An informal introduction to algebraic geometryby Enrique Arrondo(*)

Version of November 3, 2019 (rough draft, with an appendix to be completed)

0. Introduction

1. Affine sets

2. Projective sets

3. Regular functions

4. The definition of a scheme

5. Properties of morphisms

6. Dimension

7. Parameter spaces

8. Local properties

9. Vector bundles

10. Sheaves of modules

11. Cohomology and duality

12. Riemann-Roch theorem

A. Appendix: Categories and universal properties

(*) Departamento de Algebra, Geometrıa y Topologıa, Facultad de Ciencias Matematicas,

Universidad Complutense de Madrid, 28040 Madrid, Spain; E-mail: arrondo@mat.ucm.es.

Feel free to use this material, provided you quote the source.

1

0. Introduction

The scope of these notes is to introduce the reader into the scheme language for

algebraic geometry. To do this in a fast way, we will skip many proofs, concentrating

mainly in the proofs that could help to a better understanding of the topic under study.

The interested reader can find most of the missing proofs in [Ha], a text of which we will

roughly follow its structure, but with our particular taste. These notes can be regarded as

a natural continuation of my previous notes [A] (in fact, the last chapters of those notes,

which are incomplete, are a first attempt of introduction to the scheme theory). However,

the first two sections of these notes will contain the material from [A] that will be required

to follow the rest of the sections.

By definition, Algebraic Geometry is the branch of Mathematics devoted to study sets

defined by zeros of polynomials. To illustrate which kind of sets we are interested in, we

will start with some examples.

Example 0.1. Consider in the affine plane A2k the circle of equation X2 + Y 2− 1 = 0. If

you project it to its first coordinate, you get different results depending on the ground field

k you are considering. For an algebraically closed field, it is clear that you get the whole

affine line, since for any a ∈ k you find b ∈ k such that a2 + b2 − 1 = 0. If, instead, you

consider k = R, this is only true for a ∈ [−1, 1]. Therefore, in the real case, the image by the

projection (which should be a morphism in any reasonable category) of something defined

by polynomial equalities becomes something defined by inequalities. Hence, the natural

objects in real algebraic geometry are semialgebraic sets, i.e. sets defined by polynomials

equalities and inequalities. To know more about this topic, you can take a look at [BCR],

where you can find a proof that the image of a semialgebraic set is always a semialgebraic

set.

Example 0.2. A cute reader probably figured out that, if in the above example we

replace the circle X2 + Y 2 − 1 = 0 with the hyperbola XY − 1 = 0, then, independently

on the ground field k, its image under the first projection is not the whole affine line, but

the line minus the point X = 0. This can be still admissible, but it is a first sign that

affine geometry is not the best one. Observe that, if we complete the hyperbola in the

projective plane, now it has two more points (at infinity), one of them projecting to the

missing point, and the other one projecting to the infinity point of the affine line.

Example 0.3. In fact, the situation in the affine case is even worse. Consider the map

ϕ : A2k → A2

k defined by (a, b) 7→ (a, ab) (since it is defined by polynomials it would be

reasonable to consider it as a morphism). The image of any horizontal line Y = λ is

obviously the line Y = λX. Hence the image of ϕ is the union of all the lines passing

2

through (0, 0) and slope λ. Since the slope is finite, the vertical line is missing. Hence

the image of the map is the whole plane minus the vertical line (to which we previously

removed the point (0, 0)). A set like that (something defined by polynomials, to which

we remove a subset defined by polynomials, to which previously we removed a smaller

subset defined by polynomials, to which previously...) is called a constructible set. It is

also true that the image of a constructible set is a constructible set (see [H], Theorem

3.16). This shows that, in affine geometry, the right category to work with would be the

category of constructible sets, instead the category of affine set, i.e. sets defined by zeros

of polynomials.

However, in the case of projective space, we will see that the image of any projective

set –i.e. a set in the projective space defined by (homogeneous) polynomials– is again a

projective set, assuming the ground field to be algebraically closed. Hence, the right natural

ambient space space for Algebraic Geometry is the projective space over an algebraically

closed field, instead of the affine space. The underlying reason is that the projective space

is complete in the sense that there are no missing points (while in the affine space the

infinity points are missing).

We will follow essentially the steps in Differential Geometry, in which one first defines

the notion of embedded submanifold, then the notion of abstract manifold and finally

one gets a way of embedding abstract manifolds in a natural ambient space. As we have

pointed out, the analogue of embedded submanifold will be for us the notion of projective

set.

Recall that, in Differential Geometry, the notion of abstract manifold comes from the

observation that a submanifold of dimension n is locally isomorphic at any point to an open

disc of Rn. Unfortunately, one cannot do the same in Algebraic Geometry. First of all, the

natural topology one can use for an arbitrary ground field (the so-called Zariski topology)

has open sets that are quite big. This produces that, in general, no neighborhood of a

projective set is isomorphic to any open set of an affine or projective space. In particular,

it makes very difficult to define the notion of dimension of an algebraic variety (in the

case of projective sets, there is a direct way of defining the dimension using the so-called

Hilbert polynomial, and we will use that to see how to generalize the notion for an arbitrary

variety).

Remark 0.4. When the ground field is the field of complex numbers, one could use

the usual topology instead of Zariski topology, and in that case, a projective subset is

locally, around any non-singular point, (analytically) isomorphic to an open disk of some

Cn. Hence, in this situation, an analytic approach is possible, in which the notion of

dimension is automatic. We will not follow this approach, which the reader can find in

[GH]. The reader should be aware that the price to pay to follow this approach instead

3

of the algebraic approach (which requires the knowledge of Commutative Algebra) is that

it requires the knowledge, not only of Complex Analysis, but also Differential Geometry,

Algebraic Topology and Functional Analysis. The striking result in this analytic approach

is Chow’s Theorem (see [GH], page 167), stating that any analytic subvariety of a projective

space is a projective set.

Having this in mind, one needs to find the right notion of abstract variety. Since a

basis of the Zariski topology will be given by affine sets, a first notion of abstract variety

could be an object that can be covered by affine sets (as it happens for projective sets).

However, in modern Algebraic Geometry, there is a further generalization, which is a

little bit more subtle. In order to give an idea of this new generalization, recall that,when

intersecting two plane curves, we have the notion of intersection multiplicity at the different

points of their intersection. For example, a conic and a tangent line meet at the tangency

point with multiplicity two, so that in some sense their intersection is a point counted

twice, i.e. a point with some extra structure. We want to give a precise description of that

structure, in the same way that we are used to speak about a special kind of conic: double

line, i.e a line with some double structure. The notion of scheme, the most important in

algebraic geometry, will include those concepts.

Once we will have the notion of abstract scheme, our next goal will be to see when

it is possible to embed it in some projective space. This is done with the classical theory

of divisors and line bundles. To compute the possible dimension of the projective space in

which to embed the scheme, we will need the Riemann-Roch theorem and the cohomology

theory. We will end these notes particularizing all this theory to the case of curves and

surfaces.

4

1. Affine sets

Definition. An affine set X ⊂ Ank is a subset for which there exists a set of polynomials

S ⊂ k[X1, . . . , Xn] such that

X = V (S) := p ∈ Ank | f(p) = 0 for all f ∈ S.

Remark 1.1. Observe that, for any S ⊂ k[X1, . . . , Xn], one has V (S) = V (I), where

I is the ideal generated by I. In particular, since any ideal of k[X1, . . . , Xn] is finitely

generated (Hilbert Basis Theorem) it follows that any affine set can be defined by a finite

number of equations.

Exercise 1.2. Prove the following equalities:

(i) V (0) = Ank , and V (1) = ∅.

(ii) For any subsets S, S′ ⊂ k[X1, . . . , Xn], then V (S) ∪ V (S′) = V (S′′), where S′′ =

ff ′ | f ∈ S, f ′ ∈ S′.

(iii) If SjS∈J is a collection of subsets of k[X1, . . . , Xn] then⋂j∈J V (Sj) = V (

⋃j∈J Sj).

Or, in terms of ideals, prove:

(i’) V (0) = Ank , and V (k[X1, . . . , Xn]) = ∅.

(ii’) For any ideals I, I ′ ⊂ k[X1, . . . , Xn], then V (I) ∪ V (I ′) = V (II ′) = V (I ∩ I ′).

(iii’) If IjI∈J is a collection of ideals of k[X1, . . . , Xn] then⋂j∈J V (Ij) = V (

∑j∈J Ij).

Observe in particular that the sets of the form V (S) satisfy the conditions to be the closed

sets of a topology.

Definition. The Zariski topology of Ank is the topology in which the closed sets are those

of the form V (S). We will still call Zariski topology on any subset of Ank the topology

induced by the Zariski topology on Ank .

Since affine sets are defined by ideals, we can try to invert the process and get ideals

from affine sets:

Definition. We define the ideal of an affine set X (or in general of any subset X ⊂ Ank )

as

I(X) = f ∈ k[X1, . . . , Xn] | f(p) = 0 for any p ∈ X.

We now check if the analogue to properties (i) (ii) and (iii) of Remark 1.1 hold.

Exercise 1.3. Prove the following results:

5

(i) I(∅) = k[X1, . . . , Xn] and, if k is infinite, I(Ank ) = 0 (for this, use induction on n).

(ii) If Xjj∈J is a collection of subsets of Ank , then I(⋃j∈J Xj) =

⋂j∈J I(Xj).

(iii) If X,X ′ ⊂ Ank , then I(X ∩X ′) ⊃ I(X) + I(X ′).

The attentive reader will have notice that part (iii) is not the expected equality (even if

we are considering just a finite intersection), but only an inclusion. The following example

shows why we cannot expect to reach the equality.

Example 1.4. Consider L = V (Y ), C = V (Y − X2) ⊂ A2k. It is easy to check that

I(L) = (Y ) and I(C) = (Y − X2) (this is an easy consequence of the Nullstellensatz

below, but the reader is encouraged to prove it by hand). Clearly L ∩ C = (0, 0),which easily implies I(L ∩ C) = (X,Y ). However, this ideal is clearly different from

I(L) + I(C) = (X2, Y ). But observe that this other ideal keeps more information. In fact,

while a polynomial f ∈ (X,Y ) corresponds to a curve passing through (0, 0), a polynomial

f ∈ (X2, Y ) corresponds to a curve that not only passes through it, but also that the

line V (Y ) meets V (f) in (0, 0) with multiplicity at least two. Hence (X2, Y ) represents in

some sense the point (0, 0) plus an infinitely close point in the horizontal direction, which

is precisely the intersection of the parabola C with the line L.

The reason why the previous ideal cannot be the ideal of the point is that, as one can

easily check, the ideals of affine sets are radical ideals. In fact, this property characterizes

radical ideals, as the following important result (which we will not prove) shows:

Theorem 1.5 (Nullstellensatz). If k is algebraically closed, any ideal I ⊂ k[X1, . . . , Xn]

satisfies IV (I) =√I.

As a consequence, the operators V and I define a bijection between the set of affine

sets in Ank and the set of radical ideals of k[X1, . . . , Xn]. The idea of the notion of scheme

will be to consider also non-radical ideals, since they can provide more information than

just ideals of affine sets. We pass now to understand what maximals ideals correspond to

in this bijection.

Example 1.6. Let p = (0, . . . , 0). Then a polynomial f ∈ k[X0, . . . , Xn] is in I(p) if

and only it it has not independent term. This is equivalent to say that any monomial of f

contains some Xi, so that I(p) = (X1, . . . , Xn). If now p is a general point p = (a1, . . . , an),

using a translation we immediately conclude that I(p) = (X1− a1, . . . , Xn− an). Observe

also that the evaluation at p defines an epimorphism k[X1, . . . , Xn] → k whose kernel is

I(p). Hence k[X1, . . . , Xn]/I(p) is isomorphic to k, which proves that I(p) is a maximal

ideal. In the case of an algebraically closed field, the Nullstellensatz implies that all the

maximal ideal take that form:

6

Corollary 1.7 (Weak Nullstellensatz). Assume k is algebraically closed. Then, an ideal

I ⊂ k[X1, . . . , Xn] is proper if and only if V (I) 6= ∅. As a consequence, an ideal I ⊂k[X1, . . . , Xn] is maximal if and only if I = I(p) for some point p ∈ Ank .

Proof: Assume first that V (I) = ∅. Hence I(V (I)) = k[X1, . . . , Xn], so that 1 ∈ IV (I).

By the Nullstellensatz, 1 ∈√I, so that 1 ∈ I, and therefore I is not a proper ideal. Since

the reciprocal result is trivial for any ground field, this proves the first statement.

For the second statement, we already showed in Example 1.6 that any ideal I(p) is

maximal. Let now I ⊂ k[X1, . . . , Xn] be a maximal ideal. In particular, it is proper, and

we deduce as above that V (I) is not empty. Hence, if we take p ∈ V (I), it follows I ⊂ I(p),

and the maximality of I implies I = I(p), as wanted.

Remark 1.8. It is not completely honest to state the weak Nullstellensatz as a conse-

quence of the Nullstellensatz, since it is nowadays standard to prove the Nullstellensatz

from the weak Nullstellensatz using the so-called Rabinowitsch trick. It is also worth

to mention that the weak Nullstellensatz proves, without using Zorn’s Lemma, that any

nonzero finitely generated k-algebra possesses maximal ideals.

Remark 1.9. Of course the Nullstellensatz result is false if k is not algebraically closed.

For instance, (X2 + Y 2 + 1) defines the empty set in the real plane, and however is a

proper ideal (and it is even a radical ideal). We will always assume our ground field k to

be algebraically closed.

Prime ideals also have a geometrical interpretation, for which we need a first definition.

Lemma 1.10. Let X be a non-empty closed subset of a topological space. Then the

following are equivalent:

(i) If X ⊂ X1 ∪X2, with X1, X2 closed subsets then either X ⊂ X1 or X ⊂ X2.

(ii) X is not the union of two proper closed subsets.

(iii) Any two non-empty open sets of X intersect.

Proof: We will prove the implications cyclically.

(i)⇒ (ii): It is obvious.

(ii)⇒ (iii): Let U1, U2 be two open sets of X that do not intersect. Then X is the union

of the closed sets X \U1 and X \U2. By hypothesis (ii), either X = X \U1 or X = X \U2

i.e. either U1 = ∅ or U2 = ∅, which proves (iii).

(iii) ⇒ (i): Assume X ⊂ X1 ∪ X2. This implies that the intersection of the open sets

X \X1 and X \X2 is empty. By hypothesis (iii), either X \X1 = ∅ or X \X2 = ∅, i.e.

either X ⊂ X1 or X ⊂ X2.

7

Definition. A non-empty closed set X of a topological space is irreducible if it satisfies

any of the conditions of Lemma 1.10.

Lemma 1.11. If U is an open subset of a topological space and X is an irreducible closed

subset meeting U , then X ∩ U is an irreducible closed subset of U .

Proof: Let U1, U2 be two non-empty open subsets of X ∩ U . Since X ∩ U is open in X,

then U1, U2 are obviously open subsets of X, so that they meet.

The notion of irreducibility is not interesting at all for the usual topology (or any

Hausdorff topology), in which the only irreducible sets are the points. It is however im-

portant for the Zariski topology, and in fact it will play the role that connectedness plays

for the usual topology.

Proposition 1.12. An affine set X ⊂ Ank is irreducible if and only if I(X) is prime.

Proof: Assume first that X is irreducible. Then, if the product fg of two polynomials is

in I(X), this means X ⊂ V (fg) = V (f) ∪ V (g). Hence, the irreducibility of X implies

that either X ⊂ V (f) or X ⊂ V (g), i.e. either f ∈ I(X) or g ∈ I(X). Therefore I(X) is a

prime ideal.

Reciprocally, assume that I(X) is a prime ideal, and assume X ⊂ X1 ∪X2. If it were

X 6⊂ X1 and X 6⊂ X2, then we could find f ∈ I(X1) \ I(X) and g ∈ I(X2) \ I(X). Hence

fg ∈ I(X1)∩ I(X2) = I(X1 ∪X2) ⊂ I(X), contradicting the primality of I(X). Therefore

either X ⊂ X1 or X ⊂ X2, which proves the irreducibility of X.

The previous result can be considered as the geometric counterpart of the following

characterization of prime ideals:

Exercise 1.13. Prove that an ideal I of a ring A is prime if and only if any time

I ⊃ I1 · I2 necessarily either I ⊃ I1 or I ⊃ I2. In particular, if I is prime and I ⊃ I1 ∩ I2then necessarily either I ⊃ I1 or I ⊃ I2 (the converse is not true: take for example

A = k[X]/(X2) and I = (0)).

Theorem 1.14. Any non-empty affine set X ⊂ Ank can be written, in a unique way, as a

finite union X = X1 ∪ . . . ∪Xr of irreducible affine sets such that Xi 6⊂ Xj if i 6= j.

Proof: It is clear that it is enough to prove that X can be written as a finite union of

irreducible sets. Assume for contradiction that X cannot be expressed in that way. This

implies in particular that X is not irreducible, so that we can write X = X1∪X ′1 such that

X ⊆/ X1 and X ′1. Our assumption that X cannot be written as a finite union of irreducible

sets implies that the same holds for at least X1 or X ′1. We can assume, without lost of

8

generality, that X1 cannot be written as a finite union of irreducible sets. Repeating the

above reasoning for X1 we will have X1 = X2 ∪ X ′2, with X2 ⊆/ X1, X ′2 ⊆/ X1 and X2 is

not a finite union of irreducible sets. Iterating the process, we find an infinite chain

X ⊇/ X1 ⊇/ X2 ⊇/ . . .

which translates into an infinite chain

I(X) ⊆/ I(X1) ⊆/ I(X2) ⊆/ . . .

which is impossible because k[X1, . . . , Xn] is a noetherian ring, so that any infinite increas-

ing chain of ideals is necessarily stationary.

Definition. The irreducible sets X1, . . . , Xr in the above theorem are called irreducible

components of X.

Remark 1.15. The proof of Theorem 1.14 shows that the decomposition into a finite

number of irreducible components remains true for topological spaces in which any infinite

descending chain of closed subsets is stationary. A topological spaces satisfying this last

condition is called noetherian topological space.

The algebraic counterpart of the decomposition into irreducible components is the

following:

Theorem 1.16. Let I be a proper ideal of a noetherian ring. Then we can write I =

I1 ∩ . . . ∩ Ir with I1, . . . , Ir primary ideals with different radical ideals, and for each i =

1, . . . , r we have Ii 6⊃⋂j 6=i Ij . Moreover, the set of

√I1, . . . ,

√Ir is independent of

the decomposition, as well as the primary ideals Ii for which√Ii is minimal in the set

√I1, . . . ,

√Ir.

We refer to any text on Commutative Algebra for the proof, although the existence

follows the steps of the proof of Theorem 1.14:

Exercise 1.17. If we call irreducible to a proper ideal I that cannot be expressed

I = I1∩I2, with I ⊆/ I1, I2, prove that any proper ideal in a noetherian ring can be expressed

as a finite intersection of irreducible ideals. Prove also that an irreducible ideal I is primary

by showing that, for any f ∈ A, if we define In = h ∈ A | hfn ∈ I, then there exists

n ∈ N such that In = In+1; conclude that, if fg ∈ I, then I = ((fn)+I)∩h ∈ A | fh ∈ I.One could think that the whole Theorem 1.16 is not necessary for our purposes. Since

the ideal of an affine set is radical, taking radicals in the primary decomposition we get

that any radical ideal is a unique intersection of prime ideals. The next example (which

9

will illustrate the uniqueness of primary decomposition) will show that the general primary

decomposition appears naturally when working with affine sets:

Example 1.18. Let us consider in A3k the lines L = V (X,Z) and Lt = V (Y,Z − t). We

regard them as a family of lines when t varies in k, and we observe that L and Lt are not

coplanar if and only if t 6= 0. We have I(L) = (X,Z) and I(Lt) = (Y, Z − t), and if t 6= 0

we have I(L) + I(Lt) = (1), which implies I(L) ∩ I(Lt) = I(L) · I(Lt), hence

It := I(L ∪ Lt) =(XY,X(Z − t), Y Z, Z(Z − t)

).

It is a natural temptation to consider for L∪L0 the ideal I0 = (XY,XZ, Y Z,Z2). It is not

true that this is the ideal of L0 (in fact it is not even radical, since Z 6∈ I0 but Z2 ∈ I0).

However V (I0) = L0. It is another exercise to check the equality

I0 = (X,Z) ∩ (Y, Z) ∩ (X − aZ, Y − bZ, Z2)

for any a, b ∈ k. This is a primary decomposition for I0, and you cannot remove any of

these three primary ideals. Since a and b can be taken arbitrarily, this means that the

primary decomposition is not unique. The ideal (X − aZ, Y − bZ, Z2) represents, like

in Example 1.4, the point (0, 0, 0) together with the tangent direction given by the line

X = aZ, Y = bZ. When varying the values of a and b we obtain all the lines passing

through (0, 0, 0) and not contained in the plane Z = 0. The geometric interpretation

is that the ideal I0 still “remembers” that the intersection point of the lines came from

outside the plane Z = 0. Or if you prefer, you are trying to put in the same place the

point (0, 0, 0) of the line L and another point coming from Lt. Since there is no room in

(0, 0, 0) for two different points, you get two infinitely close points.

We can repeat all the above results replacing the affine space Ank with any affine

set X ⊂ Ank and replacing polynomials with polynomial functions. Observe that two

polynomials f, g ∈ k[X1, . . . , Xn] give the same polynomial function on X if and only if

f − g vanishes at all points of X, i.e. if and only if f − g ∈ I(X). Hence we can naturally

identify

A(X) := k[X1, . . . , Xn]/I(X)

with the set of all polynomial functions on X (we will usually represent with a bar the

function defined on X by a polynomial). As for the affine space, we can define

VX(S) = p ∈ X | f(p) = 0 for all f ∈ S for S ⊂ A(X)

IX(Y ) = f ∈ A(X) | f(p) = 0 for all p ∈ Y for any Y ⊂ X.

Of course, the will have that IX(Y ) is nothing but I(Y )/I(X). In particular, the maximal

ideals of A(X) are precisely the ideals IX(p) for p ∈ X, and the prime ideals of A(X)

correspond to the irreducible subsets Y ⊂ X. Hence, the k-algebra A(X) keeps all the

information about X.

10

Theorem 1.19. Let X ⊂ Ank and Y ⊂ Amk be two affine sets. For any polynomial map

ϕ : X → Y (i.e. ϕ(p) = (f1(p), . . . , fm(p)) for some f1, . . . , fm ∈ k[X1, . . . , Xn]), let

ϕ∗ : A(Y )→ A(X) be defined by ϕ∗(g) = g ϕ. Then:

(i) ϕ∗ is well-defined, and is a homomorphism of k-algebras.

(ii) The assignment ϕ 7→ ϕ∗ defines a bijection between the set of polynomial mapsX → Y

and the set of homomorphisms of k-algebras A(Y )→ A(X).

(iii) For each affine subset X ′ ⊂ X, we have ϕ∗−1(IX(X ′)) = IY (ϕ(X ′)).

Proof: It is clear that, if ϕ is defined by polynomials f1, . . . , fm, then ϕ∗(g) = g(f1, . . . , fm),

so that it is a polynomial function on X, and ϕ∗ is a homomorphism of k-algebras, so that

(i) follows.

To prove (ii), we start with a homomorphism of k-algebras ψ : A(Y ) → A(X). If we

write fi = ψ(Yi), then we will have that ψ(g) = g(f1, . . . , fm). We set ψ∗ : X → Y the

polynomial map defined by f1, . . . , fm. It is then clear that the assignements ψ 7→ ψ∗ and

ϕ 7→ ϕ∗ are inverse to each other, which proves (ii).

Finally, part (iii) is immediate, since g ∈ (ϕ∗)−1(IX(X ′)) if and only if gϕ = ϕ∗(g) ∈IX(X ′), i.e. for each p ∈ X ′ we have g(ϕ(p)) = 0, or equivalently g ∈ IY (ϕ(X ′)).

Remark 1.20. Part (iii) is saying that from the homomorphism ψ it is possible to

recover the map ϕ, since the image of a point p is the point whose ideal is ψ−1(IX(p)).

More generally, recall from Examples 0.2 or 0.3 that the image of an affine set X ′ ⊂ X

is not an affine set. However, ψ−1(IX(X ′)) = IY (ϕ(X ′)), which is the ideal defining the

Zariski closure of ϕ(X). Observe also that Theorem 1.19 implies that there is a bijective

polynomial map X → Y whose inverse is also polynomial if and only if A(X) and A(Y )

are isomorphic as k-algebras.(∗) The consequence is that the category of affine sets is

equivalent to the category of reduced (i.e. without nilpotent elements) finitely generated

k-algebras (the trivial k-algebra k corresponding to the isomorphism class of a point). The

idea of affine schemes will be to consider a wider category than the one of reduced finitely

generated k-algebras.

Example 1.21. Let us see what happens when we take a non-reduced k-algebra, for ex-

ample k[X,Y ]/(X2, Y ), corresponding to the point (0, 0) and the horizontal direction (see

Example 1.4). This can be regarded as the homomorphism ψ : k[X,Y ]→ k[T ]/(T 2) given

(∗) Notice that a bijective polynomial map does not possess necessarily a polynomial

inverse. For example, ϕ : A1k → V (Y 2 −X3) ⊂ A2

k defined by t 7→ (t2, t3) is bijective, but

the homomorphism ϕ∗k[X,Y ]/(Y 2 −X3)→ k[T ] is not an isomorphism, since T is not in

the image.

11

by X 7→ T and Y 7→ 0, whose kernel is precisely (X2, Y ). In general, a homomorphism of

k-algebras

ψ : k[X1, . . . , Xn]→ k[T ]/(T 2)

is determined by ψ(Xi) = ai + b1T , and it represents the point (a1, . . . , an) and the

direction of the vector (b1, . . . , bn). For example, the ideal (X−aZ, Y − bZ, Z2) appearing

in Example 1.18, is the kernel of k[X,Y, Z]→ k[T ]/(T 2) determined by

X 7→ aT

Y 7→ bT

Z 7→ T

and represents the direction (a, b, 1) at the point (0, 0, 0). We can wonder what happens

when we replace T 2 with a higher power. Consider, for instance, f = Y − XY − X2,

which is in (X2, Y ) (as we can understand by observing that V (f) passes through (0, 0)

in the horizontal direction). Writing f = Y (1 − X) − X2, it follows, for each m ≥ 2,

that f(X,X2 +X3 + . . .+Xm−1) is divisible by Xm, so that f is in the ideal (Y −X2 −X3 − . . .−Xm−1, Xm), which is the kernel of ψ : k[X,Y ] → k[T ]/(Tm) given by X 7→ T

and Y 7→ T 2 + T 3 + . . . + Tm−1. This corresponds to the fact that the truncated Taylor

expansion of Y in terms of X (observe that the equation of the curve can be written as

Y = X2

1−X ) is Y = X2 +X3 + . . .+Xm−1. Hence k[T ]/(Tm) corresponds to infinitesimal

data up to order m− 1. Looking at maximal ideals, the inverse image by ψ of (T ) (in fact

the only prime ideal of k[T ]/(Tm)) is the ideal (X,Y ) of the point (0, 0). Hence we miss

all the extra information only imitating Theorem 1.19(iii).

Example 1.22. We can go to the limit case (see Example A.4) of the above example,

and now consider the homomorphism ψ : k[X,Y ] → k[[T ]] defined by X 7→ T and Y 7→T 2 + T 3 + . . .. Again k[[T ]] has only one maximal ideal, (T ), whose inverse image is

(X,Y ), hence we only recover the point (0, 0). However, this time we are not only missing

some infinitesimal information, but the whole curve (which is unique) to which the Taylor

expansion corresponds. This can be recovered from the fact that ker(ψ) = (Y −XY −X2).

Hence we recover that important piece of information as the inverse image of (0), which

is the other prime ideal of k[[T ]], although is not a maximal ideal. This will be the idea

of k[[T ]] (and in general of any discrete valuation ring): it represents not only a point,

obtained from the maximal ideal, but also a (germ of) curve, obtained from (0).

We conclude the section showing the strange behavior of the Zariski topology.

Lemma 1.23. Given an affine set X ⊂ Ank , the sets of the form DX(f) := X \ VX(f)(with f ∈ k[X1, . . . , Xn]) form a basis of the Zariski topology of X. Moreover any open

set of X can be written as a finite union of sets of that form.

12

Proof: Let U ⊂ Ank be an open set. Then Ank \ U is an affine set. By Remark 1.1, we

can write Ank \ U = V (f1, . . . , fr). This means X \ U = VX(f1) ∩ . . . ∩ VX(fr), so that

X ∩ U = DX(f1) ∪ . . . ∪DX(fr). This proves the result.

Exercise 1.24. For any f ∈ k[X1, . . . , Xn], prove that the map

D(f) −→ An+1k

(a1, . . . , an) 7→ (a1, . . . , an,1

f(a1,...,an) )

induces a homeomorphism between D(f) := Ank \ V (f) and V (Xn+1f − 1).The following is the translation of the Nullstellensatz to subsets of affine sets:

Lemma 1.25. For any affine set X and polynomial functions f and fi in A(X) (with i

varying in some set), the following are equivalent:

(i) DX(f) ⊂⋃iDX(fi)

(ii) VX(fi) ⊂ VX(f)

(iii) There exist a power fs in the ideal generated by the classes fi.

Proof: It is clear that (i) and (ii) are equivalent and also that (iii) implies (ii). Finally,

observe that condition (ii) implies f ∈ I(X ∩ (⋂i V (fi))). By the Nullstellensatz, this is

equivalent to saying that some power fs is in the ideal I(X) + Σi(fi), as wanted.

We have the following striking result, showing that the write notion for completeness

cannot be compactness:

Proposition 1.26. Any open set of an affine set X is a compact set.

Proof: By Lemma 1.23, it is enough to show that any basic open set DX(f) is compact.

