.)

In other words, if U is a non-empty open subset of X, then we can cover X with translatesof X. (Translation: G× U → X is surjective.) Such an X (with a transitive G-action) is called ahomogeneous space for G. homogeneous sp

30.3.A. Exercise. Paralleling §30.3.6, show that a homogeneous space X is smooth over k.

30.3.7. The Kleiman-Bertini theorem. — Suppose X is homogeneous space for groupvariety G (over an algebraically closed field k of characteristic 0). Suppose f : Y → X andg : Z → X be morphisms from smooth k-varieties Y , Z. Then there is a nonempty open subsetV ⊂ G such that for every σ ∈ V (k), Y ×X Z defined by

Y ×X Z

// Z

g

Y

σf // X

(i.e. Y is “translated by σ”) is smooth over k of dimension exactly dimY +dimZ−dimX. Better:there is an open subset of V ⊂ G such that

(64) (G×k Y )×X Z → G

is a smooth morphism of relative dimension dimY + dimZ − dimX.(The statement and proof will carry through even if k is not algebraically closed.)The first time you hear this, you should think of the special case where Y → X and Z → X

are closed immersions (hence “smooth subvarieties”). In this case, the Kleiman-Bertini theoremsays that the second subvariety will meet a “general translate” of the first transversely.

Proof. It is more pleasant to describe this proof “backwards”, by considering how we would proveit ourselves. We will end up using generic smoothness twice, as well as many facts we now knowand love.

In order to show that the morphism (64) is generically smooth on the target, it would sufficeto apply Theorem 30.3.3), so we wish to show that (G ×k Y ) ×X Z is a smooth k-variety. NowZ is smooth over k, so it suffices to show that (G ×k Y ) ×X Z → Z is a smooth morphism (asthe composition of two smooth morphisms is smooth). But this is obtained by base changedfrom G ×k Y → X, so it suffices to show that this latter morphism is smooth (as smoothness ispreserved by base change).

This is a G-equivariant morphism G×k Yaf // X . (By “G-equivariant”, we mean that

G action on both sides respects the morphism.) By generic smoothness of the target (Theo-rem 30.3.3), this is smooth over a dense open subset X. But then by transitivity of the G action,this morphism is smooth (everywhere). (Exercise: verify the relative dimension statement.)

30.3.8. Corollary (Bertini’s theorem, improved version). Suppose X is a smooth k-variety,where k is algebraically closed of characteristic 0. Let δ be a finite-dimensional base-point-free

24Gsmooth

504

linear system, i.e. a finite vector space of sections of some invertible sheaf L. Then almost everyelement of δ, considered as a closed subscheme of X, is nonsingular. (More explicitly: eachelement s ∈ H0(X,L) gives a closed subscheme of X. For a general s, considered as a point ofPH0(X,L), the closed subscheme is smooth over k.) 25

(Again, the statement and proof will carry through even if k is not algebraically closed.)This is a good improvement on Bertini’s theorem. For example, we don’t actually need L to

be very ample, or X to be projective.

30.3.B. Exercise. Prove this!

30.3.C. Easy Exercise. Interpret the old version of Bertini’s theorem (over a characteristic 0field) as a corollary of this statement.

Note that this fails in positive characteristic, as shown by the one-dimensional linear systempP | P ∈ P1. This is essentially Example 30.3.2. 26 [Two more interesting examples,which could be starred. Hartshorne gives two exercises. Exercise III.10.7 is a linearsystem with moving singularities. Exercise III.10.8 is a linear system with movingsingularities contained in the base locus.]

[Possible exercise, either here, or at earlier Bertini statement, or something withcurves. Something where they show there is a quartic surface in P3, double along aline, but nonsingular elsewhere.]

30.4 ? Formal interpretations

At some point I should define “formal”. All “formal” things will be in starredsections. (In case I’ve forgotten to say this before: “Formally” means “in the com-pletion”.) For those of you who like complete local rings, or who want to make theconnection to complex analytic geometry, here are some useful reformulations, whichI won’t prove.

Suppose (B, n) → (A,m) is a map of Noetherian local rings, inducing an isomorphism of

residue fields, and a morphism of completions at the maximal ideals B → A (the “hat” terminologyarose first at 16.4.1). Then the induced map of schemes SpecA→ SpecB is:

• etale if B → A is a bijection. EGA IV.17.6.3. (If extension of residue fields:get Witt vectors.)Witt vectors

• smooth if B → A is isomorphic to B → B[[x1, . . . , xn]]. EGA IV.17.5.3 In otherwords, formally, smoothness involves adding some free variables.

• unramified if B → A is surjective.

Mumford’s Corollary on p. 251 is: π morphism of finite type k-schemes, where kis algebraically closed. etale iff map on formal rings at closed points is an isomorphismeverywhere. You can use this to show that algebraically etale = covering space, p.252 of Mumford. Use Hensel’s lemma.

30.4.1. Formally unramified, smooth, and etale. EGA has defines these three notionsdifferently.27 EGA IV.17.1.1. The definitions there make clear that these three definitions forma family, in a way that is quite similar to the differential-geometric definition. (You should largelyignore what follows, unless you find later in life that you really care. I won’t prove anything.)We say that π : X → Y is formally smooth (resp. formally unramified, formally etale) if for allaffine schemes Z, and every closed subscheme Z0 defined by a nilpotent ideal, and every morphismZ → Y , the canonical map HomY (Z,X)→ HomY (Z0, X) is surjective (resp. injective, bijective).This is summarized in the following diagram, which is reminiscent of the valuative criteria for

25H.C.III.10.9

26H.R.III.10.9.3

27egausedef

505

separatedness and properness.

SpecZ0//

_

nilpotent ideal

X

π

SpecZ //

?

;;wwwwwwwwwY

(Exercise: show that this is the same as the definition we would get by replacing “nilpotent” by“square-zero”. This is sometimes an easier formulation to work with.)

EGA defines smooth as morphisms that are formally smooth and locally of finite presentation(and similarly for the unramified and etale). f. sm, unr, e

30.4.2. Older stuff, not to say in class. Note that lfpr can be described in the sameway: a morphism is lfpr if HomY (·,X) commutes with projective limits of affine schemes IV.8.14.2.(Possibly remark on this when introducing lfpr morphisms.) lfpr

Remarks (not to say publicly): (i) formally unramified is precisely the same as Ωπ = 0. EGAIV.17.2.1. (ii) formally smooth implies Ωπ is locally projective. If f is lft, then this means locallyfree of finite type.

[Remark to possibly place: Assume π is lfpr. then π is unramified iff δπ is anopen immersion. EGA Cor. IV.17.4.2.]

[Definition 4 (Mumford.) Note that if Y = SpecR, and X = SpecR[x1, . . . , xn+k]/(f1, . . . , fk),where the rank of the Jacobian matrix (∂fi/∂xj) is k, then we get something smooth(check). ]

Hartshorne’s definition.

• finite type, flat,• if X′ ⊂ X and Y ′ → X are irreducible components such that f(X ′) ⊂ Y ′, then

dimX′ = dimY ′ + n.• For each point x ∈ X (not necessarily closed) dimk(x)(ΩX/Y ⊗ k(x)) = n

Useful variation I’ve used: f : X → Y morphism of schemes of finite type over k,where Y is reduced and X and Y are equidimensional. Then f is smooth of relativedimension dimX − dimY at x ∈ X if the Zariski tangent space to f−1(f(x)) is of rankdimX − dimY at x. That’s in my Hartshorne notes.

30.5 Unused

30.5.A. Possible exercise. Smooth and etale can be brought down to earth as follows. (a)Show that X = SpecR[x1, . . . , xn+k](f1, . . . , fn) → Y = SpecR is smooth of relative dimensionk at x ∈ X if rank(∂fi/∂xj)(x) = n. (b) Show that an arbitrary morphism f : X → Y of finitetype is smooth if for all x ∈ X, there are open neighborhoods U ⊂ X of x and V ⊂ Y of f(x)such that the morphism here looks like (a). A special case of this is when k = 0. 28

This almost shows that the sheaf of relative differentials is locally free of rank k. (This mightbe a handy way of getting at that Mumford proof of determining smooth morphisms via theirfibers.)

Here is a nice link to the differential-geometric picture.

30.5.1. Proposition (Geometric Hensel). — Let (R,m, k) be a complete local Noetherianring. Let f be a morphism of finite type X → Y := SpecO with f(x) = y := [m]. Assume thatwe have an isomorphism of residue fields. Then f is etale iff f is a local isomorphism near x.(Translation: there is an open neighborhood of x.) 29

[M] proof uses Hensel’s lemma, which he also proves.

28M.III.5.D.1, he takes this as a provisional definition of etale; and M.III.10.D.3 is the definition of smooth

29M.III.5.T.2

506

30.5.2. Hensel’s lemma. — Let (R,m) be a complete local ring, and let f1, . . . , fn ∈R[x1, . . . , xn]. Assume a1, . . . , an ∈ R satisfy

f1(a1, . . . , an) ≡ · · · ≡ fn(a1 , . . . , an) ≡ 0 (mod m)

and

(det ∂fi/∂xj) (a1, . . . , an) /∈ m.

Then there exist α1, . . . , αn ∈ R such that αi ≡ ai (mod m), and30

f1(α1, . . . , αn) = · · · = fn(α1 , . . . , αn) = 0.

The proof is by induction. If I use this, I may want to stick this in the algebraic glossary.When working over an algebraically closed field, then things are particularly simple. Put this

earlier.

30.5.3. Proposition. — f : X → Y morphism of nonsingular n-dimensional k-varieties. If forall closed y ∈ Y , the fiber is a finite set of reduced points (or if for all x ∈ X, mx = f∗(mf(x))Ox)then f is etale. 31

Even if k is not algebraically closed, we allow only separable extensions. Maybe we cando better: have them both just be equidimensional regular schemes of the same dimension?(Presumably do this for smooth too.)

30.5.4. Notes from Jarod: formally smooth, unramified, etale. For any surjectionφ : A′ → A with (ker φ)2 = 0, there is a natural map

hX(A)ψ // hX(A′)×hY (A′) hY (A),

where hX : (Sch)→ (Sets), S 7→ Hom(S,X).formally smooth, un-

ramified, etale Define:

• f is formally smooth provided that ψ surjective for all such A′ → A.• f is formally unramified provided that ψ surjective for all such A′ → A.• f is formally etale provided that ψ surjective for all such A′ → A.

SpecA? //

##FFF

FFFF

FFX

f

SpecA′

;;xxxxxxxxx//?

OO

Y

[that arrow should be dashed, use –]Here are some equivalent definitions of unramified EGA IV.17.4.2: f is locally of finite

presentation, and:unramified morphism

• formally unramified• ∆ : X → X ×Y X open immersion• ΩX/Y = 0• for all y, Xy formally unramified• for all y, Xy =

‘λ SpecKλ with Kλ finite separable extension of k(y).

At a point x ∈ X: write y = f(x). EGA IV.17.4.1.

• there exists a neighborhood U of x such that f |U is formally unramified• Ox/myOx is a finite separable field extension• (ΩX/Y )x = 0• SpecOx → SpecOy is formally unramified

If additionally, k(x) = k(y) and Y is locally Noetherian, then (EGA IV.17.4.4)

• Oy Ox is a surjection

Equivalent definitions of smooth: f is locally of finite presentation and:smooth morphism

• formally smooth (EGA IV.17.5.1)• f is flat and for all y ∈ Y , Xy is formally smooth (EGA IV.17.5.1)

30hensel, M.III.5 p. 247

31M.III.5.T.4

507

• f is flat and for all y ∈ Y , Xy is regular and remains so after extension of scalars upto some perfect extension of k(y). (Brian Conrad’s notes — etale cohomology noteshanded out by Andy in the etale cohomology seminar)

• f is flat and ΩX/Y is locally free with for all x rank near x equal to dimxXf(x) =

maximal dimension of an irreducible component of Xf(x) through x. (Brian Conrad’s

notes)• formally etale and for all x ∈ x, there exists and open neighborhood V of x and U off(x) with f(V ) ⊂ U such that V admits an etale U-map to some AnU . (Brian Conrad’snotes)

At a point x ∈ X (with y = f(x))

• there exists a neighborhood U of x with f |U is formally smooth (EGA IV.17.5.3)

If k(x) = k(y) and Y locally Noetherian,

• Ox ∼= Oy[[T1, . . . , Tn]]. (EGA IV.17.5.3)

Equivalent definitions of etale.: f is locally of finite presentation, and (EGA IV.17.6.2) etale morphism

• formally etale• smooth and unramified• smooth and locally quasifinite [latter is not yet defined]

• flat and unramified

Brian Conrad’s notes:

• for all x ∈ X, there is an open neighborhood V ⊂ X of x and U ⊂ Y of f(x) withf(U ′) ⊂ U such that U ∼= SpecR and U ′ is U-isomorphic to an open subscheme ofSpec(R[T ]g)g′ for some g ∈ R[T ] monic

• flat and for all y ∈ Y , Xy =‘λ Spec kλ with kλ separable finite extension of k(y).

• flat and ΩX/Y = 0

At a point x ∈ X (with y = f(x)) EGA IV.17.6.1

• exists an open neighborhood U of x such that f |U is formally etale• f is flat at x and unramified at x• f is smooth and unramified at x• f is flat and Ox/my · Ox is a finite separable field extension.

If additionally Y is locally Noetherian, (from EGA IV.17.6.3)

• Spf Oy → Spf Ox is formally etale

• Ox is a free Oy-module and Ox ⊗Oyk(y) is a finite separable field extension [should

there be a hat over that ⊗?]

If additionally Y is locally Noetherian and k(x) = k(y), (from EGA IV.17.6.3)

• OY → OX is an isomorphism

If X, Y are finite type over k (but k(x) is not necessarily k(y)) (Hartshorne Exercise III.10.4) 32

• OY ⊗k(y) k(x)→ Ox is an isomorphism

If Oy is normal: (from SGA I. Expose 1. Theorem 9.5) [Zariski’s main theorem?]

• f unramified at x and Oy → Ox injective

If x ∈ X is smooth and y ∈ Y is smooth (maybe more hypotheses needed?) (SGA I. Jarod sawthis referenced by Drezet in his paper “Luna’s Slice theorem and applications” but couldn’t findit in SGA. Drezet is working over the complex numbers.)

• Txf : TxX → TyY is an isomorphism and f is smooth.

If f is a map between algebraic C-schemes: [Conrad’s notes]

• fan : Xan → Y an etale.

][primordial:

32H.E.III.10.4

CHAPTER 31

Serre duality

[How do you define trace? t : Exti(F ,F) → Hi(O). More generally, Exti(F ,F ⊗ H) →Hi(H).

My guess as to the general statement of Serre duality:

Exti(F ,G)× Extn−i(G,F(ω)) // Extn(F ,F(ω))t // Hn(omega) // k

And this is the same with G,F “reversed”?][Serre duality in some generality is apparently discussed in SGA 2 exp. XII

Prop. 1.1.] [I can discuss this when discussing curves, because I don’t really needflatness or differentials.]

[Can I use adjunction to show the n-form result? Can I use it to deal with localcomplete intersections?] [Show that the dualizing sheaf behaves well with respect toflat base change?]

[Chad Schoen says:proof of the residue theorem (and indeed Serre duality) inchapter 2 of Serre’s book “Groupes algebriques et corps...”, “Algebraic groups andfunction fields”.]

31.1 Introduction

Recall that Serre duality arose in our section on “fun with curves” [ref ]. We’ll prove the statementused there, and generalize it greatly.

Our goal is to rigorously prove everything we needed for curves, and to generalize the state-ment significantly. Serre duality can be generalized beyond belief, and we’ll content ourselves withthe version that is most useful. For the generalization, we will need a few facts that we haven’tproved, but that we came close to proving.

(i) The existence (and behavior) of the cup product in (Cech) cohomology. For any qua-sicoherent sheaves F and G, there is a natural map Hi(X,F) × Hj(X,G) → Hi+j (X,F ⊗ G)satisfying all the properties you might hope. From the Cech cohomology point of view this isn’thard. For those of you who prefer derived functors, I haven’t thought through why it is true. Fori = 0 or j = 0, the meaning of the cup product is easy. (For example, if i = 0, the map involvesthe following. The j-cocycle of G is the data of sections of G of (j + 1)-fold intersections of affineopen sets. The cup product corresponds to “multiplying each of these by the (restriction of the)global section of F”.) This version is all we’ll need for nonsingular projective curves (as if i, j > 0,i+ j > 1).

(ii) The Cohen-Macaulay/flatness theorem. I never properly defined Cohen-Macaulay, so Ididn’t have a chance to prove that nonsingular schemes are Cohen-Macaulay, and if π : X → Y isa morphism from a pure-dimensional Cohen-Macaulay scheme to a pure-dimensional nonsingularscheme, then π is flat if all the fibers are of the expected dimension. (I stated this, however.) Thisis exercise 29.13.A.

We’ll take these two facts for granted.Here now is the statement of Serre duality.Suppose X is a Cohen-Macaulay projective k-scheme of pure dimension n. A dualizing sheaf

for X over k is a coherent sheaf ωX (or ωX/k) on X along with a trace map Hn(X,ωW ) → k, dualizing sheafωX/k

509

510

such that for all finite rank locally free sheaves F on X,1

(65) Hi(X,F)×Hn−i(X,F∨ ⊗ ωX) // Hn(X,ωX)t // k

is a perfect pairing. In terms of the cup product, the first map in (65) is the compositioncup prod

trace map

Serre dHi(X,F) ×Hn−i(X,F∨ ⊗ ωX)→ Hn(X, (F ⊗F∨)⊗ ωX)→ Hn(X,ωX).

31.1.1. Theorem (Serre duality). — A dualizing sheaf always exists. 2 3

We will proceed as follows. UPDATE THIS LATER.

• We’ll partially extend this to coherent sheaves in general: Hom(F , ωX)→ Hn(F)∨ isan isomorphism for all F .

• Using this, we’ll show by a Yoneda argument that (ωX , t) is unique up to uniqueisomorphism.

• We will then prove the Serre duality theorem 31.1.1. This will take us some time. We’llfirst prove that the dualizing sheaf exists for projective space. We’ll then prove it foranything admitting a finite flat morphism to projective space. Finally we’ll show thatevery projective Cohen-Macaulay k-scheme admits a finite flat morphism to projectivespace.

• The existence of a dualizing sheaf will be straightforward to show — surprisingly so, atleast to me. However, it is also surprisingly slippery — getting a hold of it in concretecircumstances is quite difficult. For example, on the open subset where X is smooth,ωX is an invertible sheaf. We’ll show this. Furthermore, on this locus, ωX = det ΩX .(Thus in the case of curves, ωX = ΩX . In the “fun with curves” section, we needed thefact that ΩX is dualizing because we wanted to prove the Riemann-Hurwitz formula.)

• We’ll prove the result in families (i.e. we’ll define a “relative dualizing sheaf” in goodcircumstances). This is useful in the theory of moduli of curves, and Gromov-Wittentheory. We’ll state this later.

31.1.A. Warm-up trivial exercise. Show that if h0(X,OX) = 1 (e.g. if X is geometricallyintegral [ref ]), then the trace map is an isomorphism, and conversely.

31.2 Extension to coherent sheaves; uniqueness of thedualizing sheaf

31.2.1. Proposition. — If (ωX , t) exists, then for any coherent sheaf F on X, the naturalmap Hom(F , ωX)×Hn(X,F)→ Hn(X,ωX)→ k is a perfect pairing. 4

In other words, (65) holds for i = n and any coherent sheaf (not just locally free coherentsheaves). You might reasonably ask if it holds for general i, and it is true that these other cases arevery useful, although not as useful as the case we’re proving here. In fact the naive generalizationdoes not hold. The correct generalization involves Ext groups, which we have not defined. Theprecise statement is the following. For any quasicoherent sheaves F and G, there is a natural mapExti(F ,G)×Hj(X,F)→ Hi+j(G). Via this morphism,

Exti(F , ωX)×Hn−i(X,F) // Hn(X,ωX)t // k

is a perfect pairing. 5 [Caution: we haven’t done Ext groups!] 6

Proof of Proposition 31.2.1. Given any coherent F , take a partial locally free resolution

E1 → E0 → F → 0.

1sdone

2t:sd

3serreduality, H.C.III.7.7

4extcoh

5H.T.III.7.1 is the projective space case; H.T.III.6 includes this for projective CM. Repeated below, includ-

ing label.6ext1

511

(Recall that we find a locally free resolution as follows. E0 is a direct sum of line bundles. Wethen find E1 that is also a direct sum of line bundles that surjects onto the kernel of E0 → F .)

Then applying the left-exact functor Hom(·, ωX), we get

0→ Hom(F , ωX)→ Hom(E0, ωX)→ Hom(E1, ωX)

i.e. 0→ Hom(F , ωX)→ (E0)∨ ⊗ ωX → (E1)∨ ⊗ ωX[Remember earlier to have had the exercise that Hom(E ,F) ∼= H0(E∨ ⊗ F) is E if locally free.Better: get the map from one to the other in general, and show that it is an isomorphism.]