To show that, it is enough to show that any covering of an affine set DX(f) by a colection

of sets DX(fi)i has a finite subcover. By Lemma 1.25, the inclusion DX(f) ⊂⋃iD(fi)

is equivalent to saying that some power fs can be written as a linear combination of a

finite set fi1 , . . . , fir of the given collection of polynomials. Applying again Lemma 1.25,

we have that DX(f) ⊂ DX(fi1) ∪ . . . ∪DX(fir ), as wanted.

13

2. Projective sets

In principle, the theory of projective sets seems to be the same as the one for affine sets,

with the only difference that we need to work with homogeneous polynomials. However,

this will make a big difference, as we will see. The first difference a homogeneous polynomial

does not define any function, although at least we can say whether it vanishes or not at a

point. This allows the following:

Definition. A projective set X ⊂ Pnk is a subset for which there exists a set of homogeneous

polynomials S ⊂ k[X0, . . . , Xn] such that

X = V (S) := p ∈ Pnk | F (p) = 0 for all F ∈ S

(we will use capital characters for polynomials in k[X0, . . . , Xn], and keep small characters

for polynomials in k[X0, . . . , Xn]).

One first difference is that now the ideal generated by homogeneous polynomials

contains polynomials that are not homogeneous. One possible solution to that is to regard

homogenous polynomials as defining an affine set (called the affine cone of the projective

set X) X ⊂ An+1k . This affine cone has the property that, if (a0, . . . , an) ∈ X, then

also (ta0, . . . , tan) ∈ X for any t ∈ k. This means that, for any F ∈ I(X), if we write

F = F0 +F1 + . . .+Fd (with Fi homogeneous of degree d), then, for any (a0, . . . , an) ∈ X,

0 = F (ta0, . . . , tan) = F0(a0, . . . , an) + F1(a0, . . . , an)t+ . . .+ Fd(a0, . . . , an)td

for each t ∈ k. Hence Fi(a0, . . . , an) = 0 for i = 0, 1, . . . , n, i.e. also Fi ∈ I(X). In the

particular case in which X is just one point p, we are implicitly taking the convention that

a polynomial F vanishes at p if and only if Fi(p) = 0 for all the homogeneous summands

Fi of F . This is a more general fact in graded rings, whose definition we recall now.

Definition. A graded ring is a ring S such that, as an abelian group, can be decomposed

as S =⊕

d≥0 Sd (elements in Sd are called homogeneous of degree d), with the property

that F ∈ Sd and G ∈ Se imply FG ∈ Sd+e.

Lemma 2.1. Let I ⊂ S be an ideal of a graded ring. Then the following are equivalent:

(i) I is generated by homogeneous elements.

(ii) For any F ∈ I, if F = F0 + . . .+ Fd, with Fi ∈ Si, then each Fi is in I.

Proof: It is clear that condition (ii) implies that I is generated by homogeneous elements,

so that we need to prove the converse. So let assume I is generated by homogeneous

elements and take F ∈ I. By assumption, we can write F = G1F1 + . . . + GrFr, with

14

F1, . . . , Fr homogeneous elements of I of respective degrees d1, . . . , dr, and G1, . . . , Gr ∈ S.

If Gi = Gi0 + . . .+Giei is the decomposition of each Gi into its homogeneous summands, it

is clear that the homogeneous summand of degree l of F is Fl = G1,l−d1F1+. . .+Gr,l−drFr,

which is also in I.

Definition. An ideal as in Lemma 2.1 is called a homogeneous ideal.

Remark 2.2. Observe that condition (ii) is saying that, writing Id := I ∩ Sd, the kernel

of the natural map S →⊕

d≥0 Sd/Id is I. Thus we can write S/I =⊕

d≥0 Sd/Id, and

hence S/I is also a graded ring.

Now we can repeat most of the theory of affine sets, just adding the word “homo-

geneous”. For example, observe that, with our convention, we can now define projective

sets as V (I), where I is a homogeneous ideal of k[X0, . . . , Xn]. We have the analogue of

Exercise 1.2:

Exercise 2.3. Prove the following equalities:

(i) V (0) = Pnk , and V (1) = ∅.

(ii) For any subsets S, S′ ⊂ k[X0, . . . , Xn] of homogeneous polynomials, then V (S) ∪V (S′) = V (S′′), where S′′ = FF ′ | F ∈ S, F ′ ∈ S′.

(iii) If SjS∈J is a collection of subsets of homogeneous polynomials of k[X0, . . . , Xn] then⋂j∈J V (Sj) = V (

⋃j∈J Sj).

Or, in terms of homogeneous ideals, prove:

(i’) V (0) = Pnk , and V (k[X0, . . . , Xn]) = ∅.

(ii’) For any homogeneous ideals I, I ′ ⊂ k[X0, . . . , Xn], then V (I) ∪ V (I ′) = V (II ′) =

V (I ∩ I ′).

(iii’) If IjI∈J is a collection of homogeneous ideals of k[X0, . . . , Xn] then⋂j∈J V (Ij) =

V (∑j∈J Ij).

In particular, we also have a Zariski topology in Pnk in which the close sets are now

the projective sets. We also have

Definition. We define the ideal of a projective set X as

I(X) = F ∈ k[X0, . . . , Xn] | F (p) = 0 for any p ∈ X.

If you do not like the convention of saying that F (p) = 0 if and only if Fi(p) = 0 for all

the homogeneous summands of F , you can define I(X) as the ideal generated by all the

homogeneous polynomials vanishing at all the points of X.

15

Now we have the analogue of Exercise 1.3:

Exercise 2.4. Prove the following results:

(i) I(∅) = k[X0, . . . , Xn] and, if k is infinite, I(Pnk ) = 0.(ii) If Xjj∈J is a collection of subsets of Pnk , then I(

⋃j∈J Xj) =

⋂j∈J I(Xj).

(iii) If X,X ′ ⊂ Pnk , then I(X ∩X ′) ⊃ I(X) + I(X ′).

Again the fact that in (iii) is not an equality is explained by the Nullstellensatz. To

prove it, it is enough to apply it to the affine cone in An+1k defined by the same equations.

There is a slight problem, namely that a homogeneous ideal can define the empty set

in two different ways: when the affine cone is the empty set or when it is the point

(0, . . . , 0) ∈ An+1k . Hence the result can be written in this way:

Theorem 2.5 (Projective Nullstellensatz). Let I ⊂ k[X0, . . . , Xn] be a homogeneous

ideal. Then

(i) V (I) = ∅ if and only if there is d0 such that I contains all the homogeneous polynomials

of degree d ≥ d0.

(ii) If V (I) 6= ∅, then IV (I) =√I.

(iii) In either case, if F is a non-constant homogeneous polynomial in I(V (I)), then there

exists a power Fm in I.

Remark 2.6. A first visible difference with the affine case is that the ideals of points are

not the maximal ideals. In fact, the only homogeneous maximal ideal is M = (X0, . . . , Xn),

called the irrelevant ideal, since it is defines the empty set (although we will see that it

is a very relevant ideal in all this theory). It is easy to see that any proper homogeneous

ideal is contained in M. Although not interesting for our purposes, the ideals of points

can be characterized as those homogeneous prime ideals for which there are not any other

homogeneous prime ideal between it and M. As in the affine case (Proposition 1.12), prime

ideals are in bijection with irreducible projective sets. It is useful to notice that, in the case

of homogeneous ideals (in any graded ring), properties like primality or primarity can be

characterized using only homogeneous elements. For example, a homogeneous ideal I ⊂ Sis prime if and only if for each pair of homogeneous elements F,G ∈ S such that FG ∈ Inecessarily F ∈ I or G ∈ I (the reader is invited to verify this and the corresponding

property for primary ideals). It also holds that the primary decomposition (Theorem 1.17)

of a homogeneous ideal consists of homogeneous primary ideals. For another example, see

the proof of Lemma 3.24.

Remark 2.7. If we restrict a projective set to any affine subspace we get an affine

set, so that the topology induced by the projective Zariski topology is the affine Zariski

16

topology. Indeed, let Ui := D(Xi), and let p = (a0 : . . . : an) ∈ Ui. As a point in

Ank , we can write it as (a0ai , . . . ,ai−1

ai, ai+1

ai, . . . , anai ). If X ⊂ Pnk is a projective set, then p

is in X ∩ Ui if and only if, for all F ∈ I(X) we have F (a0, . . . , an) = 0, or, in terms

of the coordinates in Ank , we have F (a0ai , . . . ,ai−1

ai, 1, ai+1

ai, . . . , anai ) = 0. Hence, using

(X0

Xi, . . . , Xi−1

Xi, Xi+1

Xi, . . . , Xn

Xi) as the coordinates of Ank , X ∩ Ui is the affine set defined by

the polynomials F (X0

Xi, . . . , Xi−1

Xi, 1, Xi+1

Xi, . . . , Xn

Xi) (called the dehomogenized polynomial of

F with respect to Xi), when F varies in I(X). Observe that, if F ∈ k[X0, . . . , Xn] is a

homogeneous polynomial of degree d, its dehomogenized with respect to Xi can be written

as FXd

i

.

Definition. Given a homogeneous ideal I ⊂ k[X0, . . . , Xn], its dehomogenized ideal with

respect to Xi is the ideal of k[X0

Xi, . . . , Xi−1

Xi, Xi+1

Xi, . . . , Xn

Xi] formed by all the dehomogenized

polynomials with respect to Xi of I.

It is not true, however, that different homogeneous ideals have always different deho-

mogenized ideals, even when considering all possible coordinates:

Proposition 2.8. Let I, I ′ ⊂ k[X0, . . . , Xn] be two homogeneous ideals. Then the follow-

ing are equivalent:

(i) For each i = 0, . . . , n the dehomogenization of I and I ′ with respect to Xi is the same.

(ii) For each F ∈ I there exist m ∈ N such that Xmi F ∈ I ′ for i = 0, . . . , n; and for each

F ′ ∈ I ′ there exist m′ ∈ N such that Xm′

i F ′ ∈ I for i = 0, . . . , n.

(iii) There is d0 ∈ N such that Id = I ′d for d ≥ d0.

Proof: We will prove the implications cyclically.

(i)⇒ (ii): By symmetry, it is enough to prove only the result for polynomials of I. Since

I is generated by homogeneous polynomial, it is also enough to prove it for homogeneous

polynomials. For each homogeneous F ∈ I, its dehomogenized with respect to Xi isF

Xdeg(F )i

. Since I and I ′ have the same dehomogenization, it follows that F

Xdeg(F )i

=F ′i

Xdeg(F ′

i)

i

for some homogeneous polynomial F ′i ∈ I ′. Taking m := maxdeg(F ′0), . . . ,deg(F ′n), the

result follows.

(ii) ⇒ (iii): Let F1, . . . , Fr be a system of homogeneous generators of I. By part (ii),

for each Fj and each i = 0, . . . , n there is some mj such that Xmj

i Fj is in I ′. Taking

d′j := (n+ 1)(mj − 1) + 1, it follows that, for any homogeneous polynomial G of degree at

least d′j , any monomial of GFj is divisible for some Xmj

i Fj , so that GFj is in I ′. Therefore

any homogeneous polynomial of I of degree at least maxd′0 + deg(F0), . . . , d′n + deg(Fn)is in I ′. Symmetrically, any homogeneous polynomial of I ′ of degree at least some precise

number is in I. Taking d0 to be the maximum of the two numbers we found, we get the

result.

17

(iii) ⇒ (i): A polynomial in the dehomogenization of I with respect to Xi is of the formF

Xdeg Fi

, with F ∈ I. Multiplying if necessary numerator and denominator with a power

of Xi, we can assume deg(F ) ≥ d0, hence F ∈ I ′, which implies that F

Xdeg Fi

is also in

the dehomogenization of I ′ with respect to Xi. Symmetrically, any polynomial in the

dehomogenization of I ′ with respect to Xi is also in the dehomogenization of I.

Definition. The saturation of a homogeneous ideal I ⊂ k[X0, . . . , Xn] is the homogeneous

ideal

sat(I) := F ∈ k[X0, . . . , Xn] | there exists m ∈ N such that Xmi F ∈ I for i = 0, . . . , n.

A homogeneous ideal is saturated if it coincides with its saturation.

Remark 2.9. It is clear that the saturation of an ideal is saturated. Proposition 2.8 is

saying that two homogeneous ideals have the same dehomogenized ideals if and only if they

have the same saturation. The saturation of an ideal is thus the maximum homogeneous

ideal having the same dehomogenizations than the original ideal. Hence the notion of

projective scheme should be equivalent to the notion of saturated ideal. In this direction,

Theorem 2.5(i) is saying that a homogeneous ideal defines the empty set if and only if its

saturation is the whole polynomial ring.

Exercise 2.10. Prove that homogeneous radical ideals are saturated.

The saturation of a homogeneous ideal is related to its primary decomposition:

Proposition 2.11. Let I ⊂ k[X0, . . . , Xn] be a proper homogeneous. Then the saturation

of I is the intersection of all the primary components of I whose radical is not M.

Proof: Write I ′ for the intersection of all the primary components of I whose radical is

not M. We will prove the equality I ′ = sat(I) by double inclusion,

Start taking F ∈ I ′ and let us prove that it is in the saturation of I. If there is no

primary component whose radical is M, then I ′ = I ⊂ sat(I) and there is nothing to

prove. If instead there is a primary component I0 with√I0 = (X0, . . . , Xn), there exists

m ∈ N such that Xmi ∈ I0 for i = 0, . . . , n. Hence each Xm

i F is in I0 (because Xmi is) and

is in I ′ (because F is). Therefore Xmi F ∈ I0 ∩ I ′ = I, i.e. F is in the saturation of I.

Reciprocally, assume F ∈ sat(I), i.e. there exists m ∈ N such that Xmi F ∈ I for

i = 0, . . . , n. We need to prove F ∈ I ′. To prove that, it is enough to prove that F is in

any primary component J of I ′. By definition, J is a component of I whose radical ideal is

not M. Therefore, there is i ∈ 0, . . . , n such that Xi /∈√J , which also implies Xm

i /∈√J .

18

But the assumption Xmi F ∈ I implies Xm

i F ∈ J , and the fact that J is primary together

with Xmi /∈

√J imply now F ∈ J , as wanted.

Ideals that are not saturated have a bad behavior:

Proposition 2.12. Let I ⊂ k[X0, . . . , Xn] be a proper homogeneous ideal, and let I =

I1 ∩ . . . ∩ Ir be a minimal primary decomposition. Then the map k[X0, . . . , Xn]/I →k[X0, . . . , Xn]/I given by the multiplication by the class of F , is injective if and only

F is not in the radical of any Ii. In particular, there exists for each d > 0 a homo-

geneous polynomial F ∈ k[X0, . . . , Xn] of degree d such that the multiplication map

·F : k[X0, . . . , Xn]/I → k[X0, . . . , Xn]/I is injective if and only if I is saturated.

Proof: Assume first that F is not in the radical of any Ii. Assume that we have G ∈k[X0, . . . , Xn] such that FG ∈ I. Then FG ∈ Ii for each i = 1, . . . , r. Since Ii is primary

and F /∈√Ii, it follows that G ∈ Ii for all i, hence G ∈ I and its class is zero. This proves

the injectivity of the multiplication map.

Reciprocally, assume that F is in the radical of some ideal Ii. Since the primary

decomposition is minimal, we can find G ∈ Ij for all j 6= i such that G /∈ Ii. On the other

hand, by assumption some power of F is in Ii, which implies that some F sG is in I. We

can take s minimum with such property, and obviously s ≥ 1. Hence F s−1G is not in I but

its product by F is, which shows that the multiplication by the class of F is not injective.

For the last part, if I is not saturated, then by Proposition 2.11, it has a primary

component whose radical is M, and hence any homogeneous polynomial of any positive

degree is in the radical of that component. As we have just proved, this implies that no

multiplication map can be injective.

If instead I is saturated,√Ii 6= M for all i = 1, . . . , r. Hence the Nullstellensatz

implies that V (Ii) is not empty. Picking a point pi ∈ V (Ii) for each i, we take F ∈k[X0, . . . , Xn]d such that V (F ) does not contain any of p1, . . . , pr (for example, the d-th

power of the equation of a hyperplane not passing through any of them). Hence F /∈√

(Ii)

for i = 1, . . . , r, and the multiplication by the class of F , is injective.

Contrary to the affine case, the ring of functions cannot characterize projective sets

up to isomorphism. In fact, we will see that, when the correct definition of regular function

–which is not immediate how to define– the only regular functions on irreducible projective

sets are the constant functions (you can think of this theorem as an analogue of Liouville’s

theorem). We can try to study the analogue of the ring that, in the affine case, was

naturally isomorphic to the ring of polynomial functions.

Definition. Given a projective set X ⊂ Pnk , the quotient S(X) := k[X0, . . . , Xn]/I(X) is

called the graded coordinate ring of X.

19

Although a homogeneous element F ∈ S(X) cannot be regarded as a function, it is

still true that one can say whether it vanishes or not at a point of X, since this does

not depend on the representative F ∈ k[X0, . . . , Xn]. We can thus define, for any subset

Σ ⊂ S(X) of homogeneous elements

VX(Σ) := p ∈ Pnk | F (p) = 0 for all F ∈ Σ

which is nothing but the projective set obtained as the intersection of X with the set

defined by representatives of Σ. We can also define the ideal IX(Z) ⊂ S(X) for any subset

Z ⊂ X, but it is not important to our purposes. Sets of the form VX(Σ) are the closed

sets in the Zarisky topology of Pnk restricted to X, and a basis is given by the open sets of

the form

DX(F ) = X \ VX(F )

where F is a homogeneous polynomial. As in the affine case (Lemma 1.25), we have the

following interpretation of the Nullstellensatz restricted to a projective set:

Lemma 2.13. For any projective set X ⊂ Pnk and homogeneous polynomials F (non-

constant) and Fi in k[X0, . . . , Xn], the following are equivalent:

(i) DX(F ) ⊂⋃iDX(Fi)

(ii) VX(Fi) ⊂ VX(F )

(iii) There exist a power F s in the ideal generated by the classes Fi.

Again contrary to the case of affine sets, S(X) will not characterizeX up to polynomial

maps or any other kind of reasonable maps:

Example 2.14. Consider the map ϕ : P1k → P2

k given by (t0 : t1) 7→ (t20 : t0t1 : t21). The

inverse over the image X = V (X0X2 −X21 ) can be defined by (X0 : X1 : X2) 7→ (X0 : X1)

when t0 6= 0 (i.e. X0 6= 0 on X), while for t1 6= 0 (i.e. X2 6= 0) the definition could

be (X0 : X1 : X2) 7→ (X1 : X2). Both definitions coincide in the intersection of the

two open sets D(X0) ∩ D(X2) ∩ X, just because the equation of X says precisely that

(X0 : X1) = (X1 : X2). And on the other hand, the two open sets cover the whole

X. As a consequence, we do not have a global definition for the projection, but it is

possible to cover X by open sets in such a way that the map has a definition on those

open sets by homogeneous polynomials. Anyway, it seems reasonable to say that the map

is an isomorphism between P1k and V (X0X2 −X2

1 ) (this is a first sign that the definition

of morphism will be a little bit subtle). However, the map ϕ induces a homomorphism

S(X) = k[X0, X1]/(X0X2 −X21 )→ k[T0, T1] = S(P1

k) that is not an isomorphism.

The main point is that S(X) does not keep track only of the isomorphism class of X,

but also about its concrete embedding in a projective space. The key tool to study S(X)

20

will be to study the dimension of each S(X)d (as a quotient of the space of homogeneous

polynomials of degree d, it is a finite-dimensional vector space over k), which will behave

nicely for high values of d:

Definition. The Hilbert function of a homogeneous ideal I ⊂ k[X0, . . . , Xn] is the function

hI : N→ N defined by hI(d) = dim k[X0, . . . , Xn]d/Id. The Hilbert function of a projective

set X ⊂ Pnk is the Hilbert function of I(X), i.e. hX(d) = dim(S(X)d

).

Observe that Proposition 2.8 says that, for big values of d, the value of hI(d) depends

only on the saturation of the ideal I. On the other hand, Theorem 2.5(i) characterizes

homogeneous ideals I ⊂ k[X0, . . . , Xn] as those for which hI is zero for big values of d. We

also have the following:

Proposition 2.15. Let X be a set of m points in Pnk . Then hX(d) = m if d ≥ m − 1.

Reciprocally, if I ⊂ k[X0, . . . , Xn] is a homogeneous ideal such that hI constant if d >> 0,

then V (I) is a finite set of points.

Proof: If X = p1, . . . , pm, we choose vectors v1, . . . , vm representing them, and define, for

each d ∈ N the evaluation map ϕd : k[X0, . . . , Xn]d → km associating to each homogeneous

polynomial F of degree d the m-uple(F (v1), . . . , F (vm)

). Then since clearly I(X)d is the

kernel of ϕd, we have that S(X)d ∼=Imϕd. Hence our first statement is equivalent to prove

that ϕd is surjective if d ≥ m − 1. It is clear that, for each i = 1, . . . ,m and j 6= i we

can find a linear form Hi ∈ k[X0, . . . , Xn] vanishing on pi but not on any other pj . Then

the product Fi = Πj 6=iHj is a homogeneous form of degree m − 1 vanishing at all the

points of X except pi. Fixing a homogeneous form G of degree d−m+ 1 not vanishing at

any of the points p1, . . . , pm, we get that the images by ϕl of the elements GF1, . . . , GFm

generate km. This proves the surjectivity of ϕd for d ≥ m − 1 and hence the first part of

the proposition.

Reciprocally, assume that dim k[X0, . . . , Xn]d/Id takes a constant value c for d >> 0.

If V (I) were not finite, we could find Z ⊂ V (I) consisting of c+ 1 points. Since I ⊂ I(Z),

there is a surjection k[X0, . . . , Xn]/I → k[X0, . . . , Xn]/I(Z) that restricts, for any d. to a

surjection k[X0, . . . , Xn]d/Id → S(Z)d. But we just proved that, for large d the dimension

of S(Z)d is c + 1 while by assumption the dimension of k[X0, . . . , Xn]d/Id is c. This

contradiction implies that V (I) is a finite number of points (and in fact at most c).

Example 2.16. In the second part of Proposition 2.15, the number of points of V (I)

is not necessarily the constant value of hI , since points could count with multiplicity.

Consider, for example, the homogenized case of Example 1.4, i.e. the ideal I = (X21 , X2) ⊂

k[X0, X1, X2]. It is clear that, for d ≥ 1, the vector space k[X0, X1, X2]d/Id is freely

generated by the classes of Xd0 and Xd−1

0 X1, hence its dimension is two. This coincides

21

with the geometric interpretation that I consists of two infinitely close points (in the

direction of the line V (X2)).

The situation of Proposition 2.15 is just a particular case of a more general result:

Theorem 2.17. For any homogeneous ideal I ⊂ k[X0, . . . , Xn], there exists a polynomial

PI ∈ Q[T ] such that, for d >> 0, hI(d) = PI(d).

Proof: If V (I) = ∅, then Theorem 2.5(i) implies the result, and PI = 0. The idea is that we

can find linearly independent linear forms H1, . . . ,Hr such that V(I + (H1, . . . ,Hr)

)= ∅

(for sure we can take as linear forms X0, . . . , Xn, but we need to chose them in a more

suitable way). First of all, it could happen that I is not saturated, but by Proposition 2.8

it is clear that we can replace I with its saturation (because if PI exists it must be the same

as Psat(I)). We now apply Proposition 2.12, and we can find a linear form H1 such that

the multiplication by its class gives an injective map k[X0, . . . , Xn]/I → k[X0, . . . , Xn]/I.

It is then easy to see that we have an exact sequence

0→ k[X0, . . . , Xn]/I → k[X0, . . . , Xn]/I → k[X0, . . . , Xn]/(I + (H1))→ 0

and, looking at homogeneous, it restricts to

0→ k[X0, . . . , Xn]d−1/Id−1 → k[X0, . . . , Xn]d/Id → k[X0, . . . , Xn]d/(I + (H1))d → 0

which implies hI(d) − hI(d − 1) = hI+(H1)(d). Now the idea is that, if hI is given by a

polynomial, necessarily hI+(H1) will be given also by a polynomial, of one less degree. We

know that, after a final number of steps, we will get a finite number of points (hence a

polynomial of degree zero) or even the empty set (hence the zero polynomial). We will

try to reverse this process, i.e., if we know that after a final number of steps we arrive to

the zero polynomial, then hI is also given by a polynomial. Observe that, in the process,

I + (H1) could be non-saturated, but again we can replace it by its saturation. We will

need to prove the following:

Claim. Let f : N → Z be a function in integres, and define ∆f : N → Z as (∆f)(m) =

f(m + 1) − f(m). If (∆r+1f)(m) = 0 for m >> 0, then there exist c0, c1 . . . , cr ∈ Z such

that f(m) = c0(m0

)+ c1

(m1

)+ . . .+ cr

(mr

)if m >> 0.

Proof of the claim. This is a sort of calculus for integer numbers, in which ∆ is the

derivative and we need to integrate. We will prove it by induction on r. The case r = 0

is immediate, because our assumption (∆f)(m) = 0 means f(m) = f(m + 1), for all

m >> 0, hence f eventually becomes a constant c0 ∈ Z. Assume then r > 0 and that

we now the result for any function f for which ∆rt is zero for large values of m. Our

22

assumption (∆r+1f)(m) = 0 can be expressed as((∆r(∆f)

)(m) = 0 for m >> 0. Hence,

our induction hypothesis implies the existence of c1, c2 . . . , cr ∈ Z such that

(∆f)(m) = c1

(m

0

)+ c2

(m

1

)+ . . .+ cr

(m

r − 1

)for m >> 0. But now, the binomial identity

(m+1i

)−(mi

)=(mi−1

)implies that also for

g : N→ Z defined by

g(m) = c1

(m

1

)+ c2

(m

2

)+ . . .+ cr

(m

r

)we also have

(∆g)(m) = c1

(m

0

)+ c2

(m

1

)+ . . .+ cr

(m

r − 1

).

This means (∆(f − g))(m) = 0 for m >> 0, so that the case r = 0 we already proved

implies the existence of c0 ∈ Z such that f(m)− g(m) = c0 if m >> 0. We thus have

f(m) = c0 + g(m) = c0

(m

0

)+ c1

(m

1

)+ . . .+ cr

(m

r

)if m >> 0, which completes the proof of the claim.

The proof of the Theorem comes now easily from the claim. Indeed we are taking

ideals I1 := sat(I + (H1)

), I2 := sat

(I1 + (H2)

)... and repeating the process as long as

V (Ij) 6= ∅, choosing Hj+1 a linear form that is not in any of the radicals of the primary

decomposition of Ij . Since V (Ij) = V (I) ∩ V (H1, . . . ,Hj), the linear form Hj+1 cannot

be a linear combination of H1, . . . ,Hj . As the intersection of n+ 1 hyperplanes defined by

linearly independent linear forms is the empty set, we will necessarily find r ≤ n+ 1 such

that V (Ir) 6= ∅ and V (Ir+1) = ∅. Since we also have hIj (d) − hIj (d − 1) = hIj+1(d), for

d >> 0, it follows (∆r+1hI)(d) = 0 for d >> 0, and the result comes now from the claim.

Definition. The Hilbert polynomial of a homogeneous ideal I ⊂ k[X0, . . . , Xn] is the poly-

nomial PI that coincides with the Hilbert function for large values of d. Similarly, the

Hilbert polynomial of a projective set X ⊂ Pnk is the Hilbert polynomial of its ideal I(X).

Exercise 2.18. Prove that the set X = (t40 : t30t1 : t0t31 : t41) ∈ P3

k | (t0 : t1) ∈ P1k is a

projective set and I(X) = (X0X3−X1X2, X31 −X2

0X2, X32 −X1X

23 , X

21X3−X0X

22 ). Show

that bad situations like in the proof of Theorem 2.17 can occur, namely I(X) + (X1−X2)

is not saturated. More precisely, prove that, for any λ 6= 0, 1,−1, a primary decomposition

23

is: I(X) + (X1 − X2) = (X0, X1, X2) ∩ (X1, X2, X3) ∩ (X0 − X1, X1 − X2, X2 − X3) ∩(X0 +X1, X1 −X2, X2 +X3) ∩ (X1 −X2, X3 − λX2, X2 − λX0, X

30 ).

Remark 2.19. The proof of Theorem 2.17 gives a very geometrical meaning to the

degree r of the Hilbert polynomial: we have seen that there exists a linear subspace of

codimension r + 1 not meeting V (I). On the other hand, any time we take a linear form

H (and in fact any homogeneous polynomial of positive degree) we always have an exact

sequence

k[X0, . . . , Xn]/I → k[X0, . . . , Xn]/I → k[X0, . . . , Xn]/(I + (H))→ 0

which implies hI+(H)(d) ≥ hI(d)−hI(d− 1), so that the intersection of V (I) with a linear

subspace of codimension r (or, more generally, with r hypersurfaces) has a nonzero Hilbert

polynomial, hence it is not empty. Observe that this geometrical interpretation of r does

not depend on the ideal I, but only on V (I) (since it can be interpreted as the maximum

codimension r such that any linear subspace of codimension r intersects V (I). When

the linear space of codimension r is well chosen (according to Proposition 2.12), we get

as Hilbert polynomial of the intersection the constant cr of the above claim, which is r!

times the leading coefficient of PI . As Proposition 2.15 and Example 2.16 suggest, this crcounts the number of points counted with multiplicity appearing in the intersection with

the linear space.

We thus have the following natural:

Definition. The dimension of a projective set X ⊂ Pnk is the degree r of the Hilbert

polynomial of any homogeneous ideal defining X. The degree of the projective set X is r!

times the leading coefficient of the Hilbert polynomial PX .

We state here the main properties of the dimension of projective sets.

Proposition 2.20. Let X ⊂ Pnk be a projective set of dimension r and degree d. Then:

(i) Any projective subset of X has dimension at most r.

(ii) If V (F ) does not contain any irreducible component of X, then dim(X∩V (F )) = r−1.

(iii) If X is irreducible, then any proper projective subset of X has dimension strictly

smaller than r.

(iv) The dimension of X is the maximum of the dimension of the irreducible components

of X.

Proof: For part (i), observe that, if Y ⊂ X is a projective subset, then there is a surjection

S(X) → S(Y ) preserving degrees. In particular, hY (l) ≤ hX(l) for any natural number

l. Since, for large values of l, such dimensions are given by polynomials PY and PX of

24

positive leading coefficient, necessarily the degree of PY is at most the degree of PX , i.e.

the dimension of Y is at most the dimension of X,

In the hypothesis of (ii), Proposition 2.12 implies that the multiplication by the class

of F is a monomorphism S(X) → S(X). If e = deg(F ), there is thus an exact sequence

for each l ≥ e:

0→ S(X)l−e → S(X)l → k[X0, . . . , Xn]l/(I(X) + (F ))l → 0

which shows that hI(X)+(F )(l) = hX(l)−hX(l−e). Since by hypothesis PX(l) = dr! lr+ . . .,

a simple computation shows PX(l)−PX(l− e) = de(r−1)! l

r−1 + . . ., hence V (I(X) + (F )) =

X ∩ V (F ) has dimension r − 1, which proves (ii).

To prove (iii), assume Y ⊆/ X. Then I(X) ⊆/ I(Y ), so that we can find a homogeneous

polynomial F ∈ I(Y ) \ I(X). Since X is irreducible, part (ii) implies that X ∩ V (F ) has

dimension r − 1. On the other hand, Y ⊂ X ∪ V (F ), so that part (i) implies that the

dimension of Y is at most r − 1, proving (iii).

For part (iv), let X = X1 ∪ . . . ∪ Xs be the decomposition of X into its irreducible

components. Then I(X) = I(X1) ∩ . . . ∩ I(Xs), which yields a natural injection S(X) ⊂S(X1) ⊕ . . . ⊕ S(Xs) preserving degrees. Hence hX(l) ≤ hX1

(l) + . . . + hXs(l). By (i) we

know that the degree of any PXi is at most r, and we deduce that above inequality shows

that at least one of the degrees is r, which proves (iv).

Exercise 2.21. Prove that degree of a projective set is the sum of the degrees of its

irreducible components of maximum dimension [Hint: use induction on the number of

components of maximum dimension and the natural exact sequence 0 → S/(I1 ∩ I2) →S/I1 ⊕ SI2 → SI1+I2 → 0 for any pair I1, I2 of ideals of a ring S].

Remark 2.22. Observe that, the prove of part (ii) of Proposition 2.20 shows that the

intersection of a projective set of dimension r and degree d with a nice hypersurface of

degree e is expected to be of dimension r−1 and degree de (this will happen if the saturation

of I(X) + (F ) to be the ideal of X ∩ V (F ); see Exercise 2.18). Iterating the process, the

intersection of X with r nice hypersurfaces of degrees e1, . . . , er must consist of de1 . . . erpoints counted with multiplicity (which is a generalization of Bezout’s Theorem for the

intersection of two plane curves).

Exercise 2.23. Prove that the map νm : Pnk → PNk , where N + 1 =(n+1m

), (called the

m-uple Veronese embedding of Pnk , i.e. the ) defined by

νm(a0 : . . . : an) = (am0 : am−10 a1 : . . . : am1 : am−1

1 a2 : . . . : amn )

25

(i.e. the coordinates in the image are all the monomials of degree m) is injective and its

image is a projective set (called Veronese variety). Compute its Hilbert polynomial and

conclude that it has dimension n and degree mn [This degree has a natural explanation:

the intersection of the Veronese variety with n general hyperplanes corresponds in Pnk to

the intersection of n general hypersurfaces of degree m, and Remark 2.22 implies that one

should get mn points. Explain also the degree identifying PNk with the set of hypersurfaces

of degree m in Pnk and regarding the Veronese variety as the set of m-fold hyperplanes].

Exercise 2.24. Prove that the map ϕn,m : Pnk × Pmk → Pnm+n+mk (called the Segre

embedding) defined by

ϕn,m((a0 : . . . : an), (b0 : . . . : bm)

)= (a0b0 : a0b1 : . . . : a0bm : a1b0 : . . . anbm)

(i.e. the coordinates in the image are all the products of a variable ai and a variable bj)

is injective and its image is a projective set (called Segre variety). Compute its Hilbert

polynomial and conclude that it has dimension n+m and degree(n+mn

).

Remark 2.25. As remarked in Example 2.14, the graded ring S(X) does not characterize

X up to isomorphism. This is due to the fact that S(X) encodes the information of X

as embedded in a precise projective space. For example, the same X can be embedded

into different projective spaces with different degrees (think of P1k embedded into any Pdk

as a rational normal curve of degree d). However, there is some intrinsic information of X

inside S(X). The main invariant is the dimension, and we will see (Proposition 6.1) that

the dimension of a projective set is obtained from its topology, hence it is invariant under

homeomorphisms.

26

3. Regular functions

We would like to define morphisms (also called regular maps) among affine or alge-

braic sets. However, this is a priori not immediate. We have seen in Example 2.14 that,

in the projective case, we expect morphisms that are not globally defined by polynomials.

On should immediately discard also to define morphisms just as continuous maps with

the Zariski topology. However, this is not a good idea, since topology will not be enough.

Indeed, observe that the Zariski topology on A1k is the topology of finite complements.

Therefore, any possible bijection from A1k to itself will be continuous, and it is not reason-

able to allow any bijection to be a morphism, because we want morphisms to be related

in some sense to polynomials.

We will imitate thus a standard technique of differential geometry. It is not the

technique of considering open sets isomorphic to an open set of an affine space, because

our open sets are so big that only few types of varieties possess that kind of open sets.

What we will do will be to define first the notion of regular function, and thus to define

morphisms as those maps preserving regular functions. The scope of this section will be

to find the right notion of regular function, and use that notion to naturally extend that

to other spaces that we will define inspired in the behavior of affine and projective sets.

We start trying to find out what a regular function on an affine set should be. The first

possibility is to define that a function on an affine set X ⊂ Ank is nothing but a polynomial

function, i.e. an element of A(X). However, this definition is not extendable to projective

sets, since elements in S(X) do not define any function, except for the constants. Even if

we will see later one that regular functions on an irreducible projective sets are constants,

it is clear that a regular function on a finite number of d points must be given by fixing a

constant for each point, hence it is not only a constant.

Another possibility is to try to define functions locally, as for example –following the

analogy with differential geometry– differentiable functions are those that are differen-

tiable at a point. As Exercise 1.24 indicates, polynomial functions on D(f) can have as

denominators powers of f . This suggests the following:

Definition. For any open subset U of an affine set X ⊂ Ank , we say that a map ϕ : U → k

is regular at a point x ∈ U if there exist polynomial funxtions f , g ∈ A(X)] and an open

neighborhood V ⊂ U ∩DX(f) of x such that ϕ restricted to V can be defined by fg . We

will write O(U) for the set of functions U → k that are regular at any point of U .

This can be extended immediately to projective sets. In fact, we recall that, when

restricting to a basic open set D(Xi), we deal with polynomials in k[X0

Xi, . . . , Xn

Xi], i.e quo-

tients of a homogeneous polynomial of some degree d over Xdi . Hence, the quotient of two

polynomial in the above definition becomes the quotient of two homogeneous polynomials

27

of the same degree. We thus have the following:

Definition. For any open subset U of a projective set X ⊂ Pnk , we say that a map

ϕ : U → k is regular at a point x ∈ U if there exist homogeneous elements F , G ∈ S(X) of

the same degree, and an open neighborhood V ⊂ U ∩DX(G) of x such that ϕ restricted

to V can be defined by FG

. We will write O(U) for the set of functions U → k that are

regular at any point of U .

Since we are using it very often, we give first the following:

Notation. Given a graded ring S, we will use brackets for the subrings of rings of fractions

consisting of quotients of homogeneous elements of the same degree. For example, if p is

a homogeneous prime ideal, we will write S(p) for the corresponding subring of Sp, and if

F is a homogeneous element of F of degree d, S(F ) will be the subring of SF consisting of

quotients GFm , where G is a homogeneous element of degree md.

The above definitions can be simplified:

Lemma 3.1. If U ⊂ X is an open set of an affine set X ⊂ Ank , the map ϕ : U → k is

regular at a point x ∈ U if and only if there exist polynomial functions f , g ∈ A(X) such

that x ∈ DX(g) ⊂ U and ϕ restricted to DX(g) can be defined by fg . Moreover, f , g can

be taken such that VX(g) ⊂ VX(f).

Proof: It is clear that in the definition we can take V = DX(h) for some h ∈ A(X). The

condition DX(h) ⊂ DX(g) is equivalent, by Lemma 1.25, to hm ∈ (g) for some m. We can

thus find p ∈ A(X) such that hm = pg. In particular, p has no zeros on DX(h). Therefore,

we can represent ϕ in DX(h) (which is the same as DX(hm)) by pfhm , and this is the wanted

form we were looking for. For the last part of the statement, we can take as representativepf hhm+1 .

Similarly, we have:

Lemma 3.2. If U ⊂ X is an open set of a projective set X ⊂ Pnk , the map ϕ : U → k is

regular at a point x ∈ U if and only if there exist homogenous polynomials of the same

degree F,G ∈ k[X0, . . . , Xn] such that x ∈ DX(G) ⊂ U and ϕ restricted to DX(G) can

be represented by FG . Moreover, F,G can be taken such that VX(G) ⊂ VX(F ) and having

degree multiple of a given d > 0.

Proof: Repeat the proof of Lemma 3.2, now using Lemma 2.13. For the part of the degree,

it is enough to replace at the end the quotient FG

by F Gd−1

Gd .

For basic open sets we can compute the set of regular functions:

28

Theorem 3.3. If X ⊂ Ank is an affine set and f ∈ A(X) is not the zero function, then

there is a natural isomorphism of k-algebras A(X)f → O(DX(f)

).

Proof: It is clear that the map is a homomorphism of k-algebras. Its injectivity (and the

fact that the definition does not depends from the representative) is easy to prove: if a

quotient gfm defines the zero map on DX(f), this is equivalent to saying DX(f) ⊂ VX(g),

i.e. the polynomial function f g is zero, which is in turn equivalent to gfm = 0 in A(X)f

(see Example A.10(ii)).

For the surjectivity, assume we have ϕ ∈ O(DX(f)

), and we have to check that it is

a polynomial function divided by a power of f . By Lemma 3.1, we can cover DX(f) with

open sets DX(gi) (a finite collection, by Lemma 1.23, although this is not strictly needed)

such that in each DX(gi) the map ϕ is defined by figi

, and we can assume that fi is zero

outside DX(gi). This means that in the whole DX(f) we have an identity giϕ = fi. On

the other hand, the inclusion DX(f) ⊂⋃iDX(gi) is equivalent, by Lemma 1.25, to the

existence of a relation fm = h1gi1 + . . .+ hr gir . Multiplying this by ϕ, we have

fmϕ = h1gi1ϕ+ . . .+ hrgirϕ = h1fi1 + . . .+ hrfir

which proves that fmϕ is polynomial, so that ϕ is in the image.

In the projective case, the result is as follows:

Theorem 3.4. If X ⊂ Pnk is a projective set and F ∈ S(X) is a homogeneous element of

degree d > 0, then there is a natural isomorphism of k-algebras S(X)(F ) → O(DX(F )).

Proof: We just repeat the proof of Theorem 3.3. The difference here is that, using Lemma

3.2, a regular function ϕ : DX(F ) → k is covered by open sets DX(Gi) on which it

is defined by Fi

Gi, and we can assume deg Fi = deg Gi = did for some di. Hence, the

condition VX(Gi) ⊂ VX(Fi) allows to write the equality of functions Gi

Fdiϕ = Fi

Fdion

DX(F ). Since, from Lemma 2.13, we can write Fm = H1G1 + . . .+ HrGr (we can clearly

take m ≥ d1, . . . , dr) we have

ϕ =(H1G1 + . . .+ HrGr)ϕ

Fm=H1F

m−d1 + . . .+ HrFm−dr

Fm.

Remark 3.5. In Theorem 3.3 we can take f = 1, so that we get that the set of regular

functions of an affine set coincides with the set of polynomial functions. We cannot do the

same in the projective case, since in Theorem 3.4 we need F to have positive degree in

29

order to apply Lemma 2.13. In fact, it will be true that regular functions on irreducible

projective sets are constant. We give a first hint in the next example.

Example 3.6. Let us see how to apply Theorem 3.4 to compute regular functions

in the projective case. For example, regular functions on D(Xi) ⊂ Pnk are elements of

k[X0, . . . , Xn](Xi) (which is nothing but an element in k[X0

Xi, . . . , Xn

Xi]). Now assume that we

have a regular function ϕ : D(X0)∪D(X1)→ k. Hence ϕ|D(X0) can be written as F0

Xd00

for

some homogenous polynomial F0 of degree d0, while ϕ|D(X1) can be written as F1

Xd11

for some

homogenous polynomial F1 of degree d1. When restricting to D(X0)∩D(X1) = D(X0X1),

ϕ|D(X0X1) must be, by Theorem 3.4, an element of k[X0, . . . , Xn](X0X1). In fact, it can

be written either asF0X

d01

(X0X1)d0or as

F1Xd10

(X0X1)d1. This means that those two elements of

k[X0, . . . , Xn](X0X1) are equal, which implies F0Xd11 = F1X

d00 . Then, necessarily F0 is

divisible by Xd00 and, since both have the same degree, the quotient is a constant c ∈ K,

which is also the quotient of F1 by Xd11 . Hence the function ϕ takes the constant value

c. This proves O(D(X0X1)) = k and in particular, also regular functions on Pnk are

constants. Notice that this proves that D(X0X1) (and in general the complement of any

codimension-two linear space) cannot be an affine set.

Remark 3.7. Having now the notion of regular function on an open set U , we can repeat

now the definition of the operators V and I. For example, given S ⊂ O(U), we can define

VU (S) = p ∈ U | f(p) = 0 for all f ∈ S

and, given Y ⊂ U , we define

IU (Y ) = f ∈ O(U) | f(p) = 0 for all p ∈ Y .

with the same properties we already know (in the case U = DX(f) for some f ∈ A(X),

this can be obtained directly from Theorem 3.3). In particular, we get that the sets of the

form VU (S) are the closed sets of the restriction of the Zariski topology, and a basis for

that topology is given by the elements of the form DU (f) = U \ VU (f).

We concentrate now in the affine case, and recall that we noticed that an affine set

X ⊂ Ank is in bijection with the set of maximal ideals of A(X), which now we have seen

that it is the ring of regular functions. From this point of view, DX(f) is in bijection with

the maximal ideals of A(X) not containing f , which in turn is in bijection with the set of

maximal ideals of A(X)f , and this ring is precisely, by Theorem 3.3, the ring of regular

functions on DX(f). This shows the importance of knowing the set of regular functions in

order to reconstruct the open set.

30

Remark 3.8. When thinking of our attempt of generalizing affine sets, by allowing not

only reduced algebras, but more generally quotients of k[X1, . . . , Xn] by arbitrary ideals

(see Example 1.22), we find some problem:

1) For example, when considering k[T ]/(T 2), even if this corresponds to only one point

(and in fact that ring possesses only one maximal ideal), a regular function should be an

element a + bT of k[T ]/(T 2), but this takes not only a value a ∈ k as the image of the

point, but there is also a second datum b (which we discussed that it is a kind of first

derivative). Hence, in order to know the whole regular map, it is not enough to know the

value it takes at the points.

2) A more striking problem comes out when we consider rings as k[[T ]]. In this case,

not only we miss an infinity of data when looking only at the value that a series takes (its

constant term), but we get another problem. We have only one maximal ideal, namely

(T ). Since T can be regarded as a regular function, we could consider the analogue of

D(T ). Of course there are not maximal ideals not containing T , so that we would expect

D(T ) to be empty. However, according to Theorem 3.3, the ring of regular functions on

D(T ) should be k[[T ]]T which is very big (in fact it is the quotient field of k[[T ]]). In

particular, it possesses a maximal ideal (the zero ideal), hence D(T ) should consist of one

point. However, that maximal ideal corresponds to the zero ideal of k[[T ]], which is not

a maximal ideal, but just a prime ideal. This gives the clue of using not only maximal

ideals, but also prime ideals, i.e. irreducible subsets, not only points. Maybe the reader

thinks that a better solution could be to exclude from our theory rings like k[[T ]]. I hope

to convince the reader throughout these notes that such rings (in general discrete valuation

rings) play a very important role.

In order to solve the first problem above, we can consider, instead of the value of a

regular function at a point, the restriction of the function to a sufficiently small neighbor-

hood (going to the idea of direct limit recalled in Examples A.7 and A.8). In order to solve

the second problem, we will perform this idea of sufficiently small neighborhoods not only

for points, but also for irreducible sets. This is the scope of the following result, valid for

affine or projective sets.

Proposition 3.9. Let Y ⊂ X be an irreducible subvariety. Consider the set of pairs

(U, f), where U ⊂ X is an open set intersecting Y and f is a regular function on U . Define

a relation ∼ of pairs by (U, f) ∼ (U ′, f ′) if and only if there exists a non-empty open subset

U ′′ ⊂ U ∩ U ′ meeting Y such that f|U ′′ = f ′|U ′′ . Then:

(i) The relation ∼ is an equivalence relation and the set OX,Y of equivalence classes is a

local ring whose maximal ideal is the set of functions vanishing at Y .

(ii) For any open set U ⊂ X meeting Y the natural restriction map OX,Y → OU,Y ∩U is

an isomorphism.

31

(iii) If X is an affine set, the map A(X)I(Y ) → OX,Y sending fg to the class of (DX(g), fg )

is an isomorphism.

(iv) If X is a projective set, the map S(X)(I(Y )) → OX,Y sending FG

to the class of

(DX(G), FG

) is an isomorphism.

(v) The ring OX,Y is an integral domain if and only if X has only one irreducible com-

ponent containg Y .

Proof: It is straightforward to check that ∼ is an equivalence relation (the main tool is

that the intersection of non-empty sets in Y is not empty because Y is irreducible). It

is also clear that the quotient set OX,Y has a natural ring structure that is well-defined.

Observe that the class of a pair (U, f) is a unit in OX,Y if and only if f is not zero on

Y ∩ U . Indeed in that case the class of (DU (f)), 1f ) is an inverse of the class of (U, f).

Hence any proper ideal is contained in the ideal of those classes vanishing on Y , which is

thus the only maximal ideal, i.e. OX,Y is a local ring, proving (i).

Part (ii) is also an easy consequence of the definition of OX,Y , because elements of

OX,Y are defined as regular functions on a sufficiently small open set meeting Y .

We prove now part (iii), starting with the injectivity (as in the proof of Theorem 3.3,

this will also prove that the map is well-defined; and once it is well-defined it is clearly a

homomorphism). Assume that the class of (DX(g), fg ) is zero. This is equivalent to say

that, when restricted to a smaller open set meeting Y , the function fg is zero. Choosing

a basic open set DX(h) meeting Y , what we are saying is that f is the zero function on

DX(h), which is equivalent to saying that f h is the zero function on X. Since h is not in

IX(Y ), this is equivalent to saying that fg is zero as en element of A(X)I(Y ), proving the

injectivity.

For the surjectivity, we can represent an element in OX,Y by a regular function on a

basic open set DX(g) meeting Y . By Theorem 3.3, such regular function is of the form fgs .

Since DX(gs) = DX(g) and gs 6∈ IX(Y ), then fgs is an element of A(X)I(Y ) whose image

is the given class.

The proof of part (iv) is completely analog to part (iii).

For part (v), assume first that OX,Y is an integral domain, and let X = X1 ∪ . . .∪Xr

be the decomposition of X into irreducible components. For each i = 1, . . . , r we take

fi ∈ I(Xi)\ I(Xj) for j 6= i. Hence the product of the classes of (X, f1), . . . , (X, fr) is zero

in OX,Y , which implies that, for example, the class of (X, f1) is zero. This means that f1

is zero when restricted to an open set U meeting Y . In particular, X = VX(f1) ∪ (X \ U)

and, since f1 is not it I(X2), . . . I(Xr), it follows that X2, . . . , Xr ⊂ X \ U . In particular,

none of them can contain Y .

Reciprocally, assume there is only an irreducible component X ′ of X containing Y ,

32

and suppose that we have the product of the classes of (U1, f1) and (U2, f2) is zero in

OX,Y . This means that the product of f1 and f2 is zero when restricted to an open set

U ⊂ U1 ∩ U2 meeting Y . In particular, X ′ ⊂ VX(f1) ∪ V(f2) ∪ (X \ U), which implies X ′

is contained in one VX(fi). Hence fi is zero when restricted to U ∩ (X \X ′′) (where X ′′

is the union of the components not containing Y ), which is an open set meeting Y . Hence

the class of (Ui, fi) is zero, as wanted.

Definition. When Y consists of a point p, the ring OX,p is called the ring of germs of

regular functions at the point p, and its points must be regarded as regular functions on

sufficiently small neighborhoods of p (thus the name germ). When X is irreducible and take

Y = X, the corresponding ring is a field (by Proposition 3.9(ii) we can assume that X is

affine, and thus Proposition 3.9(iii) shows that OX,X is the quotient field of A(X), because

IX(X) is the zero ideal), called the field of rational functions on X, and it is denoted by

K(X). A rational function must be regarded as a function defined on a sufficiently small

open set of Y .

Now we will try to organize all the information we have. We have seen that knowing

the set of regular functions on an affine set is essentially equivalent to know all the affine

set. However, for projective sets, this is not the case. But, since we can recover a projective

set from its different affine pieces, we could also be able to recover the whole projective set

if we know what its regular functions on different open sets are. This will be the main idea

of the following definition, which generalizes the idea of giving the ring of regular functions

at each open sets, and which will be the key to the definition of scheme. We will use now

rings, but one could change the category of rings with any other abelian category (groups,

k-algebras,...):

Definition. A presheaf of rings on a topological space X is a map F from the set of open

sets of X to the set of rings satisfying the following conditions:

(i) F(∅) = 0.

(ii) If V ⊂ U are two open sets of X, there is a homomorphism (which we will call

restriction map) ρUV : F(U)→ F(V ) (we will often write ρUV (s) = s|V ).

(iii) For any open set U ⊂ X, ρUU is the identity map.

(iv) If W ⊂ V ⊂ U , then ρUW = ρVW ρUV .

A sheaf of rings is a presheaf for which we also have:

(v) If an open set U is the union of open sets Ui (with i varying in an arbitrary set I),

and for each i ∈ I there is si ∈ F(Ui) such that si|Ui∩Uj= sj |Ui∩Uj

for all i, j ∈ I,

then there exists a unique s ∈ F(U) such that s|Ui= si for any i ∈ I.

33

The elements of F(U) are called sections of the (pre)sheaf F on U . The restriction of the

(pre)sheaf to an open subset U ⊂ X, is the (pre)sheaf F|U defined by F|U (U ′) = F(U ′) for

any open subset U ′ ⊂ U .

As in Proposition 3.9, we can also define germs of sections:

Definition. Given a (pre)sheaf F on a topological space X, the stalk of F at a point p ∈ Xis the ring Fp of equivalence classes of pairs (U, s), with U an open neighborhood of p and

s ∈ F(U), under the equivalence ∼ defined by (U, s) ∼ (U ′, s′) if and only if p has an open

neighborhood U ′′ ⊂ U ∩ U ′ of p such that s|U ′′ = s′|U ′′ .

Remark 3.10. The stalk of a presheaf or sheaf is not necessarily a local ring, as it

happened in Proposition 3.9. On the other hand, it is clear that, for any open set U ⊂ Xwe have a homomorphism ρU,p : F(U) → Fp, assigning to any section s ∈ F(U) its germ

sp, which is the class of (U, s). This is clearly compatible with restrictions, i.e. if V ⊂ U we

have ρU,p = ρV,p ρUV (or, in the language of restrictions, sp = (s|V )p, for any s ∈ F(U).

In other words, Fp is the direct limit of F(U) for all open neighborhoods of p.

Of course, we should think of sections as functions. Nevertheless, we are going to do

this even more explicit, giving an isomorphism between any sheaf and another sheaf whose

sections are actually functions. For this we need first to define (iso)morphisms between

sheaves:

Definition. A morphism of (pre)sheaves ϕ : F → F ′ is a collection of homomorphisms

ϕ(U) : F(U) → F ′(U) compatible with the respective restriction homomorphisms ρ, ρ′,

i.e. such that for any inclusion V ⊂ U the diagram

F(U)ϕ(U)−→ F ′(U)yρUV

yρ′UV

F(V )ϕ(V )−→ F ′(V )

is commutative. An isomorphism of (pre)sheaves is a morphism ϕ such that each ϕ(U) is

an isomorphism (hence it is possible to define a morphism of (pre)sheaves ϕ−1).

Proposition 3.11. Let ϕ : F → F ′ a morphism of presheaves on X. Then, for any

p ∈ X, there is a natural homomorphism: ϕp : Fp → F ′p compatible with the restrictions.

Moreover, if F ,F ′ are sheaves, the following are equivalent:

(i) ϕ is an isomorphism.

(ii) If Uii∈I is a basis of the topology, the maps ϕ(Ui) : F(Ui) → F ′(Ui) are isomor-

phisms for each i ∈ I.

(iii) ϕp is an isomorphism for each p ∈ X.

34

Proof: We just define ϕp by sending the class of (U, s) to the class of(U, (ϕ(U))(s)

), and it is

a straightforward exercise to prove that the definition does not depend on the representative

we take and that it is compatible with restrictions. Let us prove the equivalences cyclically:

(i)⇒ (ii): It is obvious from the definition of isomorphism.

(ii) ⇒ (iii): Assume now part that each ϕ(Ui) : F(Ui) → F ′(Ui) is an isomorphism

and, for each p ∈ X, we will prove that ϕp is an isomorphism:

–We start with the injectivity of ϕp. If the class of (U, s) in Fp maps to zero, this

means that the class of(U, (ϕ(U)(s))

)is zero in F ′p, i.e there is a smaller neighborhood

V ⊂ U of p such that (ϕ(U)(s))|V = 0. Since Uii∈I is a basis of the topolgy, we can still

find a neirghborhood Ui ⊂ V of p, and we will have 0 = (ϕ(U)(s))|Ui= ϕ(Ui)(s|Ui

). Since

ϕ(Ui) is injective, we will have s|Ui= 0, hence the class of (U, s) is zero in Fp, as wanted.

–Similarly, for the surjectivity of ϕp, take the class of (U, s′) in F ′p. Taking a basic

neighborhood Ui ⊂ U of p, we have that the class of (U, s′) is the same as the class of

(Ui, s′|Ui

). Since ϕ(Ui) is surjective, s′|Uiis the image of some s ∈ F(Ui). Therefore, the

class of (U, s′) is the image by ϕp of the class of (Ui, s) in Fp, proving the surjectivity.

(iii) ⇒ (i): Assume that all ϕp are isomorphisms, and let us prove that ϕ is an

isomorphism:

–We will prove first the injectivity of any ϕ(U) : F(U)→ F ′(U). So we take s1, s2 ∈F(U) such that

(ϕ(U)

)(s1) =

(ϕ(U)

)(s2). In particular, for each p ∈ U , the images by

ϕp of the classes in Fp of (U, s1) and (U, S2) are equal. Since any ϕp is injective, the

classes in Fp of (U, s1) and (U, S2) are also equal. This means that there exists an open

neighborhood Up ⊂ U of p such that s1|Up= s2|Up

. But now, applying to s1, s2 the

uniqueness of condition (v) in the definition of sheaf to the open covering U =⋃p∈U Up,

it follows s1 = s2, which proves the injectivity of ϕ(U).

–We prove finally the surjectivity of ϕ(U) : F(U)→ F ′(U). We thus take s′ ∈ F ′(U).

We can now use the surjectivity of each ϕp to conclude that the class of (U, s′) in F ′p is the

image of some germ of section of F . This means that, for each p ∈ U , there exists an open

neighborhood Up ⊂ U of U and a section sp ∈ F(Up) such that(ϕ(Up)

)(sp) = s′|Up

. The

idea is now to glue all the sections sp to produce a section in F(U). Since F is a sheaf,

we need to prove that, for any p, q ∈ U , we have sp|Up∩Uq= sq |Up∩Uq

. To prove that, we

will use that we already proved the injectivity of ϕ(Up ∩ Uq). Hence, we need to prove

(ϕ(Up ∩Uq))(sp|Up∩Uq) = (ϕ(Up ∩Uq))(sq |Up∩Uq

), but this is obvious, since both are equal

to s′|Up∩Uq. Finally, the fact that

(ϕ(U)

)(s) = s′ comes from the fact that F ′ is a sheaf

and that we have that, by construction, both sections coincide on each Up.

Remark 3.12. Observe that, in the above proof, we proved also that ϕ is injective if

and only if each ϕp is injective. However, this is no longer true for the surjectivity, and in

35

fact we needed to use also the injectivity. What happens is that we can define the sheaf

ker(ϕ) in the obvious way, while the natural Im(ϕ) is not a sheaf, but only a presheaf. In

the following result we will see the way of constructing a sheaf from a presheaf. This new

sheaf obtained from Im(ϕ) is what we will define as image sheaf, and with that we will

have that the image is the whole F ′ if and only if each ϕp is surjective. In general, any

definition for sheaves works better when given in terms of the stalks.

The following result reminds how we defined regular functions in a local way, and how

they became (after Theorem 3.3) global polynomial functions:

Proposition 3.13. Given a presheaf F over a topological space X, for each open set

U ⊂ X we define F+(U) as the set of functions f : U →⊔p∈U Fp such that, for each

p ∈ U , it holds f(p) ∈ Fp and there exist an open neighborhood V ⊂ U of p and s ∈ F(V )

such that f(q) = sq for each q ∈ V . Then:

(i) F+ is a sheaf of rings on X with the obvious restriction maps.