Also

Hn(E1)→ Hn(E0)→ Hn(F)→ 0

from which

0→ Hn(F)∨ → Hn(E0)∨ → Hn(E1)∨

There is a natural map Hom(H, ωX) × Hn(H) → Hn(ωX) → k for all coherent sheaves, whichby assumption (that ωX is dualizing) is an isomorphism when H is locally free. Thus we havemorphisms (where all squares are commuting)

0 //

∼

Hom(F , ω) //

(E0)∨(ω) //

∼

(E1)∨(ω)

∼

0 // Hn(F)∨ // Hn(E0)∨ // Hn(E1)∨

where all vertical maps but one are known to be isomorphisms. Hence by the Five Lemma 3.8.C,the remaining map is also an isomorphism.

We can now use a universal property argument to prove:

31.2.2. Proposition. — If a dualizing sheaf (ωX , t) exists, it is unique up to unique isomor-phism. 7 8

Proof. Suppose we have two dualizing sheaves, (ωX , t) and (ω′X , t′). From the two morphisms

(66) Hom(F , ωX)×Hn(X,F) // Hn(X,ωX)t // k

Hom(F , ω′X)×Hn(X,F) // Hn(X,ω′X)t′ // k,

we get a natural bijection Hom(F , ωX) ∼= Hom(F , ω′X), which is functorial in F . By the typicaluniversal property argument, this induces a (unique) isomorphism ωX ∼= ω′X . From (66), underthis isomorphism, the two trace maps must be the same too.

31.3 Proving Serre duality for projective space over a field

31.3.A. Exercise. Prove (65) for Pn, and F = O(m), where ωPn = O(−n− 1). (Hint: do thisby hand!) Hence (65) holds for direct sums of O(m)’s.

31.3.1. Proposition. — Serre duality (Theorem 31.1.1) holds for projective space.

Proof. We now prove (65) for any locally free F on Pn. As usual, take9

(67) 0→ K→ ⊕O(m)→ F → 0.

7uniquedualizing

8H.P.III.7.2

9sdproj

512

Note that K is flat (as O(m) and F are flat [ref ]) and coherent), and hence K is also locallyfree of finite rank (flat coherent sheaves on locally Noetherian schemes are locally free, one of theimportant facts about flatness [ref ]). For convenience, set G = ⊕O(m).

Take the long exact sequence in cohomology, and dualize, to obtain

(68) 0→ Hn(Pn,F)∨ → Hn(Pn,G)∨ → · · · → H0(Pn,H)∨ → 0.

Now instead take (67), tensor with ωPn ∼= OPn(−n−1) (which preserves exactness, as OPn (−n−1)is locally free), and take the corresponding long exact sequence

0 // H0(Pn,F∨ ⊗ ωPn) // H0(Pn,G∨ ⊗ ωPn) // H0(Pn,H∨ ⊗ ωPn)

// H1(Pn,F∨ ⊗ ωPn) // · · ·

Using the trace morphism, this exact sequence maps to the earlier one (68):

Hi(Pn,F∨ ⊗ ωPn )

αiF

// Hi(Pn,G∨ ⊗ ωPn ) //

αiG

Hi(Pn,H∨ ⊗ ωPn )

αiH

// Hi+1(Pn,F∨ ⊗ ωPn )

αi+1F

Hn−i(Pn,F)∨ // Hi(Pn,G)∨ // Hi(Pn,H)∨ // Hi+1(Pn,F)∨

(At some point around here, I could simplify matters by pointing out that H i(G) = 0 for alli 6= 0, n, as G is the direct sum of line bundles, but then I’d still need to deal with the ends, soI’ll prefer not to.) All squares here commute. This is fairly straightforward check for those notinvolving the connecting homomorphism. (Exercise. Check this.) It is longer and more tedious(but equally straightforward) to check that

Hi(Pn,H∨ ⊗ ωPn)

αiH

// Hi+1(Pn,F∨ ⊗ ωPn)

αi+1F

Hi(Pn,H)∨ // Hi+1(Pn,F)∨

commutes. This requires the definition of the cup product, which we haven’t done, so this is oneof the arguments I promised to omit. Don’t omit!

We then induct our way through the sequence as usual: α−1G is surjective (vacuously), and

α−1H and α0

G are injective, hence by the “subtle” Five Lemma 3.8.C, α0F is injective for all locally

free F . In particular, α0H is injective (as H is locally free). But then α0

H is injective, and α−1H and

α0G are surjective, hence α0

F is surjective, and thus an isomorphism for all locally free F . Thus

α0H is also an isomorphism, and we continue inductively to show that αiF is an isomorphism for

all i.

31.4 Proving Serre duality for finite flat covers of otherspaces for which duality holds

We’re now going to make a new construction. It will be relatively elementary to describe, butthe intuition is very deep. (Caution: here “cover” doesn’t mean covering space as in differentialgeometry; it just means “surjective map”. The word “cover” is often used in this imprecise wayin algebraic geometry.) [Say this when I first mention the word “cover”.]cover

513

Suppose π : X → Y is an affine morphism, and G is a quasicoherent sheaf on Y :

X

π

G

~~~~

~~~

Y.

Observe that HomY (π∗OX ,G) is a sheaf of π∗OX -modules. (The subscript Y is included toremind us where the sheaf lives.) The reason is that affine-locally on Y , over an affine open setSpecB (on which G corresponds to B-module N , and with preimage SpecA ⊂ X)

(69) A

N

B

OO

this is the statement that HomB(A,N) is naturally an A-module (i.e. the A-module structurebehaves well with respect to localization by b ∈ B, and hence these modules glue together to forma quasicoherent sheaf).

In our earlier discussion of affine morphisms, we saw that quasicoherent π∗OX-modules corre-spond to quasicoherent sheaves on X. Hence HomY (π∗OX ,G) corresponds to some quasicoherentsheaf π′G on X.

Notational warning. This notation π′ is my own, and solely for the purposes of this section.If π is finite, then this construction is called π! (pronounced “upper shriek”). You may ask whyI’m introducing this extra notation “upper shriek”. That’s because this notation is standard,while my π′ notation is just made up. π′, π!, it won’t let me

put π! in the index

upper shriek

[π! is right adjoint to Rπ!, which I think might be Rπ∗. In the case of a finite mor-phism, Rπ! = π∗. π! is one of the “six operations” on sheaves defined Grothendieck. Itis the most complicated one, and is complicated to define for general π. It is definedon the level of the derived category of sheaves, not on the category of sheaves itself.In the finite case, we can define this notion at the level of sheaves, but we can’t ingeneral.]

Here are some important observations about this notion.

31.4.1. By construction, we have an isomorphism of quasicoherent sheaves on Y

π∗π′G ∼= HomY (π∗OX ,G).

31.4.2. π′ is a covariant functor from the category of quasicoherent sheaves on Y to quasicoherentsheaves on X.

31.4.3. If π is a finite morphism, and Y (and hence X) is locally Noetherian, then π′ is a covariantfunctor from the category of coherent sheaves on Y to coherent sheaves on X.10 We show thisaffine locally, see (69). As A and N are both coherent B-modules, HomB(A,N) is a coherent B-module (17.7.1), hence a finitely generated B-module, and hence a finitely generated A-module,hence a coherent A-module.

31.4.4. If F is a quasicoherent sheaf on X, then there is a natural map11

(70) π∗HomX (F , π′G)→ HomY (π∗F ,G).

10cshriek

11shriekmap1

514

Reason: if M is an A-module, we have a natural map

(71) HomA(M,HomB(A,N))→ HomB(M,N)

defined as follows. Given m ∈M , and an element of HomA(M,HomB(A,N)), send m to φm(1).This is clearly a homomorphism of B-modules. Moreover, this morphism behaves well with respectto localization of B with respect to an element of B, and hence this description yields a morphismof quasicoherent sheaves.

31.4.5. Lemma. The morphism (70) is an isomorphism.12 [Is there an obvious reason whythe map is an isomorphism? There should be...]

Proof. We will show that the natural map (71) is an isomorphism. Fix a presentation of M :

A⊕m → A⊕n →M → 0

(where the direct sums needn’t be finite). [need better notation] Applying HomA(·,HomB(A,N))to this sequence yields the top row of the following diagram, and applying HomB(·,N) yields thebottom row, and the vertical morphisms arise from the morphism (71).

0 //

∼

HomA(M,HomB(A,N)) //

HomA(A,HomB(A,N))⊕n //

∼

HomA(A,HomB(A,N))⊕m

∼

0 // HomB(M,N) // HomB(A,N)⊕n // HomB(A,N)⊕m

(The squares clearly commute.) Be sure to convince yourself that HomB(A,N)⊕n ∼= HomB(A⊕n,N)even when n isn’t finite (and ditto for the three similar terms)! Then all but one of the verticalhomomorphisms are isomorphisms, and hence by the Five Lemma the remaining morphism is anisomorphism.

Hence π′ is right-adjoint to π∗ for affine morphisms and quasicoherent sheaves. (Also, by Ob-servation 31.4.3, it is right-adjoint for finite morphisms and coherent sheaves on locally Noetherianschemes.) In particular, there is a natural morphism π∗π!G → G.

31.4.6. Proposition. — Suppose X → Y is a finite flat morphism of projective k-schemesof pure dimension n, and (ωY , tY ) is a dualizing sheaf for Y . Then π!ωY along with the tracemorphism

tX : Hn(X, π!ωY )∼ // Hn(Y, π∗π!ωY ) // Hn(Y,ωY )

tY // k

is a dualizing sheaf for X.(That first isomorphism arises because X → Y is affine.)

Proof.

Hn−i(X,F∨(π!ωY )) ∼= Hn−i(Y, π∗(F∨ ⊗ π!ωY )) as π is affine

∼= Hn−i(Y, π∗(Hom(F , π!ωY )))

∼= Hn−i(Y,Hom(π∗F , ωY )) by 31.4.5

∼= Hn−i(Y, (π∗F)∨(ωY ))

∼= Hi(Y, π∗F)∨ by Serre duality for Y

∼= Hi(X,F)∨ as π is affine

At the third-last and second-last steps, we are using the fact that π∗F is locally free, and it ishere that we are using flatness!

[The example of a doubled plane to A2 gives an example of a coherent sheafwhere cohomology disagrees with Ext. This is worth mentioning.]

12H.E.III.6.10ab

515

31.5 All projective Cohen-Macaulay k-schemes of puredimension n are finite flat covers of Pn

We conclude the proof of the Serre duality theorem 31.1.1 by establishing the result in thetitle of this section.

Assume X → PN is projective Cohen-Macaulay of pure dimension n (e.g. smooth).First assume that k is an infinite field. Then long ago in an exercise that I promised would be

important (and has repeatedly been so) [ref ], we showed that there is a linear space of dimensionN − n − 1 (one less than complementary dimension) missing X. Project from that linear space,to obtain π : X → Pn. Note that the fibers are finite (the fibers are all closed subschemes ofaffine space), and hence π is a finite morphism (see 15.6.C, or possibly Theorem 23.8.3). I’vestated the “Cohen-Macaulay/flatness theorem” 29.13.A that a morphism from a equidimensional

Cohen-Macaulay scheme to a smooth k-scheme is flat if and only if the fibers are of the expecteddimension. Hence π is flat.

31.5.A. Exercise. Prove the result in general, if k is not necessarily infinite. Hint: show thatthere is some d such that there is an intersection of N − n− 1 degree d hypersurfaces missing X.Then try the above argument with the dth Veronese of PN .

31.6 Serre duality in families

31.6.A. Exercise: Serre duality in families. Suppose π : X → Y is a flat projectivemorphism of locally Noetherian schemes, of relative dimension n. Assume all of the geometricfibers are Cohen-Macaulay. (Possibly just the fibers over closed points will suffice.) Thenthere exists a coherent sheaf ωX/Y on X, along with a trace map Rnπ∗ωX/Y → OY such that,for every finite rank locally free sheaves F on X, each of whose higher pushforwards are locallyfree on Y , 13

(72) Riπ∗F ×Rn−iπ∗(F∨ ⊗ ωX) // Rnπ∗ωXt // OY

is a perfect pairing. (Hint: follow through the same argument!)14

Note that the hypothesis, that all higher pushforwards are locally free on Y , is the sort ofthing provided by the cohomology and base change theorem. (In the solution to Exercise 31.6.A,you will likely show that Rn−iπ∗(F∨ ⊗ ωX) is a locally free sheaf for all F such that Riπ∗F is alocally free sheaf.)

You will need the fibral flatness theorem.Joe Rabinoff on the fibral flatness theorem: EGA IV(3).11.3.10–11. Specifically,

Corollary 11.3.11 says that if g : X → S, h : Y → S are locally of finite presentation,and f : X → Y is an S-morphism, then TFAE: (a) g is flat and fs : Xs → Ys is flat forall s ∈ s, (b) h is flat at all points of f(X) and f is flat. The proof looks really long,but a lot of that seems to be elimination of Noetherian hypotheses, so you might beable to extract a shorter proof using only local flatness.

[Less important. Possible exercise, once I think it through: interpret duality forfinite flat morphisms as a special case of this. Make sure to say what I mean byduality for finite flat morphisms — I currently don’t do so!]

31.7 What we still want

There are three or four more facts I want you to know.

13sdtwo

14sdf

516

• On the locus of X where k is smooth, there is an isomorphism ωX/k ∼= det ΩX/k. (Note

for experts: it isn’t canonical!) We define det ΩX/k to be KX . [Is this bad?] We

used this in the case of smooth curves over k (proper, geometrically integral). This issurprisingly hard, certainly harder than the mere existence of the canonical sheaf!

• The adjunction formula. If D is a Cartier divisor on X (so D is also Cohen-Macaulay[ref ]), then ωD/k =

`ωX/k ⊗OX(D)

´|D .

On can show this using Ext groups, but I haven’t established their existence or properties.So instead, I’m going to go as far as I can without using them, and then I’ll tell you a little aboutthem.

But first, here are some exercises assuming that ω is isomorphic to det Ω on the smooth locus.

31.7.A. Exercise (Serre duality gives a symmetry of the Hodge diamond). Suppose X isa smooth projective k-variety of dimension n. Define ΩpX = ∧pΩX. Show that we have a natural

isomorphism Hq(X,Ωp) ∼= Hn−q(X,Ωn−p)∨. Add a reference to this when we computeHodge diamondthings on projective space. Remark on other symmetry. 15

ΩpX

31.7.B. Exercise (adjunction for smooth subvarieties of smooth varieties). SupposeX is a smooth projective k-scheme, and and D is a smooth effective Cartier divisor. Show thatKD ∼= KX(D)|D . 16 Hence if we knew that KX ∼= ωX and KD ∼= ωD , this would let us computeωD in terms of ωx. We will use this in §where?. [Move this to Ω section, along with thedefinition of K.]

31.7.C. Exercise. Compute K for a smooth complete intersection in PN of hypersurfaces ofdegree d1, . . . , dn. Compute ω for a complete intersection in PN of hypersurfaces of degree d1,. . . , dn. (This will be the same calculation!) Find all possible cases where K ∼= O. These areexamples of Calabi-Yau varieties (or Calabi-Yau manifolds if k = C), at least when they havedimension at least 2. If they have dimension precisely 2, they are called K3 surfaces.K3, CY

31.8 The dualizing sheaf is an invertible sheaf on the smoothlocus

A simpler proof in the case where X is a curve is given in §31.9.) 17

We begin with some preliminaries.(0) If f : U → U is the identity, and F is a quasicoherent sheaf on U , then f ′F ∼= F .(i) The ′ construction behaves well with respect to flat base change, as the pushforward does.

In other words, if

X′h //

g

X

e

Y ′

f // Yis a fiber diagram, where f (and hence h) is flat, and F is any quasicoherent sheaf on Y , thenthere is a canonical isomorphism h∗e′F ∼= g′f∗F .

(ii) The ′ construction behaves well with respect to disjoint unions of the source. In otherwords, if fi : Xi → Y (i = 1, 2) are two morphisms, f : X1 ∪X2 → Y is the induced morphismfrom the disjoint union, and F is a quasicoherent sheaf on Y , then f ′F is f ′1F on X1 and f ′2F onX2. The reason again is that pushforward behaves well with respect to disjoint union.

Exercise. Prove both these facts, using abstract nonsense.Given a smooth point x ∈ X, we can choose our projection so that π : X → Pn is etale at

that point. Exercise. Prove this. (Hint: We need only check isomorphisms of tangent spaces.)[Refer back to why!]

15H.C.III.7.13

16adjunctionsmooth

17harderdc

517

So hence we need only check our desired result on the etale locus U for X → Pn. (This is anopen set, as etaleness is an open condition, [ref ].) Consider the base change.

X ×PnkU h //

g

X

e

U

f // Pnk .

There is a section U → X ×Pn U of the vertical morphism on the left. Exercise. Show thatit expresses U as a connected component of X ×Pn U . (Hint: Show that a section of an etalemorphism always expresses the target as a component of the source as follows. Check that s is ahomeomorphism onto its image. Use Nakayama’s lemma.) The dualizing sheaf ωPn

kis invertible,

and hence f∗ωPnk

is invertible on U . Hence g!f∗ωPnk

is invertible on s(U) (by observation (0)). By

observation (i) then, h∗g∗ωPnk∼= h∗ωX is an invertible sheaf.

We are now reduced to showing the following. Suppose h : U → X is an etale morphism.(In the etale topology, this is called an “etale open set”, even though it isn’t an open set in anyreasonable sense.) Its image is an open subset of X (as etale morphisms are open maps). SupposeF is a coherent sheaf on X such that h∗F is an invertible sheaf on U . Then F is an invertiblesheaf on the image of U .

(Experts will notice that this is a special case of faithfully flat descent.)Exercise. Prove this. Hint: it suffices to check that the stalks of F are isomorphic to the

stalks of the structure sheaf. [ref ]. Hence reduce the question to a map of local rings: suppose(B, n)→ (A,m) is etale, and N is a coherent B-module such that M := N⊗BA is isomorphic to A.We wish to show that N is isomorphic to B. Use Nakayama’s lemma to show that N has the sameminimal number of generators (over B) as M (over A), by showing that dimB/n N − dimA/m M .

Hence this number is 1, so N ∼= B/I for some ideal I. Then show that I = 0 — you’ll use flatnesshere.

[Hope to show that for finite etale morphisms, f ! = f∗. Then show that we canproject so that we avoid the singular locus. At this point we show that we get aninvertible sheaf. Next, in codimension 1, show that “Riemann-Hurwitz” holds.

What I might like: is there a natural map π∗G → π′G? I can’t quite see it. (Itmight be the following. HomA(M,A⊗B N) → HomB(M,N), sending (m 7→P

ai ⊗ ni) 7→(m 7→P

aini).) If this is true, then we would then take our nonsingularity fact to getan isomorphism above the smooth locus on the target, between the dualizing sheaf

downstairs and the dualizing sheaf upstairs.I might then hope to see how this fails for a discrete valuation ring.]

31.9 An easier proof that the dualizing sheaf of a smoothcurve is invertible

18Here is another proof that for curves, the dualizing sheaf is invertible. We’ll show that itis torsion-free, and rank 1.

First, here is why it is rank 1 at the generic point. We have observed that f ! behaves well withrespect to flat base change. Suppose L/K is a finite extension of degree n. Then HomK(L,K) isan L-module. What is its rank? As a K-module, it has rank n. Hence as an L-module it has rank1. Applying this to C → P1 at the generic point (L = FF (C), K = FF (P1)) gives us the desiredresult. (Side remark: its structure as an L-module is a little mysterious. You can see that somesort of duality is relevant here. Illuminating this module’s structure involves the norm map.)

Conclusion: the dualizing sheaf is rank 1 at the generic point.Here is why it is torsion free. Let ωt be the torsion part of ω, and ωnt be the torsion-free

part, so we have an exact sequence

0→ ωt → ω → ωnt → 0.

18easierdc

518

31.9.A. Exercise. Show that this splits: ω = ωt⊕ωnt. (Hint: It suffices to find a splitting mapω→ ωt. As ωt is supported at a finite set of points, it suffices to find this map in a neighborhoodof one of the points in the support. Restrict to a small enough affine open set where ωnt isfree. Then on this there is a splitting ωnt → ω, from which on that open set we have a splittingω→ ωt.)

Notice that ωnt is rank 1 and torsion-free, hence an invertible sheaf. By Serre duality, forany invertible sheaf L, h0(L) = h1(ωnt⊗L∨) and h1(L) = h0(ωnt⊗L∨)+h0(ωt⊗L). SubstituteL = OX in the first of these equations and L = ωX in the second, to obtain that h0(X,ωt) = 0.But the only skyscraper sheaf with no sections is the 0 sheaf, hence ωt = 0.

31.10 The sheaf of differentials is dualizing for a smoothprojective curve

One can show that the determinant of the sheaf of differentials is the dualizing sheaf usingExt groups, but this involves developing some more machinery, without proof. Instead, I’d liketo prove it directly for curves, using what we already have proved. (Note again that our proof ofSerre duality for curves was rigorous — the cup product was already well-defined for dimension 1schemes.)