(ii) There is a natural morphism α : F → F+ and αp : Fp → F+p is an isomorphism for

each p ∈ X.

(iii) α is an isomorphism if and only if F is a sheaf.

Proof: Part (i) is straightforward. The usual difficult part to prove (property (v) in the

definition) is now trivial since sections are functions.

For part (ii), we define each α(U) : F(U) → F+(U) by assigning to each s ∈ F(U)

the function U →⊔p∈U Fp that sends each p ∈ U to the germ of s at p. It is a simple

exercise to check that α is a morphism. Now fix p ∈ X, and let us check that αp is an

isomorphism.

For the injectivity, assume that the germ of (U, s) at p goes to the class of zero. This

means that, in a smaller open set V ⊂ U containing p such that the map V →⊔q∈V Fq

given by q 7→ sq is zero. In particular, the germ sp of s at p is zero, as wanted.

For the surjectivity, consider the germ at p of a function f : U →⊔q∈U Fp. By

definition of F+, there exists a neighborhood V ⊂ U of p and s ∈ F(V ) such that f(q) = sq

for each q ∈ V . This means that the class of (U, f) (which is also the class of (V, f|V )) is

the image of the class of (V, s).

Finally, part (iii) follows from Proposition 3.11. Indeed, if F is a sheaf, then α is an

isomorphism by that proposition and part (ii). Reciprocally, if α is an isomorphism, then

F is a sheaf because F+ is.

Definition. The sheaf F+ constructed in Proposition 3.13 is called the sheaf associated

to the presheaf F .

36

Remark 3.14. Observe that the above construction, together with Proposition 3.11 says

that, in order to determine a sheaf we only need to know its behavior on a basis of open

sets. Specifically, assume that we have an assignment U 7→ F(U) only for the open sets U

of a basis, but that this satisfies all the properties of the definition of a sheaf. From this

data we can still define the stalks at each point, and hence construct the sheaf F+ and

for each basic open set we will have a homomorphism α(U) : F(U)→ F+(U), compatible

with the restrictions, thus inducing isomorphisms αp : Fp → F+p . In the same way as in

Proposition 3.11, we will get that this implies that all the maps α(U) are isomorphisms (it

is here that we will need that the assignment U 7→ F(U) behaves as a sheaf, even if this

is only defined for basic open sets). This also shows that all the sheaves constructed from

the given assignment are isomorphic to each other, because Proposition 3.11 says that it

is enough to have isomorphisms for basic open sets.

For example, with this point of view, the sheaf or regular sections on an affine set

X could be define as follows. First, for each f ∈ A(X) we define O(DX(f)) := A(X)f .

About the restriction maps, DX(f) ⊂ DX(g) is equivalent, by Lemma 1.25, to gp = fs for

some s ∈ N and some p ∈ A(X), hence we can define A(X)g → A(X)f by sending hgr to

hpr

frs . All this set-up satisfies the sheaf condition because of Theorem 3.3 (see its proof).

All these ideas give the hint to reproduce for any ring what we did to interpret it as

the ring of regular functions of a suitable topological space. We start with a couple of

examples, which essentially rephrase Remark 3.8.

Example 3.15. We could find non-zero functions that however vanish at any point. This

is the case, for example, when we take A = k[X,Y ]/(X2, Y ) (corresponding to the point

(0, 0) plus an infinitely close point in the horizontal direction, as discussed in Example 1.4)

or A = k[X,Y ]/(X2) (corresponding to a double line). In both cases, the class of X is

a non-zero element vanishing everywhere (at the point (0, 0) in the first case, and at the

vertical line in the second one). Of course, the reason is that we are taking classes modulo

a non-radical ideal, so that we get nilpotent elements like the class of X. However, we will

see next that we could have the same situation even when there are no nilpotent elements.

Example 3.16. Consider now the ring A = k[[T ]]. If we use the fact that, in the

affine case, points correspond to maximal ideals, we have only one point, namely the one

corresponding to the ideal (T ). Hence any element in this ideal vanishes at the point, and

now the reason cannot be due to nilpotent elements, since there are now. The clue is given

in Example 1.22, which shows that also the zero ideal (the other prime ideal of A) plays an

important role, and in fact it keeps most of the geometrical information. This motivates

the following definiton.

37

Definition. The spectrum of a ring A is the set X = Spec(A) consisting of all the prime

ideals of A.

In order to make it a topological space, we need to define the analogue of Zariski

topology. For affine sets, a polynomial function vanishes at a point (or, more generally, on

an irreducible set) if and only if the polynomial is in the maximal ideal of the point (or,

respectively in the prime ideal of the set). This suggests the following:

Definition. Given a subset S ⊂ A of a ring, we define

V (S) = p ∈ Spec(A) | f ∈ p for all f ∈ S = p ∈ Spec(A) | p ⊃ S.

It is clear that V (S) = V (I), where I is the ideal generated by S, and we have the

same properties as for the operator V of affine sets (see Exercise 1.2, which we reproduce

now in this new context):

Exercise 3.17. Given any ring A, prove the following equalities:

(i) V (0) = Spec(A), and V (1) = ∅.(ii) For any subsets S, S′ ⊂ A, then V (S)∪V (S′) = V (S′′), where S′′ = ff ′ | f ∈ S, f ′ ∈

S′.(iii) If SjS∈J is a collection of subsets of A then

⋂j∈J V (Sj) = V (

⋃j∈J Sj).

Or, in terms of ideals, prove:

(i’) V (0) = Spec(A), and V (A) = ∅.(ii’) For any ideals I, I ′ ⊂ A, then V (I) ∪ V (I ′) = V (II ′) = V (I ∩ I ′).(iii’) If IjI∈J is a collection of ideals of A then

⋂j∈J V (Ij) = V (

∑j∈J Ij).

Definition. We calle the Zariski topology of Spec(A) to the topology in which the closed

sets are the sets of the form V (S).

The main difference with the Zariski topology of affine sets is that not all points are

closed:

Lemma 3.18. The closure of the set p is V (p). In particular, p is closed if and only

if p is a maximal ideal.

Proof: The closure of any set is the intersection of all closed sets containing it. Since

p ∈ V (I) if and only if I ⊂ p, then the closure of p is the intersection of all the V (I)

for which I ⊂ p. Obviously, one of such I is precisely p, and since for any other such I we

have V (p) ⊂ V (I), it follows that the closure is indeed V (p).

Example 3.19. To give a first idea of the new kind of points that we get when using

schemes, let us compute all the points of A2k regarded as Spec(k[X,Y ]). First of all, we

38

have the closed points, which are the maximal ideals. A maximal ideal is of the type

(X − a, Y − b), and it corresponds to the point (a, b) in the classical affine plane. On the

opposite side, we have the prime ideal (0), whose closure is the whole plane. Take now a

prime ideal p ⊂ k[X,Y ] that is none of the above. In particular, p must contain a nonzero

polynomial. Since p is prime, some of its irreducible factors is in p, so that we conclude

that p contains an irreducible polynomial f . If p contained a polynomial g that is not

a multiple of f , then V (f, g) would consist of a finite number of points (in the classical

sense). Since V (p) ⊂ V (f, g) and V (p) is irreducible, then V (p) would be just one point,

i.e. p would be a maximal ideal. Summing up, the points of the scheme A2k are:

–The closed points, corresponding to points in the classical affine plane.

–The principal ideals generated by an irreducible polynomial, corresponding to an

irreducible curve.

–The zero ideal, corresponding to the whole plane.

To complete the analogy with the affine case, we need to define the notion of ideal of

a subset. We do so and recover the main properties –including the Nullstellensatz!– in the

following (compare with Exercise 1.3):

Lemma 3.20. Let A be a ring and set X = Spec(A). For any subset Z ⊂ X we write

I(Z) = f ∈ A | f ∈ P for each P ∈ Z (i.e. I(X) =⋂P∈X P ).

(i) I(∅) = A and I(X) =√

(0) (see Example 3.15).

(ii) I(Z1 ∪ Z2) = I(Z1) ∩ I(Z2).

(iii) I(Z1 ∩ Z2) ⊃ I(Z1) + I(Z2).

(iv) For any ideal I ⊂ A, it holds IV (I) =√I.

Proof: Parts (i), (ii), (iii) are left as an exercise, while part (iv) is a consequence of the

fact that⋂p⊃I p =

√I.

Remark 3.21. The above property⋂p⊃I p =

√I is proved using Zorn’s Lemma, which

is equivalent to the Axiom of Choice. If the reader has conscience problems with that,

the Nullstellensatz implies that the radical of any ideal is the intersection of all maximal

ideals containing it, in case the ring is a finitely generated k-algebra (even if k is not

algebraically closed). This property also encloses the other important fact that is needed

from commutative algebra: the existence of maximal ideals (which also follows from the

Nullstellensatz in the case of finitely generated k-algebras). In fact, we could have the fact

in which I does not contain any prime, so that⋂p⊃I p becomes the whole A; therefore√

I = A, and equivalently I = A. In other words, this property is saying that, if I 6= A,

39

there is always a prime ideal containing I (i.e. if A/I is an honest ring, with 0 6= 1, then

it contains a prime ideal).

We can also recover the results about a basis of the topology (Lemmas 1.23 and 1.25):

Lemma 3.22. Let A be a ring and set X = Spec(A). Then

(i) The sets D(f) := X \ V (f) with f ∈ A form a basis of the Zariski topology on X.

(ii) D(f) ⊂⋃iD(fi) if and only if a power of f is in the ideal generated by the fi.

(iii) D(f) =⋃iD(fi) if and only if a power of f is in the ideal generated by the pi, where

pi is an element such that pif = fsii for some si.

Proof: Part (i) is straightforward, and proved as in Lemma 1.23(∗)

Theorem 3.23. Let A be a ring, and set X = Spec(A). Then:

(i) The assignment D(f) 7→ Af defines a sheaf OX (see Remark 3.14).

(ii) For each p ∈ Spec(A), the map Ap → OX,p that assigns to each gf ∈ Ap the germ

class of (D(f), gf ) is an isomorphism.

Proof: It is analogue to that for affine sets (Theorem 3.3 and Proposition 3.9(iii)), with

lots of technical details that do not provide any light. The reader is thus allowed to skip

that proof, although we will write it down completely.

For part (i),we need to follow the steps given in Remark 3.14. First of all, the restric-

tion maps are given in the same way. Indeed, D(f) ⊂ D(g) is equivalent to f ∈ IV (g),

i.e., by part (ii) f ∈√

(g); thus there exist s ∈ N and p ∈ A such that fs = gp. We

therefore define the restriction map Ag → Af by hgr 7→

hpr

fsr . Next we need to check that

this assignment satisfies the sheaf properties, namely part (iv) of the definition, i.e. that,

given a covering D(f) =⋃iD(fi), compatible elements on each Afi extend in a unique

way to an element of Af .

We will split that property into two different parts, uniqueness and existence of the

extension. Before that, we translate the condition D(f) =⋃iD(fi) using (ii). First of

all, each inclusion D(fi) ⊂ D(f) is equivalent to fi ∈ IV (f), i.e. to a relation fsii = pif ,

which means D(fi) = D(f)∩D(pi). Therefore the equality D(f) =⋃iD(fi) is equivalent

to D(f) ⊂⋃iD(pi), i.e f ∈ IV (pii∈I).

–For the uniqueness, assume that an element gfr ∈ Af restricts to zero to each Afi .

This means thatgp

sii

fssii

is zero as an element of Afi . Therefore there exists mi such that

(∗) In fact, if A is noetherian, it also holds that any open set is a finite union of open

basic sets.

40

fmii gpsii = 0 in A. In particular also fmisi

i gpsii = 0, i.e. pmi+sii gfmi = 0. Since f ∈

IV (pii∈I) = IV (pmi+sii i∈I) there is a relation, by (ii), of the type

fs = hi1pmi1+si1i1

+ . . .+ hirpmir+sirir

(in particular this implies D(f) ⊂ D(pi1)∩ . . .∩D(pir ), i.e. D(f) = D(fi1)∪ . . .∪D(fir )).

Hence, if m = maxmi1 , . . . ,mir, we have

gfs+m = hi1pmi1+si1i1

gfm + . . .+ hirpmir+sirir

gfm = 0

which implies that gfr is the zero element of Af , as wanted.

–The existence part will be analogue to the proof of Theorem 3.3. Assume that for each

i we have an element gifnii

∈ Afi . We have already seen that D(f) = D(fi1) ∪ . . . ∪D(fir ).

Let us see first that it is enough to find gfs that restrict to

gij

fnijij

on each Afij . Indeed

for any other i we need to see that the restriction of gfs coincides with gi

fnii

in Afi . For that

we decomposes D(fi) = D(fifi1) ∪ . . . ∪D(fifir ) and, by the uniqueness part we proved

we need to see that the restriction to each D(fifij ) = D(fi)∩D(fij ) of gfs and gi

fnii

are the

same. But this is so because we are assuming that gfs restricts to

gij

fnijij

on D(fij ), and our

compatibility condition says thatgij

fnijij

and gifnii

have the same restriction on D(fi)∩D(fij ).

Let us finally prove that there exists gfs ∈ Af that restrict to

gij

fnijij

on each Afij .

The point now is that we have to deal with a finite number of open sets, so that we can

take common exponents in the computations. We thus start assuming ni1 = . . . , nir = n,

and this can also be the exponent in the relations fnij = pijf . Hence the compatibility

condition thatgijfnij

andgikfnik

coincide on D(fijfik) meansgij f

nik

(fij fik )n =gikf

nij

(fij fik )n as elements in

Afij fik and thus there exists m such that

(fijfik)mgijfnik

= (fijfik)mgikfnij

in A, which implies

(fijfik)mngijfnik

= (fijfik)mngikfnij

and thus

pmij pm+1ik

f2m+1gij = pm+1ij

pmikf2m+1gik . (∗)

Finally, since D(f) ⊂ D(pi1) ∩ . . . ∩D(pir ) (see above), we have f ∈ IV (pi1 , . . . .pir) =

IV (pm+1i1

, . . . .pm+1ir), and by (ii) there is a relation of the type

fs = hi1pm+1i1

+ . . .+ hirpm+1ir

41

Multiplying this equality by each f2m+1pmij gij and applying (*) for k = 1, . . . , r, we get

fs+2m+1pmij gij = f2m+1pm+1ij

(hi1pmi1gi1 + . . .+ hirp

mirgir ).

Writing g := hi1pmi1gi1 + . . .+ hirp

mirgir , multiplying by pm+s+1

ij and using again fnij = pijf

we get fn(s+2m+1)ij

gij = fn(2m+1)ij

ps+1ij

g, which implies the equality

ps+1ij

g

fn(s+1)ij

=gijfnij

i.e. the restriction to Afij of gfs+1 is

gijfnij

. Hence all thegijfnij

glue together to produce gfs+1 ,as

wanted.

Finally we prove part (ii) following the steps of Proposition 3.9(iii). For the surjectiv-

ity, a germ at p is some class (D(f), gfs ), with p ∈ D(f), i.e. f /∈ p. Thus this germ is the

image of gfs ∈ Ap.

For the injectivity, the germ (D(f), gf ) is zero if and only if there exists an open

neighborhood D(h) ⊂ D(f) of p such that the restriction of gf is zero. Recall that D(h) ⊂

D(f) is equivalent to hs = pf for some s ∈ N and some p ∈ A, and that the restriction ofgf to D(h) is thus the element gp

hs ∈ Ah. Now gphs is zero in Ah if and only if gphr = 0 for

some r. Since phr /∈ p, this means gf is zero as an element of Af .

Definition. An affine scheme is the spectrum X of a ring A together with the sheaf of

rings OX given in Theorem 3.23. Although we have seen it is useless for practical purposes,

the precise description of the structure sheaf is that OX(U) consists of the set of maps

s : U →∐p∈U Ap such that for each p ∈ U s(p) ∈ Ap and there exists a neighborhood V

of p and elements f, g ∈ A satisfying that for each p′ ∈ V it holds s(p′) = fg (in particular

g 6∈ p′).

We can now repeat the same procedure for graded rings, trying to imitate the situation

of projective sets and their homogeneous rings. As it happened there, the problems will

come with the analogue of the maximal homogeneous ideal.

Definition. Given a graded ring S, let S+ :=⊕

d>0 Sd. Then we define Proj(S) to be the

set of homogeneous prime ideals of S not containing S+. For any homogeneous ideal I,

we define V (I) = p ∈ Proj(S) | p ⊃ I. We have the same properties as in Exercise 2.3

or Exercise 3.17, so that Proj(S) possesses a Zariski topology in which the closed sets are

those of the form V (I).

We have now the analogue of Theorem 3.23:

42

Theorem 3.24. Let S be a graded ring, then:

(i) The sets D(F ) := Proj(S) \ V (F ), with F ∈ S homogeneous of positive degree, form

a basis of the Zariski topology.

(ii) If I ⊂ S a homogeneous ideal and F ∈ S a homogeneous element of positive degree,

then V (I) ⊂ V (F ) if and only if there exist a power Fm ∈ I.

(iii) The assingnment D(F ) 7→ S(F ) defines a sheaf OX on Proj(S).

(iv) For each p ∈ X, there is an isomorphism S(p) → OX,p sending each GF ∈ S(p) to the

germ of (D(F ), GF ).

Proof: For (i), let U be an open set and take p ∈ U . Hence p is not in Proj(S) \U , which

takes the form V (I) for some homogeneous ideal I ⊂ S. Therefore, there is a homogeneous

element F ∈ I that is not in p. In case deg(F ) = 0, we can replace it by its product by

another homogeneous polynomial of positive degree that is not in p (which exists, since p

does not contain S+). Hence p ∈ D(F ) ⊂ U , proving (i).

We prove now (ii). It is clear that, if Fm ∈ I, then V (I) ⊂ V (Fm) = V (F ). Recipro-

cally(∗), assume F /∈√I. Consider the set

Σ := I ′ ⊂ S | I ′ is a homogeneous ideal such that I ⊂ I ′ and F /∈√I ′.

This is a non-empty set, since I ∈ Σ and, with the order given by the inclusion it is clear

that any chain in Σ has a supremum (use the standard trick that the union of a chain of

ideals is still an ideal). Hence, by Zorn’s Lemma, the set Σ has a maximal element p. If

we prove that p is prime we are done, because then p ∈ Proj(S) (the fact F /∈ p implies

p 6⊃ S+, because F ∈ S+) and we have p ∈ V (I) \ V (F ), a contradiction.

For the primality of p, let G,H ∈ S homogeneous elements not in p. Then p+(G) and

p + (H) are homogeneous ideals strictly containing p. By the maximality of p in Σ, those

ideals cannot be in Σ. Hence F is in the radical of both, i.e there exist powers F r ∈ p+(G)

and F s ∈ p + (H). But then, multiplying those relations, we get F r+s ∈ p + (GH). Since

F /∈ √p, it follows GH 6∈ p. This proves that p is prime.

For the proofs of (iii) and (iv), just imitate, respectively, the proofs of Theorem 3.4

and Theorem 3.9(iv), as we did in the proof of Theorem 3.23.

Definition. A projective scheme is the topological space X = Proj(S) of a graded ring S

together with the sheaf of rings OX given in Theorem 3.24. Specifically, OX(U) consists

of the set of maps s : U →∐p∈U S(p) such that for each p ∈ U s(p) ∈ S(p) and there exists

(∗) We will essentially repeat the proof that the radical of an ideal is the intersection of

all the primes containing it, but in the homogenoeus case

43

a neighborhood V of p and homogeneous elements F,G ∈ S of the same degree satisfying

that for each p′ ∈ V it holds s(p′) = FG (in particular G 6∈ p′).

44

4. The definition of a scheme

In this section we will study the notion of scheme, which is the generalization of

abstract variety. This will be an object that locally is isomorphic to an affine scheme.

For this, we need to define first what an isomorphism is (the notion of morphism is more

delicate, and we will leave it for the second half of this section). We will arrive to that

definition by looking closely at the example of Veronese varieties:

Example 4.1. Consider the m-uple Veronese embedding of Pnk (see Exercise 2.23), which

we expect to be an isomorphism onto its image, which is the Veronese variety Vn,m ⊂ PNk .

We can take coordinates ZI in PNk , where I varies in the ser of (n+1)-uples (i0, . . . , in) with

i0 + . . .+ in = m (representing the monomial Xi00 . . . Xin

n ). If S is the polynomial ring with

indeterminates XI , then it is clear that the ideal of the Veronese variety is the kernel of

the homomorphism of k-algebras ψ : S → k[X0, . . . , Xn] determined by ZI 7→ Xi00 . . . Xin

n .

Therefore, the graded ring of the Veronese variety is S(Vn,m) = S/I(Vn,m) ∼= Im(ψ). The

image of ψ is clearly the subring of k[X0, . . . , Xn] consisting of those polynomials such

that all its monomials have degree a multiple of m. Hence S(Vn,m) can be identified with

that subring, in which we homogeneous part of degree d is the space of homogeneous

polynomials of degree md. This shows that, contrary to what happens for affine sets, the

coordinate ring does not characterize the projective set up to isomorphism, since any two

Vn,m with the same n are expected to be isomorphic, while not two S(Vn,m) are isomorphic.

Proposition 4.2. Let S be a graded ring, fix m > 0 and define a new graded ring

S′ =⊕

d S′d≥0, where S′d := Smd. Then:

(i) The assignment p 7→ p ∩ S′ defines a homeomorphism ϕ : Proj(S) → Proj(S′) such

that, for any F ′ ∈ S′ homogeneous of positive degree, ϕ−1(D(F ′)) = D(F ′).

(ii) For any F ′ ∈ S′ homogeneous of positive degree, the inclusion map S′(F ′) → S(F ′) is

an isomorphism.

Proof: To prove (i), we first prove that the map ϕ is well defined. Since ϕ(p) is the inverse

image of p by the inclusion S′ ⊂ S, it is a prime ideal. It is clearly homogeneous, since

p is. Finally, since p does not contain S+, there is F ∈ S homogeneous of positive degree

not in p, hence Fm is not in p′, which proves that p′ is in Proj(S′).

Let us define an inverse for ϕ. For any p′ ∈ Proj(S′), set ψ(p′) =F ∈ S | Fmi ∈

p′ for all homogeneous components Fi

. We first observe that ψ(p′) is easily seen to be a

homogeneous prime ideal. Since p′ ∈ Proj(S′), there exists a homogeneous G of positive

degree not in p′, hence not in ψ(p′), which proves ψ(p′) ∈ Proj(S).

Let us prove first ψ(ϕ(p)) = p. We take a homogeneous element F ∈ S. Then,

F ∈ ψ(ϕ(p)) if and only if Fm ∈ ϕ(p), i.e. Fm ∈ p, which is equivalent, by the primality

45

of p, to F ∈ p.

Similarly, we prove ϕ(ψ(p′)) = p′. A homogeneous F ′ ∈ S′ is in ϕ(ψ(p′)) if and only

if it is in ψ((p′), i.e. F ′m ∈ p′. By the primality of p′, this is equivalent to F ′ ∈ p′.

Once we know ϕ is a bijection, let us prove that is a homeomorphism. To prove

that it is continuous, ley us prove that, for any F ′ ∈ S′ homogeneous of positive degree,

ϕ−1(D(F ′)) = D(F ′). Indeed, p ∈ ϕ−1(D(F ′)) if and only if F ′ /∈ ϕ(p), i.e. F ′ /∈ p or,

equivalently p ∈ D(F ′).

Applying this to F ′ = Fm, where F ∈ S is any homogeneous element of positive

degree, and using that ϕ is bijective, we get ϕ(D(F )) = ϕ(D(Fm)) = D(Fm). Hence ϕ is

a homeomorphism.

Finally, part (ii) is obvious.

Example 4.3. Let us apply Proposition 4.2 to a generalization of a projective space. The

weighted projective space P(a0, . . . , an) is defined to be Proj(k[X0, . . . , Xn]), but with the

peculiarity that we set deg(Xi) = ai. Of course, if a0 = . . . = an = 1 we obtain the usual

projective space (and more generally, if a0 = . . . = an = m, the above proposition implies

that we again get something isomorphic to Pnk ). In general, weighted projective spaces are

singular. For example, let us study closely the weighted projective plane P(1, 1, 2). This

is Proj(S), with S = k[U, V,W ], in which deg(U) = deg(V ) = 1 and deg(W ) = 2. Taking

m = 2 in Proposition 4.2, we get a graded epimorphism k[X0, X1, X2, X3]→ S′ defined by

X0 7→ U2

X1 7→ UV

X2 7→ V 2

X3 7→W

whose kernel is (X0X2 − X21 ). Therefore P(1, 1, 2) can be identified, by Proposition 4.2,

with Proj(k[X0, X1, X2, X3]/(X0X2 −X2

1 )), which represents a quadratic cone in P3

k.

Observe that Proposition 4.2 is giving an isomorphism of sheaves between the structure

sheaf OProj(S′) and the sheaf defined by U ′ 7→ OProj(S)(ϕ−1(U ′)). We thus arrive to the

following definitions.

Definition. Given a continuous map ϕ : X → X ′ and a sheaf F on X, we define the direct

image sheaf ϕ∗F on X ′ by (ϕ∗F)(U ′) = F(ϕ−1(U ′)), with the natural restriction maps

ρ′U ′,V ′ := ρϕ−1(U ′),ϕ−1(V ′).

46

Remark 4.4. It is not evident what the stalk of ϕ∗F at a point p′ ∈ X ′ is. However,

given a point p ∈ X, for each open neighborhood U ′ of ϕ(p), since ϕ−1(U ′) is an open

neighborhood of p, there is a homomorphism (ϕ∗F)(U ′) = F(ϕ−1(U ′)) → Fp. We thus

get a homomorphism (ϕ∗F)ϕ(p) → Fp.

Proposition 4.5. Let S be a graded ring and let F ⊂ S be a homogeneous element of

strictly positive degree. Then there is a homeomorphism ϕ : D(F )→ Spec(S(F )) inducing

an isomorphism of sheaves OSpec(S(F )) → ϕ∗OX|D(F ).

Proof: If F has degree m, Proposition 4.2 allows us to replace S with S′, so that we can

assume that F has degree one. We thus define ϕ : D(F )→ Spec(S(F )) by:

ϕ(p) = GF s| G ∈ p homogeneous of degree s

and we will see that ϕ has an inverse ψ : Spec(S(F ))→ D(F ) defined by

ψ(p′) =G ∈ S | Gi

F i∈ p′ for any homogeneous component Gi of i of degree i

.

Let us see, for example, ψ(ϕ(p)) = p for any p ∈ D(F ). By definition of ψ, a homogeneous

element G ∈ S of degree s is in ψ(ϕ(p)) if and only if GF s ∈ ϕ(p), which is equivalent to

G ∈ p.

Similarly , ϕ(ψ(p′)) = p′ for any prime ideal p′ ⊂ S(F ). Indeed, an element GF s ∈ S(F ),

with G homogeneous, is in ϕ(ψ(p′)) if and only if G ∈ ψ(p′), i.e GF s ∈ p′.

Take now a basic open set D( F′

F s ) (with s = deg(F ′)). Then ϕ(p) ∈ D( F′

F s ) if and only

if F ′

F s /∈ ϕ(p), i.e. F ′ /∈ p. Hence ϕ−1(D( F′

F s )) = D(F ′) ∩D(F ), so that ϕ is a continuous

map. But the fact that ϕ is bijective implies also ϕ(D(F ′) ∩ D(F )) = D( F′

F s ) for any

homogeneous F ′ ∈ S, which implies that ϕ is a homeomorphism.

We finally construct the isomorphism between the structure sheaf of Spec(S(F )) and

ϕ∗(OProj(S)|D(F )). For that, we just need to define compatible isomorphisms between

O(D( F′

F s )) and O(ϕ−1(D( F′

F s ))) = O(D(FF ′)), i.e. isomorphisms(S(F )

)F ′Fs→ S(FF ′).

But it is enough to define them as

GFm

( F′

F s )r7→ F srF ′

m−rG

(FF ′)m

(it is clear that multiplying numerator and denominator of GFm by powers of F we can

assume m ≥ r).

Remark 4.6. The same proof as Proposition 4.5 shows that, if A is a ring and f ∈ A is

not nilpotent, there is a homeomorphism ϕ : D(f) → Spec(Af ) inducing an isomorphism

of sheaves OSpec(Af ) → ϕ∗OSpec(A)|D(f).

47

Proposition 4.5 inspires the following general definition (showing that Proj(S) is a

scheme):

Definition. A scheme is a topological space X, together with a sheaf of rings OX , called,

the structure sheaf such that there exists an open covering X =⋃i Ui such that each

for each i there is a homeomorphism ϕ : Spec(Ai) → Ui for which the restriction of the

structure sheaf OX |Uiis isomorphic to ϕ∗(OSpec(Ai)).

The last part of the above definition can be re-written in terms of isomorphisms of

schemes, by giving the following:

Definition. An isomorphism between two schemes is a homeomorphism ϕ : X → Y

together with an isomorphism of sheaves OY → ϕ∗OX .

Remark 4.7. If U ⊂ X is an open subset of the scheme X covered by affine sets

Ui ∼= Spec(Ai), then we can cover U by the open sets U ∩ Ui. Since Spec(Ai) has a basis

consisting of elements of the form D(fj), Remark 4.6 implies that U can be covered by

affine schemes, so that U with OX|U has a natural scheme structure (and we will say that

U is an open subscheme).

Example 4.8. Proposition 4.5 says in particular that Pnk can be regarded as different affine

pieces glued together. We will analyze first the case n = 1, in which we only need to glue

two different pieces. So let S = k[X0, X1] and write X = X1

X0and Y = X0

X1. Then Proj(S)

consists of glueing D(X0), i.e Spec(k[X]) and D(X1), i.e. Spec(k[Y ]). The intersection

of these two pieces is D(X0X1) ⊂ Proj(S), which corresponds to D(X) in Spec(k[X], i.e

Spec(k[X]X) and to D(Y ) in Spec(k[Y ]), i.e Spec(k[Y ]Y ). The way of glueing Spec(k[X])

and Spec(k[Y ]) is to identify Spec(k[X]X) and Spec(k[Y ]Y ) by means of the isomorphism

k[X]X ∼= k[Y ]Y determined by sending X to 1Y .

Example 4.9. You can wonder what happens if, in the above example, the way of

identifying Spec(k[X]X) and Spec(k[Y ]Y ) is what seems to be the natural one, i.e. using

the isomorphism k[X]X ∼= k[Y ]Y determined by sending X to Y . In that case, we are

glueing together two affine lines, and the way we are doing is to identify each point of the

first one, except 0, with the same point of the other line. As a result, what we get is an

affine line in which we have replaced the point 0 with two different points.

Example 4.10. The general procedure to glue schemes is the following. Assume you

have a family Xii∈I of schemes and that, for each i, j ∈ I there is an open subscheme

Xij ⊂ Xi and an isomorphism ϕij : Xij → Xji with the conditions:

(i) Xii = Xi and ϕii = idXi .

(ii) For any i, j, k ∈ I, ϕij(Xij ∩ Xik) = Xji ∩ Xjk and ϕik|Xij∩Xik= ϕjk|Xji∩Xjk

ϕij|Xij∩Xik

.

48

Then it is possible to glue all those Xi along the Xij by defining first the topological space

X :=∐iXi/ ∼ in which p ∼ q if and only if p ∈ Xi, q ∈ Xj and ϕij(p) = q (observe that

conditions (i) and (ii) imply that ∼ is an equivalence relation). We can endow X with a

topology in which a basis is given by the basic open affine sets of the Xi’s, and this also

allows to define the structure sheaf.

Exercise 4.11. Prove that the scheme obtained by glueing Spec(A) and Spec(B) without

any restriction (i.e. the disjoint union, in which we choose the empty set as open subset

along which to glue) is isomorphic to Spec(A×B).

Since now we have points that represent irreducible subvarieties, we can read (and

generalize) Proposition 3.9 in terms of germs:

Lemma 4.12. Let X be a scheme. Then:

(i) The map p 7→ p defines a bijection between X and the set of irreducible subsets of

X.

(ii) For any sheaf F on X and any p ∈ X, the stalk Fp is the set of equivalence classes

of pairs (U, s), with U ∩ p 6= ∅ and s ∈ F(U), under the equivalence relation

(U, s) ∼ (U ′, s′) if and only if there exists U ′′ ⊂ U ∩ U ′ meeting p such that

s|U ′′ = s′|U ′′ .

Proof: To prove (i), we first observe that the closure of a point is always an irreducible

set (because p ⊂ Z1 ∪ Z2 if and only if p ∈ Z1 ∪ Z2, for Z1, Z2 closed subsets). On the

other hand, let us see that any irreducible set Z ⊂ X is the closure of a point. For that,

take any affine open set U meeting Z. It is clear that any point whose closure is Z must

be in U and its closure in U must be Z ∩ U . Since U is affine and Z ∩ U is an irreducible

subset of U (see Lemma 1.11), this subset corresponds to a prime ideal, and hence there

is only one p ∈ U (namely that prime ideal) whose closure in U is Z ∩ U . Let us see that

p = Z:

The inclusion p ⊂ Z is clear since p ∈ Z. To prove the other inclusion we need to

prove that any neighborhood of any point of Z contains p. We thus take any point q ∈ Z,

and any open neighborhood V of q. Since Z is irreducible and meets U and V , it follows

that Z ∩U ∩V is not empty. Since Z ∩U is the closure of p in U , then its non-empty open

subset Z ∩ U ∩ V contains p, so that p ∈ V , as wanted.

Part (ii) is an easy consequence of the fact that an open set U meets p if and only

if p ∈ U . Indeed, p /∈ U if and only if p is in the closed set X \ U , which is equivalent to

p ⊂ X \ U , i.e. U does not meet p.

Definition. The generic point of an irreducible subset Z of a scheme X is the unique point

whose closure is Z.

49

According to Theorem 3.23, the stalk OX,p of the structure sheaf is always a local

ring (depending on the context of schemes or varieties, we will right OX,p or OX,Y , where

Y = p). It can be computed by restricting to any affine open neighborhood of p of the

type Spec(A), since OX,p is canonically isomorphism to Ap, where p is the prime ideal

corresponding to p. We can thus make the following:

Definition. Let X be a scheme, p a point and Mp the maximal ideal of OX,p. The residual

field of the point p is the quotient k(p) := OX,p/Mp.

Remark 4.13. The above definition allows to regard the sections of the structure sheaf

as functions. Indeed a section in an open set U of a scheme X is a map that sends any

point p ∈ U to its stalk OX,p, and we can quotient by Mp to get an element of k(p). Hence

a section can be regarded as a function that sends any point p to a value in k(p). Of

course the field k(p) is different for each point p. If we want to have a common field k, at

least for closed points, one possibility would be to have X covered by the spectra of finitely

generated k-algebras. Since we will need those k-algebras to be compatible with each other

in their intersections, we can have that structure “from the top”, i.e. we can assumeOX(X)

to be a k-algebra, which means that we have a homomorphism k → OX(X). Therefore the

composition with the restriction map k → OX(X)→ OX(U) induces a k-algebra structure

for any open set U ⊂ X, and with this structure all restriction maps OX(U) → OX(V )

are homomorphisms of k-algebras.

Definition. Given a field k, a k-scheme is a scheme X such that OX(X) has a structure of

k-algebra. If X can be covered by open sets Spec(Ai) such that each k-algebra Ai (with the

structure explained in Remark 4.13) is finitely generated, then X is said to be a k-scheme

of locally finite type. If moreover the covering is finite X is called a k-scheme of finite type.

A standard example of k-scheme is Proj(S), where S is a graded k-algebra (if S is a

finitely generated k-algebra, then Proj(S) is a k-scheme of finite type). If k is algebraically

closed, then for any closed point p of a k-scheme of locally finite type the residual field

k(p) is isomorphic to k. Of course it is not enough to have a k-scheme. For example,

Spec(k(T )) is a k-scheme with only one point, and its residual field is k(T ).

Example 4.14. When k is not algebraically closed, we still can do a lot of geometry

with schemes. Take for example k = R Then the closed points of A1R are the maximal

ideals of R[X], which are of the form (X − a) (thus corresponding to the real point a)

or ((X − a)2 + b2) (corresponding to the pair of conjugate imaginary points a ± bi). In

general, in an R-scheme of locally finite type there are two kind of closed points: the

real points for which the residual field is R and the pairs of conjugate imaginary points,

whose residual field is C. For example, Spec(R[X,Y ]/(X2 + Y 2 + 1)), which represents

the imaginary conic V (X2 + Y 2 + 1) ⊂ A2R, has in fact a lot of points as a scheme, all

50

of them imaginary. In general, the generalization of the Nullstellensatz implies that the

closed points of a k-scheme of locally finite type have residual field a finite extension of

k, i.e. they represent conjugate points with coordinates in that extension. In particular

taking k = Q, the theory of schemes reveals to be a powerful tool for number theory.

In order to repeat the proof of Theorem 1.14 we need (see Remark 1.15) the following:

Definition. A noetherian topological space is a topological space such that any descending

chain of closed sets is stationary.

If X = Spec(A) then X is noetherian if A is a noetherian ring. The converse is

not true; for example, it is an easy exercise (left to the reader) to prove that the ring

A = k[X1, X2, . . . , ]/(X1, X22 , . . .) is not noetherian but Spec(A) is a noetherian topological

space (it consists of only one point).

Definition. A noetherian scheme is a scheme that it is a finite union of open spectra of

noetherian rings.

Proposition 4.15. Any closed set in a noetherian scheme can be written in a unique way

as an irredundat finite union of irreducible sets.

Proof: As shown in the proof of Theorem 1.14, it is enough to see that any noetherian

scheme X is a noetherian topological space. But this is immediate because X is the finite

union of open noetherian sets U1, . . . , Us. Hence, for any descending chain Z1 ⊃ Z2 ⊃ . . .

of closed sets of X, the restriction to any Ui is stationary, i.e. there exists ni such that

Zn ∩ Ui = Zni∩ Ui for any n ≥ ni. Hence, for n ≥ n0 := maxn1, . . . , ns it follows

Zn = Zn0.

We try to understand now the right notion of morphism of schemes. Since a the

structure of a scheme is given by its structure sheaf, we need morphisms to be compatible

with structure sheaves. The idea in fact comes from differential geometry, in which a

differential map is a map that sends differential functions to differential functions. The

first temptation would be to imitate the notion of isomorphism and define a morphism of

schemes as a continuous map ϕ : X → Y together with a morphism of sheaves OY →ϕ∗OX . However this will not be enough:

Example 4.16. Let us see that it is not as easy as it could seem to construct a constant

morphism. For that, the most trivial example of a scheme with only one point, the spec-

trum of a field. So let us try to define the constant map ϕ : X = Spec(k(T )) → A1k that

sends the unique point of Spec(k(T )) to the point p = (T ) ∈ Spec(k[T ]). The sheaf ϕ∗OX

51

is defined by

(ϕ∗OX)(U) =

k(T ) if p ∈ U

0 if p /∈ U.

We can thus define a morphism of sheaves ψ : OA1k→ ϕ∗OX by defining, for any f ∈ k[X],

the map ψ(D(f)) : OA1k(D(f)) → (ϕ∗OX)(D(f)) as the natural inclusion k[T ]f → k(T )

if f 6∈ p, and as the zero map if f ∈ p. But there is something wrong in this definition,

since one should expect that the image on any section, regarded as a function on D(f), is

the constant map whose image is the value of the function at the point represented by p.

Even taking germs, the germ map ψp : OA1k,p → (ϕ∗OX)p is given by the natural inclusion

k[T ]p → k(T ). But this is still not correct, since this sends each fg (a germ whose value at

k(p) is f(0)g(0) ) to f

g , whose value at the residual field is fg itself. The reason why this does

not work is that we would like the germ map OA1k,p → (ϕ∗OX)p to factor through the

residual fields, i.e. that it sends the maximal ideal to the maximal ideal. In other words,

the correct germ map must be defined as the composition k[T ]p → k → k(T ), where the

first map is the evaluation at 0.

Thus the best way is to change the morphism of sheaves OA1k→ ϕ∗OX by defining,

for any f ∈ k[X] not in p, the map OA1k(D(f)) → (ϕ∗OX)(D(f)) as the composition

k[T ]f → k → k(T ), where the first map is the evaluation at 0.

Definition. A morphism of schemes X and Y is a continuous map ϕ : X → Y between

two schemes, together with a morphism of sheaves ψ : OY → ϕ∗OX such that, for each

p ∈ X, the composition map OY,ϕ(p)

ψϕ(p)−→ (ϕ∗OX)ϕ(p) → OX,p (see Remark 4.4 for the

second homomorphism) sends Mϕ(p) into Mp.

Proposition 4.17. Given a ring homomorphism ψ : A→ B there is a natural morphism

of schemes ϕ : Spec(B)→ Spec(A). Reciprocally, any morphism of schemes from Spec(B)

to Spec(A) is induced by a (unique) ring homomorphism A→ B.

Proof: Given ψ : A → B, we define the map ϕ : Spec(B) → Spec(A) by ϕ(q) = ψ−1(q).

For any f ∈ A we have that q ∈ ϕ−1(D(f)) if and only if ψ−1(q) ∈ D(f), i.e. ψ(f) /∈ q.

This means that ϕ−1(D(f)) = D(ψ(f)). Hence, in order to define a morphism of sheaves

OSpec(A) → ϕ∗OSpec(B) we need to give, for any f ∈ A, a homomorphism Af → Bψ(f),

and we take the natural one induced by ψ. This gives the map ϕ : Spec(B) → Spec(A)

the structure of morphism of schemes. Observe that we can recover ψ by taking f = 1 in

the morphism of sheaves.

Reciprocally, assume that we have a morphism of schemes ϕ : Spec(B) → Spec(A).

From the corresponding morphism of sheaves OSpec(A) → ϕ∗OSpec(B), taking global sec-

tions, we get a homomorphism ψ : A → B, and we need to see that ϕ comes from ψ

52

according to the above construction. For any q ∈ Spec(B), we have a commutative dia-

gram when taking stalks at ϕ(q):

Aψ−→ B

↓ i ↓ jAϕ(q)

ϕq−→ Bq

The maximal ideal of Bq is qBq, and we have j−1(qBq) = q. Also the maximal ideal

of Aϕ(q) is ϕ(q)Aϕ(q), and we have i−1(ϕ(q)Aϕ(q)) = ϕ(q). By hypothesis we have

ψq(ϕ(q)Aϕ(q)) ⊂ qBq, so that ϕ(q)Aϕ(q) ⊂ ψ−1q (qBq) and, since ϕ(q)Aϕ(q) is a maxi-

mal ideal we have an equality. Therefore we have

ϕ(q) = i−1(ϕ(q)Aϕ(q)) = i−1(ψ−1q (qBq)) = ψ−1(j−1(qBq)) = ψ−1(q)

and hence the map ϕ is defined from ψ as in the above construction. As we have seen

in the first part, this implies ϕ−1(D(f)) = D(ψ(f)) for any f ∈ A. Therefore the map

(OSpec(A))(D(f))→ (ϕ∗OSpec(B))(D(f)) is the bottom map of a commutative diagram

Aψ−→ B

↓ ↓Af −→ Bψ(f)

and there is only one such map. Hence the morphism of sheaves is the one defined in the

first part, which concludes the proof.

Remark 4.18. Observe that Proposition 4.17 implies that two affine schemes Spec(A) and

Spec(B) are isomorphic if and only if their corresponding rings A and B are isomorphic.

In particular, given two non-isomorphic fields k, k′, the schemes Spec(k) and Spec(k′)

are not isomorphic, despite of both consisting of only one point. The same happens for

Spec(k[T ]/(T 2)

), which consists only of one point (although it represents also a tangent

direction, as we have seen in Example 1.21), but is not isomorphic to Spec(k). One more

example that, in the theory of schemes, the structure sheaf plays a more important role

than the set of points is Example 1.18. If we consider Spec(k[X,Y, Z]/I0

), where

I0 = (XY,XZ, Y Z,Z2) = (X,Z) ∩ (Y,Z) ∩ (X − aZ, Y − bZ, Z2)

then we miss the point represented by (X − aZ, Y − bZ, Z2), not only as an element of

Spec(k[X,Y, Z]/I0

), but also in its decomposition into irreducible components, since there

are only two irreducible components, corresponding to the ideals (X,Z) and (Y,Z). This

is why a primary component whose radical is not minimal is called embedded component.

We can still strengthen Proposition 4.17:

53

Proposition 4.19. If X,Y are schemes and Y is affine, there is a bijection between the

set of morphisms from X to Y and the set of homomorphisms of rings from OY (Y ) to

OX(X).

Proof: As in Proposition 4.17, to any ϕ : X → Y we associate the homomorphism

OY (Y )→ OX(X) obtained when taking global sections in the morphism of sheaves OY →ϕ∗OX . Reciprocally, assume that we have a homomorphism of rings ψ : OY (Y )→ OX(X),

and let us see how to produce from it a morphism of schemes X → Y . We will construct

it by glueing affine pieces. Specifically, let X =⋃i Ui a cover by affine open sets. Thus

for each i we have a homomorphism of rings ψi : OY (Y ) → OX(X) → OX(Ui) and, by

Proposition 4.17, it corresponds to a morphism ϕi : Ui → Y . We need to check that,

for any i, j, the morphisms ϕi and ϕj coincide on Ui ∩ Uj . For that, we now take an

affine open covering Ui ∩ Uj =⋃k Vij,k. Since the composition of ψi with the restriction

map OX(Ui) → OX(Vij,k) coincides with the composition of ψj with the restriction map

OX(Uj) → OX(Vij,k), it follows that ϕi|Vij,k= ϕj |Vij,k

as morphisms, for each k, hence

also ϕi|Ui∩Uj= ϕj |Ui∩Uj

.

Obviously both operations are inverse to each other, which completes the proof.

For our geometrical purposes, morphisms coming from general ring homomorphisms

could not desirable. For example the homomorphism C[X] → C[X] that conjugates the

coefficients of the polynomials define the conjugation map in A1C, and we do not want to

consider it a morphism. To avoid that kind of situation, we would need the homomorphisms

to be the extra structure of being homomorphisms of C-algebras. This leads again to the

notion of k-schemes. Observe that, as a corollary of Proposition 4.19, we have that a

scheme X is a k-scheme if and only if there is a morphism of schemes X → Spec(k). More

generally:

Definition. Given a scheme S, an S-scheme is a scheme X endowed with a morphism

of schemes X → S. A morphism of S-schemes is a morphism ϕ : X → Y such that, if

p : X → S and p : X ′ → S are the morphisms providing the S-scheme structure, then

p′ = p ϕ.

Remark 4.20. We can generalize Remark 4.13. By Proposition 4.19, if S = Spec(B),

then X is an S-scheme if and only if each OX(U) is a B-algebra and the restriction maps

OX(U)→ OX(V ) are homomorphisms of B-algebras. And Proposition 4.19 now gives that

giving a morphism of S-schemes X → SpecA is equivalent to giving a homomorphism of

B-algebras A→ OX(X).

We give a list of examples:

54

Example 4.21. Given finitely generated k-algebras k[X1, . . . , Xn]/I and k[Y1, . . . , Ym]/J ,

a k-morphism Spec(k[X1, . . . , Xn]/I) → Spec(k[Y1, . . . , Ym]/J) is equivalent to a homo-

morphism of k-algebras k[Y1, . . . , Ym]/J → k[X1, . . . , Xn]/I, and this is univoquely de-

termined by the image of the classes of Y1, . . . , Yr. Therefore, giving a k-morphism

Spec(k[X1, . . . , Xn]/I) → Spec(k[Y1, . . . , Ym]/J) is equivalent to giving the classes of m

polynomials f1, . . . , fm ∈ k[X1, . . . , Xn], exactly as it happened for affine sets. More gen-

erally, a k-morphism X → Ank is determined, by Proposition 4.19, by a homomorphism

of k-algebras k[X1, . . . , Xn] → OX(X), i.e. it is determined by f1, . . . , fr ∈ OX(X). In

particular, there is a one-to-one correspondence between the set of k-morphism from X to

A1k and the k-algebra OX(X).

Definition. A regular function on a k-scheme X is a k-morphism of schemes X → A1k.

Remark 4.22. Observe that now we can interpret k-schemes in a more pleasant way.

First, the structure sheaf associates to each U ∈ X the k-algebra of regular functions U →A1k. Now, a continuos map of k-schemes ϕ : X → Y is a morphism if and only if, for each

open set V ⊂ Y and any regular function V → A1k, the composition ϕ−1(V )→ V → A1

k is

a regular function.

Proposition 4.23. If X ⊂ Pn is a projective subscheme, then a map ϕ : X → Pmk is a

k-morphism if and only if for each p ∈ X there exist an open set U 3 p such that ϕ|U is

determined by m + 1 homogeneous polynomials of the same degree F0, . . . , Fm satisfying

U ∩ V (F0, . . . , Fm) = ∅.

Proof: Let us see first that such a map is a morphism. As usual, we can take each U

to be a basic open set DX(G), and we need to prove that any ϕ|DX(G) is a morphism.

For each i = 0 . . . ,m we have a map DX(GFi) → D(Yi), i.e from Spec(S(X)(GFi)) →Spec(k[Y0, . . . , Ym](Yi)), and by assumption this map is given by the homomorphism of

k-algebras k[Y0, . . . , Ym](Yi) → S(X)(GFi) defined byYj

Yi7→ GFj

GFi, so that it is a morphism.

Reciprocally, given a morphism ϕ : X → Pmk , in particular it is a continuous map,

hence ϕ−1(D(Yi)) is an open set, which we could cover by a finite number of basic open

sets. Thus we have our morphism splits in pieces DX(F )→ D(Yi), i.e. Spec(S(X)(F ))→Spec(k[Y0, . . . , Ym](Yi), which are necessarily given by a homomorphism of k-algebras k[Y0, . . . , Ym](Yi) →S(X)(F ). The image of each

Yj

Yiis a quotient Fi

F s , with Fi homogeneous of degree sdeg(F ).

Theorem 4.24. A map ϕ : Pnk → Pmk is a morphism if and only if it is determined

by homogeneous polynomials F0, . . . , Fm ∈ k[X0, . . . , Xn] of the same degree such that

V (F0, . . . , Fm) = ∅.

Proof: By Proposition 4.23, we know that morphisms ϕ : Pnk → Pmk are those given

locally by homogeneous polynomials of the same degree, and we need to prove that it is

55

always possible to choose the same polynomials. Indeed, assume that we have F0, . . . , Fm

determining ϕ on U and F ′0, . . . , F′m determining ϕ on U ′. This implies F0, . . . , Fm and

F ′0, . . . , F′m determine the same morphism on some D(F ), in particular on some D(FF0F

′0).

This meansFF ′0Fi

FF0F ′0=

FF0F′i

FF0F ′0on k[X0, . . . , Xn](FF0F ′0) for i = 1, . . . , n, which implies F0F

′i =

F ′0Fi. Since F0 cannot have a common factor with all F1, . . . , Fm it follows that any factor

of F0 is also a factor of F ′0, and reciprocally. Therefore there exists a nonzero constant

λ ∈ k such that F ′0 = λF0, and clearly this also implies F ′i = λFi for all i.

Given an open set U of a scheme X, with the structure of open subscheme given in

Remark 4.7, the inclusion map U → X becomes a morphism. For any other subset of X

there is no natural way of inducing a scheme structure, and in fact we can have several

structures. For example, a point in P2k can have multiple structures: as a simple point, as

a point together with a tangent direction,... This is in fact the interest of the schemes. In

the next examples we will give ways of providing scheme structure to different subsets.

Example 4.25. Given an epimorphism of rings A→ B, if I is its kernel, we can factorize

it as A → A/I→B, which yields a composition of morphisms Spec(B)→Spec(A/I) →Spec(A). This gives an isomorphism between Spec(B) and the closed subscheme V (I).

Definition. A closed embedding is a morphism of schemes ϕ : X → Y such that Y can be

covered by open affine schemes Y =⋃i Vi such that each ϕ−1(Vi) is an affine open set and

the induced homomorphism OY (Vi)→ OX(ϕ−1(Vi)) is surjective.

Remark 4.26. Given a closed embedding ϕ : X → Y , we can define, for each open

set V ⊂ Y , the ideal I(V ) to be the kernel of OY (V ) → OX(ϕ−1(V )). In that way, we

get what is called a sheaf of ideals on X. Reciprocally, given a sheaf of ideals I on Y ,

for each affine open set V ⊂ Y we can consider the affine scheme Spec(OY (V )/I(V )

).

Glueing together all these affine schemes in the natural way, we get a scheme X and a

closed embedding X → Y .

Example 4.27. A particular example is when B = A/√

(0) and we take the natural

projection A→ B; in that case we get a morphism that is a homeomorphism (but not an

isomorphism unless A is a reduced ring). For example, if A = k[X1, . . . , Xn]/I, then B =

k[X1, . . . , Xn]/√I, and its spectrum represents just an affine set whose ideal is

√I, and we

forget about any possible extra structure. For any scheme X we can repeat that operation

for its affine sets, and everything glues together to give a closed embedding Xred → X,

which is a homeomorphism. The new scheme Xred is called the reduced structure of X.

Given a morphism X → Y , there is a natural induced morphism Xred → Yred which, as a

map of topological spaces, is the same.

56

Example 4.28. We can also give a (reduced) scheme structure to any point p of a

scheme X, even if the point is not closed. For that purpose, we take any open affine set

U = Spec(A) containing p. If p corresponds to the prime ideal p, we can consider the

composition A→ Ap = OX,p → OX,p/Mp = k(p), which yields a morphism Spec(k(p))→Spec(OX,p) → U ⊂ X, whose image is the point p, and this composition is independent

of the choice of U . The above composition provides the same map as the composition

A → A/p → k(p), in which now k(p) is identified with the quotient field of the domain

A/p. If p is a closed point, p is a maximal ideal, so that the map A → k(p) is an

epimorphism, and hence the morphism Spec(k(p))→ X is a closed embedding.

Example 4.29. Similarly as in the above examples, we can provide a natural structure of

affine scheme to the subset p ∈ Spec(A) | p∩S = ∅ of Spec(A), where S is a multiplicative

set of the ring A. Indeed, that subset can be identified with the image of the morphism

Spec(S1A) → Spec(A) induced by the natural homomorphism A → S−1A. When f is a

not nilpotent element of A, this identifies the open subscheme D(f) with Spec(Af ). And

when S = A \ p, where p is a prime ideal, we get the map Spec(Ap) → Spec(A) of the

previous example.

57

5. Properties of morphisms

In this section we will study properties of morphisms, mainly those characterized by

universal properties. We will mainly get inspired in the properties of morphisms among

projective sets.

One of the main tool to study morphisms is to study their graphs. For that, we need

the notion of product of schemes. This is not as easy as one can think at a first glance.

For example, it is clear that we want the product of A1k by itself to be A2

k. But then we

cannot just take the cartesian product, since we will miss many points of A2k, namely the

irreducible curves that are not horizontal or vertical lines (see Example 3.19). For the

precise definition, we need to use the standard universal property in category theory:

Definition. The product of two S-schemes X,Y is a scheme X ×S Y endowed with two

S-morphisms (called projections) p : X ×S Y → X and q : X ×S Y → Y satisfying the

following universal property: For any S-scheme Z with two S-morphisms p′ : Z → X and

q′ : Z → Y there exists a unique S-morphism ϕ : Z → X ×S Y such that p′ = p ϕ and

q′ = q ϕ.

Since it is defined by a universal property, the product is unique up to isomorphism.

Using glueing techniques (see Example 4.10) it is enough to construct the product in the

affine case. We will construct the affine product in detail, and skip the details of how to

glue the different affine pieces to obtain general products, although we will study some

particular case.

Proposition 5.1. Given two C-algebras A,B, let S = Spec(C), X = Spec(A) and Y =

Spec(B). Then Spec(A ⊗C B) is the S-product of X and Y , where p : Spec(A ⊗C B) →Spec(A) is given by the homomorphism i : A → A ⊗C B defined as a 7→ a ⊗ 1 and

j : Spec(A ⊗C B) → Spec(B) is given by the homomorphism B → A ⊗C B defined as

g : b 7→ 1⊗ b.

Proof: By Proposition 4.19 (see also Remark 4.20), it is enough to prove that A ⊗C Bsatisfies the following universal property: For any C-algebra D and homomorphisms of

C-algebras f : A→ D and g : B → D there exists a unique homomorphism of C-algebras

ψ : A⊗C B → D such that f = ψ i and g = ψ j.

The first observation is that A ⊗C B has indeed a structure of C-algebra, in which

the multiplication as a ring is determined by (a ⊗ b)(a′ ⊗ b′) = (aa′) ⊗ (bb′). Now the

uniqueness of ψ comes immediately from

ψ(a⊗ b) = ψ((a⊗ 1)(1⊗ b)

)= ψ(a⊗ 1)ψ(1⊗ b) = ψ(i(a))ψ(j(b)) = f(a)g(b).

58

The existence is a consequence of the universal property defining the tensor product, since

the map A×B → D defined by (a, b) 7→ f(a)g(b) is bilinear.

Remark 5.2. The above proof shows that the tensor product of algebras has its own

universal property, different from the one giving only the module structure (which for-

gets about the product operation). For example, let B be an A-algebra, i.e. we have a

homomorphism of rings ϕ : A → B. Let us figure out the tensor product as A-algebras

of A[X1, . . . , Xn]/I and B[Y1, . . . , Ym]/J , where I and J are ideals in their corresponding

polynomial rings. Assume we have homomorphisms of A-algebras

f : A[X1, . . . , Xn]/I → C

g : B[Y1, . . . , Ym]/J → C.

While f is determined by giving the images c1 = f(X1), . . . , cn = f(Xn), the map g is

determined by the images c′1 = g(Y1), . . . , c′m = g(Ym) and a homomorphism of A-algebras

g : B → C. We can thus build a homomorphism of A-algebras

ψ : B[X1, . . . , Xn, Y1, . . . , Ym]→ C

by g and mapping each Xi to ci and each Yj to c′j . Obviously, for each p ∈ I, if ϕ(p) is the

polynomial obtained by mapping by ϕ the coefficients of p,we have (ϕ(p))(c1, . . . , cn) = 0.

Also, for each q ∈ J , we have q(c′1, . . . , c′m) = 0. Hence ψ factorizes through a map

ψ : B[X1, . . . , Xn, Y1, . . . , Ym]/I ′ → C

where I ′ is the ideal generated by all the polynomials ϕ(p) with p ∈ I and all the polyno-

mials q ∈ J . Moreover, there are natural maps

i : A[X1, . . . , Xn]/I → B[X1, . . . , Xn, Y1, . . . , Ym]/I ′

j : B[Y1, . . . , Ym]/J → B[X1, . . . , Xn, Y1, . . . , Ym]/I ′

and it is clear that ψ is the only homomorphism of A-algebras such that f = ψ i and

g = ψ j. This shows that we can take

A[X1, . . . , Xn]/I ⊗A B[Y1, . . . , Ym]/J = B[X1, . . . , Xn, Y1, . . . , Ym]/I ′.

We list here the most characteristic examples we are going to use:

1) When A = B = k, we get Spec(k[X1, . . . , Xn]/I) ×k Spec(k[Y1, . . . , Ym]/J) ∼=Spec(k[X1, . . . , Xn, Y1, . . . , Ym]/(I, J)), where (I, J) is the ideal generated by the polyno-

mials of I and J . In particular, Ank ×k Amk ∼= An+m. As Example 3.19 shows, A2k is not

59

the cartesian product of A1k with itself. Hence, in the category of schemes, the categorical

product is not the cartesian product.

2) Let ϕ : A → B be a ring homomorphism and let I ⊂ A[X1, . . . , Xn] be an ideal.

Then A[X1, . . . , Xn]/I ⊗AB is canonically isomorphic to B[X1, . . . , Xn]/J , where J is the

ideal generated by the polynomials of I after mapping their coefficients by ϕ.