I’ll do this in a sequence of exercises.Suppose C is an geometrically irreducible, smooth projective k-curve.We wish to show that ΩC ∼= ωC . Both are invertible sheaves. (Proofs that ωC is invertible

were given in §31.8 and §31.9.)Define the genus of a curve as g = h1(C,OC). By Serre duality, this is h0(C, ωC). Also,

h0(C,OC) = h1(C, ωC) = 1.Suppose we knew that h0(C,ΩC) = h0(C, ωC), and h1(C,ΩC) = h1(ωC) (= 1). Then

deg ΩC = degωC . Also, by Serre duality h0(C,Ω∨C ⊗ ωC) = h1(ΩC) = 1. Thus Ω∨C ⊗ ωC is adegree 0 invertible sheaf with a nonzero section. We have seen [ref ] that this implies that thesheaf is trivial, so ΩC ∼= ωC .

Thus it suffices to prove that h1(C,ΩC) = 1, and h0(C,ΩC) = h0(C, ωC). By Serre duality,we can restate the latter equality without reference to ω: h0(C,Ω) = h1(C,OC). Note that we

can assume k = k: all three cohomology group dimensions hi(C,ΩC), h0(C,OC) are preserved byfield extension [ref ].

Until this point, the argument is slick and direct. What remains is reasonablypleasant, but circuitous. Can you think of a faster way to proceed, for example usingbranched covers of P1?

31.10.A. Exercise. Show that C can be expressed as a plane curve with only nodes as sin-gularities. (Hint: embed C in a large projective space, and take a general projection. TheKleiman-Bertini theorem, or at least its method of proof, will be handy.)

Let the degree of this plane curve be d, and the number of nodes be δ. We then blow up P2

at the nodes (let S = Bl P2), obtaining a closed immersion C → S. Let H be the divisor classthat is the pullback of the line (O(1)) on P2. Let E1, . . . , Eδ be the classes of the exceptional

divisors.

31.10.B. Exercise. Show that the class of C on P2 is dH−2PEi. (Reason: the total transform

has class dH. Each exceptional divisor appears in the total transform with multiplicity two.)

31.10.C. Exercise. Use long exact sequences to show that h1(C,OC) =`d−1

2

´− δ. (Hint:

Compute χ(C,OC) instead. One possibility is to compute χ(C ′,OC′ ) where C′ is the image ofC in P2, and use the Leray spectral sequence for C → C ′. Another possibility is to work on Sdirectly.)

31.10.D. Exercise. Show that ΩC = KS(C)|C . Show that this is

(−3H +X

Ei) + (dH −X

2Ei).

519

Show that this has degree 2g − 2 where g = h1(OC). [Use long exact sequences. Or elseintersection theory.]

31.10.E. Exercise. Show that h0(ΩC) > 2g − 2− g + 1 = g − 1 from

0→ H0(S,KS)→ H0(S,KS(C))→ H0(C,ΩC).

31.10.F. Exercise. Show that ΩC ∼= ωC .

31.11 Trying to prove Serre duality for curves, followingForster

I still need two facts. First, h1(Ω) = 1. Second, H0(Ω(−D)) → H1(O(D))∨. Thelatter looks potentially hopeful: given any element of H0(Ω), cook up an element ofH1(O) that produces a nonzero residue.

Goal: Prove Serre duality for curves. There are two possibilities. (1) Prove it for line bundles,without using more general duality theory, or (2) Show that H0(Ω) ×H1(O) → H1(Ω) → k is aperfect pairing. The following argument shows both.

Suppose we are given a map H1(Ω)→ k, possibly using residues. (NEEDED)Then

H0(Ω(−D)) ×H1(O(D))→ H1(Ω)→ k

inducesiD : H0(Ω(−D))→ H1(O(D))∨.

It is injective. (NEEDED)We now show surjectivity. Suppose λ ∈ H1(O(D))∨ is non-zero. We wish to show that it is

in the image of iD. We may assume λ 6= 0.Here is our strategy. Choose any point P ∈ C. We will find some n 0, non-zero ψ ∈

H0(O(nP )), and ω ∈ H0(Ω(−D + nP )) so that in the following diagram19

(73) 0

0

0 // H0(Ω(−D))

×ψ

iD // H1(O(D))∗

×ψ

0 // H0(Ω(−D + nP ))

iD−nP // H1(O(D − nP ))∗

λ×ψ (take λ in the upper right, and multiply by ψ to obtain an element of the lower right) equalsiD−nP (ω) (take ω in the middle bottom, and apply iD−nP to obtain an element of the lowerright).

Exercise.

(a) Verify that the rows and columns of (73) are exact, i.e. verify the four injectivities.(b) Verify that this diagram commutes.(c) Show that the surjectivity follows from the existence of such a diagram.

We’ll get this by ensuring that the upper right and bottom middle have dimensions addingto more than that of the lower right, by cranking up n.

31.11.1. The map j. Consider j : H0(O(nP ))×λ // H1(O(D − nP ))∨ defined as follows. The

image of ψ ∈ H0(O(nP )) is determined by the image of λ in H1(O(D))∨ → H1(O(D − nP ))∨.If ψ is non-zero, this map is injective from the long exact sequence associated to

0 // O(D − nP )×ψ // O(D) // O(D)nP // 0,

so j is an inclusion.

19sdproof

520

By Riemann-Roch,

dim(im j) = h0(O(nP )) = n+ 1− h1(O).

31.11.2. The map iD−nP . For n 0, using Serre vanishing,

dim im iD−nP = h0(Ω(−D + nP )) = (− degD + n) + 1− h1(O).

Next, for n 0, h0(O(D − nP )) = 0, so

h1(O(D − nP )) = − degD + n− 1 + h1(O).

As n 0, dim im j + dim im iD−nP > h1(O(D − nP )), so we can find something in theirintersection. Thus we have created a diagram (73) as desired.

(In other news, I fear that ΩC/B is not ωC/B , and that in particular R1π∗ΩC/B is not

congruent to OB .)

31.12 Ext groups, and adjunction

[At some point mention the exact sequence

0→ H1(Ext0(F ,G))→ Ext1(F ,G)→ H0(Ext1(F ,G))→d2 H2(Ext0(F ,G)).]

Let me now introduce Ext groups and their properties, without proof. Suppose i is aExtnon-negative integer. Given two quasicoherent sheaves, Exti(F ,G) is a quasicoherent sheaf.Ext0(F ,G) = Hom(F ,G). Then there are long exact sequences in both arguments. In otherwords, if

0→ F ′ → F → F ′′ → 0

is a short exact sequence, then there is a long exact sequence starting

0→ Ext0(F ′′,G)→ Ext0(F ,G)→ Ext0(F ′,G)→ Ext1(F ′′,G)→ · · · ,and if

0→ G′ → G → G′′ → 0

is a short exact sequence, then there is a long exact sequence starting

0→ Ext0(F ,G′)→ Ext0(F ,G)→ Ext0(F ,G′′)→ Ext0(F ,G′)→ · · · .Also, if F is locally free, there is a canonical isomorphism Exti(F ,G) ∼= Hi(X,G ⊗ F∨).

For any quasicoherent sheaves F and G, there is a natural map Exti(F ,G) × Hj(X,F) →Hi+j(G).

For any coherent sheaf on X, there is a natural morphism (“cup product”) Exti(F ,G) ×Hj(X,F)→ Hi+j(X,G).

31.12.A. Exercise. Suppose X is Cohen-Macaulay, and finite type and projective over k (soSerre duality holds). (This was stated earlier in §31.2.1.) Via this morphism, show that

Exti(F , ωX)×Hn−i(X,F) // Hn(X,ωX)t // k

is a perfect pairing. 20 Feel free to assume whatever nice properties of Ext-groups you need (as wehaven’t proven any of them anyway). [To get from here to the general case, 0→ K→ O(?)⊕? →F → 0. Take the long exact sequence for Ext(·, ω) (NEEDED). Dualize. We need our pairing tobe respect this. use the sophisticated 5-lemma repeatedly.]

Hence Serre duality yields a natural extension to coherent sheaves. This is sometimes calledSerre duality as well. This more general statement is handy to prove the adjunction formula.

31.12.1. Adjunction formula. — If X is a Serre duality space (i.e. a space where Serreduality holds), and D is an effective Cartier divisor, then ωD = (ωX ⊗O(D))|D .

We’ve seen that if X and D were smooth, and we knew that ωX ∼= det ΩX and ωD ∼= det ΩD ,we would be able to prove this easily (Exercise 31.7.B). ref

20H.T.III.7.1 is the projective space case; H.T.III.6 includes this for projective CM.

521

But we get more. For example, complete intersections in projective space have invertibledualizing sheaves, no matter how singular or how nonreduced. Indeed, complete intersections inany smooth projective k-scheme have invertible dualizing sheaves.

A projective k-schemes with invertible dualizing sheaf is so nice that it has a name: it is saidto be Gorenstein. (Gorenstein has a more general definition, that also involves a dualizing sheaf. GorensteinIt is a local definition, like nonsingularity and Cohen-Macaulayness.)

31.12.B. Exercise. Prove the adjunction formula. (Hint: Consider 0 → ωX → ωX(D) →ωX(D)|D → 0. Apply HomX(F , ·) to this, and take the long exact sequence in Ext-groups.) Asbefore, feel free to assume whatever facts about Ext groups you need. [We need that Ext canbe computed on a “bigger” scheme. We need that duality respects the resultingexact sequence.]

The following exercise is a bit distasteful, but potentially handy. Most likely you should skipit, and just show that ωX ∼= det ΩX using the theory of Ext groups.

31.12.C. Exercise. We make a (temporary) definition inductively by definition. A k-variety is“nice” if it is smooth, and (i) it has dimension 0 or 1, or (ii) for any nontrivial invertible sheaf Lon X, there is a nice divisor D such that L|D 6= 0. Show that for any nice k-variety, ωX ∼= det ΩX .(Hint: use the adjunction formula, and the fact that we know the result for curves.)

31.12.2. Remark. You may wonder if ωX is always an invertible sheaf. In fact it isn’t, forexample if X = Spec k[x, y]/(x, y)2 . I think I can give you a neat and short explanation of thisfact. If you are curious, just ask. [Use the “prime” construction. X → A2

Y → Y .]

31.13 Older notes

Here is a puzzle. Given an element of H1(P1,ΩP1 ), what is the trace map? We write this asa cocycle using the usual open cover. Then we could take the residue at 0, or we could take theresidue at ∞. Which is it? Answer: either! [This is great: place this! This seems to mean that ωis not Ω in families of curves.]

31.13.1. Approach to ω = det Ω. Here are some useful facts. If L is an invertible sheaf, thenπ!(G ⊗L) ∼= (π!G)⊗ π∗L. Reason: by the pushforward formula, π∗ plays well with π∗, and hencewith π!. Hence if Y is Gorenstein, wX = π!ωY = π∗ωY ⊗ ωX/Y is Y .

Then perhaps show that for etale π, ωX/Y = 0, and for a branched cover, ωX/Y = O(R).

Actually, because π! commutes with flat base change, it suffices to prove this when the target isa discrete valuation ring. So we need to deal only with finite flat extensions of discrete valuationrings!

Another possibility: show that this works for the desingularization of a node.f ! is right-adjoint to Rf∗, see Brian Conrad’s duality book §3.3.

31.13.2. I should mention that in Exercise 22.6.E they have seen Serre duality in aninteresting special case, for the structure sheaf on hyperelliptic curves, and the sameidea — leveraging from a finite flat over of projective space — will be used.

Hartshorne requires Ext groups to make this work. Also, he hasn’t defined the cup product,so his section isn’t well-founded.

Hartshorne then shows that if X is projective, then it has a dualizing sheaf, and that it is infact Extr

PN (OX , ωPN ); a key lemma is that ExtiPN (OX , ωPn) = 0 for all i < codimPn X. 21

He then proves duality for a projective scheme. More generally: there are natural functorialmaps θi : Exti(F , ωX) → Hn−i(X,F)∨ such that θ0 is the map given in the definition of the

dualizing sheaf, and furthermore, the following are equivalent:22

(1) X is CM and equidimensional(2) for any F locally free on X, Hi(X,F(−q)) = 0 for i < n and q 0

21H.L.III.7.3, statement of L.7.4 and P.7.5

22H.T.III.7.6

522

(3) all θi are isomorphisms for all coherent FI’m not sure why we care about these added bonuses. But he gets our version, Theorem 31.1.1,as a corollary.

31.13.3. Serre duality apparently gives that Ext1(E,E) ∼= Ext2(E,E) for a Calabi-Yau threefold.

31.14 Weil’s proof of Serre duality for curves

Look at my file from MIT. Page 1 has a little history. I think this is over a field. Jump topage 4, then go to the top of page 8. It is about 4 pages long. I likely don’t want to give this inthe notes.

][primordial:

CHAPTER 32

Formal functions

In this section, we’ll prove a series of important theorems (Theorem on cohomology andbase change; Zariski’s Main Theorem; Stein factorization). These proofs will require some newtechniques, involving understanding differential techniques. They key new technique will be thetheorem on formal functions.

You should probably skip ahead to the three punchlines first, before reading the chapter.Second, read over the facts about inverse limits; we will use Corollary A.1.5. Note that thisrequires Noetherian rings; hence this will be an unavoidable hypothesis.

32.1 Notes on the structure of what I need

We’ll need II.9.1 (Mittag-Leffler) and II.9.1.2 (if we have an inverse system of finite-dimensionalvector spaces over a field, or more generally, an inverse system of modules with dcc over a ring,then they satisfy ML), and Krull 3.1A. Also Theorem II.9.3A, on completions of Noetherian ringsalong ideals.

These will buy us the theorem on formal functions. From this we get Zariski’s ConnectednessLemma (III.11.3), then Stein and Zariski Main theorem (III.11.5 and III.11.4).

Also from the theorem on formal functions, we get (using II.9.1.2 and II.9.1) III.12.10, fromwhich cohomology and base change III.12.11.

32.2 The theorem on formal functions

Let’s cut to the chase. Here’s the key technical step.Explain the morphisms in the statement.

32.2.1. Theorem on formal functions. — 1 X → Y projective morphisms of Noetherianschemes, F coherent on X, y ∈ Y . Then Rif∗(F )y → lim←−H

i(Xn,Fn) is an isomorphism for alli ≥ 0.

As usual, we will apply this in the case i = 0, but prove it in general, not just because wecan, but also because our argument will be by descending argument on i.

32.2.2. Remark. H points out that this is true for proper morphisms [EGA, III.4]. Grothendieck

does this by showing that both are the same as Hi(X, F), which I won’t introduce. 2

[One consequence, we’ve already seen: higher direct image sheaves vanish in degree higherthan the dimension of the maximal fiber (Theorem 23.8.2). But our earlier argument worked forquasicoherent sheaves, and non-Noetherian schemes...]

Proof. [I am really following [H] here!] First, it is local on Y . Hence we can reduce to the casewhere X → PnY . Next, we can reduce to the case where X = PnY . [Give ref for this: we computecohomology on closed immersions. Awkwardly, I now have 2 n’s in the proof.]

1t:ff H.T.III.11.1

2tffproper, H.R.III.11.1.1

523

524

As usual, we write

0→ K→ ⊕O(m)→ F → 0.

The result is clear for ⊕O(m): we just compute it in the same way. Let N = ⊕O(m) forconvenience.

We prove the statement by descending induction on i. Clearly true for i > n by dimensionalvanishing [ref].

We have

Kn →Nn → F → 0.

Write this as two exact sequences:

0→An → Kn → Bn → 0

and

0→ Bn →Nn → Fn → 0.

Verify that A and B form inverse systems. [Do this myself first!][Now I’ll just use the same names for the morphisms as [H]. Re-arrange this argument, so

that we don’t have such a big diagram.]

Hi(X,K)∨ //

α1

Hi(X,N )∨

α2

// Hi(X,F)∨

α3

// Hi+1(X,K)∨ //

α4

Hi+1(X,N )∨

α5

lim←−Hi(Xn)

β1

lim←−Hi(Xn)

β2

lim←−H

i(Xn,Bn) // lim←−Hi(Xn,Nn) // lim←−H

i(Xn,Fn) // lim←−Hi+1(Xn,Bn) // lim←−(Xn,Nn)

The top row is exact, because completion is an exact functor for Mittag-Leffler reasons again; herewe need hypotheses that Y is Noetherian.completion is an exact

functor The trickiest thing to note is that β1 and β2 are both isomorphisms. We want to show thatlim←−H

i(Xn,An) = 0. We first show that for any n, there is an m > n such that Am → N is thezero map. By quasicompactness, we may reduce to the case X affine, say X = SpecB. Thenthere is a brief argument on [H, p. 279] using Krull’s theorem A.1.6. This needs to be explainedmore (and ideally extirpated).

Now α2 and α5 are isomorphisms because we understand N , and α4 is an isomorphism bythe inductive hypothesis. Then by the 4 1

2-lemma (3.8.C), α3 is surjective. Hence α1 is surjective

(by doing the same construction for it), and hence α3 is an isomorphism by the Five Lemma 3.8.C,hence we are done.

32.3 Zariski’s Main Theorem and Stein Factorization

The hypothesis of projectivity can be replaced by properness, by Remark 32.2.2. [In fact, Iwonder if it works with much weaker hypotheses.]

Proof. (old proof of Connectedness lemma) Assume otherwise. Then f−1y = X1 ∪X2. Then the

nth order neighborhood is also disconnected. Let e1 be the function 1 on X1 and 0 on X2, and lete2 be the function 1 on X2 and 0 on X1. (This makes sense for any order neighborhood.) Then

they have natural images in the inverse limit. Thus we get elements e1, e2 ∈ Oy with e1 + e2 = 1,

e1e2 = 0. However, Oy is a local ring (Exercise A.1.B). e1e2 = 0, so neither is a unit, so both arein the maximal ideal, hence e1e2 6= 1. Contradiction.

32.3.1. Zariski’s Main Theorem. — This is now at 29.14.3.

Mumford gives other statements in [M, III.9]:

• I. original form• II. topological form (complex geometry)

525

• III. power series form• IV. Grothendieck form• V. Connectedness form: variation of H’s version. Work over k. Birational proper

morphism. Work with a specific normal point y ∈ Y , without requiring y to be normal.

He invokes Z-S for III, and shows III implies I and II. He refers to EGA III.4.3 for Grothendieck’scohomological proof of V. In my EGA notes, for IV, I have EGA III.4.4, and a reference also toEGA IV.8.12.

Jason Starr has a proof of ZMT in his notes.

32.3.2. [An exercise from H that I like: III.11.4 Principle of connectedness. Xt flat family ofclosed subschemes, parametrized by an irrreudcible curve T of finite type /k. Suppose there is a

nonempty open U ⊂ T such that for all closed points, Xi is connected. Then Xt is connected forall t ∈ T . Another (III.11.8): f : X → Y projective morphism, F coherent sheaf on X flat overY , Hi(Xy ,Fy) = 0 for some i and some y ∈ Y . Then Rif∗F = 0 in a neighborhood of y.] 3

semicontinuity ideas32.3.3. Semicontinuity ideas. If X → Y is a proper surjective morphism of irreducible

varieties, and Y is normal, then the number of geometric preimages is a lower semicontinuousfunction. (An extended discussion may be in order here. We ask the question for varieties oversome algebraically closed field, such as C, and just count preimages. We ask the question moregenerally, and then realize that we will need the hypotheses. We clearly want to now do this overany point. Then we move to geometric preimages, motivated both by Spec Z, and generic points(consider double covers). Then we show that the number is a constructable function (definethis earlier!). Then we are reduced to a question about local rings, which we can hope to prove. constructable functionUsing Stein factorization, we can even consider only finite morphisms. (I’d discussed this withDaniel Allcock and Martin Olsson, and Daniel knew the relevant algebraic fact.)

Can I do something similar with number of components?The dimension of the Zariski tangent spce is an uppersemicontinuous function on the set of

closed points of Y . Here Y is finite type over an algebraically closed field (or something weaker?).4

This is dumb.

32.4 Cohomology and base change (still to place)

32.4.1. Cohomology and base change.

F is a quasicoherent sheaf on X, and π : X → Y is separated and quasicompact (so push-forward sheaves are defined). Say Y ′ → Y . We want to compare base change of the pushforwardand the pushforward of the base change. Important special case: Y ′ is a closed point of Y , inwhich case we want to compare higher pushforward sheaf with cohomology.

Here’s the question:

0→ C0 →d0 C1 → · · · → Cn → 0

is a complex of R-modules, and we have a ring morphism R→ S.

32.4.2. Theorem. — 5

(a) (cohomology and base change) There is a natural morphism Hi(C•)⊗RS → Hi(C•⊗RS).

(b) This morphism is an isomorphism if R→ S is flat.

Perhaps this will be an exercise. [now done earlier]Example: if y → Y is the generic point. [now done earlier]Remark: in the i = 0 case, this is just the statement that functions pull back. This says that

cohomology classes pull back. [now done earlier]

3H.E.III.11.4, H.E.III.11.8

4H.E.III.12.1

5t:cohbasechange (a) is H.R.III.9.3.1, (b) is H.P.III.9.3. He has the additional condition that the morphism

is finite type of Noetherian schemes

526

Proof. We will use0→ im di−1 → ker di → Hi(C•)→ 0.

We have natural maps

(im di−1)⊗R S //

(ker di)⊗R S //

Hi(C•)⊗R S // 0

0 im(di−1 ⊗R S) // ker(di ⊗R S) // Hi(C• ⊗R S) // 0

(a little thought is needed for the vertical homomorphisms) from which we get a natural morphism

Hi(C•) ⊗R S → Hi(C• ⊗R S).(b)

0 0

0 // ker(di ⊗R S) // Ci ⊗R S

OO

// im(di ⊗R S)

OO

// 0

(ker di)⊗R S //

OO

Ci ⊗R S

OO

// (im di)⊗R S

OO

// 0

0

OO

Using a spectral sequence [make this a spectral sequence exercise in glossary?] or by a slightlyweaker form of the Snake Lemma 3.8.5, or by diagram chase, we get that the following sequenceis exact:6

(74) (ker di)⊗R S → ker(di ⊗R S)→ (im di)⊗R S → im(di ⊗R S)→ 0.

Now using flatness of S, we get a slightly better version of that diagram in the proof of (a):

A 0

B

0 (im di−1)⊗R S //

(ker di)⊗R S //

Hi(C•)⊗R S // 0

0 im(di−1 ⊗R S) // ker(di ⊗R S) // Hi(C• ⊗R S) // 0

0 C D

Using the Snake Lemma 3.8.5, we see first that A and B are 0, and then by (74) that C is 0, andhence that D is 0, from which we are done.

]

6blick

CHAPTER 33

? ? Comparing complex analytic and complex

algebraic geometry: Serre’s “GAGA” Theorem

[S] 1

[For compact complex spaces, there is at most one algebraic structre. This is proved on p.34 of [MAV], ans use Chow’s theorem. If X is a complete algebraic variety and Y is a closedanalytic subset of Xhol, then Y is Zariski-closed in X. For something noncompact, this is false.Mumford p. 35 mentions an example of Serre, but I think there is ane easier example: C2. Theny = ex and y = x can’t both be algebraic.]

[Notes about analytification functor. Do it by universal property. Black box: OanX is coherent.Proposition: an: coh → coh. Proof: an: fpr → fpr.Easy exercise: vb → vb.Exercise: Zariski-connected implies analytically connected. for projective things. [If affine

case can be done, then Zariski-connected implies analytically connected always.][Earlier notes:] Mention at some point Chow’s theorem, that complex subvarieties of Pn are

algebraic. This is proved in [MuCPV, Ch. 4]. Use Riemann-Roch to show that all complexcurves are projective.

This is Ravi’s summary of Jarod and Nikola’s explanation of the proof of GAGA, followingKiran’s roadmap (May 15, 2005). (Reference: J.-P. Serre, Geometrie algebrique et geometrieanalytique, Ann. Inst. Fourier (Grenoble) 6 (1955–56), 1–42.)

I later want to add: analytification to normalization chapter.Theorem (GAGA, by Serre). There is a natural equivalence of categories between coherentalgebraic sheaves and coherent analytic sheaves on Pn. Their cohomology groups are naturally iso-morphic. Every analytic subvariety of Pn is algebraic (Chow’s Theorem). The previous statementsapply with Pn replaced by any projective analytic/algebraic variety.

Analytic Background. We will need to know some analytic facts, and we’ll collect themhere. A coherent analytic sheaf is something which is may be locally written as the cokernel of twofree finite rank sheaves. Coherent analytic sheaves on X form an abelian category, so for examplecokernels and kernels of morphisms of coherent analytic sheaves are also coherent analytic sheaves.We need sheaf cohomology.

We will assume that we know: (1) Hialg(P

n,O) = Hian(Pn,O) for all i. (2) Hi

an(Pn,F) is a

finite-dimensional C-vector space for i = 1. (References: [GH], Grauert and Remmert) We willassume that the ideal sheaf of any analytic subvariety is coherent. (3) OanPn,x is Noetherian. (4)

There is some N (a function only of n) such that Hi(Pn,F) = 0 for any coherent analytic sheaffor i > N .

Analytification. Given a complex algebraic variety X, we can construct a complex analyticvariety Xan along with a morphism π : Xan → X of local-ringed spaces. The slickest way to dothis is via a universal property construction.

If F is a coherent algebraic sheaf on X, then define

Fan := π∗F = π−1F ⊗π−1OXOXan .

This gives a covariant functor from coherent algebraic sheaves on X to analytic sheaves on Xan.Note that the analytification of invertible sheaves are also invertible.

Claim. Suppose that X is quasiprojective. Then Fan is coherent.

1gaga

527

528

Proof. Write

⊕O(m′)→ ⊕O(m)→ F → 0.

Apply the exact functor π−1, and then the right-exact functor ⊗π−1OXOXan , to obtain

⊕O(m′)an → ⊕O(m)an → Fan → 0.

In the above proof, we used the fact that ⊗π−1OXOXan was right exact. In fact it is exact:

Claim. OXan is a flat π−1OX -module.In particular, analytification is an exact functor from coherent algebraic sheaves on X to

coherent analytic sheaves on X.

Proof. We will use the following algebraic fact: if A is a local Noetherian ring, then A → A isfaithfully flat.

We can check this at stalks at C-points. The stalk of π−1OX at x is just OX,x (as pullbackpreserves stalks). We have maps of local rings:

OX,x →OanX,x → OX,x.

Here we are using the fact that the completion of a local ring is (naturally isomorphic to) the

completion of its analytification; this requires a little argument, that OX,x/mm ∼→ OanX,x/m

m.

Now OX,x → OX,x and OanX,x → OX,x are both faithfully flat. Thus if M1 → M2 → M3 is any

complex of OX,x-modules, then

M1 →M2 →M3 exact

⇔ M1 ⊗OX,xOX,x →M2 ⊗OX,x

OX,x →M3 ⊗OX,xOX,x exact

⇔ M1 ⊗OX,xOanX,x ⊗Oan

X,xOX,x →M2 ⊗OX,x

OanX,x ⊗Oan

X,xOX,x →M3 ⊗OX,x

OanX,x ⊗Oan

X,xOX,x exact

⇔ M1 ⊗OX,xOanX,x →M2 ⊗OX,x

OanX,x →M3 ⊗OX,x

OanX,x exact

so OX,x → OanX,x is also faithfully flat, and in particular flat. [We use here the “black box” that

OanX,x is Noetherian.

Claim. Analytification and twisting by O(n) commute.

Proof.

Fan ⊗O(n)an = π−1F ⊗π−1OXOXan ⊗OXan π

−1O(n) ⊗π−1OXOXan

=“π−1F ⊗π−1OX

π−1O(n)”⊗π−1OX

OXan

= π−1F(n)⊗π−1OXOXan . (say something here?)

Observe that we have natural maps Hi(Pn,F) → Hi(Pn,Fan), for example by taking Cechcohomology. Here is a better description: we have natural maps H i(Pn,F) → Hi(Pnan, π

−1F),and since ⊗π−1OX

OanX is exact, we have a natural map Hi(Pnan, π

−1F) → Hi(Pnan,Fan). These

maps behave well with respect to long exact sequences for example. [The first is a universal δ-functor, the second is a δ functor (this uses the fact that analytification is exact), and there is amap on H0.]

(This description should be improved. One possibility is to bring in derived functor coho-mology, and we have to be careful, because the category of coherent algebraic sheaves doesn’t

have enough injectives, and we have to work a larger category, of OX-modules. Then we need toextend the analytification functor to OX -modules.)Claim. The natural morphism Hi(Pn,F) = Hi(Pn,Fan) is an isomorphism.

Proof. We first show that this is true for F = OPn(m), by by induction on n. The result is clearfor n = 0. Using 0 → OPn (m − 1) → OPn(m) → OPn−1 (m) → 0, and from the known fact form = 0, using the five-lemma, we are done.

Next, write

0→ K→ ⊕O(m)→ F → 0

529

where K is coherent. Take the long exact sequence in the algebraic category, and the long exactsequence in the analytic category. (Note: 0→ Kan → ⊕O(m)an → Fan → 0 is exact by exactnessof analytification.) The first exact sequence maps to the second. We have isomorphisms atHi(Pn,⊕O(m)). For i > N,n (where N = N(n) is the number given by Cech vanishing in theanalytic setting, see analytic background section), Hi(Pn,K) = Hi(Pn,Kan) = 0. Let’s look atthe long exact sequence near there:

HN (Pn,⊕O(m)) //

∼

HN (Pn,F) //

HN+1(Pn,K) //

∼

HN+1(Pn,⊕O(m))

∼

HN (Pn,⊕O(m)an) // HN (Pn,Fan) // HN+1(Pn,Kan) // HN+1(Pn,⊕O(m)an)

By the 4 12-lemma (Lang’s Algebra p. 169), the remaining vertical arrow is surjective, i.e.HN (Pn,F)→

Hn(Pn,Fan) is surjective for all (coherent algebraic) F . But then by examining

HN (Pn,K) //

surj.

HN (Pn,⊕O(m)) //

∼

HN (Pn,F) //

surj.

HN+1(Pn,K)

∼

HN (Pn,K) // HN (Pn,⊕O(m)an) // HN (Pn,Fan) // HN+1(Pn,Kan)

and using the 4 12-lemma, the central vertical arrow is injective as well, i.e.HN (Pn,F)→ Hn(Pn,Fan)

is injective for all (coherent algebraic) F .Repeating this argument downward inductively gives the desired result.

Claim. If F and G are coherent algebraic sheaves, then Hom(F ,G)an → Hom(Fan,Gan) is anisomorphism.

(As an added bonus, we could see that the same result is true for Ext’s, but we won’t needthat here.)

Proof. We need only check this at stalks at C-points. Let x be a C-point. Then

(Hom(F ,G)an)x ∼=“π−1Hom(F ,G)⊗π−1OX

OXan

”x

∼= Hom(F ,G)x ⊗π−1OX,xOXan,x (π−1 preserves stalks)

∼= Hom(Fx,Gx)⊗π−1OX,xOXan,x

∼= Hom(Fx ⊗π−1OX,xOXan,x,Gx ⊗π−1OX,x

OXan,x)

(Reason: B is A-flat, M is finitely presented implies HomA(M,N)⊗AB = Hom(M⊗AB,N⊗AB).This is in Matsumura, but is best taken as an exercise.)

∼= Hom(π−1F ⊗π−1OXOXan , π−1G ⊗π−1OX

OXan)x

∼= Hom(Fan ,Gan)x.

(Note that this isomorphism is natural, i.e. agrees with the natural map from the left to theright.)

Corollary. Hom(F ,G)an → Hom(Fan,Gan) is an isomorphism.Thus the analytification functor is fully faithful.

Proof. We apply H0 to both sides of the previous claim. Note that the right side is coherent, asthe left side is.

We’re now most of the way there. We now need to show that, given a coherent analyticsheaf G on Pn, there is a coherent algebraic sheaf F with G ∼= Fan. We will say that a coherentanalytic sheaf is algebraic if it is the analytification of an algebraic sheaf. We want to show that

any analytic coherent sheaf G on Pn is algebraic.Claim. Every coherent analytic sheaf G satisfied G(m) generated by global sections for m 0.Nikola’s proof.

Claim. Every coherent analytic sheaf was algebraic.

530

Proof. (This is the proof we’ve seen before for coherent algebraic sheaves.) Say G(m) is generatedby ` global sections. Then we have a surjection O⊕` → G(m) from which O(−m)⊕` → G → 0 is

exact. If K is the kernel, then K is coherent, and similarly we can find a surjection O(−m′)⊕`′ →K→ 0. Then we have a presentation

O(−m′)⊕`′ φ→O(−m)⊕` → G → 0.

But φ is an algebraic morphism by the previous claim, from which its cokernel is algebraic. Morecompletely: let F be the cokernel of φ in the algebraic category: in the category of coherentalgebraic sheaves,

O(−m′)⊕`′ φ→ O(−m)⊕` → F → 0.

Applying the exact functor π−1 and the right-exact functor ⊗π−1OXOanX , we have

O(−m′)⊕`′ φ→O(−m)⊕` → Fan → 0

in the analytic category, i.e. Fan = cokerφ in the analytic category. As G = cokerφ, we aredone.

[This is as far as the notes are complete. We now want to show that a coherent analytic sheafis “eventually generated by global sections”.]

Here are some earlier notes.

Proof. We proceed by induction on n. The n = 0 result is immediate.Write 0→ K → G(−1) → G → L → 0. By our inductive hypothesis, K and L are algebraic.

Split this into two short exact sequences: 0→ K→ G(−1)→M→ 0 and 0→M→ G → L→ 0.Twist by m 0, and use Serre vanishing for K, to get Hi(Pn,G(m− 1)) ∼= Hi(Pn,M(m)). Thenfrom the long exact sequence for the second short exact sequence (using Serre vanishing for L), wehave H1(Pn,G(m − 1))→ H1(Pn,G(m)) → 0. Now (by our analytic preliminary) Hi(Pn,G(m′))is finite-dimensional for each m′, so for m 0, H1(Pn,G(m−1))→ H1(Pn,G) is an isomorphism,from which

H0(Pn,G(m))→ H0(Pn−1,L(m))

is surjective.This was supposed to give us something.

Chow’s Theorem. Suppose X ⊂ Pn is an analytic subvariety. Then X is algebraic.

Proof. Coming soon. Basically, let IX is the ideal sheaf of X. Show that it is coherent. ...

Here are some other notes: stalkifying commutes with analytification. Pushforward of coher-ent is coherent. The assumptions: (1) OX,x is Noetherian.

(2) If F is coherent, then Hi(Pn,F) is finite-dimensional over C (11.10.2 in some book, Cartan-Serre theorem.)(3) Hi(Pn,OPn) is what you expect C if i = 0, and 0 otherwise.(4) F coherent, i 0, Hi(Pn,F) = 0. Cartan B: X Stein, S coherent then S is acyclic. Also, C∗

and C are Stein. (something about p. 223, 9.3.3 in some book)(5) Ideal sheaf coherent.(6) Kernel of coherent is coherent, i.e. coherent sheaves form an abelian category. That’s theorem9.5.4 in some book. Cokernels of coherent to coherent being coherent: that’s needed in Nikola’sproof.

Joseph L. Taylor: Several complex variables with connections to algebraic geometry and Liegroups.

][primordial:

Part VIII

? Further topics to explore

CHAPTER 34

? Germs of topics

34.1 ? Quotients

Group quotients.If G is an algebraic group, then a closed subgroup P ⊂ G is said to be parabolic if G/P is

projective. This is equivalent to: P contains a Borel. (Partial explanation: Suppose P containsa Borel B. Now G/B is projective (describe it as a tower of projective bundles), and G/P isquasiprojective (Jack explained why), and G/P is the image of G/B, so G/P is projective. Next,suppose G/P is projective. then... hmm... try to build up the borel as follows. Find a fixed pointby P . Then find a fixed point under ... hmmm.)

34.2 ? Geometric invariant theory

34.3 ? Classifying surfaces

Introduction to the problem.

34.3.1. Introduction to the minimal model program. Translation: Minimal modelprogram for surfaces

34.4 Deformation theory

intro to deformation theory; deformations of complete intersections; deformations of smoothvarieties.

If A = k[[x1, . . . , xn]]/I where I ⊂ m2, then (I/mI)∗ is an obstruction space, and indeed theminimal one. (Barbara Fantechi’s informal lecture in Cetraro.)

34.5 ? Complex varieties

Presumably just state facts.They are smooth in real codimension 1. This tells you something which isn’t I think that

obvious. Moreover, you can “stratify” using generic smoothness.Milne Remark 21.9 and p. 120 (presumably his online etale nots). Connnected in the zariski

topology implies connected in the analytic topology.Refer to GAGA.

533

534

Apparently Munoz has a recent example of a non-compact complex analytic space with twodifferent algebraic structures. Someone told me this at GAEL.

34.5.1. Topology. There exists a Whitney stratification. (See notes inside front cover ofmy copy of [KS].)

Normal implies topologically normal (in Shafarevich). The converse is not true (cusp).

34.5.2. Analytic methods.

34.6 Grobner bases

In a polynomial ring. Choose a term order. Say you have an ideal. You want to checkideal membership, and you have generators of the ideal. You could try the division algorithm. Itdoesn’t always work (x100, x100 − y). It will work if the initial ideal is generated by the initialelements of the generators (exercise). Such a set of a generators is called a Grobner basis.

How do you make one? Given any two generators f1, f2, you check if the S-polynomial passesthe division algorithm check. Basically, you cancel the leading terms of f1 and f2 (by multiplyingboth by an appropriate monomial), and see what you get.

If the S-polynomial doesn’t pass the division algorithm check, then you don’t have a Grobnerbasis obviously. Exercise: if it does, then you do have a Grobner basis. Proof: say h ∈ I, soh =

Pgifi. Look at initial elements of the fi (i.e. if the initial ideal of h comes from initial

elements of some of the fi). If there is any cancellation, use the S-polynomial to get rid of all butone of those fi’s where there is cancellation.

Buchsberger algorithm.

CHAPTER 35

? Toric varieties

1

35.0.1. Toric varieties. M character lattice of torus T = Hom(T, k∗).N dual lattice = 1-parameter subgroup = Hom(M,Z) = Hom(k∗, T ).toric variety with torus T corresponds to fan ∆ in NR: rational polyhedral cones meeting

along shared faces.T -invariant affine open sets are in one-to-one correspondence with cones σ ∈ ∆. Denote such

an open Uσ . Then Uσ∩σ′ = Uσ ∩ Uσ′ .Example: P2, Pn. (Note: dimensions are opposite those of N .)Ask them: When is it singular?Compact?Examples: Fn. An singularities.

35.0.2. Line bundles.Toric line bundle corresponds to piecewise linear function ψL on Nσ .Cartier and Weil divisor. When is Weil = Cartier?Example of non-Cartier: lattice = 〈e1, e2, (e1 + e2)/2〉, cone = 〈e1, e2〉. Then no integral

linear functional on the cone restricts to this.Some later notes: a divisor (Cartier) corresponds to picking one vertex for each affine open

set. Adjacent open sets (sharing a divisor) are connected by a line in the direction of that divisor.Globally generated means that each of these lines points in the “correct” direction. Ample meansthat in addition, they have positive length. The number of global sections is what you’d think.

35.0.3. Morphisms. Morphisms of toric varieties. Resolving An-singularities. The smallresolution of the cone over the quadric surface.

35.0.4. Sam Payne’s example.From Sam Payne, May 20, 2006:”Here is an example with Picard rank 1 and no effective Cartier divisors. (I like to picture

the effective cone in the space of classes of R-Weil divisors, which in this case is 5-dimensional,with the line of R-Cartier classes going through the vertex but missing the rest of the cone.) Theexample referred to is the example in math.AG/0501204 – so the lattice is N = Z3, and the raysare generated by

v1 = (1, 1, 1), v2 = (−1, 1, 1), v3 = (−1,−1, 1), v4 = (1,−1, 1),

v5 = (1, 0,−1), v6 = (0, 1,−1), v7 = (−1, 0,−1), v8 = (0,−1,−1).

Presumably one can construct such examples with arbitrarily large Picard rank, but I don’tknow how to do it off the top of my head. Also, I don’t know if one can control the codimension ofthe space of R-Cartier classes — for example, can the space of R-Cartier classes be a hyperplanecutting out the vertex of the effective cone?”

35.1 June 2007 notes from Batyrev

1toric

535

536

Suppose ∆ is a lattice polytope.Define P∆ := ProjS∆. S∆ = ⊕i≥0Si∆.Better definition: Let M := Hom(Tn,Gm), χ : Tn → Cn. Then M ∼= Zn. [forme:

Mnemonic: “meeting”.]N := Hom(Gm,Tn), ν : C∗ → Tn. N = Zn [forme: Mnemonic: “nugget”]There is an obvious pairing 〈·, ·〉 : M ×N → Z, χ× ν : C∗ → C∗.∆ → MR. Then we have a fan [forme: mnemonic: “faN”] N ⊂ NR defined by the

equivalence relation y ∼ y′ in NR if min∆〈x, y〉 and min∆〈x, y′〉 are obtained on the same face of∆.

Example: small triangle.P∆ =

‘θ<∆ Tθ . dimTθ = dim θ.

Flow interpretation: ν ∈ N gives a one-parameter family C∗ → T, which yields a limit ast→ 0, and each equivalence class gives a different fixed point.

CHAPTER 36

? Intersection theory

36.1 Intersection theory on surfaces

36.2 Intersection theory in general

Earlier notes: Introduction to intersection theory? Flat pullback, proper pushforward, inter-section with effective Cartier. State GRR.

Definition: cycles. 2 points on P1 are isomorphic. Flat pullback and proper pushforward.Degree map (integral notation). Chern classes, excess in-

tersectionMost important facts to know: Gysin pullback. Can pull back under lci morphism.Chern classes: how to think of them. Properties: where they live. Vanishing. Chern roots.

Behavior in exact sequences.Chow groups of projective bundles. Degree of a vector bundle.Intersections on something smooth: transverse; multiplicities.Excess intersection.Operational Chow classes (“Chow cohomology”). Commutative in genus 0.Maps to Borel-Moore homology. But I’m not talking about non-algebraic things.Useful calculations:

• Degree of the discriminant• Is every fibration by Pn’s a Pn-bundle in the Zariski topology? No; but yes in the etale

topology. (Hence maybe talk about Brauer group, see discussion of gerbes in 37.0.10.