To give a hint on how to glue together the affine pieces to obtain a product, we give

some examples of geometric flavor.

Example 5.3. If A is a k-algebra and we consider S = A[X0, . . . , Xn], then Proj(S) ∼=Pnk×kSpec(A), where the projections are given by the maps p : Proj(S)→ Proj(k[X0, . . . , Xn])

defined by p(p) = p ∩ k[X0, . . . , Xn] (since A is a k-algebra, there is an inclusion k ⊂ A)

and q : Proj(S)→ Spec(A) defined by q(p) = p∩A. In fact, Pnk ×k Spec(A) is obtained by

glueing together the affine pieces D(Xi)×k Spec(A) = Spec(k[X0

Xi, . . . , Xn

Xi])×k Spec(A) =

Spec(A⊗k k[X0

Xi, . . . , Xn

Xi]) = Spec(A[X0

Xi, . . . , Xn

Xi]), and this is the corresponding D(Xi) in

Proj(S).

Observe that in the above example, when taking A = k[Y1, . . . , Ym], we get Proj(S) ∼=Pnk ×k Amk , and the equations we use there are polynomials in k[X0, . . . , Xn;Y1, . . . , Ym, ]

that are homogeneous in the variables X0, . . . , Xn. If we want to glue together several

products Pnk ×k Amk to get Pnk ×k Pmk , then we will need to use in the final Pnk ×k Pmkequations that are polynomials in k[X0, . . . , Xn;Y0, . . . , Ym] that are bihomogeneous (of

some bidegree (a, b)), i.e. homogeneous (of degree a) in the variables X0, . . . , Xn and

homogeneous (of degree b) in the variables Y0, . . . , Ym. This can be done using the Segre

embedding (see Exercise 2.24), that allows to identify Pnk ×k Pmk with a projective subset

of Pnm+m+nk . Let us see how this works in a concrete example:

Example 5.4. Consider the curve X ⊂ P3k of Example 2.18. The first generator of I(X)

is X0X3 − X1X2, which is precisely the equation of the image of the Segre embedding

ϕ := ϕ1,1 : P1k × P1

k → P3k defined by ϕ

((t0 : t1), (s0 : s1)

)= (t0s0 : t0s1 : t1s0 : t1s1).

Hence, performing that substitution in the other generators of I(X), we get bihomogeneous

equations:

X31 −X2

0X2 = T 30 S

31 − T 2

0 T1S30 = T 2

0 (T0S31 − T1S

30)

X32 −X1X

23 = T 3

1 S30 − T0T

21 S

31 = T 2

1 (T1S30 − T0S

31)

X21X3 −X0X

22 = T 2

0 T1S31 − T0T

21 S

30 = T0T1(T0S

31 − T1S

30)

As a conclusion, ϕ−1(X) is defined by just one equation, namely T0S31 − T1S

30 , which is of

bidegree (1, 3). It is clear that, in the same way, any projective subvariety of the image of a

Segre embedding of Pn×Pm in Pnm+n+mk can be identified with a subset of Pn×Pm defined

by bihomogeneous polynomials (in fact any equation of degree d produces a bihomogeneous

60

equation of bidegree (d, d)). What this example shows is that sometimes we can reduce

the bihomogeneous polynomials we get, and get smaller polynomials, that now can have

bidegree (a, b) with a 6= b. Precisely this observation gives the hint of how the converse

is true: any subset of Pn × Pm defined by bihomogeneous polynomials corresponds to a

projective subset of the image of the Segre variety in Pnm+n+mk . Indeed, it is clear how to

get a homogeneous equation of degree d in Pnm+n+mk from any bihomogeneous equation of

bidegree (d, d) in Pn × Pm. If instead we start with a bihomogeneous equation of bidegree

(a, b) with a 6= b, we can reverse the above procedure in the example. For example, if

we start with the bihomogeneous equation T0S31 − T1S

30 , since is degree in the variables

T0, T1 is one, strictly smaller than the degree three in the variables S0, S1, we multiply that

equation with all possible monomials of degree two in the variables T0, T1, so that we now

get a series of equivalent equations, now all of them of bidegree (3, 3), which come now

from homogeneous polynomials of degree three in X0, X1, X2, X3. Observe that, in order

to get equivalent equations it is enough to multiply by powers of all the variables instead

of by all monomials of the required degree. For example, if we forget the last of the above

three equations (obtained when multiplying by T0T1 the equation T0S31 −T1S

30), we would

get, instead of I(X) the ideal

(X0X3 −X1X2, X31 −X2

0X2, X32 −X1X

23 )

whose saturation is precisely I(X) (check it as an exercise).

As a conclusion, there is in general a bijection among the projective subsets of the im-

age of the Segre embedding of Pnk×Pmk and the subsets of Pnk×Pmk defined by homogeneous

polynomials. In fact, we can extend all the theory we did for graded rings to bigraded

(or in general multigraded) rings T =⊕

a,b Ta,b. For example, for subsets of Pnk × Pmk we

would also have Hilbert polynomials (now in two variables), and for any bigraded ring T

we could define an analogue of Proj, now consisting on the sets of bihomogeneous prime

ideals (i.e. generated by bihomogeneous elements) not containing neither T ′ =⊕

a>0 Ta,0nor T ′′ =

⊕b>0 T0,b; the scheme structure is given in a similar way. Then, for a pair of

graded k-algebras S, S′, the schematic product Proj(S) ×k Proj(S′) would be the scheme

we would get from the bigraded ring S ⊗k S′.

Observe that we can regard the product of S-schemes from other points of view:

Definition. The fiber product of two morphisms f : X → S and g : Y → S is a scheme

X ×S Y together with two morphisms p : X ×S Y → X and q : X ×S Y → Y such that

f p = g q, and satisfying the following universal property: For any other scheme Z with

morphisms p′ : Z → X and q′ : Z → Y such that f p′ = g q′, there exists a unique

morphism ϕ : Z → X ×S Y such that p′ = p ϕ and q′ = q ϕ.

61

Definition. If X is a scheme over S and S′ → S is a morphism, then the fiber product

X ×S S′ is a scheme over S′ (with the second projection) called the base extension of X

from S to S′.

Many properties are preserved by base extension:

Proposition 5.5. Let X → S be a closed embedding and let S′ → S be a morphism.

Then the projection X ×S S′ → S′ is a closed embedding.

Proof: By definition, X → S is locally a morphism Spec(A/I) → Spec(A) induced by

the projection A → A/I, while the morphism S′ → S is locally a morphism Spec(B) →Spec(A) induced by a ring homomorphism ψ : A → B. By part 2) of Remark 5.2 (when

n = 0), the projection X ×S S′ → S′ is locally the morphism Spec(B/J) → Spec(B)

induced by the natural projection B → B/J , where J is the ideal generated by ψ(I). As

a consequence, X ×S S′ → S′ is a closed embedding.

Example 5.6. If X is a scheme over a field k and we have a field extension k ⊂ k,

then the base extension corresponding to Spec(k) → Spec(k) produces a scheme over

k. For instance, if X = Spec(k[X1, . . . , Xn]/I), part 2) of Remark 5.2 implies that the

base extension is X = Spec(k[X1, . . . , Xn]/I), where I is the ideal generated by a set

of generators of I, but now regarded as polynomials in k[X1, . . . , Xn]. For example, the

base extension induced by the inclusion R ⊂ C produces the complexification of real

schemes. This kind of base extensions is very used in arithmetic geometry. If k is a field of

characteristic p > 0, then it is typical to consider the extension produced by the Frobenius

map k → k defined by a 7→ ap. Also, if X is a scheme over Z (for example, a variety

defined by polynomials whose coefficients are integers), then its reduction modulo p is its

base extension determined by the natural quotient Z→ Zp.

Example 5.7. We can apply Proposition 5.5 when we have two closed embeddings

X → Z and Y → Z, in which we expect the fiber product to be the intersection in

Z of the subschemes X and Y . As in that proof, the local situation corresponds to

two subschemes Spec(A/I) and Spec(A/J) of an affine scheme Spec(A). Since the ideal

generated by the image by A → A/J of I is (I + J)/J , and the quotient of A/J by it is

canonically isomorphic to A/(I + J), we conclude that the intersection of the two closed

subschemes is Spec(A/I+J). As a consequence, the intersection of two closed subschemes

of a scheme is defined by the sum of their corresponding ideal sheaves. This corresponds

to the original idea we got in Example 1.4.

Another application of these new point of view of fiber products is the following:

Definition. The fiber of a morphism ϕ : X → Y over a point q ∈ Y is the fiber product

of ϕ and the morphism Spec(k(q))→ Y defined in Example 4.28.

62

Observe that then, properties of morphisms preserved by base extension can be re-

garded as properties on the fibers as schemes over a field.

Example 5.8. Consider the natural homomorphism k[T ] → k[X,Y, T ]/(X2 − TY ),

defining a morphism Spec(k[X,Y, T ]/(X2 − TY )

)→ A1

k. Geometrically, this can be

regarded as a fibration over A1k in which the fiber over a point t ∈ A1

k is the conic V (X2−tY ). Observe, that for a general t (precisely when t 6= 0), the fiber is a smooth conic

(namely a parabola), while for the special value t = 0 we get a double line. Let us see

that, using schemes, we get exactly that situation. Indeed, for a point pt = (T − t), we

have k(pt) = k[T ]/(T − t) ∼= k (and the isomorphism is given by evaluating polynomials in

t), hence the fiber over pt = Spec(k[T ]/(T − t)

)is (see Remark 5.2)

Spec( k[X,Y, T ]

(X2 − TY )⊗k[T ]

k[T ]

(T − t)) ∼= Spec

( k[X,Y ]

(X2 − tY )

)which is in fact a parabola if t 6= 0 and a double line if t = 0. But there exists another

point in A1k, namely the zero ideal p = (0). In that case, k(p) = k(T ), and its fiber is now

Spec(k(T )[X,Y ]

(X2−TY )

). This is now a parabola in the affine space A2

k(T ). Hence, the fiber at this

point represents what happens to a general point.

Example 5.9. We can put Example 5.8 in a different perspective. We can start with the

double line Spec(k[X,Y ]/(X2)

), and we wonder whether it is possible to find a smooth

deformation of it, i.e. a morphism to some space in which one of the fibers is the double

line but any general fiber is a smooth conic. A first attempt would be what is called a

first-order infinitesimal deformation, i.e. to consider a morphism to Spec(k[T ]/(T 2)

), and

of course we can take the morphism Spec(k[X,Y, T ]/(T 2, X2 − TY )

)→ Spec

(k[T ]/(T 2)

)defined by the natural homomorphism k[T ]/(T 2)→ k[X,Y, T ]/(T 2, X2 − TY ). This mor-

phism still gives few information, since Spec(k[T ]/(T 2)

)has only one point, and its fiber

is again Spec(k[X,Y ]/(X2)

). The same happens if we take higher order deformations

Spec(k[X,Y, T ]/(Tm, X2 − TY )

)→ Spec

(k[T ]/(Tm)

). We can, however, take the limit

case and consider the morphism Spec(k[[T ]][X,Y ]/(X2−TY )

)→ Spec(k[[T ]]). This time

Spec(k[[T ]]) has two different points, the closed point (T ), whose fiber is again the double

line, but we also have the generic point (0), and its fiber is the parabola V (X2 − TY ) in

the affine plane over k((T )). Hence, in this formal deformation, even if we do not have

actual smooth fibers (in the sense of fibers of a closed point), we get a smooth parabola as

a generic fiber.

Remark 5.10. The interpretation of the above example is related with the theory of

moduli spaces. Roughly speaking, a moduli space is a set parametrizing all the objects of

the same type. These objects can be, for example, conics (we considered them in the affine

63

plane, but it works better in the projective plane). Of course we want moduli spaces to

have a nice structure (for example, the set of conics in P2k can be identified with a projective

space P5k, if we also allow double lines). The way of giving a nice structure is to ask to

be a scheme M satisfying the following universal property: giving a morphism X → T

such that the fibers of the closed points are objects of the type we want to parametrize is

equivalent to give a morphism T →M in which the image of each closed point is precisely

the point parametrizing the fiber of the morphism. Then Example 5.8 shows that, if we

want the moduli space M of plane conics to be complete, we actually need to consider

in it double lines, since otherwise we would get a map A1k \ p0 → M that cannot be

extended to the point p0. On the other hand, the universal property for M shows that

Example 5.9 consists of giving maps Spec(k[T ]/(Tm) → M . The case m = 2 consists of

giving a point of M (the one corresponding to the double line) and a tangent vector at it.

For arbitrary m, this is just giving infinitesimal data at the point up to order m. The final

part of Example 5.9 is now giving a map Spec(k[[T ]])→M , i.e. a germ of a curve passing

through the point corresponding to the double line. The process of deformation is usually

like that: start with infinitesimal deformations of first, second,... order, and eventually a

formal deformation, which then one needs to integrate in order to get an actual curve in

the moduli space.

The fact that a moduli space has a scheme structure and is not only an algebraic set

is due to the fact that it is defined by a universal property. For example, the fiber of a

morphism is defined as a fibered product, hence by a universal property, and Example

5.8 shows that in fact fibers of a morphism can be non-reduced. And indeed also moduli

spaces can also be non-reduced.

We briefly recall now how to prove that the image of a projective set is a projective set,

and we will see the ingredients we need in order to generalize it to the general framework

of schemes.

Theorem 5.11. Let ϕ : X → Y be a morphism of projective sets. Then ϕ is a closed

map, i.e., it maps projective sets to projective sets.

Proof: It is clear that we can replace Y by a projective space. On the other hand, since

ϕ is locally defined by polynomials, it is clear that locally, its graph is locally defined

by bihomogeneous polynomial, and hence (see Example 5.3) it is locally closed. Since

being closed is a local property, the graph itself is closed. It is thus sufficient to prove

that the second projection Pnk × Pmk → Pmk is a closed map. Again, since being closed

is a local property, we will prove it, because it simplifies the writing, for the projection

q : Pnk × Amk → Amk , and in fact the proof works for any projection Pnk × Y → Y .

We thus take a closed set Z ⊂ Pnk×Amk , and let F1, . . . , Fr ∈ k[X0, . . . , Xn, Y1, . . . , Ym]

64

a set of polynomials defining Z (they are homogeneous as polynomials in the indeterminates

X0, . . . , Xn, according again to Example 5.3). Now, a point (b1, . . . , bm) ∈ Amk is in the

image of Z if and only if the set of homogeneous polynomials

S :=F1(X0, . . . , Xn, b1, . . . , bm), . . . , Fr(X0, . . . , Xn, b1, . . . , bm)

vanish at some point of Pnk . By the projective Nullstellensatz, this is equivalent to say that

the ideal generated by X does not contain all the homogeneous polynomials of large enough

degree. Therefore, we can express the image of Z as the set of points (b1, . . . , bm) ∈ Amksuch that the homogeneous part of degree d >> 0 of the ideal generated by S is not

the whole k[X0, . . . , Xn]d. In other words, Z =⋂d>>0 Zd, where Zd is the set of points

(b1, . . . , bm) ∈ Amk such that the homogeneous part of degree d of the ideal generated by

S is not the whole k[X0, . . . , Xn]d. Therefore, it is enough to see that, for d >> 0, the set

Zd is closed. We can also describe Zd as the set of points (b1, . . . , bm) ∈ Amk such that the

set

S′ :=MFi(X0, . . . , Xn, b1, . . . , bm) | i = 1, . . . , r and M is a monomial of degree d− di

(where di is the degree of Fi as a polynomial in X0, . . . , Xn) do not generate k[X0, . . . , Xn]das a vector space. But the coordinates of the elements of S′ with respect to the basis of

k[X0, . . . , Xn]d consisting of the set of all monomials of degree d are just zero or the coeffi-

cients of some F1(X0, . . . , Xn, b1, . . . , bm), which are polynomial expressions in b1, . . . , bm.

Since the condition that S′ does not generate k[X0, . . . , Xn]d is equivalent to the vanishing

of the maximal minors of its matrix of coordinates, then a point (b1, . . . , bm) ∈ Amk is in

Zd if and only if some polynomial expressions in b1, . . . , bm are zero. This shows that Zdis closed, completing the proof.

To try to imitate the above proof, we need first the right definition of graph:

Definition. The graph morphism of an S-morphism ϕ : X → Y is the unique morphism

γϕ : X → X ×S Y such that p γϕ = idX and q γϕ = ϕ. The diagonal morphism of an

S-scheme X is the graph map δX : X → X ×S X of idX .

Contrary to what one could think, the graph is not always a closed set in the product.

Example 5.12. Consider the diagonal map of the scheme constructed in Example 4.9.

Since it is formed by glueing two pieces, Spec(k[X]) and Spec(k[Y ]), the product will be

formed by glueing four pieces. We concentrate in the piece Spec(k[X]) ×k Spec(k[Y ]) =

Spec(k[X,Y ]). Since Spec(k[X]) and Spec(k[Y ]) are glued along Spec(k[X]X) ∼= Spec(k[Y ]Y )

(sending X to Y ), the restriction of the image of diagonal to Spec(k[X,Y ]) is V (X − Y ) \V (X,Y ), which is not closed.

Definition. An S-separated scheme is an S-scheme X such that the diagonal map δX :

X → X ×S X is a closed embedding.

65

Proposition 5.13. If Y is an S-separated scheme, then the graph map of any morphism

ϕ : X → Y of S-schemes is a closed embedding.

Proof: It is easy to check that the graph map is a fibered product

Xγϕ−→ X ×S Yyϕ yϕ×idY

YδY−→ Y ×S Y

Since, by assumption, the diagonal map δY is a closed embedding, the result follows now

from Proposition 5.5.

Exercise 5.14. Show that, if ϕ,ψ : X → Y are two morphisms of S-schemes that coincide

on a non-empty open set U ⊂ X, X is irreducible and Y is separated, then ϕ = ψ. Show

that the result is not true if we do not assume Y separated.

The second ingredient that we used in the proof of Theorem 5.11 was that the second

projection Pnk × Y → Y was closed for any Y (obviously, the same holds when replacing

Pnk with any projective set).

Definition. An S-scheme X is universally closed if for any S-scheme Y the projection

X ×S Y → Y is closed. A proper scheme over k is a universally closed scheme of finite

type over k.

Separation and properness are closely related:

Theorem 5.15. Let X be a scheme of finite type over k. Then

(i) X is separated over k if and only if for each DVR A with quotient field K, any

morphism Spec(K)→ X extends to Spec(A) in at most one way.

(ii) X is proper if and only if for each DVR A with quotient field K, any morphism

Spec(K)→ X extends uniquely to Spec(A).

Proof: See [Ha], Theorem 4.3, Theorem 4.7 and Exercise 4.11 of Chapter II.

We can use this result to prove that general projective schemes are proper (observe

that proper schemes over k are separated):

Proposition 5.16. Let I ⊂ k[X0, . . . , Xn] be a homogeneous ideal, and consider S =

k[X0, . . . , Xn]/I. Then Proj(S) is a proper k-scheme.

Proof: We will prove it applying Theorem 5.15. Let A be a DVR and let K be its quotient

field. Given a morphism ϕ : Spec(K)→ Proj(S), the image of ϕ is one point, say p, so that

66

it will be in one D(Xi), and we will assume for simplicity i = 0. We thus have a morphism

ϕ : Spec(K)→ D(X0), which is equivalent to a k-homomorphism S(X0) → K, and this is

determined by the images α1, . . . , αn of, respectively, X1

X0, . . . , Xn

X0. If the valuation of all

the nonzero α1, . . . , αn are non-negative, then α1, . . . , αn are in A and hence ϕ extends to

Spec(A). Otherwise, let αi be the nonzero element of minimum valuation. In particular,Xi

X0is not in the kernel of S(X0) → K, which is the prime ideal corresponding to p; hence p

is also in D(Xi). Regarding now ϕ as a morphism ϕ : Spec(K)→ D(Xi), it is determined

by sending X0

Xito 1

αiand

Xj

Xito

αj

αi. All these images are now in A, so that ϕ extends to

Spec(A).

The following result shows that we cannot recover a proper scheme (in particular a

projective scheme) from its ring of regular functions.

Proposition 5.17. Let X be a proper irreducible k-scheme. Then the only regular func-

tions on X are the constant maps.

Proof: A regular function on a scheme of finite type over k is, by definition, a morphism

ϕ : X → A1k, which we will regard as a morphism ϕ : X → P1

k whose image is contained

in D(X0). Since P1k is separated, Proposition 5.13 implies that the graph of ϕ is closed in

X ×k P1k. Since X is proper, this implies that the image of this graph, i.e. the image of ϕ

is closed. It is also irreducible (because X is), and it is not the whole P1k, hence necessarily

the image is one single point, as wanted.

Even if a general morphism is not closed, we can still get much information.

Proposition 5.18. Let ϕ : X → Y be a morphism of schemes and let X ′ ⊂ X be an

irreducible closed subset corresponding to the point p ∈ X. Let Y ′ ⊂ Y be the closure of

ϕ(X ′). Then Y ′ is irreducible and its generic point is ϕ(p).

Proof: We need to prove the equality ϕ(p) = ϕ(p). For that, it will suffice to prove

that a closed set of Y contains ϕ(p) if and only if it contains ϕ(p). But obviously, ϕ(p) ∈ Zif and only if p is in the closed set ϕ−1(Z), and this is in turn equivalent to p ⊂ ϕ−1(Z),

i.e. ϕ(p) ⊂ Z, as wanted.

Definition. A dominant morphism is a morphism whose image is dense, or equivalently,

the inverse image of any non-empty open set is a non-empty open subset.

Example 5.19. In general, it is not true that the image contains an open set. For

example, we can consider the morphism ϕ : Spec(k[[T ]]) → Spec(k[T ]) = A1k induced by

67

the inclusion k[T ] ⊂ k[[T ]]. Its image is the set (0), (T ), which is dense in A1k, hence ϕ

is dominant, but the image of ϕ does not contain any non-empty open subset.

Let ϕ : X → Y be a dominant morphism of irreducible schemes. Then there is a

natural inclusion K(Y ) → K(X). In fact, for any class (V, g), with g ∈ OY (V ) we can

associate the corresponding (ϕ−1(V )), g), where g is the image by g of the homomorphism

OY (V )→ OX(ϕ−1(V )).

Theorem 5.20. Let X,Y be two k-irreducible reduced schemes of locally finite type.

Then K(X) and K(Y ) are isomorphic k-algebras if and only if there exist non-empty open

sets U ⊂ X and V ⊂ Y that are isomorphic to each other.

Proof: It is obvious that if X,Y contain isomorphic open subsets U, V , then K(X) =

K(U) ∼= K(V ) = K(Y ), so that we only need to check the converse.

Take open subsets U ⊂ X and V ⊂ Y such that U ∼= Spec(k[X1, . . . , Xn]/I) and V ∼=Spec(k[Y1, . . . , Ym]/J). Since, by assumption, K(U) = K(X) and K(V ) = K(Y ) are iso-

morphic k-algebras, there is an isomorphism ψ between the quotient fields of k[X1, . . . , Xn]/I

and the quotient field of k[Y1, . . . , Ym]/J . By taking common denominators of the images,

we can assume ψ(Xi) = gig for i = 1, . . . , n, and ψ−1(Yj) =

fjf

for j = 1, . . . ,m. If

ψ(f) = qgs and ψ−1(g) = p

fr , then ψ restricts to an isomorphism (k[X1, . . . , Xn]/I)f p →(k[Y1, . . . , Ym]/J)gq. This defines an isomorphism between the open subsets DX(f p) ⊂U ⊂ X and DY (gq) ⊂ V ⊂ Y .

68

6. Dimension

In this section we want to extend to any scheme the definition (and properties) of

dimension that we got for projective sets. That definition was based on the existence of

the Hilbert polynomial, which is a specific property of projective sets. However we will see

that it is possible to give an alternative definition that preserves the main properties that

we have seen. The key point is the following:

Proposition 6.1. The dimension of a projective set X ⊂ Pnk is the maximal length r of

a chain

Z0 ⊆/ Z1 ⊆/ . . . ⊆/ Zr

of irreducible projective subsets of X.

Proof: Let r′ be the dimension of X. By Proposition 2.20(iv), X possesses an irreducible

component Zr′ of dimension r′. By 2.20(ii),, the intersection of Zr′ with a hypersurface

not containing Zr′ has dimension r′−1, so that it contains an irreducible component Zr′−1

of dimension r′ − 1. Iterating the process as long as we get non-empty intersections, we

get a chain

Z0 ⊆/ Z1 ⊆/ . . . ⊆/ Zr′

of irreducible sets, in which each Zi has dimension i. This shows dim(X) ≤ r.On the other hand, take a chain of maximum length. By 2.20(iii), the strict inclusion

Zi−1 ⊆/ Zi implies dim(Zi−1) < dim(Zi). Hence we have a chain of inequalities

dim(Z0) < dim(Z1) < . . . < dim(Zr).

Since dim(Z0) ≥ 0, it follows dim(Zr) ≥ r, hence dim(X) ≥ r, so that we get the wanted

equality.

This allows to define the notion of dimension for any topological space. Of course

it will make sense only for topologies like Zariski topology. For example, the following

definition will give dimension zero for any topological space based on the usual topology

(in which the only irreducible sets are the points).

Definition. The dimension of a topological space X is the maximum length r of a chain

Z0 ⊆/ Z1 ⊆/ . . . ⊆/ Zr

of closed irreducible sets of X. Similarly, the codimension of an irreducible set Z ⊂ X is

the maximum length s of a chain of

Z = Z0 ⊆/ Z1 ⊆/ . . . ⊆/ Zs

69

of closed irreducible sets of X.

Remark 6.2. If X = Spec(A) then dim(X) is the Krull dimension of the ring A, which

is defined as the maximal length of a chain of prime ideals of A. If A is an integral finitely

generated k-algebra, then dim(A) = transdegk(K), where K is the quotient field of A. For

example, the dimension of Ank is n. On the other hand, if Z ⊂ Spec(A) is the irreducible

set corresponding to a prime ideal p. i.e. Z = V (p), then the codimension of Z is the height

of the prime ideal p, i.e. the maximum length of a chain of prime ideals of A contained in

p or, equivalently, the dimension of Ap.

With this definition we can readily generalize the following properties of the dimension

of projective sets (compare with Proposition 2.20):

Exercise 6.3. Let X be a topological space. Prove:

(i) If Y ⊂ X, then dim(Y ) ≤ dim(X) and, if Y is irreducible, dim(Y ) + codim(Y,X) ≤dim(X).

(ii) If X = X1∪. . .∪Xs, with each Xi a closed set, then X = maxdim(X1), . . . ,dim(Xs).(iii) If X is irreducible, then any proper closed subset Y ⊆/ X has dimension strictly smaller

than the dimension of X.

In the context of schemes, we can find several unexpected situations:

Example 6.4. Let X = Spec(k[[T ]]). Then X consists only of two points, p = (T ) and

M = (0), hence there are only two irreducible subsets, namely the closed point p and

the closure of M, which is X. Hence X has dimension one, even if it consists only of two

points. Moreover, we can consider the (dense) open set U = X \p, and now dim(U) = 0.

Observe also that the map Spec(k((T ))⊕ k

)→ X determined by k[[T ]]→ k((T ))⊕ k, in

which the first map is the inclusion and the second one is the evaluation at 0, is a surjective

map even if Spec(k((T )) ⊕ k

)(which is the disjoint union of Spec

(k((T ))

)and Spec(k),

according to Exercise 4.11) has dimension zero.

Since, for studying schemes, we typically restrict to affine open subschemes, we will

need to compare dimensions when restricting to open sets. We have seen in the above

example that strange things must happen, but let us see what we can say in general:

Proposition 6.5. Let U ⊂ X be a non-empty open subset of an irreducible topological

space. Then the restriction to U and the closure in X define inverse bijections among the

set of irreducible closed subsets of X meeting U and the set of irreducible closed subsets

of U . As a consequence, dim(U) ≤ dim(X).

Proof: We know from Lemma 1.11 that the restriction to U of an irreducible subset of

Z ⊂ X meeting U is irreducible in U .

70

Next we prove that the closure of an irreducible closed set W of U is also irreducible.

Indeed, if W ⊂ Z1 ∪ Z2, then also W ⊂ (Z1 ∩ U) ∪ (Z1 ∩ U), which implies W ⊂ Zi ∩ Ufor some i = 1, 2, thus W ⊂ Zi and hence W ⊂ Zi.

We now need to prove that both operations are inverse to each other.

Let us prove first that, if Z is an irreducible set of X meeting U , then Z ∩ U = Z.

From the decomposition Z = (Z ∩ U) ∪ (Z \ U) we get Z = Z ∩ U ∪ (Z \ U). Since Z is

irreducible and does not coincide with Z \ U , we get Z ∩ U = Z.

Finally, let us see that, for any closed set W of U (not necessarily irreducible), we

have W ∩ U = W . This is so because any element of W ∩ U is clearly in the closure of W

in U , and this is W , since W is closed in U .

The final part of the statement is immediate, since any strict chain of irreducible

closed sets W0 ⊆/ W1 ⊆/ . . . ⊆/ Wr in U produces, by taking closures, a strict chain of

irreducible closed sets W0 ⊆/ W1 ⊆/ . . . ⊆/ Wr in X.

The key tool to have equality in the case of schemes with more geometrical flavor (such

as those of finite type over a field) will be to generalize the following result for projective

spaces:

Proposition 6.6. A projective set X ⊂ Pnk can be written as X = V (F ) for some non-

constant homogeneous polynomial F ∈ k[X0, . . . , Xn] if and only if all the irreducible

components of X have dimension n− 1.

Proof: If X = V (F ) and F = F a11 . . . F arr is the decomposition of F into irreducible factors,

it is clear that X = V (F1) ∪ . . . ∪ V (Fr) is the decomposition of X into its irreducible

components, and each of these components has dimension n− 1 by Proposition 2.20(ii).