537

CHAPTER 37

? Moduli spaces, moduli functors, and stacks

[Maps of functors: locally of finite presentation over S if for all filtered projective systemsof affine schemes Zλ ∈ Schemes/S (directed systems of rings, “cofinal”), we have lim→ F (Z) =limF (lim←Zλ

). This is handy if you can prove something for Noetherian rings, and you want toprove it for all rings. Brian C says: that’s not the best example.] The equivalence in the case ofmaps of rings is from EGA.

[This section is currently an idea pit]Ideas from deformation theory workshop.

• outline: first, category nonsense. Then stack. Then Grothendieck topologies.• category: never saying “same”. (Equivalence of categories is an example of this —

essentially the same category.)• functor of points: there is no moduli space of curves; nor moduli of vector bundles.

But it is kind of geometric.• What is a stack? Notice that to glue vector bundles together, we need additional

information. (Use vector bundles instead of sheaves, because you don’t want to saythat “sheaves form a sheaf”.)

• Another example of this: moduli of curves.• Towards a stack in a topology, motivated by moduli of curves or vector bundles or

sheaves. First approximation: groupoid for each base. But you want restriction maps,aka pullback. Groupoid over the base category (couple of axioms: pullbacks exist, andthat triangle axiom). Exercise: show that these two things make sure that the fibercategory is a groupoid.

• Then add gluing facts to make it a stack. Isom should be a sheaf. (This is a prestack,according to kresch et al.) Objects glue given isomorphisms, just like vector bundles.Example: moduli of vector bundles form a stack. Moduli of sheaves form a stack. BG:locally trivial G-bundles. A group quotient. Scheme is a stack (even a sheaf).

• Harder: moduli of curves form a stack.• Grothendieck topologies. (Explain why we need etale topology, and indeed fppf.)• Deligne-Mumford stack.

Motivation for moduli of curves: why is there a natural line bundle associated to the nodalcurve locus?

37.0.1. November 20, 2006: Toward stacks.Motivations:3 motivations: orbifolds, moduli space, quotients (subcase: point / Gm).counting things with fractions.enumerative geometry: elliptic curve count. how many cubic elliptic curves with given j-

invariant pass through 9 general points? Better question: given any curve C, how many cubiccurves with a given isomorphism to C pass through 9 general points? (side question: “generalcurve”? what does it mean?)

Moduli spaces Intuition Path in the space. Loop. (Hence π1(GR(1, n)) surjects onto Z/2.Hence nontrivial. G(k, n)?)

Psueudofunctor as spaces. Definition of moduli space. Yoneda. Examples to keep in mind.

Mg (define rigorously: flat families). X/G. Then BG: do G = Z/2.Line bundle on M. PicM.

Then cohomology class. Chern classes of vector bundles.Geometric point.

539

540

Zariski tangent space (e.g. Mg) — caution here, as we don’t know what closed points are.Map of one space to another. Example: Mg,1 →M. Getting the cotangent line bundle. κ-classesetc.

Harder: fibered product. Then representable (schematic?) functor.Then define stack?Coarse moduli space. Best approximation. Mg,1 →M gives Mg,1 → Mg . Line bundle on

M giving line bundle on M is harder!

37.0.2. May 9, 2006.Functors are like spaces. Maps between functors. Functor category; Yoneda says that objects

are fully faithful subcategory.Invertible sheaves on a functor. Hodge bundle on Mg .Open and closed subfunctor. Seesaw theorem gives closed subfunctor. Locus where something

is a section is a locally closed subfunctor. Mg,n → Mg,n−1; but do this where you have noautomorphisms.

Tangent space to a functor “over k”. Kodaira-Spencer map.The functor category of schemes is kind of large. The ones we care about will be geometrically

natural.Representable functors. Tangent space/Kodaira Spencer map. Moduli space (or “fine moduli

space”). Representable functors have a universal object. Remark: They are sheaves.Grassmannian (several definitions); prove. (Ex: show that Pn is Grassmannian) Why is

it projective? (Interesting alternative definition of Grassmannian: G = GL(n) acts on ∧kKn.Consider the vector v = e1 ∧ · · · ∧ en. Consider Gv. The stabilizer is a parabolic. How do youknow the orbit is closed?

Moduli of genus 0 curves. (Needed: they have no automorphisms.) M0,3 = pt: alwaystrivial. Give proof: map to P1, canonical.

M0,4∼= P1. The idea is this. Given π : C → B. The four sections are Cartier (why?).

π∗ωπ(s1 + s2 + s3 + s4) is locally free of rank 2. Relative very ample (why?). Hence C → aP2-bundle. We can trivialize it (needed: those 4 points aren’t on the same line) as follows: x isthe section on which s1 and s2 lie, etc.; this gives a one-dimensional subspace.

Hilb and possibly Quot (state).Exercise: flag. Mor. Isom. Aut. Pic (state; already discussed)What to do with a functor that is not representable? First, an example. Moduli of etale

double covers, or of an object with an automorphism. Do this over a banana curve. Then youhave 2 open sets on which it is trivial, yet glued together it is not trivial.

Two answers.(i) Coarse moduli spaces. (are not moduli spaces in our sense of the word). GIT.Mg .(ii) Stacks. Example: Modular forms: sections of an invertible sheaf over a stack. Cusp

forms. Advantage: modulo p.Pseudofunctor.

37.0.3. X ⊂ Pn, Z ⊂ Pm. The locus where Z ⊂ X is closed. Hom(X, Y ). Locus where itis a graph of a morphism is open. Hence Isom(X, Y ), where it is a graph of morphism in bothdirections.

(The Picard variety and the moduli space of curves made an appearance in Section 25.16.3.Perhaps define the moduli space of stable maps.)Picard variety, moduli

space of curves and sta-

ble maps

Representable functors.

representable functors

Open subfunctor. Line bundles on a functor. If we can cover a functor with open subfunctors,

open subfunctor

each of which is representatble, then the functor in question is representable. We have used thisin constructing fibered products.

Pn, in terms of sheaves. G(k, n) in terms of sheaves. Show they are both representable.How do we know it is projective? Pn, because it is a proj. G(k, n): valuative criterion I

guess. Or describe it as a proj, using Plucker coordinates, which are already in the picture.Existence of the Hilbert scheme. (Remark: Hilbert scheme of points. Smooth for smooth

things of dimension 1 and 2. Show that it is not even irreducible in dimension three.) Quotscheme. Projective space and Grassmannian re-done in this way. Reference for existence of theHilbert scheme, Quot

scheme, Isom scheme,

aut scheme, algebraic

space, representable

morphisms

Hilbert scheme: Kollar’s rational curve book, in ten pages. He refers to Mumford.Exercise: Isom, Aut.

541

37.0.4. Coarse moduli spaces. [Rough notes]We need to work over an algebraically closed field.Example: finite group quotient. x = −x.Example: 6 points in P1. Maybe cross ratio.Ring of invariants: tell you when two are isomorphic.Compactify but get a scheme.Example: j-line.Remark: M→M′ implies M →M ′. Moduli space of curves: show Mg,n+1 →Mg,n.I could refer back to the Hodge bundle, which may have been described earlier (except for a

family of curves). 1 Hodge bundle 2

37.0.5. Toward stacks. [Category = space, functor = map, equivalence of categories =homotopy equivalence. homotopy product = homotopy fibration.]

[Remark about orbifolds.] orbifoldThe problem with considering things up to automorphism. Gluing doesn’t always hold.

(Translation: we don’t always have descent.) Motivating example: Think of objects with auto-morphisms. Say we have families over the two open sets covering a circle. Trivial on both opensets. Then we still need the gluing data. (“Clutching function”.) Algebraic version: over thebanana curve.

We can algebraize that without too much trouble; consider families over a banana curve.A more algebraic version: a line bundle is locally trivial.Descent notes are a bit below, in 37.0.7. Why do we use the etale topology? Reason: We like descent

schemes. Each scheme is a sheaf in the Zariski topology: if X = U1 ∪U2, and we have morphismsU1, U2 → Y , then we can construct X → Y . We need more information if we want families to bea sheaf: we need a choice of isomorphism of families over U1 and families over U2; this requiresextending our notion of sheaf. Unhappily, our families aren’t sheaves in the Zariski-topology.Thanks to Grothendieck, they are sheaves in the flat topology (fppf, or fpqc). So we can make astack in this topology. We’d like to get this topology as close to Zariski as possible, and etale is

pretty close. The question is: can we find a flat cover? If so, can we find a smooth cover? Thenetale cover? Then Zariski cover? If the last one, we would have a scheme.

etale also makes us happy because the local rings are essentially the same. (Somewhat moreprecisely: we allow finite separable extensions of residue fields.) So local behavior, e.g. differentialinfo, is still there.

Instead: we describe all families.Pseudofunctor. pseudofunctorWe toss in some properties that we want, as part of the definition.The notion of homotopy fibered product, or 2-fibered product. homotopy fibered prod-

uct, 2-fibered productIf it has a Zariski-cover by a scheme, we would have a scheme.Define the etale topology? This is mentioned when introducing etale morphisms. Grothendieck

Grothendieck topol-

ogy (etale, smooth,

fppf=flat), etc.

topology, etale, smooth, fppf=flat, fpqc. 23

Define Deligne-Mumford stack? Artin stack? Algebraic space?

algebraic space [defined

elsewhere], DM stack,

Artin stack

Examples: BG. Note that BGm is a moduli space for line bundles. (Remark: BC∗ has

BG, classifying space

dimension −1! Topologists would be unhappy.)Quotient stacks more generally. A1/Gm is a moduli stack of line bundles with sections.[Place: Homotopy fibered product = 2-fibered product. Hom’s are categories. They are only

defined up to equivalence of categories.You can never use the word “same” or else you will get confused, if there are things with

automorphisms. E.g. BZ2 is not a scheme.TpX ∩ TpY = Tp(X ∩ Y ) (false for varieties!). Move this back into “scheme-theoretic

intersection” discussion!Hom(X, Y ) = Hom(X1, Y ) ×Hom(X1∩X2,Y ) Hom(X2, Y ) is true for schemes, but false for

stacks.]

The moduli space of curves. Mg . Mg,n. Mg,n. Gesture toward its construction, as a moduli space of curvesquotient.

This is an active area of research, see for example Harris-Morrison.

1hodge2

2Gtopology

3Gtopology

542

Place later: nodal locus is cut out by a single equation. It is given by a Fitting idealconstruction. (Heard from Dan Abramovich in Cetraro.)

37.0.6. Defining stacks? algebraic space. Example of an algebraic space but not a scheme:algebraic space (also de-

fined elsewhere) Take Hironaka’s example 29.3.7 and quotient by Z/2. Better: my curve example, which I showedwasn’t a scheme.

If X is an algebraic space, then it has a biggest open subset that is a scheme (easy).Define Deligne-Mumford and Artin stack? I don’t like “algebraic stack”. (Votes: Max saysDM and Artin stacks

alg is different from the two. DM called their stacks algebraic. Max uses “algebraic stack” tomean Artin.)

Morphisms of stacks.Warning: each stack is a category. Morphisms of stacks are functors. What is an isomorphism

of stacks? Not F G = I. (You’re never allowed to say equals!) Hence there should be a naturaltransformation to the identity. Define equivalence of categories.equivalence of categories

Representable morphism “relative scheme”. Example: Mg,n+1 →Mg,n.reprsentable morphismOther properties of (representable) morphisms: anything behaving well under descent. That’s

why we work in some Grothendieck topology, which was defined in 37.0.5.Proper morphism of stacks: slightly different definition, using valuative criterion. In partic-

ular, base change allowed.proper morphism of

stacks Nonsingular stacks. Dimension of a stack (in both DM and Artin case).Coarse moduli space of a stack: exists under nice circumstances. If it exists, it is unique

up to “unique isomorphism”? Pushforward on coherent sheaves is an exact functor (Dan A inCetraro).

Fibered product of stacks. You’re not allowed to say “=”. So we don’t do the naive thing.fibered product of stacksThis is the 2-fibered product, or homotopy fibered product. Well defined up to equivalence.

Perhaps give as an example: the inertia stack. Is it I = M×M M? Example: M = BG.inertia stackI(M) = [G/G] where G is acting on G by conjugation. Then I(M) =

‘(g) B(C(g)).

37.0.7. Descent. 4 Here are some versions. From [M, p. 160]:5 we have a functor F :(Rings) → (Sets) that is the functor of points. This is compatible with faithfully flat desent.More precisely: let q : A → B be a faithfully flat ring homomorphism, i.e. for all ideals I ⊂ A,the map I ⊗A B → I ·B is an isomorphism, and q−1(I ·B) = I. Then if p1, p2 : B → B⊗AB, theinduced diagram of sets (2nd should be doubled):

F (A)F (q) // F (B)

F (p1)

F (p2)// F (B ⊗A B)

is exact, i.e. is an equalizer exact sequence. Mumfored doesn’t give a proof.

37.0.8. Etale cohomology. We immediate have etale cohomology (in good situations).We can calculate it in a Cech way. (Why?) Caution: things meet thmselves.

Even for schemes, this is nice; behaves like analytic cohomology of complex things. Vanishesin dimension bigger than 2 dim.

Example: etale cohomology of BG where G is a finite group. Get group cohomology.

37.0.9. Intersection theory on stacks. The same sort of formalism works out. Takerational coefficients. Quote Vistoli [V]. Mention Kresch for Z-coefficients [K]. Get isomorphismwih that of coarse moduli space. Bonus: stack may be smooth, but coarse space singular.

Hence we get a degree map (proper pushforward to point). Each point is counted with size1/ automorphism group. Examples. Moduli space of curves: get fractions.

Cohomology on stacks: by Kai (Jim says), in some C.I.M.E. volume.

37.0.10. Gerbes. (Thaddeus in Cetraro) 6 F a sheaf of abelian groups on X. An F-torsorgerbeis a sheaf of F -modules, locally free of rank 1. F-torsors form a sheaf of categories over X.

An F-gerbe (or gerbe banded by F is a stack over X) locally isomorphic to T (F).Just as any two trivialazations of an F-torsor differ by a section of F , any two trvializations

of an F-gerbe differ by an F-torsor.

4descent

5Mp160b

6gerbe

543

Example: O∗X = line bundles, U(1) = locally constant functions, values in U(1). Get line

bundle with flat U(1) connection (= local system).F-gerbes / F-torsors / F-sections.Example: the gerbe of liftings. Pn → P → X, locally trivial in etale topology. Liftings means

choice of vector bundles. Let BU = E → U,PE ∼= P|U is a lifting of P to a vector budnle.That’s a gerbe! Any two liftings differ by a line bundle.Cech interpretation: trivialize gerbe on Uα, Uβ , Uγ . Compare the trivliazations on the 3

overlaps, to get 3 torsors.On the triple overlap, you’ll get Lαβ |Uαβγ

→ Lβγ |Uαβγ→ Lγα|Uαβγ

→ Lαβ |Uαβγwhich

gives you a section of F on each Uαβγ . This arises from the category-theoretical, and is a bitsubtle. This is a Cech cochain, hence you get:

Theorem. The isomorphism classes of F-gerbes are classified by H2et(F). (Milne, Brylinski,

Giraud)Sections classified by H0, torsors by H1, and this is the next in the sequence.If F = O∗, we get the Brauer group. This would be good to work out.

37.0.11. Existence of the relative Picard scheme.Suppose C/k is a smooth geometrically irreducible projective curve. Describe the functor

PicC/k. C has a point. PicC/k =‘

Picd C/k. We’ll construct Picd C/k for d > 2g − 2, andhence PicC/k.

Effective Cartier divisors of degree d, X = Symd C. Idea: Cd/Sd. Likely do this via theHilbert scheme.

Question to figure out.

Dcl. imm.//

textfiniteflat

""FFF

FFFF

FFC × T

T

Why is D Cartier?Then by the Seesaw theorem 29.7.2, we get a closed immersion R → X×X such that R→ X

is a Pd−g-bundle. This also uses the cohomology and base change theorem 29.8.2.We then need that X/R exists, and is smooth of finite type. (Quotients are discussed soon.)

This als generales without change to C → B families of curves, flat, smooth, projective, geomet-rically irreducible.

Claim: X/R represents Pic.Reason: Given an invertible sheaf L on C × T → T . π∗L is free, hence we get a Pd−1 by

cohomlogy and base change. P(π∗L) is a Pd−g-bundle. We then get a canonical section of L overC×P(π∗L)→ T , which gives a Cartier divisor by that earlier fact. We get map T → X/R locally.Any two differ by an element of R. Conversely, given a map T → X/R, we get a line bundle onC × Ti → Ti. (How do we glue them together? Do we need a section of C?)

37.0.12. The meaning of X/R. Y → X/R, Yi → X, Yi ∩ Yj → X ×X lies in R.So what we want is not quite Hom(Y,X)/Hom(Y,R). We have to sheafify this. (It already

satisfies “identity”, i.e. it is a separated presheaf. Reason: given 2 elements of Hom(Y,X)/Hom(Y,R)that restrict to the same Hom(Yi,X)/Hom(Yi, R). They are the same. Reason: Hom(·, R) is asheaf, hence satisfies gluability. Another reason: the presheaf quotient of one sheaf by anotheris a separated presheaf. Perhaps: the quotient of a separated presheaf by a sheaf is a separatedpresheaf.)

37.1 ? Grothendieck topologies (and descent)

][primordial:

CHAPTER 38

Things still to place, and things not to include

Cone over a smooth quadric (at least over C) is called a “conifold singularity” in the physicsliterature. I should mention this when I first introduce the notion. conifold singularity

scheme-theoretic support: this works for any quasicoherent sheaf, but works best for a co- scheme-theoretic sup-

portherent sheaf.

38.0.1. Issues about coherent sheaves: finitely generated modules (= finite type) vs.finitely generated finitely presented modules.

Suppose M →M ′′ → 0. If M is finitely generated, clearly so is M ′′.Suppose M ′ → M → M ′′ → 0. If M is finitely generated and finitely presented, and M ′ is

finitely generated, then M ′′ is finitely generaed and finitely presented. Reason: take the generatorsof M as the generators of M ′′; the relations are the relations of M and the generators of M ′.

Suppose

0→M ′ →M →M ′′ → 0.

Then ifM ′ andM ′′ are finitely generated, so isM . Proof: let Rn′ →M ′ → 0 and Rn

′′ →M ′′ → 0

be generators for M ′ and M ′′. Lift the latter to Rn′′ →M . Then use the Snake Lemma 3.8.5 on

0 // Rn′

// Rn′+n′′

// Rn′′ //

0

0 // M ′

//M //M ′′

// 0

0 0

Next, if M ′ and M ′′ are finitely generated and finitely presented, so is M . Proof: use theSnake Lemma 3.8.5 as above, look at the “top” of the snake, and use the previous finite generationresult there.

Here now is the tricky part. (a) If M and M ′′ are finitely generated, is the same true of M ′?(b) If they are finitely generated and finitely presented, is the same true of M ′?

The answer to (a) is NO: let R be a ring, and I a non-finitely-generated ideal (e.g. R =k[x1, . . . ] and I the ideal of the origin). Then consider 0→ I → R→ R/I → 0.

Certainly if our base ring is Noetherian, then (i) finite generation implies finite presentation,and (ii) any submodule of a finitely generated module is also finitely generated.

38.0.2. Uppersemicontinuity of dimension. From Mike Roth (summer ’05):I think that there’s unfortunately a flaw (probably fatal) in the plan for proving upper semi-

continuity by using the normal cones.I don’t see an obvious way to fix it, but it would be nice if there were a way around the

problem since I like the idea of using the fibre product a lot (it would force people to grapple withthe fibre product for instance).

If I understand the argument correctly the outline was like this:f Suppose that X —¿ Y is your morphism, then

(1) Let N be the normal cone to the diagonal embeddingD X —-¿ X x X. YI agree that by projectivizing and using the easy upper semi-continuity argument for projec-

tive morphisms you get that the function

545

546

x — –¿ dim(N ) x is an upper semi-continuous function of the point x in X (Where N meansthe fibre of N over x). x

(2) Fix a point x in X, let y=f(x), let X be the fibre over y, and set yM = the normal cone to x in X . x yI agree that dim(M ) = the local dim. of X at x. x y(3) To finish the proof of upper semi-continuity you want to claimM = N , x xor even just that dim(M ) = dim(N ), x xand I think that the argument breaks down here.There is a natural map M —¿ N which is a closed x x embedding, and even an isomorphism

on tangent spaces at the vertex, but this map is not an isomorphism in general, and the conesdon’t even have to have the same dimension.