Reciprocally, let X = X1 ∪ . . . ∪ Xr be the decomposition of X into its irreducible

components, and assume that each Xi has dimension n− 1. If we could write X = V (Fi),

then X = V (F1 . . . Fr). Hence it is enough to prove that an irreducible subset X ⊂ Pnk of

dimension n− 1 can be written as X = V (F ).

Since X ⊆/ Pnk , then I(X) 6= 0. We can thus find a nonconstant polynomial in I(X)

and, since I(X) is a prime ideal (because X is irreducible), some irreducible factor F of it

is also in I(X). Hence X ⊂ V (F ), which immediately implies the equality.

Remark 6.7. The reader could (and should) wonder why we forgot for a while of

schemes. The reason is that Proposition 6.6 is no longer true for projective schemes. For

example, we could consider the ideal I = (X1) ∩ (X21 , X2) ⊂ k[X0, X1, X2]. The scheme

Proj(k[X0, X1, X2]/I) represents the union of the line V (X1) plus the nonreduced scheme

71

consisting of the point (1 : 0 : 0) and the direction of the line V (X2). However, as a

topological space, Proj(k[X0, X1, X2]/I) is homeomorphic to Proj(k[X0, X1, X2]/(X1)

),

because any prime ideal containing I contains (X1). Hence, Proj(k[X0, X1, X2]/I) is irre-

ducible of dimension one, but I cannot be generated by a single homogeneous polynomial

F . We state the right result for subschemes of the projective space.

Proposition 6.8. Let I ⊂ k[X0, . . . , Xn] be a homogeneous ideal.

(i) If I is primary and dim(V (I)) = n − 1, then there exists a homogeneous polynomial

F ∈ k[X0, . . . , Xn] of positive degree and a positive integer m such that I = (Fm).

(ii) I can be generated by a homogenous polynomial of positive degree if and only if, for

each primary component Ii of I, the projective set V (Ii) has dimension n− 1.

Proof: To prove (i), observe that the fact that I is primary implies that V (I) is irreducible

and, since it has dimension n−1, Proposition 6.6 implies that√I = IV (I) = (F ) for some

irreducible homogeneous polynomial. Observe that√I cannot contain any polynomial G

not divisible by F , since otherwise V (I) = V (√I) ⊂ V (F,G), and V (F,G) has dimension

n − 2 by 2.20(ii) (because V (G) does not contain the irreducible set V (F ) of dimension

n − 1). Since√I = (F ), there exists m such that Fm ∈ I. Fix the minimum such m,

and let us see that I is in fact generated by Fm. For any homogeneous H ∈ I, we write

H = F sG, with G not divisible by F . Hence G is not in√I, and the primarity of I implies

that F s is in I. Now the minimality of m implies s ≥ m, so that H ∈ (Fm), as wanted.

Part (ii) is now an easy consequence of (i). First, if I = (F ) for some homogeneous

polynomial of positive degree, we consider its decomposition F = Fm11 . . . Fmr

r into irre-

ducible factor. Then I = (Fm11 )∩ . . .∩ (Fmr

r ) is the primary decomposition of I, and each

V (Fmii ) has dimension n− 1.

Reciprocally, if I = I1 ∩ . . . ∩ Ir is the primary decomposition and each V (Ii) has

dimension n− 1, we obtain from (i) that each Ii is generated by some Fmii . Since the Fi’s

are coprime to each other, it follows I = (Fm11 . . . Fmr

r ), as wanted.

Remark 6.9. Trying to translate directly Proposition 6.6 to the affine space shows the

kind of difficulties we should overcome. For example, given an affine set X ⊂ Ank and re-

garding Ank as the open subset D(X0) of Pnk , we can consider the Zariski closure of X in Pnk ,

defined by the homogenization of I(X) (the homogenization of an ideal I ⊂ k[X1, . . . , Xn]

is the ideal I ⊂ k[X0, . . . , Xn] consisting of all polynomials F ∈ k[X0, . . . , Xn] such that

all their homogeneous components Fi satisfy Fi(1, X1, . . . , Xn) ∈ I). Now, if all the ir-

reducible components of X have dimension n − 1, by Proposition 6.5, all the irreducible

components of X have dimension at least n− 1. Since they cannot have dimension n (by

Proposition 2.20(iii)), then all the irreducible components of X have dimension exactly

72

n − 1 and thus X = V (F ) for some homogeneous F ∈ k[X0, . . . , Xn]. Hence X = V (f),

where f = F (1, X1, . . . , Xn). However, if we try to prove that, when f ∈ k[X1, . . . , Xn]

has positive degree, V (f) has dimension n− 1, we find problems: the natural way(∗) to do

so would be to use that V (F ) (where F is the homogenization of f) has dimension n− 1,

so that there exists a chain

Z0 ⊆/ Z1 ⊆/ . . . ⊆/ Zn−1

of irreducible open sets in V (F ). But, when intersecting with D(X0), some of the closed

sets could become empty. We would need to know that we can find such a chain all of

whose members meet D(X0). In other words, we would need the chain to have the point

Z0 in our original V (f). In fact, we can proof much more: when looking for a chain

determining the dimension of a projective set, we can find one containing any irreducible

subset we want. For that, we will need the following generalization of Proposition 2.20(ii)

and Proposition 6.6:

Theorem 6.10. Let X ⊂ Pnk be an irreducible projective set of dimension r. Then, if

F is a nonconstant homogeneous polynomial not in I(X), any irreducible component of

VX(F ) has dimension r − 1.

Proof: See [A] Theorem 7.21.

Remark 6.11. We could have stated also the affine version of it ([Sh], Chapter I, 6,

Theorem 5), which is in fact the most useful for our purposes. However, its proof is much

more algebraic than the one given in [A], and requires a more sophisticated definition of

dimension. The algebraic statement for these results is the so-called Krull’s Hauptidealsatz

([AM], Corollary 11.17): Let A be a noetherian ring, and let f ∈ A be an element which

is neither a zero divisor nor a unit. Then every minimal prime ideal p containing f has

height 1. However, we will need first to find conditions for which the height of a prime

ideal has the right geometric interpretation (i.e. when the codimension has the expected

meaning). We will check this for finitely generated k-algebras (using precisely Theorem

6.10).

Example 6.12. Theorem 6.10 deals only with non-embedded components. Consider, for

example, X = Spec(k[X0, X1, X2, X3]/I), where I = (X0X3 − X1X2, X31 − X2

0X2, X32 −

X1X23 , X

21X3 − X0X

22 ). In other words, X is the affine cone in A4

k of the curve given in

Exercise 2.18. As that exercise shows, VX(X1−X2) is, as a scheme, the union of four lines

(∗) If we want to use Commutative Algebra, we could use the fact that, when taking f

irreducible, the transcendence degree of the quotient field of k[X1, . . . , Xn]/(f) is n− 1.

73

plus a scheme supported in (0, 0, 0, 0). As in Remark 6.8, that last part is not actually an

irreducible component, but its presence is a strange phenomenon.

We can improve now Proposition 6.1:

Theorem 6.13. Let Y ⊆/ X ⊂ Pnk be two irreducible projective sets of respective dimen-

sions s, r. Then there exists a chain

Y = Zs ⊆/ Zs+1 ⊆/ . . . ⊆/ Zr−1 ⊆/ Zr = X

of irreducible closed sets.

Proof: Since I(X) ⊆/ I(Y ), we can take a homogeneous polynomial F ∈ I(Y ) \ I(X).

Therefore we have Y ⊂ X ∩ V (F ), and all the components of X ∩ V (F ) have dimension

r − 1, by Theorem 6.10. Since Y is irreducible, it is contained in at least one component

Zr−1 of X ∩ V (F ). If Y ⊆/ Zr−1 (which is equivalent to s < r − 1), we repeat the process,

which we will finish when we will arrive to Zs, which will be necessarily Y .

With this, we can now prove the equality in the last statement of Proposition 6.5

when passing from affine sets to their Zariski closures in the projective space and thus we

can also rephrase Theorem 6.13 for affine sets:

Corollary 6.14. If X ⊂ Ank is an irreducible affine set of dimension r, then:

(i) If X ⊂ Pnk is the Zariski closure of X, dim(X) = r.

(ii) If f /∈ I(X) is a non-constant polynomial, X ∩V (f) is either empty or any irreducible

component of it has dimension r − 1.

(iii) If Y ⊂ X is irreducible of dimension s, there exists a chain

Y = Ws ⊆/ Ws+1 ⊆/ . . . ⊆/ Wr−1 ⊆/ Wr = X

of irreducible closed sets. In other words, codim(Y,X) = dim(X)− dim(Y ).

Proof: For part (i), take p ∈ X. If r = dim(X), we know from Theorem 6.13 that we

can find a chain Z0 ⊆/ Z1 ⊆/ . . . ⊆/ Zr of irreducible subsets of X in which Z0 = p and

Zr = X. Restricting to X, we get a strict chain of length r of irreducible subsets of X, so

that dim(X) ≥ r, and Proposition 6.5 implies that we have equality.

In the hypothesis of (ii), the homogenization F of f is now a non-constant polynomial

not in I(X), so that, by Theorem 6.10, any component of X ∩ V (F ) has dimension r − 1.

Now, by Proposition 6.5, the irreducible components of X ∩ V (f) are the restriction to

D(X0) of the components of X ∩ V (F ) meeting D(X0) and, by part (i), the dimension of

74

these restrictions have still dimension r−1 (because their closures are, by Proposition 6.5,

the corresponding components of X ∩ V (F )). This proves (ii).

For part (iii), we can either repeat the proof of Theorem 6.13 or, as we will do, use

it. So we take the Zariski closures of Y and X, which we will have respective dimensions

s and r, by part (i). We now apply Theorem 6.13 and get a chain

Y = Zs ⊆/ Zs+1 ⊆/ . . . ⊆/ Zr−1 ⊆/ Zr = X.

Restricting to X we get the wanted chain on X.

Remark 6.15. Theorem 6.13 and Corollary 6.14(ii) are thus saying that all maximal

chains of irreducible closed subsets from an irreducible closed set Y to another irreducible

closed set X have the same length (a catenary topological space is a topological space with

that property). What is important is that we are saying is that such length (which, by

definition, is the codimension of Y in X) is precisely dim(X) − dim(Y ), so that we have

equality in the second statement of Exercise 6.3(i). We did that for projective and affine

subsets, but it obviously holds for subschemes, since their underlying topological space are

affine sets.

Observe that, in the affine case, we can interpret this in the following way. Take any

ideal I and consider the ring A = k[X1, . . . , Xn]/I. Two irreducible subsets Y ⊂ X of

Spec(A) correspond to two prime ideals p ⊂ q. Corollary 6.14(ii) says that the maximal

length of a strictly ascending chain of prime ideals from p to q is constant (such A is

called a catenary ring), and the important part is that this maximal length is precisely

dim(A/p)− dim(A/q). In the particular case in which A is a domain and p = 0, then we

have that the height of the prime ideal q is dim(A)− dim(A/q).

We can now generalize the above results for more general schemes. Since we need to

have decomposability into irreducible components, finite type is the natural assumption.

Theorem 6.16. Let X be an irreducible k-scheme of finite type. Then:

(i) For each irreducible set Y ⊂ X and any open set U meeting Y , we have dim(Y ∩U) =

dim(Y ).

(ii) For each p ∈ X, dim(OX,p) = codim(p, X) = dimX − dim(p).(iii) If f : X → A1

k is a non-constant regular function, either f−1(0) is empty or every

irreducible component of it has dimension equal to dim(X)− 1.

Proof: If dim(Y ) = r, let Z0 ⊆/ Z1 ⊆/ . . . ⊆/ Zr be a chain of irreducible closed sets of Y . In

particular, Z0 consists of a closed point p ∈ Y . Take an affine open neighborhood of p of

75

the form V = Spec(A), where A = k[X1, . . . , Xn]/I. Since p ∈ V , the above chain restricts

to a chain in Y ∩V , and hence dim(Y ∩V ) = r. Since Y is irreducible, its non-empty open

subsets Y ∩ U and Y ∩ V meet. Taking now a closed point q ∈ Y ∩ U ∩ V , we know from

Corollary 6.14(iii) (taking s = 0), that we can find a chain of irreducible closed sets in V

starting in q and having length r. We can restrict that chain to U ∩ V , showing that

still dim(Y ∩ U ∩ V ) = r. Hence, also dim(Y ∩ U) = r, proving (i).

For (ii), take now an affine open neighborhood of p of the form U = Spec(A), where

A = k[X1, . . . , Xn]/I. If p corresponds to the prime ideal p, then OX,p = Ap, whose

dimension is codim(p ∩ U,X ∩ U). Since A is a finitely generated k-algebra, we know

from Corollary 6.14(iii) that codim(p∩U,X∩U) = dim(X∩U)−dim(p∩U). Therefore

(ii) follows now from (i).

For part (iii), it is enough to restrict affine open sets and apply Corollary 6.14(ii),

having in mind part (i).

The above result is saying that, for schemes of locally finite type, the notion of di-

mension is essentially local. We can find many useful applications of this.

Example 6.17. If we want to generalize Proposition 6.6 to a product of projective

spaces (we do it for a product of two projective spaces, but it is clear that it can be

generalized to an arbitrary number of factors), we need to prove that, for any irreducible

bihomogeneous F ∈ k[X0, . . . , Xn;Y0, . . . , Yn], the set V (F ) ⊂ Pnk ×k Pmk is irreducible of

dimension n+m− 1. Let us consider its dehomogenization f ∈ k[X1, . . . , Xn, Y1, . . . , Ym]

with respect to X0 and Y0. It cannot be a constant polynomial (unless F = X0 or

F = Y0, in whose case the result is clear), and clearly f is irreducible (since a nontrivial

decomposition of f would produce another for F ). Hence V (f) is irreducible and, since its

closure in Pnk ×k Pmk is V (F ), the result follows.

Since the above result is saying that, for schemes of locally finite type, the notion of

dimension is essentially local, we will use it to prove several results abut the behavior of

the dimension in morphisms. The first basic result is the following.

Lemma 6.18. Let ϕ : X → Y be a dominant k-morphism of irreducible k-schemes of

finite type of respective dimensions r and s. Then, for any closed point q ∈ Y in the image

of ϕ, each component of the fiber of q has dimension at least r − s.

Proof: If V ⊂ Y is a non-empty open set, then we know from Theorem 6.16(i) that

dim(X) = dim(ϕ−1(V )) and dim(Y ) = dim(V ), so that we can assume that Y is affine.

On the other hand, applying s times Corollary 6.14(ii), we can find s regular functions

f1, . . . , fs ∈ O(Y ) such that V (f1, . . . , fs) is a finite number of points including q. Re-

stricting Y to an open set containing q but not the other points, we can assume that

76

V (f1, . . . , fs) = q. Hence the fiber of q is the intersection of s regular functions on

X, and therefore any component of it has dimension at least r − s, applying recursively

Theorem 6.16(iii). This proves the result.

To get stronger results, we need to use the properties of projective sets:

Lemma 6.19. Let X ⊂ Pnk ×k Y be an irreducible closed subset of dimension r such

that Y is an irreducible k-scheme of finite type of dimension s and the second projection

ϕ : X → Y is surjective. Then:

(i) For each c, the set Yc := q ∈ Y | dim(ϕ−1(q)) ≥ c is closed.

(ii) The open set Yr−s \ Yr−s+1 is not empty.

Proof: To prove (i), recall from Remark 2.19 that Y ⊂ Pnk has dimension at least c if and

only if meets any linear subspace Λ ⊂ Pnk of dimension n−c. Hence, if p : X → Pnk denotes

the projection onto the first factor, q ∈ Yc if and only if p(ϕ−1(q))∩Λ 6= ∅ for any Λ. This

is equivalent to ϕ−1(q) ∩ p−1(Λ) 6= ∅, i.e. q ∈ ϕ(p−1(Λ)). Therefore, Yc is the intersection

of all the closed sets ϕ(p−1(Λ)) and hence it is closed.

For (ii), we know from part (i) that Yr−s \ Yr−s+1 is an open set, so that it is enough

to prove that it is not empty. We will use induction on s, the case s = 0 being trivial.

We can reduce to the case in which Y is affine. We take a non-constant regular function

f : Y → A1k and we can assume, after a possible translation, that f−1(0) is not empty.

Let Y ′ be an irreducible component of f−1(0), which has dimension s − 1, by Theorem

6.16(iii). Since f ϕ is a non-constant regular function on X, again by Theorem 6.16(iii)

all the irreducible components of ϕ−1(f−1(0)) have dimension r − 1. In particular, if

ϕ−1(Y ′) = X ′1 ∪ . . .∪X ′m, each X ′i has dimension r− 1. Since ϕ is a closed map and Y ′ is

irreducible, at least one of the ϕ(X ′i) is Y ′. For each i = 1, . . . ,m, we define a non-empty

open set V ′i ⊂ Y ′ as follows:

–If ϕ|X′i

: X ′ → Y ′ is not surjective, then V ′i = Y ′ \ ϕ(X ′i).

–If ϕ|X′i

: X ′ → Y ′ is surjective, by induction hypothesis, there is a non-empty open

set V ′i ⊂ Y ′ of points such that the fiber at each q ∈ V ′i has dimension r − s.Now, the intersection V ′1 ∩ . . . ∩ V ′m (which is not empty, because Y ′ is irreducible) is

contained in Yr−s \ Yr−s+1, proving (ii).

Theorem 6.20. Let ϕ : X → Y be a dominant k-morphism of irreducible k-schemes of

finite type of respective dimensions r and s. Then the image of ϕ contains a non-empty

open set V ⊂ Y such that dim(ϕ−1(q)) = r − s for all q ∈ V (in particular, r ≥ s).

Proof: We can clearly assume that Y is affine. Moreover, if X =⋃i Ui is a finite cover

by affine sets, each Ui has dimension r and the restriction of ϕ is still dominant (because

77

the generic point of each Ui is the generic point of X, which maps to the generic point of

Y ). If for each i we find a non-empty open set Vi ⊂ Y such that dim(ϕ−1(q)∩Ui) = r− sfor all q ∈ Vi, then the non-empty open set V =

⋂i Vi satisfies the required condition. In

other words, we can also assume X to be affine.

Since Y is separated (because it is affine), the graph of ϕ is closed, so that we can

assume X is a closed subset of Ank ×k Y and ϕ is the second projection. Let X ⊂ Pnk ×k Ybe its Zariski closure and let ϕ denote the second projection. Since ϕ is a closed dominant

map, it is surjective. The image of ϕ consists of the points that are image by ϕ of some

point of X that is not in V (X0). Set X∞ := X ∩ V (X0). If X∞ = X1 ∪ . . . ∪Xm is the

decomposition into irreducible components, we know from Theorem 6.16(iii) that each Xi

has dimension r − 1. For each i = 1, . . . ,m, we distinguish two possibilities:

–If ϕ(Xi) ⊆/ Y , then we take the non-empty open set Vi = Y \ ϕ(Xi).

–If ϕ(Xi) = Y , we can apply Lemma 6.19(ii) to the restriction of ϕ to Xi and find

a non-empty open set Vi ⊂ Y such that, for each q ∈ Vi, the fiber of ϕ|Xiof q, i.e.

ϕ−1(q) ∩Xi, has dimension r − s− 1.

We also consider, using Lemma 6.19(ii), a non-empty open set V0 ⊂ Y such that the

fiber of ϕ at each point of q has dimension r−s. Since Y is irreducible, V := V0∩V1∩. . .∩Vmis a non-empty open subset of Y , and any q ∈ V1 ∩ . . . ∩ Vm satisfies that ϕ−1(q) has all

its components of dimension r − s and is not contained in X∞, hence q is in the image of

ϕ and its fiber is ϕ−1(q) \X∞, which still has dimension r − s, by Theorem 6.16(i). This

proves the result.

Theorem 6.21. Let ϕ : X → Y a dominant morphism of k-schemes of finite type. Assume

that Y is irreducible of dimension s and that, for any closed point q ∈ ϕ(X), the fiber

ϕ−1(q) is also irreducible of dimension c. Then X is irreducible of dimension s+ c.

Proof: Let X =⋃iXi the decomposition of X into irreducible components. For each i,

we construct a non-empty open set Vi ⊂ Y as follows:

–If ϕ(Xi) 6= Y , we take Vi = Y \ ϕ(Xi).

–If ϕ(Xi) = Y , by Theorem 6.20(i) applied to ϕ|Xi, we can take Vi such that

dim(ϕ−1(q) ∩Xi) = dim(Xi)− s for all q ∈ Vi.Since Y is irreducible,

⋂i Vi is not empty, hence there is a point q0 ∈

⋂i Vi. Decom-

posing ϕ−1(q0) =⋃i(ϕ−1(q0) ∩ Xi) and using the irreducibility of ϕ−1(q0), we get that

ϕ−1(q0) = ϕ−1(q0) ∩Xi for some i and, necessarily, ϕ(Xi) = Y . In particular,

c = dim(ϕ−1(q0)) = dim(ϕ−1(q0) ∩Xi) = dim(Xi)− s

hence dim(Xi) = s + c. But now, for any q ∈ ϕ(X) we have, by Lemma 6.18 applied to

ϕ|Xi, dim(ϕ−1(q) ∩Xi) ≥ c. And since ϕ−1(q) is irreducible of dimension c and contains

78

ϕ−1(q) ∩ Xi, necessarily ϕ−1(q) = ϕ−1(q) ∩ Xi. In other words, all the fibers of ϕ are

contained in Xi, and therefore X = Xi, as wanted.

Remark 6.22. It is clear that all the hypotheses are needed in the above result. For in-

stance, it is not valid in the affine case (in which maps are nor closed in general). For exam-

ple, the projection onto the first coordinate of X = V (XY −1)∪V (X,V ) ⊂ A2k satisfies all

the hypotheses except being a closed map. Similarly, the condition that all the fibers have

the same dimension is necessary; one could consider the map Proj(k[X0, X1, X2]/(X1X2)

)→

Proj(k[X0, X1]) that is the natural isomorphism on the component V (X2) and sends the

other component V (X1) to V (X1, X2).

Remark 6.23. One case in which we could apply Theorem 6.21 is when X is proper and

Y is separated. For example, X ×k Y is irreducible of dimension dim(X) + dim(Y ). In

fact, by the local nature of dimension, we get that dim(X ×k Y ) = dim(X) + dim(Y ) for

any k-schemes of finite type. For example, if X,Y ⊂ Ank , then X ∩ Y is isomorphic to the

intersection of X ×k Y ⊂ Ank ×k Ank and the diagonal ∆ ⊂ Ank ×k Ank . Since the diagonal is

defined by n equations, we get from Corollary 6.14(ii) that each component of X ∩ Y has

dimension at least r + s− n. Due to the local nature of the dimension, the same holds if

instead X,Y ⊂ Pnk .

79

7. Parameter spaces

The contents of this section can also be found in section 13 of [A], which should be

the reference until this part is completely written in an organized and complete way.

We start with the first non trivial parameter space, the one parametrizing linear spaces

in the projective space.

Example 7.1. The space parametrizing hyperplanes in Pnk is known to be another

projective space, the dual space Pnk∗, where a point of coordinates (u0 : . . . : un) ∈ Pnk

∗

represents the hyperplane of equation u0X0 + . . .+unXn. Observe that, if the hyperplane

is generated by the points whose coordinates are the rows of the matrix

M =

a00 . . . a0n...

...an−1,0 . . . an−1,n

then the equation of the hyperplane is∣∣∣∣∣∣∣∣

X0 . . . Xn

a00 . . . a0n...

...an−1,0 . . . an−1,n

∣∣∣∣∣∣∣∣so that the coordinates (u0 : . . . : un) as a point of Pnk

∗ are given, up to sign and order,

by the maximal minors of the matrix M . This motivates the following definition for the

general case:

Definition. The Grassmannian G(k, n) is the set of linear subspaces of dimension k in

Pnk . The Plucker embedding is the map ϕk,n : G(k, n) → P(n+1k+1)−1

k that associates to the

subspace Λ ⊂ Pnk generated by the rows of the matrix

M =

a00 . . . a0n...

...ak0 . . . akn

the point in P(n+1

k+1)−1

k with Plucker coordinates pi0...ik0≤i0<...<ik≤n, where

pi0...ik =

∣∣∣∣∣∣∣a0i0 . . . a0ik

......

aki0 . . . akik

∣∣∣∣∣∣∣80

Proposition 7.2. The Plucker embedding is an injective well-defined map whose image

is a projective set.

Proof: First of all, since the rows of M span a k-dimensional linear subspace, M has max-

imal rank, so that at least one maximal minor is not zero, so that the Plucker coordinates

define a point in P(n+1k+1)−1

k . Moreover, this point does not depend on the choice of the

matrix M . Indeed, a different choice of M , i.e. a different choice of generators of Λ is

given by a new matrix M ′ = PM where P is a (k+1)×(k+1) invertible matrix. Therefore

all the maximal minors of M ′ are the maximal minors of M multiplied by det(P ), hence

the corresponding point in P(n+1k+1)−1

k is the same.

For the injectivity, assume a point of Plucker coordinates pi0...ik0≤i0<...<ik≤n is in

the image of ϕk,n. Some of the coordinates is different from zero, and we can assume

for simplicity p0...k 6= 0. We will thus use affine coordinates pi0...ik

p0...k0≤i0<...<ik≤n for

D(p0...k). For any possible matrix M whose whose maximal minors give the point of

coordinates pi0...ik0≤i0<...<ik≤n, we can take the equivalent matrix PM , where P is the

inverse of the matrix a00 . . . a0k...

...ak0 . . . akk

(which is invertible because p0...k 6= 0). Therefore, any subspace Λ ∈ G(k, n) whose image

by ϕk,n is in D(p0...k) can be represented by a unique matrix of the type

M ′ =

1 . . . 0 a′k+1,0 . . . a′0n...

. . ....

......

0 . . . 1 a′k,k+1 . . . a′kn

.

Observe, for example, that, for any j = k + 1, . . . , n, one hasp0...k−1,j

p0...k= a′kj . Similarly,

any a′ij = (−1)k−ip0...i−1,i+1,...j

p0...k. Hence, any point in the image is in the image of only one

element of G(k, n).

Moreover, a point of D(p0...k) is in the image of ϕk,n if and only if eachpi0...ik

p0...kis

the corresponding minor of M ′. Since the entries of M ′ are a′ij = (−1)k−ip0...i−1,i+1,...j

p0...k, it

follows that the intersection of Im(ϕk,n with D(p0...k) (and in general with any D(pi0...ik)

is closed in the Zariski topology. Therefore, Im(ϕk,n) is a closed subset in P(n+1k+1)−1

k .

Remark 7.3. A natural way of given a scheme structure to G(k, n) (besides using the

Proj construction) would be to use the glueing construction (see Example 4.10) to glue to-

gether the(n+1k+1

)affine pieces corresponding to the different D(pi0...ik). The previous proof

shows that the intersection of Im(ϕk,n) with D(pi0...ik) is isomorphic to the affine space

81

of dimension (k + 1)(n − k) (its coordinates being the Plucker coordinates corresponding

to the entries of M ′). In particular, this proves that G(k, n) is irreducible of dimension

(k + 1)(n− k).

Example 7.4. Let us study the first example of Grassmannian that is not a projective

space, namely G(1, 3). In this case, the matrix M ′ of the above proof becomes

M ′ =

(1 0 −p12p01

−p13p01

0 1 p02p01

p03p01

)and the only equation we get is

p23 =

∣∣∣∣−p12p01−p13p01

p02p01

p03p01

∣∣∣∣i.e. p01p23 − p02p13 + p03p12 = 0. The different matrices we obtain when varying the open

set are:

M ′ =

(1 p12

p020 −p23p02

0 p01p02

1 p03p02

)on D(p02)

M ′ =

(1 p13

p03

p23p03

0

0 p01p03

p02p03

1

)on D(p03)

M ′ =

( p02p12

1 0 −p23p12

−p01p120 1 p13

p12

)on D(p12)

M ′ =

( p03p13

1 p23p13

0

−p01p130 p12

p131

)on D(p13)

M ′ =

( p03p23

p13p23

1 0

−p02p23−p13p23

0 1

)on D(p23).

In all cases we always obtain the same equation p01p23 − p02p13 + p03p12 = 0, while the

points generating the line can be collected as the rows (or columns) of the skew-symmetric

matrix

A =

0 p01 p02 p03

−p01 0 p12 p13

−p02 −p12 0 p23

−p03 −p13 −p23 0

.

Observe that the determinant of A is precisely (p01p23−p02p13 +p03p12)2 (the determinant

of a skew-symmetric matrix is always a perfect square of an expression called the Pfaffian

of the matrix), so that the quadratic equation is saying that the rank of A is smaller than

four. Since the rank of a skew-symmetric matrix is always even, this means that the rows

of A span a line.

82

Example 7.5. Another way of representing an element Λ ∈ G(k, n) is as the intersection

of n−k linearly independent hyperplanes, i.e. as a linear subspace of dimension n−k−1 in

Pnk∗. In other words, we can identify G(k, n) with G(n−k−1, n). The Plucker coordinates

of Λ as an element of G(n− k− 1, n) are called dual Plucker coordinates, and are actually

the original Plucker coordinates given in a different order and with possible different sign.

For example, for G(1, 3), in the open set D(p01), the line with Plucker coordinates (p01 :

p02 : p03 : p12 : p13 : p23) can be described as the intersection of the planes

p12

p01X0 −

p02

p01X1 +X2 = 0

p13

p01X0 −

p03

p01X1 +X3 = 0

Varying the affine open set, the line can be obtained as the intersection of the planes whose

coefficients are the rows of the matrix

B =

0 p23 −p13 p12

−p23 0 p03 −p02

p13 −p03 0 p01

−p12 p02 −p01 0

.

If we want all these planes to contain all the points of the rows of the matrix A of Example

7.4, we need to impose ABt = 0. Since ABt is the product of p01p23−p02p13 +p03p12 with

the identity matrix, we recover the known quadratic equation of G(1, 3).