——————————————————–I think that the easiest example where things go wrong is this:1 1 Pick X = A , Y = A , and the map X —¿ Y to be a 2-1 cover branched over 0.Then the fibre product X x X is isomorphic to the Y 2 two axes in A, (i.e., given by uv=0 in

some coordinates) with the diagonal the u-axis. It follows that the normal cone N to the diagonalis 1-dimensional over the origin.

i.e., if x = the origin, then N is one dimensional and M zero dimensional. x x

——————————————————–There may be some way to get around it by first stratifying X, but I couldn’t think of one.MikeHey Ravi,I was a bit surprised that the normal cone idea didn’t work out (at least as stated), since it

seemed like exactly the right kind of thing.I also hadn’t thought about base change issues with the normal cone before, so I had to think

a bit to see that (3) wasn’t actually true. [The confusion for me was that the degree 1 piece ofthe graded algebra for the normal cone is the cotangent sheaf, and computing that is compatiblewith base change].

Anyway, the ”usual” proof of the upper semi-continuity, which doesn’t involve any advancedmethods, (you factor the morphism through a series of morphisms where the fibre dimension iseither 0 or 1 each time), is still more complicated than I had expected, and I’d also like to see anice construction that made it more straightforward.uppersemicontinuity

38.1 Other stuff

When discussing the category of k-schemes, say that morphisms in this category are preciselygiven by polynomials with k-coefficients. This is the same with a ring R of course.

Koszul cohomology stuff. bij(X) = dimToriS(S(X), k)i+j , that Tor is called Kij(X,L), aKoszul cohomology group. You can use this to find the minimal resolution. Properties: Lefschetzhyperlane theorem. X ⊂ Pr, Y hyperplane section, then Kij(X,L) isomorphic to Kij(Y,K|Y ).Duality: Kij(C,KC) = Kg−2−i,3−j (C,KC)∨.

Eventual polynomiality of Hilbert function: mention Hilbert syzygy theorem, and possiblygive alternate proof.

Mention confusion about the word “points”: closed points. points. T -valued points. geomet-ric points. Advantage of geometric points: a finite etale degree d map, all fibers are d points.

Bend-and-break type argument: if you have a 1-parameter family of curves through 2 points,then it breaks. This could go in the “Cartier divisors pairing with curves” section.

Grothendieck’s theorem that all vector bundles on P1 split.Exercise: normal bundle to line in Pn.For finite morphisms, Restriction of the fiber is the fiber of the restriction. I think I’ve proved

this.For a coherent sheaf on a nonsingular curve, torsion-free = locally free = flat (=projective?).Suppose π : X → Y is smooth of relative dimension 1, and s is a section. Show that s(Y ) is

an effective Cartier divisor on X.

547

38.1.1. Vanishing theorems. Kawamata-Viehweg implies Grauert-Riemannschneider vanish-ing (proof heard from Mihnea). KV vanishing

38.1.2. Grauert-Riemannschneider vanishing theorem. — Suppose f : X → Y isa generically finite morphism where X is nonsingular (or has rational singularieties). ThenRif∗ωX = 0 for all i > 0. GR vanishing

Better: Rif∗(ωX × L) = 0 for i > 0 where L is nef.

Proof. Let A 0 on Y . Hi((Rjf∗ωX)⊗A) abuts to Hi+j(ωX ⊗ f∗A), and the latter is big andnef, so hence is 0 for i+ j > 0. Our goal is to show that Hi(Rjf∗ωX ⊗A) = 0 for i+ j > 0 (neattrick!). Now if i > 0, then Hi(Rj ⊗ A) = 0 by Serre. Thus H0(Rjf∗ωX ⊗ A) = 0, from whichRjf∗ωX = 0.

38.1.3. Fourier-Mukai transform (heard from Mihnea).Suppose we have P over X × Y . Then the Fourier-Mukai transform is pullback from X,

tensor with P , and push forward to Y . The inverse transform is to do the same in reverse, excepttensor instead with P∨ ⊗ p∗Y ωY [g] where the last is a degree sheaf.

38.1.4. Good exercises.

38.1.A. Exercise. Suppose there is a dominant rational map Pn 99K X. Show that there isa dominant rational map PdimX 99K X. Mention unirationality. Perhaps do this over an unirationalityalgebraically closed (or infinite) field.

38.1.B. Exercise on representability. Suppose we have a proper or projective morphismZ → B, and two subschemes X and Y . Show that the locus of B where X meets Y is naturallya closed subscheme. Ditto for the locus where X is contained in Y . Ditto for the locus for whereX = Y .

38.1.5. Image schemes. Remark that the image of scheme to scheme can be anything.Exercise: image scheme. Bad case: image of generic point; underlying set of image scheme is not(closure of) image. [Later: equal if proper?] Image of image is image. image scheme

38.1.6. Base of sheaf comments. I don’t know of a “reasonable” criterion for whenSpecR → Spec S is an open immersion, and I’m not convinced that we should care. Distin-guished subcategory of a topology. In the case of something affine, we have something with goodintersections. For a general scheme, we’ll have something more like “quasiseparated”.

Sheaf on a base ↔ sheaf. Morphism between sheaf ↔ morphism between sheaves on a base.A ring homomorphism A→ B induces a morphism of ringed spaces SpecB → SpecA. (Use

“morphisms on a base”.) The original ring homomorphism can be recovered from the map onglobal sections. Those morphisms of ringed spaces SpecB → SpecA of this form are calledschematic.

A morphism of schemes is a morphisms in the category of ringed spaces that is schematic forall SpecB → SpecA.

Show that a schematic morphism of affine schemes is indeed a morphism of schemes.Some problems: why do morphisms glue together? Why do closed immersions correspond to

X → Y as topological spaces, along with surjections OY → f∗OX?Exercise: it induces a map of stalks.Definition of a quasicoherent sheaf: we have a module on each affine, which does what you

think for each distinguished morphism. Then we have identity and gluability on this base, so weget a sheaf. It suffices to check on an affine cover. (There is some logic missing here. We needsome fancier gluability for modules.)

38.1.7. Other topics.Serre’s criterion for affineness. X → Y finite. X affine implies Y affine. (Warning: quasifinite

doesn’t suffice, consider curve.)

548

38.2 Group schemes (and abelian varieties, INCORPORATEEARLIER)

GL(n). Gm. Ga. Torsion: comes together, show this for roots of unity. We get this forabelian varieties too.

Linear algebraic group. Abelian varieties.

38.3 Things not to include

Formal schemes: because we don’t use them. But here are some interesting facts. Forexample, if A is a topological ring that is appropriately “nice” (pre-adic? pre-admissible? adic?admissible?), with ideal of definition I then we define SpfA, and we have an open set SpfAf . Thetopology is the induced one, which is also the If -adic topology. The structure sheaf is a bit weirdthough, because we want Af to be Hausdorff and complete. So we complete, and the restriction

map is A → Af . Thus to pursue the same programme here, we will need to re-do identity and

gluability, and anything to do with them. Also ⊗.Serre duality: ωB = Extn−rA (B,A).

38.3.1. Topology of complex projective varieties. Hironaka 1975: Any complex alge-braic/analytic variety of dimension n has a triangulation. (Heard from Young-Hoon Kiem, May5 ’06.)

][primordial:

APPENDIX A

Glossary

A.1 Category theory

Words they should know. Bold: in advance. Italics: will be defined in notes. Plain: nice toknow, but not essential.

Recommend [W] as a reference. monomorphismcategory, object, morphism, identity morphism, composition of morphisms, iso-

morphisms, monic morphism = monomorphism, epi morphism = epimorphism (cancellationrules), subobject, kernel, cokernel, initial object, final object, zero object, opposite or dualcategory, functor, faithful functor, full functor, forgetful functor, sum product, fibered product,concrete category, subcategory, Yoneda embeding is fully faithful [W, A.3.4], covariant and con-travariant functor, natural transformation of functors, equivalence of categories, inverse limit(=limit), direct limit (=colimit), filtereed direct limits, adjoint functors, Ab-category (homs areabelian groups), additive category (0, A×B), abelian category (every morphisms has kernel andcokernel; every monic is kernel of its cokernel, every epi is cokernel of its ernel).

[forme: Handy diagram to remember some definitions:

kernel monic

F

∃!

%%KKKKKKKKKKK D

≤1

$$HHHHHHHHH

K // A //

≤1$$

B //

%%JJJJJJJJJJ

C

∃!

E G

epi cokernel

]Monomorphism is defined in §3.3.9, and epimorphism is defined in Remark 3.3.N.Cokernel is defined in Exercise ??, and also in §3.7.In §3.7, the following terms are defined: initial object, final object, 0-object, additive category,

kernel, cokernel, abelian category. initial, final, and 0-

obj; (co)kernel; add and

abelian categoryA.1.1. Various limits. Direct limits (category-theoretic limit), and how to think of them: havea representative. Canonical example: stalk. This is sometimes called a colimit by topologists.

direct and inverse limitsInverse limits are sometimes just called limits by topologists. Inverse limits lim←− can be

defined in the same way (with the arrows reversed). The reader may ponder this. We will need aspecial case: where the “index category” is the non-negative integers. Motivating example p-adics.

We have groups A0, A1, . . . , and maps φmn : Am → An where m > n such that φnpφmn =φmp.

549

550

A.1.A. Exercise. Show that the inverse limit is isomorphic to the subset ofQAi where

φmn(am) = an. (Alternatively, the reader may just take this as a preliminary definition.)Note how the p-adics work.Thus such inverse limits exist for groups for example.Useful example: completion of a ring along an ideal lim←−R/I

n.

A.1.B. Easy exercise. The completion of a local ring along its maximal ideal is also a maximalideal. 1 (We will use this in the proof of Zariski’s Main Theorem 29.14.3, and Stein FactorizationTheorem 29.14.2.) [Idea: just show that anything not in the maximal ideal is invertible!]

A.1.2. Exactness properties for inverse limits.

A.1.C. Straightforward but important exercise. [Do this with the diagram correspondingto the natural numbers.] Suppose we have an abelian category, and a diagram in the categorysuch that inverse limits exist. Then an exact sequence of diagrams in that category is defined inthe obvious way. Show that lim←− is a left-exact functor from our auxiliary category to our ooriginalcategory. When might it be right-exact?

We say that an inverse system (An, φmn) satisfies the Mittag-Leffler condition if for eachML conditionn, the decreasing family φmnAm ⊂ An | m ≥ n is eventually stationary.

A.1.3. Key example. If the An are finitely generated modules over a Noetherian ring,then they satisfy the Mittag-Leffler condition. (A special case of this is: finite-dimensional vectorspaces over a field.) 2 [Perhaps this section should be put after algebraic preliminaries?]

Suppose An satisfies the Mittag-Leffler condition. Define the stable image A′n ⊂ An.

A.1.D. Easy and important exercise. A′n form an inverse system, and the inverse limits arecanonically isomorphic.

A.1.4. Proposition. — If An satisfies the Mittag-Leffler condition, then the left-exact sequenceis exact. (Perhaps better: in any category with the descending chain condition, the inverse limitfunctor is exact.)3

Proof. Let’s set notation that should have been set before. Suppose 0 → An → Bn → Cn → 0.Fix cn ∈ Cn. Let B′n be the preimage of the cn’s. They are each An-torsors, and they satisfythe Mittag-Leffler condition. Then we can build a preimage by taking an element of the stableset of B′0, and lifting to an element of the stable set of B′1, etc. (This requires the axiom ofchoice.)

[Unimportant remark: H points out that if Bn surjects onto Cn, and Bn satisfies ML, thenso does Cn. I don’t think we use this.4. This, along with the previous Prop, suggests that lim←−

2 is

always 0. Perhaps check in Weibel [W]]Combining this with Example A.1.3, we have:

A.1.5. Corollary. Suppose An (awkwardly) is a system of Noetherian rings. Then we have anexact functor from finitely generated rings to blah blah. 5

We will use this to prove the theorem on formal functions 32.2.1.We’ll need another result too (that I would like to excise):

A.1.6. Krull’s theorem. — A noetherian ring, M ⊂ N finitely generated A-modules, I ⊂ Aan ideal. Then the I-adic topology on M is induced by the I-adic topology on N . Equivalently,for any n > 0, there is m ≥ n such that InM ⊃M ∩ ImN . 6Krull’s Thm

1completitionoflocalring

2ML

3H.P.II.9.1b

4H.P.II.9.1a

5H.E.II.9.1.2, c:inverselimitexact

6Krull, H.P.III.3.1A. I would like to excise this. He refers to [AM] [1,10.11] or ZS [1, vol. II,Ch. VIII,

Thm. 4]. This is H’s wording.

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A.2 Algebraic glossary

Galois theory.Prime ideals. Localization (is exact). Quotients. Tensor products.Transcendence theory.Short things: Hilbert basis theorem; Going Up (easy).

Noether normalization

uukkkkkkkkkkkkkkk

))SSSSSSSSSSSSSSSS Nakayama

weak Nullstell. Krull (trans. version) finiteness of i.c.

Hartogs

wwnnnnnnnnnnnn

Krull (ft/k)

Know about rings, prime ideals. Localizing is exact. Tensor is left-exact. Prime, irreducibleelements. UFD. Hilbert basis theorem. Transcendence theory.

Extensions: algebraic, separable, purely inseparable (implicitly algebraic; everything is a pthroot); separable closure.

To place:

A.2.1. Theorem. — Theorem [H, I.4.8A]. If k is a perfect field (i.e. any algebraic extension

is separable), in particular if k = k, then any finitely gnerated field extension K/k is separablygenerated. Apparently Matsumura Ch. 10 Cor. p. 194. Used in proof of Proposition 13.3.1, andin a couple more places. 7

A.2.2. Theorem (Fundamental Theorem of Elimination Theory). — Let f1, . . . , fr behomogeneous polynomials in x0, . . . , xn, having indeterminate coefficients aij . Then there is aset g1, . . . , gt of polynomials in the aij , with integer coefficients, whic are homongeeousl in thecoefficients of each fi separately, with the following property: for any field k and for any set ofspecial values of the aij ∈ k, a necessary and sufficient condition for the fi to havce a common

zero different from (0, 0, . . . , 0) is that the aij are a common zero of the polynomials gj .8 Fundamental Theorem

of Elimination TheoryThat’s Hartshorne’s wording, and he quotes Van der Waerden. I want a more recent reference.Also to work over a ring, and also to get the scheme-theoretic image. Check out Mumford forexample (see 21.4.2), and Eisenbud-Harris.

That’s used to show that projective morphisms are universally closed.

A.2.3. Lemma. — Let K be a field. A local ring R contained in K is a valuation ring of K ifand only if it is a maximal element of the set of local rings contained in K, with respect to therelation of domination. Every local ring contained in K is dominated by some valuation ring ofK. 9

Hartshorne credits Bourbaki, and also [AM, Ch. 5, p. 65, and exercises, p. 72]. Use Atiyah-Macdonald’s wording.

That’s used to prove the valuative criterion for separatedness in its general form.

A.2.4. Noetherian rings. R Noetherian implies S−1R and R/I Noetherian. Hilbert basis N rings, H basis thmtheorem: R Noetherian implies R[x] noetherian. (Give ref.) I now state this in the text.

A.2.5. Finiteness of integral closure. — 10 [H, Thm. I.3.9A, p. 20]; refers to Zariski-Samuel[1, vol. 1, Ch. V, Thm. 9, p. 267]; Mumford also refers here [M, p. 278]. Reword. Let A be adomain which is a finitely generated k-algebra. Let L be a finite extension of the quotient fieldof A. Then the integral closure A′ of A in L is a finite[ly generated] A-module (and hence afinitely-generated k-algebra).

7sepgen

8FTET, H.T.I.5.7A

9VRdom, H.T.I.6.1A

10finitenessofintegralclosure

552

A.2.6. Nakayama’s Lemma. 11 This is now now done in the text. References: [M-CRT,Nakayamap. 8-9] [AM, p. 21]. Shafarevich BAGv1, Prop. 3, p. 283. Mumford [M] has three versions in p.III.2, p. 212–3. 12

Not yet in text: Geometric Nakayama (see p. 124 Eisenbud): X Noetherian, F coherent.If Fx ⊗ k(x) = 0 then there exists V such that F|V = 0. [Currently in discussion of coherentsheaves!]

A.2.7. Discrete valuation rings. 13 This is now done in the text. References: [AM, 9.2]discrete valuation ringIf R contains its residue field, then the completion with respect to m is k[[t]].Observe: valuation of an element a ∈ R is l(R/a).(4) PID. (Exercise.)

A.2.8. Factoriality. Now in text.

A.2.9. Flatness.Exercises: Everything is flat over a field. Flatness is preserved by base change. Flatness is

stalk-local, i.e. if it is true at all stalks then it is true for the entire ring.flatFlat and finite type implies open [AM, Ex. 7.25].

A.2.A. Exercise. [cut if not used.] (a) Show that if M ′ ⊕M ′′ is flat, then M ′ is flat. (b) Show

that if M = ⊕Γ(X,F(n)) is flat, then M is flat. [Answer: localize, and then take 0th gradedpiece, using (a).]

A.2.10. Hilbert’s Nullstellensatz.

A.2.11. Theorem (Nullstellensatz). — Let k be an algebraically closed field, let I ⊂k[x1, . . . , xn] be an ideal, and let f ∈ k[x1, . . . , xn] be a polynomial vanishing at all closed pointsof Z(α). Then fr ∈ I for some r > 0. [Copied from [H], from [AM, p. 85].] 14Nullstellensatz

Give different versions, either here, or where first discussed in the text.

A.2.12. Noether normalization. in text.

A.2.13. Koszul complex. in text.

][primordial:

11nakayama

12Mnakayama

13DVR

14nullstellensatz

APPENDIX B

Crib notes of facts

This used to be properties.tex in my ag directory.

B.1 Properties of schemes X

Here p denotes prime ideal, m denotes maximal.W (k) is the initial object in the category of complete local Noetherian algebras with residue

field k.

B.1.1. Topological properties. Dim, codim. irreducible, connected.qc = quasicompact = finite union of affines = finite subcover propertyNoetherian space (⇔ every open subset is qc).obscure definition: Sober = each irreducible closed subset has a unique generic point. Zariski

space = sober N space [H, Ex. II.3.17]. Also called Kolmogorof space [EGA, 0.2.1.3]. T0 axiomof separatedness.

B.1.2. Nontopological properties.List degrees of “locality”. (local ring; Zariski-local = affine-local)reduced = all open sets = covering open = p. m?

• Constructing Xred: same space, structure sheaf is sheafification of U →Ox(U)red [H].Reduction necessary: consider infinite disjoint union of fatter and fatter points.

integral = reduced + irreducible = all U OX(U) is int domainlN = locally Noetherian ([dJ] Mar. 8 ⇒ qs)N = Noetherian = lN+qc

• N → N space, but not vice versa. Example: k[xε](xε)

• SpecR Noetherian ⇔ R Noetherian [H, II.3.2] (This doesn’t seem right, see previousexample.)

regular local ring (A,m) r-dimensional N local ring, m generated by r elementsregular = lN + all local rings are regularregular ring

• N + Ap regular (sufficient to check m)• ⇒ normal• SpecR is regular

normal ring = integrally closed domain

• Ap int closed domain for all p

• Spec of int closed domains are normal• ⇒ completion is int closed domain [?, ZMT III.9]

normal = on affine opens, normal rings = R1+S2 = check p = check m.

• Normalization is finite if X is reduced excellent. (e.g. A int. domain ft over k,K = K(A), L finite over K. Then integral closure A′ of A in L is finite A-mdule, andft over k, [H, Thm I.3.9 A]). Counterexample if X not reduced: Q[ε]/(ε2) is a subsetof Z[ε]/(ε2). Integral closure in a + bε|a ∈ Z is Z⊕Qε. Fix this.

UFD ⇔ normal + (Cl= 0).locally factorial = all local rings are UFD

553

554

Dedekind domain: noeth int closed dimension 1 domain.Cohen-Macaulay

• motivation: degree of projective scheme• depth: longest regular sequence; ≤ dim; independent of order• local coho criterion• open condition

X quasisep ⇐⇒ X → Spec Z isaffine (Serre’s criterion [EGA, II.5.2])lN = locally Noetherian (fppf-local) (lN ⇒ qs easy) = all affines are NoetherianN = Noetherian = lN + qc

B.2 Properties of morphisms f : X → Y

Locality type 0 (where from?): P (X → S) ⇒ P (X ′ → S) if X′ → X is an isomorphism. Ppreserved under base change, and Zariski-local on the base. e.g. surjective etale flat proper qcisomorphism qf finite ft lfpr sm sep, cl imm, open imm. Representable morphisms of DM stackscan have these properties.

[dJ, Mar. 10]. Locality type 1: stable under base change, fppf-local on base.Locality type 2: stable under etale base change, etale local on base. e.g. open immersion

with dense image. dominant. birational. finite fibers. reduced fibers.Locality type 3: etale local on source and bazse, e.g. flat smooth etale unr lft lfpr.The six properties of morphisms. Property P : closed immersions have P , closed under

composition and base extension [H, Ex. II.4.8]. E.g. ft, proper, sep, closed immersion, proj,affine, qc [EGA, I.6.6.4], lft [EGA, I.6.6.6]. Universally closed ([Milne, 20]; check [EGA])?.Then products have P ; if f : X → Y , g : Y → Z, g f has P , ∆g has P (e.g. g sep), then f hasP . fred has P . In the case of ft, g sep can be replaced by f qc [H, Ex. 3.13 (f)]; is this true ingeneral? Generalizations in [EGA, I.5.5].

B.2.1. Tool: Affine Induction. On schemes: reduced, lN, integral. On morphisms (on base) ft,lft, qc, affine, etc. On sheaves: Γ(M ) = M , coh.

B.2.2. Topological properties.qc = quasicompact = pre-image of affines are finite unions of affines.

• suffices to check for a cover of the base (easy; [EGA, I.6.6.1])• f qc ⇒ f(X) closed iff stable under specialization ([H, Lemma II.4.5], [EGA]).

• (Stacks: for all qc scheme mapping into Y , the tensor product is qc.)

surjectiveclosed, openuniversally closeddominant

B.2.3. Non-topological properties.In each case, open-local means: i) exists a covering ⇔ ii) for all opens.closed immersion (diff geo: embedding), open immersion, locally closed immersion (sometimes

just immersion — bad b/c of diff geo meaning)lft = locally of finite type (fppf-local)

• X → Y → Z lft ⇒ X → Y lft.

Q: lN morphism? N morphism? check [EGA].ft = finite type = lft + qc

B.2.4. Affine.affine = pre-image of affine is affine = X = SpecY (A) for some qcoh OY -alg A. (⇒ sep)qa = quasi-affine. qa scheme = quasicompact open subscheme of affine qa morphism:

• qa morphism: If X is qa, preimage of affine is quasiaffine. What did I mean here?

555

• X → Spec(Γ(OX )) is a qc open immersion [EGA, II]; hence ⇒ qc, sep, but notnecessarily qprof (as qproj requires ft)

• [EGA, II.5.2], lots of equivalent versions• Fact: [EGA, IV.18.12.12] qf+sep ⇒ qa (if Y is lN, [EGA, II.6.2.7]).

B.2.5. Finiteness.If Y irreducible with generic point η, generically finite = f−1(η) is finite set ([H, Ex. II.3.7]).finite = f affine and for all V ⊂ Y affine open, O(V )→O(f−1V ) is finite. (⇒ proper (easy),

affine) = qf + proper ([dJ, Mar. 8]; Milne too; or G’s form of ZMT [?, III.9])

• Y irr f dom generically finite ft then exists U ⊂ Y with f−1(U) → U finite [H,Ex. II.3.7].

• proper + qa = finite (pf of ⇒: proper + qa ⇒ proper + affine easily. Then π∗OX iscoherent. Hence ft, so X = SpecOX is finite over Y )

qf = quasifinite = f ft and for all y ∈ Y the scheme Xy = f−1(y) is finite over Spec(κ(y))

• ft + form unr ⇒ qf

All immersions into Noetherian are ft.

B.2.6. Finite presentation.lfpr = locally of finite presentation (fppf-local) = “limit-preserving”

• On rings: TFAE (i) A → B fpr (ii) Whenever (Ci, φij)i∈I is a directed system ofA-algebras, HomA(B, lim→i Ci) = lim→i HomA(B, Ci). (iii) Same, where Ci are fpr.[dJ, Oct. 27 ’98].

• If (Ti, φi,j) inverse (indirect) system of affine schemes over Y , then T = lim←i Ti existsas a scheme, and T = Spec(lim→i Γ(Ti,O)).

• Then f lfpr iff all T = lim←−Ti as above, Hom(T,X) = lim−→Hom(Ti,X) (ref [dJ, Oct. 27

’98], [EGA, IV.8.14]). This is what you use to check lfpr in practice; can also definelfpr for contravariant functors. Algebraic spaces are lfpr (by def). For this reason lfpris sometimes called limit-preserving.

• Lemma. X → Y → Z lfpr, then X → Y lft. Pf. Algebra A → B → C =A[x1, . . . , xn]/(f1, . . . , fr), so C is a finite-type B-algebra. [dJ, Feb. 25 ’99]

• lfpr morphism between affines is actually fpr (easy).

fpr = lfpr + qc + qs [EGA, IV.1.6.1]lci, ci [careful!] (on rings: A → B is a c.i. if B ∼= A[x1, . . . , xn]/(f1, . . . , fc), c ≤ n, B flat

over A, and dimB ⊗A κ(p) = n− c for all y ∈ SpecA.)

B.2.7. Monomorphism. ⇔ diagonal map X → X ×Y X is an isomorphism. Examples: closedimmersions, open immersions, mapping Spec of a local ring in. Then tensors over X and Y arethe same. See [EGA].

• ⇒ qs clearly.

B.2.8. Separatedness.sep = separated = ∆ is closed immersion. Valuative criterion valuation ring (when Y noeth?

[dJ]: only need dvr when ft over Z). [EGA, I.5.5.6] criterion that’s more useful.

• A prevariety is separated iff the intersection of any two open sets SpecR and Spec Sis affine, and (naturally) SpecRS inside K(R) = K(S). ([?, p. 39 Exer.]; likely that isthe EGA criterion above) Can I fix this for general schemes?

• Sufficient condition: if for every x and y there is an open affine containing both, thenX is sep. (What is an example of something failing this condition?)

• Useful application: Given two morphisms f, g : Y → X with separated target, thelocus where they agree is closed. (Pf: Consider Y ×X×X X.) Works over base tooobviously.

• birational + sep + et = open immer ([EGA, 18.10.18] (III or IV?), [dJ, Apr. 4])

lsep = locally separated = ∆ is locally closed immersion (only interesting for alg spaces. allschemes are lsep, any stack with a “stacky point” is not)

qs = quasi-separated = ∆ : X → X ×Y X is qc.

• equivalently, any 2 affines intersect in finite union of affines (or affine replaced by qc)

556

• then (non-trivially) ∆ is ft + qa• for any scheme, ∆ is sep + lft (Not nec qc. But see [dJ, Apr. 4]: for DM stacks, ∆ is

sep+qc.)• If |X| is lN, then X → Y is qs [EGA, IV.1.2.8].• immersions are mono ⇒ qs, so by 6 properties, X qs ⇒ X → Y qs; If qs then ⇐ as

well.

B.2.9. Properness.proper = ft + sep + uc = qc + lft + sep + uc. Valuative criterion: dvr (trait := Spec dvr)

if Noetherian ([dJ] ft over Z); vr otherwise.Chow’s Lemma. If f is ft + sep, then there exists P → X → S quasi-projective, and P → X

surjective proper and isomorphic over dense open set. This basically characterizes proper. ([EGA,II.5.6]; copy in)

An injective proper morphism of varieties is affine [Newstead, Lemma 2.5].Is proper + affine = finite? ⇐ certainly true.

B.2.10. Projective.projective (following [EGA], not [H]) locally (on Y ) = ([dJ] Mar. 2 ’99 handout) X isom to

ProjY (S), where S is a qcoh graded OY module generated in degree 1 such that S1 is a finite-typeOY -module [what now?:] an closed subscheme of PnY

qproj = quasi-projective = [EGA, II.5.3] ft and f-ample sheaf (This means for all V ⊂ Yaffine open, can find sections of some high power of L over f−1(V ) which gives an immersionf−1(V ) → PNV over V .) = open subscheme of projective (qproj + proper ⇐ proj; ⇒ if base isqc+qs [EGA, II.5.5.3] )

B.2.11. Lifting properties (unr, et, sm) [EGA, IV.17.3].Notation caution: lcimm ⇒ unr ⇒ etale ⇒ smooth. Differential geometry terminology:

embedding ⇒ local embedding ⇒ immersion⇒ submersion. What’s an imbedding? embed-ding?

I ⊂ A ideal, I2 = 0, ψ : Spec(A/I)→ X, SpecA→ Y . Then ψ : SpecA→ X.formally unramified = no vertical tangent vectors = (via [EGA]) Ω1

X/Y= 0 = at most one

ψ.

• X → Y → Z formally unr then X → Y formally unram (Pf. algebra A → B → C.Ω1C/A

= 0⇒ Ω1C/B

= 0. [dJ, Feb. 25 ’99])

unramified = formally unr + lfpr

• e.g. of something form unr but not unr: Spec Q → Spec Q. Also, if Spec(A/I) →SpecA is formally unr, then it is unr if I can be generated by finite number of elements.

• can check over alg closed fields• Question in case sep, ft / k, is it = scheme-theoretic fiber is reduced finite set? See

etale below.

• Diff geo: “immersion”

formally etale: there exists a unique ψetale (⇒ open) = formally et + lfpr = (non-trivially) flat + unr

• In case of sep, ft over k, it is equivalent to– flat + scheme-th fiber is reduced finite set ([MuPicMod, p. 43])– flag + geo fiber is reduced finite set [?, II.p. 303]– sm + scheme-th fiber is finite set (from previous case, [MuPicMod, p. 52])

– isomorphism of completions (clear): see [H, Ex. III.10.4] for k 6= k– flat + ΩX/Y = 0 [H, Ex. III.10.4]

• Diff geo: “covering space”

formally smooth = exists a ψsmooth = formally sm + lfpr = (non-trivially) lfpr + flat + smooth fibers

• [H, III.10]: smooth of relative dimension n of ft schemes equiv to each of following– (definition) flat + relative dimension n on irr comps + for each point (not nec.

cl.) dimk(x)(ΩX/Y ⊗ k(x)) = n

557

– flat + fibers are geometrically regular of equidimension n (=for each point,

Xy ⊗k(y) k(y) is regular and equidimensional of dimension n)

– [?] gives local Jacobian condition version• ⇒ open by “flat+lft ⇒ open” below• Infinitesimal lifting property (lfpr or something required?) SpecA→ S, and Spec(A/I)→X → S with I2 = 0, then SpecA→ X [dJ, Nov. 3 ’98]

• Translation to functors: A transformation F → G of covariant functors on C is smoothif for any surjection B → A in C we have that F (B)→ F (A)×G(A)G(B) is surjective.

• Lemma [Sch, p. 211]. If A1 → A2 is R-algebra hom between complete local Noether-ian rings with residue fields k, then hA2

→ hA1is smooth ⇐⇒ A2

∼= A1[[x1, . . . , xr]]for some r. Presumably if residue fields give separable extension, the obvious statementholds too.

• [LS, p. 55]: formally smooth + ft + lN source + lN target; see [SGA]2

• for varieties, presumably for ft / k:– flat + scheme-theoretic fibers are ns varieties [MuPicMod]

– Ox = Of(x)[[x1, . . . , xn]] [MuPicMod]

– ΩX/Y is locally free of same rank as relative dimension [H, Prop. III.10.4] (“same

rank” needed, witness Spec k[ε]/(εp)→ Spec k, or Spec k[tp]→ Spec k[t])– for every closed point, the induced map on Zariski tangent spaces is surjective

[H, Prop. III.10.4]

• [SGA, Thm 3.1?] A fd local k-algebra, I ⊂ A,

SpecA/I → X↓ ↓

SpecA → Y

Then exists lifting SpecA → X. Think of this as “homotopy lifting property”, sosmooth morphisms are loosely fiber spaces.

B.2.12. Flatness. In progress: What you need to know about flatness: trivial: base change,transitivity, localization. Tor1 fact?

• [H, III.9.1] (+ more comments there):– base change, transitivity, localization (check only m)– iff for all fg ideals I⊗M →M . Hence A domain: flat⇒ torsion-free; A PID: flat⇔ torsion-free; A = k[ε]/(ε2)], M/εM → εM is an isomorphism.only one ideal tocheck.

– M fg / A lN “flat ⇔ [locally?] free” (must use torsion-free ⇔ free in light ofprevious comment) [This can’t be right... what do I mean here?]

– 0→M ′ →M →M ′′ → 0, “M,M ′′ ⇒M , M,M ′′ ⇒M ′”, pf by Tor criterion.• Let R be a ring, I an ideal, I2 = 0, M an R-module. Then M flat ⇐⇒ M/IM is flat

over R/I and IM ∼= I ⊗RM [dJ, Nov. 5 ’98]• good summary [Milne, Ch. 2]• flat + lft ⇒ open [AM, Ex. 7.25] via Chevalley’s theorem (f(constr) = constr) and

“going down theorem” for flat ring maps.• f : X → Y varieties, Y reg, X cm, fiber equidimensional, then f flat, [H, Ex. III.10.9]

fp = faithfully flat = surjective and flat; affine description.

• The “descent exact sequence” [Milne, Prop. 2.18], comes up in faithfully flat descent.

B.2.13. fpqc. Main example [BC]. Spec A→ SpecA when A → A maps Noetherian local ring

to its completion, or a field to its algebraic or separable closure. Then A×A A is not Noetherian.When doing descent, we need to make sure algebraic geometry works over not-necessarily (locally)Noetherian schemes.

(Also, if Y is Noeth, and G finite, Y/G need not be. Need Y not ft over field for this towork.)

Theorem (Grothendieck). ([Neron] or [FGA]). Let p1, p2 be projections from X ×Y Xto X. If f is fpqc then for F ′ a qcoh sheaf on X, descent data on F ′ is α : p∗1F

′ ∼= p∗2F′ so that

558

cocycle condition holds. Then

(qcoh on X) → ( qcoh on Y with descent data)

F 7→ (p∗F,α : p∗1p∗F ∼= p∗2p

∗F )

is an equivalence of categories. ([dJ, Oct. 15, 1998]) One says that descent data is effective.

B.2.14. Factorization theorems.

B.2.15. Zariski’s Main Theorem. The first few are from [?, III.9].

• Original form: f birational with finite fibers, Y normal. Then f is an isomorphism ofX with an open subset of Y .

• Topological form: Y normal over C, y ∈ Y closed point, S singular locus of Y . Thenthere is a basis of complex neighbourhoods of y such that Ui −Ui ∩ S is connected forall i.

• Power series form: Y normal variety over k, y a point. Then the completion of thestalk is an integral domain, integrally closed in its quotient field.

• Grothendieck’s form: f finite fibers, X and Y varieties. Then Xu→ X′

f ′→ Y , u openimmersion, f ′ finite. ([dJ, Mar. 2 handout]: f qf+sep, Y qc+qc. Then same result.)

• Connectedness theorem. Y noetherian integral scheme over k normal at y. f birationalproper. Then f−1(y) is connected. (Combination of Mumford and [H, Cor. 11.4].)

• Form from Mumford’s [MAV]: f proper bijective (X, Y varieties?) then f is finite.

• Application: f qf. Then Xu→ X′

f ′→ Y , u open immersion, f ′ finite. Presumably fshould be sep too.

• Moduli application: SupposeM is a moduli stack with a morphism to a normal schemeM that is a bijection on geometric points. Then M is the coarse moduli space.

B.2.16. Stein Factorization. f projective, X, Y Noeth. Then Xu→ X′

f ′→ Y where u isprojective with connected fibers, and f ′ is finite. [EGA]: projective replace by proper; some lNhypotheses required in [EGA] 3 but [EGA]4, [SGA] removes them.

B.2.17. Nagata’s Theorem. f ft+sep, Y Noeth. Then there is a factorization Xu→ X′

f ′→ Y , uopen immersion, f ′ proper. (Johan: presumably holds with Y qc+qs.)

B.3 Properties of sheaves

• = presheaf (functor) + identity (injectivity) + gluability (surjectivity).• stalk. Can check inj, surj at stalks. Can check inj. on opens.• Sheafification has universal properties.

• Image, quotient, cokernel, direct limits (Ex. II.1.12) require sheafification; kerneldoesn’t. Direct limits don’t unless X is a Noeth space (Ex. II.1.10–11) (This iswrong!).

• f−1F is odd ([H, p. 65]).• Restriction F|Z to topological subspaces. Support. Sheaf Hom.• Adjoint property: f−1 left-adjoint of f∗. HomX(f−1G,F) = HomY (G, f∗F).• If U open in X, j inclusion, then j! is extension by 0. 0→ j!(F|U )→ F → i∗F|X\U →

0 [H, Ex. II.1.19]• subsheaf with supports [H, Ex. II.1.20].• ringed space: morphisms (f, f#) : (X,OX)→ (Y,OY ) with f# : OY → f∗OX .• π∗ qcs = qcs if π is qs [EGA, IV.1]• π∗ cs = cs if π is proper. (cs = fg + fr in [EGA]?)

Sheafifying: 0 →? → P → S →?→ 0. Do the thing once to get identity, i.e. separatedness:0→ P∨ → S →?→ 0. Do it again to get gluability: 0→ P∨∨ → S → 0. This should be thoughtof as descent data.

ample vector bundle

559

B.4 Miscellaneous notes

Philosophy: Everything is relative. Descent. Functors.

B.4.1. Nakayama.

• A local, M an A-module. M/IM generated by k elements. Then M can be generatedby k elements [H, p. 175, 178].

• M finite A module. If M/IM = 0 then exists a ∈ A such that aM = 0, a ≡ 1 (mod I).I ⊂ radA implies a unit of A, so M = 0 (generic interpretation is clear). [M-CRT,p. 8]

• Cor. I ⊂ radA, N ⊂M , M/N finite over A. Then M = N + IM ⇒M = N .

• Variation 1. If N = M and M/mM ≡ N/mN then M = N .• Variation 2. If M/mM = (A/m)n then M = An. Pf. Let u1, . . . , un ∈ M be

generators of M/mM , N = (u1, . . . , uk). Use Variation 1 or [M-CRT, p. 9 Thm. 2.3].Then 0 → K → Am →M → 0. Show (using Nak) that K = 0. (How? This seems togive K/mK ∼= Tor1(A/m,M).

B.4.2. Hilb. Hilbert schemes exist for projective morphisms. For proper, get Hilbert algebraicspaces [dJ, Oct. ’98].

B.4.3. Quot. [FGA] X → S proper fpr, OX(1) ample on X rel to S, F sheaf of fpr over OX .Then Quotp

F/X/Sis representable. Application: MorS(X, Y ).

B.4.4. Hensel’s Lemma. Equivalence of categories between separable extensions and completions... (?). [dJ, Dec. 3].

B.4.5. Grothendieck Topologies. See [dJ, Oct. 15, Feb. 2]. site = topology, topos = sheaveson that site. stack in groupoids [dJ, Feb. 2]. etale sites associated to a DM stack, [dJ, Mar. 8].Good ref: [MuPicMod]

B.4.6. Algebraic spaces. pre-eq relations, equiv relations, groupoid, etale equiv rels,Theorem (Knutson). X alg sp with quasifinite map to scheme Y . Then X is scheme [dJ,

Dec. 1]Theorem. (Artin; [dJ, Dec. 8]) Pic is representable in the category of alg spaces.

B.4.7. Grothendieck’s big theorem. [dJ, Nov. 24] f : X → Y = Spec(A) proper. Then coherent

OX-modules = coherent OX

-modules, where X and Y are completions. This relates to Gr’stheorem on algebraization of formal schemes [dJ, Dec. 3]. What precisely is this?

[BC]: “Formal GAGA”; lift from directed systems to formal schemes. Then often use Artinapproximation to lift from formal schemes to actual schemes, up to any desired degree of accuracy.

B.4.8. Descent. fppf descent for qcs is effective, compatible with tensor product, and morphisms.(Check! get ref.) Note in [dJ, Mar. 2]: qa needed for some reason, [SGA] I.

B.4.9. Stacks.Example of an algebraic space that is not a Deligne-Mumford stack? Jarod says something

like: take the axes, glue together the y-axis minus the origin with the x-axis minus the origin.This is not locally separated.

Isom is a sheaf. Objects satisfy descent.Diagonal is representable

• ⇒ maps of schemes into it are representable (⇐? I think so: [dJ, Mar. 10])• (⇔ Isom is a sheaf?)

[dJ, Mar. 8], DM stack S is said to be separated if ∆ proper ⇔ ∆ finite ⇔ for all X, x, y ∈ SX ,scheme I representing IsomX(x, y) is proper over X ⇔ for all X, and all x, y, I → X is finite.

Poll “algebraic” = DM: no one. “algebraic = Artin”: no one. Abstain: no one.

560

B.4.10. Useful diagrams. [EGA] 5.3.5 (which vol?! likely I):

X ×S Y → X ×T Y↓ ↓S → S ×T S

Application: If U → S is a monomorphism, and S → T is separated, then U → S is too.Application: A special case is in the hint to [H, Ex. II.4.8(e)]. If we let X = Y , we get a

version that appears in [EH, p. 243]:

X ×Y X → Y↓ ↓

X ×S X → Y ×S YHence if Y → S is sep, then the left vertical arrow is a closed immersion. Also, if f is a finite mapof irreducible varieties, then it is unramified iff X → X ×Y X is an isomorphism [EH, p. 244–5],i.e. f is a monomorphism.

The following square is Cartesian.

X × Y → X ×S Y↓ ↓

S × S → S

Caution!! The above looks bogus!If M is a stack, and X → M is a cover (etale, say) the following Cartesian square will give

you a presentation.R → X ×X↓ ↓M → M ×M

Also,R → X↓ ↓

X ×M X → M

What was that?! That can’t be right.

]

Bibliography

[AM] Atiyah and Macdonald[BC] Brian Conrad[dJ] Johan de Jong’s 1998-99 course notes.[E] Eisenbud, Commmutative Algebra with a View to Algebraic Geometry.[EH] Eisenbud and Harris.[EGA] EGA.[FAC] Serre, Faisceaux Algebriques Coherents.[FGA] Grothendieck, A., Fondements de la geometrie algebrique, in Collected Bourbaki

Talks, mimeographed by the Secretariat Mathematique, Paris, 1962. I think this isGrothendieck, Sem. Bourbaki 1960/61.

[GH] Griffiths and Harris.[Gr] A. Grothendieck, Sur quelques points d’algebre homologique, Tohoku Math. J. (2)

9 1957, 119–221. [forme: cited in category chapter][H] Hartshorne.[KS] M. Kashiwara and P. Schapira, Sheaves on Manifolds.[K] A. Kresch, Invent. Math.[L] S. Lang, Algebra (rev. 3rd ed.), GTM 211.[LS] Lichtenbaum-Schlessinger.[M-CRT] Matsumura’s Commutative Ring Theory.[Milne] Milne.[MilneEC] Milne, presumably his book on etale cohomology; this is a reference in [MilneLET].[MilneLET] Milne, Lectures on etale cohomology, version 2.01.[MuPicMod] Mumford’s Picard groups of moduli problems.[M] Mumford’s red book.[MAV] Mumford Abelian Varieties

[MuCPV] Mumford, Complex Projective Varieties I[Neron] Neron models.[Newstead] Newstead.[Sch] Schlessinger.[S] J.-P. Serre, GAGA.[SGA] SGA.[V] A. Vistoli, Invent. Math.[W] Weibel

561

Index

(−1)-curve, 461

(−2)-curve, 461

(I : x), 147

0-object, 549

0-ring, 11

2-fibered product, 541

A-scheme, 138

A-valued points, 206

A1, 97, 98

A1Q bold, 99

A1R bold, 99

A2k, 100

An, 100

AnX , 344

BG, 541

FF (X), 134

I(S), 113

L mathcal with bars, 340

N mathfrak, 104

O(a, b) oh, 211, 340

OX-modules [fix O], 301

P 1k , 125

PnA , 127

PnX , 346

Pnk , 127

Q-Cartier divisor, 446

Q-divisor, 446

T -valued points, 206

V (S), 105

X(R), 206

X(T ), 206

Aut(·), 32

Γ∗(F ) cal, 329

Mor, 31

Ω, 354

ΩpX , 516

ΩA/B , 371

SpecZ bold, 160

SpecA, 97

SpecZ bold, 98, 123

SuppF mathcal, 88

, 211, 340, 342

A, 289

ωX/k, 509PicX, 306, 309Proj underline, 344, 347

Spec underline, 343⊕, 50

⊗, 36π-ample, 441π′, 513√I, 106

M , 120×k, 195×Spec k, 195

d-tuple embedding, 339, 340, 364, 389, 422,426

d-uple embedding, 193

f−1, 86f−1, inverse image sheaf, 86

f∗, 73mathcalOX,p, 94n-plane, 190

pa, 390etale, 496, 497

etale morphism, 507etale topology, 496

abelian category, 31, 50, 549

absolute product, 195additive category, 49, 549

additive functor, 50adjoint, 44, 329adjoint functors, 44

adjoint matrix, 252adjoint pair, 44

adjunction formula, 374, 375affine line, 98affine communication lemma, 135

affine cone, 187, 188affine line, 97

affine line with doubled origin, 125affine morphism, 165affine morphisms as Spec underline, 344

affine morphisms separated, 219affine open, 121

affine plane, 100

563


affine scheme, 97, 118affine space, 100affine topology/category, 301affine variety, 138affine-local, 135Algebraic Hartogs’ Lemma, 123, 127, 139algebraic set, 109algebraic space, 541, 542algebraic space, DM stack, Artin stack, 541ample, 440ample cone, 445arithmetic genus, 390arrow, 31Artin stack, 541Artinian, 266Artinian module, 266Artinian ring, 269Artinian scheme, 269ascending chain condition, 110

associated point, 148associated prime, 147assumptions on graded rings, 184Auslander-Buchsbaum theorem, 292Aut scheme, 540automorphism, 32axiom of choice, 104axiom of choice, 11, 85

base, 205base change, 205base change diagram, 196base locus, 340base of a topology, 88base scheme, 205, 340base-point, 340base-point-free, 340, 439Bertini’s theorem, 364Bezout’s theorem, 400, 418big divisor, 446big invertible sheaf, 445birational, 230birational (rational) map, 230blow up, 452blow-up, 233blowing up, 453boundary, 51branch divisor, 367branch points, 366

Calabi-Yau varieties, 516Calabi-Yau variety, 25Cancellation Theorem for morphisms, 221canonical bundle, 374canonical bundle K, 374canonical curve, 425, 428

Cartesian diagram, 196Cartesian diagram/square, 39category, 31category of open sets on X, 70

Cech cohomology, 381

change of base, 205Chern classes, 537

Chevalley’s theorem, 259

Chevalley’s Theorem, 176Chinese Remainder Theorem, 128

Chow’s lemma, 350

class group, 309class group in number theory, 307

classifying space, 541closed point, 108, 131

closed immersion, 168

closed immersion affine-local, 168closed map of topological spaces, 239

closed morphism, 243

closed subscheme, 168closed subscheme exact sequence, 316

coarse moduli space, 541

codimension, 248cofinal, 44, 302

cofinal system, 122Cohen-Macaulay, 292, 389

Cohen-Seidenberg Going-up theorem, 252

coherent sheaf, 311, 312cohomology of a double complex, 57

cokernel, 50, 549

colon ideal, 147complete (k-scheme), 239

complete intersections in projective space,292

complete linear series, 340completion, 288, 550

completion is an exact functor, 524

completition, 550complex, 51

complex intersections in projective space,400

composition series, 268

cone over quadric surface, 257

cone over smooth quadric surface, 128cone over the quadric surface, 141

conic, 171

conifold singularity, 545connected, 112, 131

connected component, 113

conormal sheaf, 358constant presheaf, 72

constructable function, 525constructable set, 259

constructable subset of a Noetherianscheme, 176

convergence of spectral sequence, 59coproduct, 41, 43

cotangent sheaf, 353

cotangent vector = differential, 276covariant, 34

cover, 512

covering space, 496

Bibliography 565

Cremona transformation, 233, 413, 461cross-ratio, 369cubic, 171cup product, 510curve, 248curves of low genus, 373cusp, 281, 459cycle, 51

Dedekind domain, 237degenerate, 340degree of a finite morphism, 180degree of a projective scheme, 389degree of a vector bundle on a smooth

projective curve, 437degree of invertible sheaf on Pnk , 322Deligne-Mumford stack, 541depth, 292derivation, 357derived functor, 471derived functor cohomology, 379descending chain condition, 109, 266descent, 199, 541desingularization, 452determinant, 241determinant bundle, 310devissage, 405, 446diagonal morphism δ, 214diagonalizing quadrics, 142differential = cotangent vector, 276dimension, 247, 272dimensional vanishing, 400direct limit, 43direct image sheaf, 73direct limit = colimit, 549discrete topology, 79discrete valuation, 283discrete valuation ring, 552distinguished affine base, 301distinguished open set, 103, 115domain of definition of rational map, 228dominant, 229dominant rational map, 229dominating, 229double complex, 56dual numbers, 99, 104dual numbers, 370dual of a quasicoherent sheaf, 309dual variety, 365dualizing sheaf, 509

effective Cartier divisor, 171, 317, 319effective divisor, 446effective invertible sheaf, 445effective Weil divisor, 320

elimination of quantifiers, 261elliptic curve, 307embedded points, 144embedding, 496

enough injectives, 471

enough projectives, 471

epi morphism, 50epimorphism, 40

equidimensional, 248

equivalence of categories, 542

Euler characteristic, 327, 386

Euler exact sequence, 361Euler test, 280

exact, 51

exceptional divisor, 233, 453

excess intersection formula, 537exponential exact sequence, 82

Ext, 520

extending the base field, 199

extension of an ideal, 203exterior algebra, 310

factorial, 141, 322

faithful functor, 35, 38faithfully flat, 468

Faltings’ Theorem (Mordell’s Conjecture),16, 425

fiber above a point, 205

fiber diagram, 196

fibered diagram/square, 39fibered product of schemes, 195

fibered product of stacks, 542

fibral flatness theorem, 493

final object, 37, 549finite presentation, 180

finite extension of rings, 166

finite implies projective, 348

finite module, 166

finite morphism is closed, 243finite morphism is quasifinite, 180

finite morphisms are affine, 166

finite morphisms are projective, 347

finite morphisms separated, 219finite presentation, 311

finite type, 178

finite type A-scheme, 138

finite type (quasicoherent) sheaf, 312finitely generated, 311

finitely generated modules over discretevaluation rings, 286

flasque sheaf, 85

flat, 392, 465, 552

flat A-module, 466flat lft morphisms are open, 475

flat limit, 474

flat morphism, 467

flat quasicoherent sheaf, 467flat quasicoherent sheaf over a base, 468

flat ring homomorphism, 466

flex line, 431

forgetful functor, 34

formal schemes, 12


formally etale, 505, 506formally smooth, 505, 506formally unramified, 505, 506free sheaf, 304Freyd-Mitchell Embedding Theorem, 51full functor, 35, 38function field FF (·), 134function field, 134, 144, 257functor, 34functor category, 38Fundamental Theorem of Elimination

Theory, 551

Gaussian integers mathbbZ[i], 280, 285, 368

generalization, 108generated by global sections, 327, 439generated in degree 1, 184generated in degree 1, 190generic point, 131generic fiber, 205generic point, 109generically separable morphism, 366generization, 131genus, 373genus of nodal curve, 425geometric fiber, 210geometric genus, 373, 374geometric point, 210geometrically connected, 210geometrically integral, 210, 394geometrically irreducible, 210geometrically nonsingular fibers, 500geometrically reduced, 210gerbe, 542germ, 70germ of function near a point, 122gluability axiom, 71gluing along closed subschemes, 477gluing two schemes along closed

subschemes, 237gluing two schemes together along a closed

subscheme, 237Going-Up theorem, 252Gorenstein, 293, 521graph morphism, 221graph of rational map, 229Grassmannian, 363Grauert-Riemannschneider vanishing, 547Grothendieck topology, 496, 542Grothendieck’s theorem on bundles over P 1

bold, 400group scheme, 161group schemes, 433groupoid, 32

Hard Lefschetz theorem, 375Hartogs’ Theorem, 307Hausdorff, 213, 215height, 248

higher direct image sheaf, 391

higher pushforward sheaf, 391

Hilbert basis theorem, 110, 551Hilbert function, 387

Hilbert polynomial, 387

Hilbert scheme, 540

Hilbert syzygy theorem, 399Hilbert’s Nullstellensatz, 101

Hironaka’s example, 349, 461, 477

Hodge bundle, 485, 541

Hodge diamond, 375, 516Hodge theory, 375

homogeneous ideal, 184

homogeneous space, 503

homology, 51homotopic maps of complexes, 469

homotopy fibered product, 541

hypercohomology, 57

hyperelliptic, 426hyperelliptic curve, 372, 425

hyperelliptic curves, 419

hyperelliptic involution, 426

hyperplane, 171, 190hyperplane class, 321

hypersurface, 171, 248

ideal denominators, 265

ideal of denominators, 140

ideal sheaf, 169identity axiom, 71

image scheme, 547

immersion, 496

index category, 41induced reduced subscheme structure, 177

induced reduced subscheme structure, 177

inertia stack, 542

infinite-dimensional Noetherian ring, 250initial object, 37, 549

injective object, 407

injective object in an abelian category, 471

integral, 133integral extension of rings, 251

integral morphism, 167

integral morphism of rings, 251

inverse image, 86

inverse image ideal sheaf, 205inverse image scheme, 205

inverse image sheaf, 86

inverse limit, 42, 549

invertible ideal sheaf, 171invertible sheaf, 307

irreducible, 107, 131

irreducible component, 111

irreducible components, 131irrelevant ideal, 185

Isom scheme, 540

isomorphism, 32

isomorphism of schemes, 121

Bibliography 567

Jacobian, 356Jacobian criterion, 278

Jacobian matrix, 497Jacobson radical, 255

K3 surface, 25K3 surfaces, 516Kawamata-Viehweg vanishing, 547

kernel, 50, 549Kleiman’s criterion for ampleness, 445Kleiman-Bertini theorem, 374Koszul complex, 292, 293

Krull, 261Krull dimension, 247Krull dimension, 247Krull’s Principal Ideal Theorem, 261Krull’s Theorem, 550

Luroth’s theorem, 419Lefschetz hyperplane theorem, 375Lefschetz principle, 261left-adjoint, 44left-exact, 54

left-exactness of global section functor, 84Leibniz rule, 354, 371Leray spectral sequence, 384, 394limit, 42

line, 190line in projective space, 262linear space, 190linear series, 340

linear systsem, 340local complete intersection, 368local complete intersections are

Cohen-Macaulay, 292local-ringed space, 157locally ringed spaces, 158

locally closed immersion, 173locally constant sheaf, 72locally free sheaf, 307locally integral (temp.), 282

locally Noetherian scheme, 136locally of finite type A-scheme, 138locally of finite presentation, 180locally of finite presentation (morphisms),

505

locally of finite type, 178long exact sequence, 53long exact sequence of higher pushforward

sheaves, 392

mapping cone, 62, 383minimal primary decomposition, 146

minimal prime, 108, 111Mittag-Leffler condition for inverse

systems, 550module of Kahler differentials, 354module of relative differentials, 354

moduli space of curves, 436, 540, 541

moduli space of stable maps, 540monic morphism, 50

monomorphism, 40, 549Mordell’s conjecture, 16, 425morphism, 31morphism of (pre)sheaves, 75morphism of ringed spaces, 156

morphism of schemes, 159multiplicity of a singularity, 459

Neron-Severi group, 444Nagata, 250, 323Nagata’s Lemma, 323

Nakai’s criterion for ampleness, 445Nakai-Moishezon criterion for ampleness,

401Nakayama’s Lemma, 242, 254, 255, 552natural transformation of functors, 38

nef, 445nice base of a topology, 88nilpotents, 104, 132nilradical, 104, 104, 106node, 281

Noetherian induction, 112Noetherian ring, 110, 110, 551Noetherian rings, important facts about,

110Noetherian scheme, 131, 136

Noetherian topological space, 108, 109non-degenerate, 340non-zero-divisor, 103nonsingular, 275, 279nonsingularity, 275

normal, 123, 139normal = R1+S2, 290normal exact sequence, 95, 368normal sheaf, 358

normalization exact sequence, 425notation, 13Nullstellensatz, 101, 138, 552number field, 237

numerically effective, 445

object, 31octic, 171Oka’s theorem, 312open subscheme, 121open immersion, 164, 164

open subfunctor, 540open subscheme, 164orbifold, 541orientation of spectral sequence, 57

page of spectral sequence, 57

partially ordered set, 33partition of unity, 383Picard group, 306, 309Picard number, 444

Picard scheme, 483


Picard variety, 436, 540

plane, 190

plane in projective space, 262pole, 284

poset, 33

presheaf, 70

primary decomposition, 146

primary ideal, 145prime avoidance (temp. notation), 147

product, 35

projection formula, 394

projective space, 127

projective A-scheme, 185projective and quasifinite implies finite, 395

projective cone, 188

projective coordinates, 127, 183

projective implies universally closed, 349projective line, 125

projective morphism, 347, 348

projective object in an abelian category,470

projective variety, 187

projectivization of a locally free sheaf, 346proper, 239

proper morphism of stacks, 542

proper non-projective surface, 476

proper transform, 452, 453pseudofunctor, 541

pullback diagram, 196

pullback for [locally?] ringed spaces, 337

pullback of cohomology of quasicoherentsheaves, 341

pure dimension, 248pushforward of coherent sheaves, 341, 351

pushforward of quasicoherent sheaves, 333

pushforward sheaf, 73, 73

quadric, 171quadric surface, 257

quadric surface, 141, 193

quartic, 171

quasiaffine, 446

quasicoherent sheaf, 120, 302quasicoherent sheaves: product, direct sum,∧, Sym, cokernel, image, ⊗, 310

quasicompact, 131

quasicompact morphism, 163

quasicompact topological space, 108

quasifinite, 179quasiprojective, 446

quasiprojective scheme, 186

quasiprojective is separated, 220

quasiseparated, 218, 334quasiseparated morphism, 163

quasiseparated scheme, 132

quintic, 171

Quot scheme, 540

quotient object, 51

radical, 106

radical ideal, 106

radiciel morphisms, 162ramification divisor, 367, 419

ramification points, 366

rank of finite type quasicoherent sheaf, 315

rank of locally free sheaf, 307rational map, 227

rational function, 144

rational normal curve, 193, 339, 373

rational normal curve take 1, 109

rational section of invertible sheaf, 307reduced, 133, 137

reduced locus (closed in locally Noetheriancase), 178

reduced ring, 104

reduced scheme, 132reducedness is stalk-local, 133

reduction, 177

Rees algebra, 452

reflexive sheaf, 309, 358regular, 275

regular scheme, 279

regular function, 145

regular implies Cohen-Macaulay, 292regular local ring, 279

regular point, 275

regular ring, 279

regular section of invertible sheaf, 307

regular sequence, 293, 461relative canonical bundle, 374

relatively ample, 440, 441

relatively base point free, 345

relatively generated, 439relatively generated by global sections, 345

representable functor, 161

representable functors, 540

representable morphism, 542residue field of a scheme at a point, 122

resolution of singularities, 452

restriction of a quasicoherent sheaf, 335

restriction of sheaf to open set, 76resultant, 241

Riemann-Hurwitz formula, 419, 436

Riemann-Roch, 417

right exact, 36

right-adjoint, 44right-exact, 54

ring of fractions, 143

ring of integers in a number field, 237

ring scheme, 434ringed space, 74, 93

rulings on the quadric surface, 193

S2, 290

Sard’s theorem, 501

saturated module, 330

scheme over A, 138

Bibliography 569

scheme, definition of, 120

scheme-theoretic inverse image, 205scheme-theoretic pullback, 205

scheme-theoretic support, 545

scheme-theoretic support of quasicoherentsheaf, 405

sections over an open set, 70

Segre, 440

Segre embedding, 211, 340Segre product, 211

Segre variety, 211semicontinuity ideas, 525

separable morphism, 366

separated, 125, 214separated over A, 215

separated presheaf, 71

separatedness, 121septic, 171

Serre duality, 436, 510Serre vanishing, 380, 400, 442

Serre’s criterion for affineness, 392

Serre’s criterion for normality, 290sextic, 171

sheaf, 70

sheaf Hom (Hom underline), 76sheaf Hom (Hom underline) of

quasicoherent sheaves, 309

sheaf Hom (underline), 76sheaf determined by sheaf on base, 302

sheaf of ideals, 169

sheaf of relative differentials, 353sheaf on a base, 88, 89

sheaf on a base determines sheaf, 89

sheaf on affine base, 301sheafification, 77, 79

singular, 275, 279skyscraper sheaf, 72

smooth, 275, 496–498

smooth morphism, 374, 506smooth quadric surface, 142

Snapper’s theorem, 446

specialization, 108, 131spectral sequence, 55

spectrum, 97stack, 90

stack (Deligne-Mumford, Artin), 541

stalk, 70stalk-local, 135

Stein factorization, 492

strict transform, 453structure morphism, 159

structure sheaf, 93structure sheaf (of ringed space), 74

structure sheaf on SpecA, 117

submersion, 496subobject, 51

subsheaf, 81

support (of a sheaf, or a section thereof), 85

support of a sheaf, 88

support of a Weil divisor, 320

surface, 248

surjective morphism, 208

symbolic power of a prime ideal, 271

symbolic power of an ideal, 267

symmetric algebra, 310

tacnode, 281

tame ramification, 367

tangent line, 431

tangent vector, 276

tautological bundle, 344

tensor algebra, 310

tensor product, 35, 36

tensor product of sheaves, 85

topology (etale, fppf, etc.), 541

torsion section of a sheaf, 311

torsion sheaf, 311torsion-free quasicoherent sheaf, 310

total fraction ring, 144

total transform, 453

trace map, 510

trigonal curve, 425

twisted cubic, 171

twisted cubic, 193, 257, 400

twisted cubic curve, 109

two planes meeting at a point, 293, 400

ultrafilter, 113

underline S, 72underline Spre, 72

uniformizer, 282

unirationality, 547

universal property, 30

universal property of blowing up, 453

universally closed, 239, 349

unramified, 496, 497, 500

unramified morphism, 506

upper shriek, 513

uppersemicontinuity, 546

uppersemicontinuity of rank of finite typesheaf, 315

valuative criterion for separatedness, 294

value of a quasicoherent sheaf at a point,315

value of function at a point, 122

vanishing set, 105

vanishing theorems, 386

variety, 213, 217

vector bundle, 344

Veronese, 340

Veronese embedding, 193, 339, 340, 364,389, 422, 426

Veronese surface, 193

vertical (co)tangent vectors, 353

very ample, 440


Weierstrass normal form, 431weighted projective space, 194Weil divisor, 320wild ramification, 367Witt vectors, 504

Yoneda embedding, 38Yoneda’s Lemma, 37

Zariski tangent space, 275

Zariski topology, 105, 105Zariski’s Main Theorem, 492zero ring, 11zero object, 37, 49zero ring, 71zero-divisor, 103, 147

foundations of algebraic geometry ... - stanford university

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