Example 7.6. The situation for a general G(k, n) is as in the case of G(1, 3). Any

subspace Λ can be spanned by the points whose coordinates of a matrix A whose entries

are Plucker coordinates. When k = 1, we can take A to be a general skew-symmetric

matrix. Dually, we can also describe Λ as intersection of the hyperplanes whose coefficients

are the rows of a matrix B whose entries are Plucker coordinates (and, when k = n − 2,

B can be taken to be skew-symmetric). Imposing the vanishing ABt, we get quadratic

equations that can be proved (not in an easy way) to generate the ideal of G(k, n) as a

projective set.

Example 7.7. One of the main uses of Grassmannians (and parameter spaces in general)

is to give structure of variety to the so-called incidence varieties. One typical example is

the incidence variety of points and linear spaces of dimension k in the projective space,

namely

I = (p,Λ) ∈ Pnk ×k G(k, n) | p ∈ Λ.

Since this is subset of Pnk ×k G(k, n) ⊂ Pnk ×k P(n+1k+1)−1, we need to find bihomogeneous

equations to describe I. First of all, we need to insure the second element of each pair

83

to be in G(k, n). Since these equations (we remarked in Example 7.6 that they can be

taken to be quadratic in the Plucker coordinates) only depend on the Plucker coordinates,

they are bihomogeneous of bidegree (0, 2). Secondly, to impose the point of coordinates

(x0 : . . . : xn) to be in the subspace Λ of Plucker coordinates pi0...ik0≤i0<...<ik≤n is

equivalent to impose that (x0 : . . . : xn) belongs in a set of hyperplanes whose intersection

is Λ. Therefore, the incidence condition consists of B

x0...xn

=

0...0

, where B is the

matrix of Example 7.6, and these are equations of bidegree (1, 1). For example, when

k = 1, n = 3, the equations of the incidence variety of points and lines in P3k are

p01p23 − p02p13 + p03p12

p23X1 − p13X2 + p12X3

−p23X0 + p03X2 − p02X3

p13X0 − p03X1 + p01X3

−p12X0 + p02X1 − p01X2.

Using incidence varieties we can be more precise about how many “good” linear spaces

intersect a projective set in the right dimension.

Proposition 7.8. Let X ⊂ Pnk be an irreducible projective set of dimension r. Then,

for any k ≥ n − r − 1, the set of linear spaces of dimension k intersecting X in a set of

the expected dimension r + k − n is a non-empty open set of G(k, n). Moreover, in the

case k = n− r − 1, the complementary closed set of the subspaces of dimension n− r − 1

intersecting X is irreducible of codimension one in G(n− r − 1, n).

Proof: Consider the incidence variety

I := (p,Λ) ∈ X ×k G(n− r, n) | p ∈ Λ

and the second projection p2 : I → G(n− r, n). The fiber of a linear subspace Λ under p2

is the intersection of X with Λ. Hence, by Lemma 6.19(i) the subset of all subspaces Λ

whose intersection with X has dimension at least r+ k− n+ 1 is closed, an we know it is

not the whole Grassmannian.

If k = n − r − 1, we observe that the fiber of any p ∈ X is the set of subspaces

of dimension n − r − 1 passing through p. It is a simple exercise to prove that, fixing

a hyperplane H ⊂ Pnk and assigning to each subspace passing through p its intersection

with H, we get an isomorphism among the fiber of p and the Grassmannian of linear

84

subspaces of dimension n− r− 2 in H. Hence the fibers of p1 are irreducible of dimension

(n−r−1)(r+1). Then Theorem 6.21 implies I is irreducible of dimension r+(n−r−1)(r+

1) = (n − r)(r + 1) − 1. Since there are linear subspaces of dimension n − r − 1 meeting

X only in a finite number of points, this implies that p2(I) is irreducible of dimension

r + (n − r − 1)(r + 1) = (n − r)(r + 1) − 1, which means that has codimension one in

G(n− r − 1, n).

Example 7.9. If we want to get something interesting with incidence varieties we will

need more sophisticated examples. For instance, let identify P(k[X0, . . . , Xn]d) with the

space of hypersurfaces of degree d in Pnk . A natural set of coordinates is given by the

coefficients of the polynomial defining the hypersurface. Now we would like to prove that

I = (Λ, X) ∈ G(k, n)×k P([X0, . . . , Xn]d) | Λ ⊂ X

is a projective set. However, it is not so easy to write the precise equations for the inclusion

Λ ⊂ X in terms of the Plucker coordinates of Λ and the coefficients of the hypersurface

X. We can use the trick of considering a bigger incidence variety

I ′ = (p,Λ, X) ∈ Pnk ×k G(k, n)×k P([X0, . . . , Xn]d) | p ∈ Λ, p ∈ X.

The condition p ∈ Λ can be written, as observed in Example 7.7, using trihomogeneous

equations, of tridegree (0, 2, 0) to impose Λ to be in G(k, n) inside P(n+1k+1)−1

k and of tridegree

(1, 1, 0) to conclude that the point p is in the subspace Λ. It is even easier to write the

condition p ∈ X, since this is given precisely by the equation of X, which should be

regarded as a trihomogeneous polynomial of tridegree (d, 0, 1). Consider now the projection

ϕ : I ′ → G(k, n) ×k P(k[X0, . . . , Xn]d). The fiber of ϕ at a closed point (Λ, X) can be

naturally identified with Λ∩X. Therefore, I is the set of pairs whose fiber has dimension

at least k. Hence, by Lemma 6.19(i), I is a closed set, i.e. a projective set.

Example 7.10. Let us give a concrete application of the incidence varieties that will

inspire how to use them in general. We will specialize Example 7.9 to the case of lines in

surfaces of degree d in P3k. We consider thus the diagram

I = (L, S) ∈ G(1, 3)×k P(k[X0, X1, X2, X3]d) | L ⊂ Sp1 p2

G(1, 3) P(k[X0, X1, X2, X3]d)

and we wonder about the surjectivity of p2, i.e. we want to know when any surface of

degree d contains a line or not. To study this, we will use the part of the diagram we

understand. On the left side, we have an irreducible projective set G(1, 3) of dimension

85

4 and a map p1 that is surjective. Moreover, the fiber of p1 at each line L ⊂ P3 is

naturally identified with P(I(L)d

)⊂ P(k[X0, X1, X2, X3]d). Since there is an isomorphism

k[X0, X1, X2, X3]/I(L) ∼= k[T0, T1] (given, for example, by a parametrization P1k → L),

we get that p−11 (L) is a projective space of dimension

((d+3

3

)− 1)− (d + 1). Hence, by

Theorem 6.21, I is irreducible of dimension((d+3

3

)− 1)− (d− 3). Therefore, when d ≥ 4,

we have dim(I) < dim(P(k[X0, X1, X2, X3]d)

), which implies that p2 cannot be surjective

(see Theorem 6.20). As a consequence, a general surface of degree d ≥ 4 does not contain

lines (the word “general” is used in Algebraic Geometry in the sense that there is a non-

empty open set for which the property holds. In this case, any surface in the open set

P(k[X0, X1, X2, X3]d) \ ϕ(I) does not contain lines). Observe also that the set of surfaces

containing a line forms a projective set, since p2 is a closed morphism (Theorem 5.11).

We can check what happens when d ≤ 3, in which dim(I)−dim(P(k[X0, X1, X2, X3]d)

)=

d− 3 ≥ 0, so that we expect the general fiber of p2 to have dimension d− 3, i.e a general

surface of degree d ≤ 3 is expected to contain a (d− 3)-dimensional family of lines.

–When d = 1, any plane contains a two-dimensional family of lines (what is called the

dual plane).

–When d = 2, any smooth quadric contains two one-dimensional families of lines. Also

quadrics of rank three contain a one-dimensional family of lines. It is only for the closed

set of quadrics of rank at most two (pairs of planes or double planes) that they contain

two-dimensional families of lines.

–When d = 3, one should expect a general cubic surface to contain a finite number of

lines. If this were not the case, i.e. if p2 were not surjective, the fibers of p2 : I → p2(I)

would have strictly positive dimension, so that they would be infinite. To exclude that

case, it is enough to find a cubic surface with a finite number of lines. The standard

example is the Fermat cubic V (X30 + X3

1 + X32 + X3

3 ) ⊂ P3k. For any of the 3 possible

partitions (up to order) 0, 1, 2, 3 = i, j ∪ k, l and any choice of the 9 possible pairs

(ω, ω′) of cubic roots of −1, the line V (Xi − ωXj , Xk − ω′Xl) is contained in the Fermat

cubic. So that we find 27 lines in the cubic, and it is possible to prove that there are no

more lines. Hence a general cubic contains a finite number of lines. The strong result is

that one can be much more precise and prove that any smooth cubic surface in P3k contains

exactly 27 lines (for an elementary proof, see [R], Cap. III §7).

Remark 7.11. It is clear that, if we can parametrize not only lines or hypersurfaces,

the ideas of Example 7.10 could provide much more general results. One of the keys to

generalize that result is hidden in its proof. Indeed the main part of the proof in Example

was to compute the dimension of the fiber of p1. That depended on the dimension of I(L)d,

and that was equivalent to compute implicitly to compute the dimension of S(L)d. This

dimension is given, in general for large d, by the value of the Hilbert polynomial. This

86

is why the best families to parametrize are those consisting of schemes sharing the same

Hilbert polynomial.

We state next the existence theorem of Hilbert schemes, but we will not prove it, since

a complete proof would require tools that we did not develop. We will give just the main

ideas of the construction. The interested reader can try to take a look at [Se].

Theorem 7.12. Fix P ∈ Q[T ] a polynomial taking integral values. Then, for any n ∈ N,

there exists a scheme HilbP (Pnk ) parametrizing all schemes X ⊂ Pnk having P as their

Hilbert polynomial.

Idea of the proof: Recall (Theorem 2.17) that, given a homogeneous ideal I ⊂ k[X0, . . . , Xn]

with Hilbert polynomial P , there is d0 such that hI(d) = P (d) for any d ≥ d0. Moreover,

if I is generated by F1, . . . , Fr and we choose the above d0 bigger than the degrees of

F1, . . . , Fr, then, for any d ≥ d0 we have that each Fi multiplied by all monomials of

degree d − deg(Fi) is in Id. This means that F1, . . . , Fr are in the saturation of the ideal

generated by Id. The first key observation (this is a hard theorem) is that we can find the

same d0 for all saturated ideals I with the same Hilbert polynomial P , i.e. if d ≥ d0 we

have hI(d) = P (d) and that the saturation of the ideal generated by Id is I. This allows

to define an injective map

HilbP (Pnk )→ G((n+ d

d

)− P (d)− 1,

(n+ d

d

)− 1)

that maps a scheme defined by the saturated ideal I to the linear subspace P(Id) of

P(k[X0, . . . , Xn]d). The point now is to see that the image of the map is closed, at least

inside an open set of the Grassmannian (another question that we will not treat here would

be to check that all the images we get from different values of d are isomorphic to each

other). So we need to characterize when, given a linear subspace W ⊂ k[X0, . . . , Xn]d of

codimension P (d), the saturation I of the ideal generated by W has Hilbert polynomial P

and Id = W . If we prove that the ideal generated by W has Hilbert polynomial P , then

this is also the Hilbert polynomial of I. Hence hI(d) = P (d) and, since W ⊂ Id, it follows

W = Id. Therefore we need to characterize when the ideal generated by W has Hilbert

polynomial P . For this, we need to impose that, for each degree e ≥ 0, the image of the

multiplication map

W × k[X0, . . . , Xn]e → k[X0, . . . , Xn]d+e

has dimension(n+d+ed+e

)− P (d + e). Since a system of generators of the image can be

expressed in terms of the Plucker coordinates of W , it follows that imposing that the

dimension of the image is at most(n+d+ed+e

)− P (d + e) defines a closed set in G

((n+dd

)−

P (d) − 1,(n+dd

)− 1

). The intersection of all these closed sets for e ≥ 0 determines a

87

projective set. However, to impose that the dimension of the image is exactly(n+d+ed+e

)−

P (d+ e), we need to remove the closed set of those W for which the dimension is at most(n+d+ed+e

)−P (d+ e)− 1. This is a slight problem, because this means that we need to take

now an infinite intersection of open sets. However, one can remove that difficulty by a

trick based on the fact that a polynomial of degree d can be determined by its values at

d+ 1 points.

As a consequence, HilbP (Pnk ) can be regarded as an open set of a projective subvariety

of G((n+dd

)−P (d)−1,

(n+dd

)−1). A hard task that we will not approach is to show that this

algebraic structure is independent of the choice of d (i.e. the subvarieties corresponding to

different values of d are isomorphic).

Remark 7.13. The reason why the set HilbP (Pnk ) is a scheme and not just a simple

variety is because it is defined through a universal property. The idea is that one expects

to have an incidence variety I ⊂ Pnk ×k HilbP (Pnk ) consisting of pairs (p,X) such that p is

a point of the scheme X. The first projection I → HilbP (Pnk ) has the property that the

fiber at each point is a scheme with Hilbert polynomial P . The universal property that

this satisfies is that, for any other H ′ with a subscheme I ′ ⊂ Pnk ×k H ′ such that all the

fibers of the second projection I ′ → H ′ are projective subschemes of Hilbert polynomial P ,

there exists a unique morphism H → H ′ for which I ′ is the fiber product I ×H′ H. It can

be proved that there exists a projectve scheme HilbP (Pnk ) satisfying the above universal

property, so that it is unique up to isomorphism. It is a suitable “completion” of the

subvariety defined in the proof of Theorem 7.12.

Example 7.14. Let us study explicitly a concrete example, namely Hilb2T+1(P3k).

This means that we deal with conics lying in some plane of P3k. Any saturated ideal

I ⊂ k[X0, X1, X2, X3] with PI(T ) = 2T + 1 is generated by a linear form H and a

quadratic form Q not divisible by H. The value d0 of the proof of Theorem 7.12 can be

taken to be d0 = 2, since I2 is generated, as a vector space, by X0H,X1H,X2H,X3H,Q,

which is enough to recover I. So we can embed Hilb2T+1(P3k) into G(4, 9) by map-

ping each I to P(I2) ⊂ P(k[X0, X1, X2, X3]2). The way of identifying when a subspace

P(W ) ⊂ P(k[X0, X1, X2, X3]2) in in the image is when there exists [H] ∈ P3k∗

such that

X0H,X1H,X2H,X3H ∈W . In other words, if we consider the incidence variety

I := ([H],P(W )) ∈ P3k∗ ×k G(4, 9) | X0H,X1H,X2H,X3H ∈W

then Hilb2T+1(P3k) is the image of I under the second projection map. Since P3

k∗

is irre-

ducible of dimension 3, and the fiber under the first projection of any [H] ∈ P3k∗

is naturally

isomorphic to the 5-dimensional projective space P((k[X0, X1, X2, X3]/H)2

), then Theo-

rem 6.21 implies that I is irreducible of dimension 8. The fiber under the second projection

88

of any conic is just one point, namely the only plane containing the conic (in fact, this

second projection is an isomorphism), so that Hilb2T+1(P3k) is irreducible of dimension 8.

Observe that we also derived at once that Hilb2T+1(P3k) is a projective scheme.

Example 7.15. We are now in position to imitate Example 7.10 and study, depending

on d when a general surface of degree d in P3k contains some conic. We thus start with the

incidence variety

I := (C, S) ∈ Hilb2T+1(P3k)×k P(k[X0, X1, X2, X3]d) | C ⊂ S.

Now we know that Hilb2T+1(P3k) is irreducible of dimension 8, while the fiber under the

second projection of any C ∈ Hilb2T+1(P3k) is P

(I(C)d

)⊂ P(k[X0, X1, X2, X3]d), which is

irreducible of dimension((d+3

3

)−1)−(2d+1). Therefore, by Theorem 6.21, I is irreducible

of dimension((d+3

3

)−1)− (2d−7). This implies again, as in Example 7.10, that a general

surface of degree d ≥ 4 contains no conics (this is a particular case of Noether-Lefschetz

theorem, stating that the only curves on a general surface of degree d ≥ 4 in P3k are those

obtained as the intersection of the surface with another surface; in particular, by Bezout’s

theorem, the degree ofany possible curve on a general surface of degree d is a multiple of

d). We analyze now the cases d ≥ 3, for which we expect the second projection to be

surjective with general fiber of dimension 7− 2d:

-For d = 1, indeed any plane contains a 5-dimensional family of conics.

-For d = 2, the set of conics on an irreducible quadric surface is obtained intersecting

the quadric with all possible planes in P3k, so that we get a 3-dimensional family. When

the quadric is a pair of planes (or a double plane) we instead get a 5-dimensional family

of conics on the quadric.

-For d = 3, if a plane conic is contained in an irreducible cubic surface, then the plane

containing the conic intersects the surface in the plane cubic consisting of the conic plus a

line. Reciprocally, any time we fix a line inside the surface and a plane containing the line,

that plane must intersect the surface in a plane cubic consisting of the line plus a conic.

Since a general cubic surface contains 27 lines, the set of conics contained in the surface is

in correspondence with the 27 pencils of planes defined by the lines.

We go on now with another parameter space, based on Proposition 7.8, saying that

the set of linear spaces of dimension n−r−1. Since we understand codimension one better

in products of projective spaces (Example 6.17) than in Grassmannians, we start with the

following:

Proposition 7.16. Let X ⊂ Pnk be an irreducible subvariety of dimension r and degree

d. Let U ⊂ Pnk∗ × r+1). . . ×Pnk

∗ the open set of u-ples (H1, . . . ,Hr+1) such that dim(H1 ∩

89

. . . ∩Hr+1) = n − r − 1. Then the subset of WX ⊂ U of u-ples (H1, . . . ,Hr+1) such that

H1 ∩ . . . ∩Hr+1 intersects X is irreducible of codimension one. Moreover, the closure of

WX in Pnk∗ × r+1). . . ×Pnk

∗ is the zero locus of an irreducible multihomogeneous polynomial

FX of multidegree (d, r+1). . . , d).

Proof: Consider the incidence variety I ⊂ X × U consisting of u-ples (p,H1, . . . ,Hr+)

such that p ∈ H1 ∩ . . . ∩Hr+1. Considering the first projection p1 : I → X, let us study

the fiber of any p ∈ X. Writing Ap ⊂ Pnk∗ for the set of hyperplanes of Pnk containing

p (Ap is a hyperplane in Pnk∗), the fiber of p is the open set of Ap × r+1). . . ×Ap of u-ples

(H1, . . . ,Hr+1) such that dim(H1 ∩ . . . ∩ Hr+1) = n − r − 1. Hence all the fibers are

irreducible of dimension (r + 1)(n− 1). From Theorem 6.21, I is irreducible of dimension

r+(r+1)(n−1) = (r+1)n−1. Since WX is the image of the second projection p2 : I → U , it

follows that WX is irreducible. Moreover, since there are clearly subspaces of codimension

r + 1 meeting X in just one point, it follows that the dimension of a general fiber of a

point in WX is zero, which shows that WX has codimension one.

As a consequence, the closure of WX in Pnk∗×r+1). . . ×Pnk

∗ is also irreducible of codimen-

sion one, so it is defined by an irreducible multihomogeneous polynomial FX . To check

its multidegree, we need to fix r general hyperplanes in r of the factors of the product

(by symmetry we can assume they are the first ones) and let the other hyperplane (that

we will take in the last factor) to vary in a pencil. The intersection of the first r hyper-

planes H1, . . . ,Hr will intersect X in d points (we will see in Remark 8.22 that they can

be obtained to be different). If we want the last hyperplane Hr+1 such that the u-ple

(H1, . . . ,Hr+1 is in (the closure of) WX , then Hr+1 should contain one of the above d

points. Imposing also that Hr+1 is in a general pencil implies that we have exactly one

possibility for each of the d points. This proves that the multidegree of FX is (d, . . . , d).

Definition. The above polynomial FX is called the Chow form of X and WX . When X is

equidimensional (i.e. all its irreducible components have the same dimension), the Chow

form of X is the product of the Chow forms of its irreducible components.

This idea of Chow forms allow to give a nice parameter space for the set of irreducible

subsets X ⊂ Pnk of dimension r and degree d (with a little bit care, this can be easily

extended to equidimensional subsets):

Theorem 7.17. Let P the projectivization of the space of multihomogeneous forms of

multidegree (d, r+1). . . , d) in (r+ 1)(n+ 1) variables. Then the assignment that maps to each

irreducible subset X ⊂ Pnk of dimension r and degree d to (the class of) its Chow form in

P is injective, and its image is an open set of a projective set.

Proof: Let us start with a simple observation (keeping the notation of the proof of Theorem

90

7.16). Consider any point p ∈ Pnk . Then, the dimension of the open subset (Ap×r+1). . . ×Ap)∩U of all u-ples (H1, . . . ,Hr+1) ∈ U such that p ∈ H1 ∩ . . .∩Hr+1 is (r+ 1)(n− 1). For any

X ⊂ Pnk of dimension at most r, this subset (Ap× r+1). . . ×Ap)∩U is contained in WX if and

only if p ∈ X (since, if p /∈ X, there is a subspace of codimension r + 1 passing through p

and not meeting X).

Let FX be a Chow form. We consider now the incidence variety I ⊂ Pnk×kU consisting

of the u-ples (p,H1, . . . ,Hr+1) such that F (H1, . . . ,Hr+1) = 0 and p ∈ H1 ∩ . . . ∩Hr+1.

Then, by the above observation, X can be characterized as the subset of points of Pnk such

that the fiber of the first projection p1 : I → Pnk has maximal dimension (r + 1)(n − 1).

Hence X is univocally determined by its Chow form.

Let us prove now that the subset of P consisting of Chow forms is an open set of a

projective set. For that, we first consider the incidence variety I ′ ⊂ Pnk×kU×kP consisting

of u-ples (p,H1, . . . ,Hr+1, [F ]) such that F (H1, . . . ,Hr+1) = 0 and p ∈ H1 ∩ . . . ∩Hr+1.

Define subset I ′′ ⊂ Pnk ×k P as the subset of pairs (p, [F ]) such their fibers under the

projection I ′ → Pnk ×k P has maximal dimension (r + 1)(n− 1), or equivalently, such that

(Ap×r+1). . . ×Ap)∩U ⊂ V (F ). We now consider the closed subset Σ ⊂ P of the set of classes

[F ] such that their fiber under the second projection I ′′ → P has dimension at least r, and

consider its open subset Σ0 consisting of irreducible forms.

Assume that we have a class [F ] ∈ Σ0, i.e. F is irreducible and there is a subset

X ′ ⊂ Pn of dimension at least r such that, for all p ∈ X ′, (Ap× r+1). . . ×Ap)∩U ⊂ V (F ). We

can take an irreducible set X ⊂ X ′ of dimension r. Since we are assuming V (FX) ⊂ V (F )

and F is irreducible, it follows that F = FX .

Definition. The closure of the above set Σ0 is called Chow variety and denoted by

Chr,d(Pnk )

Remark 7.18. The new points appearing in the Chow variety are not only the Chow

forms of reducible varieties, but also Chow forms with multiplicity. To be precise, one

can consider a cycle of irreducible varieties n1X1 + . . . + nsXr of irreducible varieties

X1, . . . , Xs ⊂ Pnk of dimension r (i.e. we are putting multiplicities n1, . . . , ns to the sub-

varieties). If n1 deg(X1) + . . . + ns deg(Xr) = d, then the class of Fn1

X1. . . Fns

Xsis also in

Chr,d(Pnk ).

It would have been more natural to give a notion of Chow form in such a way that,

in the case of hypersurface, it coincides with the equation of the hypersurface. The cor-

responding result (which allows to construct the Chow variety in the same way as above)

needs a proof that has some slight difficulty, as we will see next:

Proposition 7.19. Let X ⊂ Pnk be an irreducible subvariety of dimension r. Consider

91

the open subset U ⊂ Pnk × n−r). . . ×Pnk of n − r independent points. Then the subset of

W ⊂ U consisting of points whose span intersects X is irreducible of codimension one in

U . Moreover, the closure of W in Pnk × n−r). . . ×Pnk is the zero locus of a multihomogeneous

polynomial of multidegree (d, n−r). . . , d).

Proof: The proof follows the lines of the one of Theorem 7.16. We first consider the

incidence variety I ⊂ X × U of u-ples (p, q1, . . . , qn−r) such that p is in the span of

q1, . . . , qn−r. We want to prove now that the fibers of the first projection p1 : I → X are

all irreducible of the same dimension. For this we first consider the set Ω(p) ⊂ G(n−r−1, n)

consisting of all the linear subspaces of dimension n− r− 1 passing through p. Since Ω(p)

is isomorphic to G(n− r− 2, n− 1) (prove it as an exercise), it is irreducible of dimension

(r+ 1)(n− r− 1). On the other hand, the fiber of p under p1 has a natural surjective map

p−11 (p) → Ω(p) (mapping any u-ple of points (q1, . . . , qn−r) to its linear span). The fiber

under such map of a linear space Λ is the open subset of Λ× n−r). . . ×Λ consisting of u-ples

of linearly independent points, hence it is irreducible of dimension (n− r)(n− r − 1). As

a consequence, the fibers of p1 are irreducible of dimension (n+ 1)(n− r− 1) and hence I

has dimension (n+ 1)(n− r − 1) + r = n(n− r)− 1.

Remark 7.20. As the above proof shows, there is a more intrinsic way of defining the

Chow form using the Grassmannian G(n − r − 1, n) instead of any of the products of

projective spaces that we just used. The hard part here is to show that an irreducible

subset of a Grassmannian of codimension one is the intersection of the Grassmannian with

a hypersurface. Giving this as granted, another construction of the Chow variety as an

open set of a projective set of P(S(G(n−r−1))d

)is easy to find. To put an easy example,

take X ⊂ Pnk to be the r-dimensional linear space X = V (Xr+1, . . . , Xn). Then, it is easy

to check that the subset of G(n− r− 1, n) consisting of the linear subspaces meeting X is

determined by the vanishing of pr+1,...,n. Of course, not any linear form in P is the Chow

form of a linear subspace. For example, the reader can check that G(1, 3) ∩ V (p01 + p23)

is not the set of lines meeting a given line of P3k.

92

8. Local properties

We have seen in Theorem 6.16(ii) that the dimension of an irreducible scheme of

locally finite type can be computed from its local ring OX,p at a closed point. It is thus

natural to wonder how much information we can get from that ring. The first geometric

local notion at a point is the tangent space, and we will see that this can be obtained in a

canonical way from OX,p.

Theorem 8.1. Let X be a scheme of localy finite type over k and let p ∈ X be a closed

point. Then the set of morphisms ϕ : Spec(k[T ]/(T 2))→ X whose image is p is a k-vector

space naturally isomorphic to (Mp/M2p)∗.

Proof: We take an open set containing p of the form Spec(k[X1, . . . , Xn]/I), and let

(X1 − a1, . . . , Xn − an) be the maximal ideal corresponding to p. Then a morphism

ϕ : Spec(k[T ]/(T 2))→ X whose image is p is the same as a homomorphism of k-algebras

ψ : k[X1, . . . , Xn]/I → k[T ]/(T 2) such that ψ(Xi) = ai+biT . The map will be well-defined

if and only if, for each f ∈ I, f(a1 + b1T, . . . , an + bnT ) ∈ (T 2). Writing f in terms of

X1 − a1, . . . , Xn − an (using that f(p) = 0 for all f ∈ I),

f =∂f

∂X1(p)(X1 − a1) + . . .+

∂f

∂Xn(p)(Xn − an) + . . .

we get that the condition becomes ∂f∂X1

(p)b1 + . . . + ∂f∂Xn

(p)bn = 0 for all f ∈ I. We thus

get that the set of morphisms ϕ : Spec(k[T ]/(T 2)) → X is in bijection with the linear

subspace of vectors (b1, . . . , bn) subject to the above linear relations when f varying in I.

Observe that the map ψ factorizes through the localization of k[X1, . . . , Xn]/I at the

maximal ideal (X1−a1, . . . , Xn−an), i.e. through OX,p → k[T ]/(T 2), and it is completely

determined by its restriction to Mp, which maps to (T )/(T 2) ∼= k. Again, this new map

factorizes through Mp/M2p → (T )/(T 2) ∼= k, and any such map defines a map ψ. This

proves the result.

Definition. The embedded tangent space of an affine scheme X = Spec(k[X1, . . . , Xn]/I)

at a point p = (a1, . . . , an) ∈ X is the linear space

TpX :=⋂

f∈I(X)

V( ∂f∂X1

(p)(X1 − a1) + . . .+∂f

∂Xn(p)(Xn − an)

).

Exercise 8.2. Prove that, if the ideal I is generated by f1, . . . , fm, then the embedded

tangent space of X = Spec(k[X1, . . . , Xn]/I) at a point p = (a1, . . . , an) ∈ X is the linear

space

TpX :=

m⋂i=1

V( ∂fi∂X1

(p)(X1 − a1) + . . .+∂fi∂Xn

(p)(Xn − an)).

93

Example 8.3. Set C = Spec(k[X,Y ]/(Y 2−X3)

). Then the embedded tangent space of

C at the origin is the whole affine plane. This is due to the existence of a singularity at

the point. But the presence of a singularity only asserts that the embedded tangent space

is bigger than expected. For example, if C ′ = Spec(k[X,Y, Z]/(Y 2 − X3, Z)

), now the

embedded tangent space of C ′ at the origin is V (Z). In fact, C and C ′ are isomorphic,

hence their embedded tangent spaces must be isomorphic, since Theorem 8.1 is saying

that there is an intrinsic tangent space, and this is isomorphic to (Mp/M2p)∗. The reader

with a knowledge of Commutative Algebra will recognize in this intrinsic expression what

is called the Zariski’s tangent space, and that this dimension is, by Nakayama’s Lemma,

also the minimum number of generators of Mp, which is at least the dimension of the local

ring. We will prove directly these results in the geometric case.

Proposition 8.4. Let X be a scheme of finite type over k and let p ∈ X be a closed

point. Set r = dim(TpX). Then there exists a non-empty open neighborhood U ⊂ X of p

and regular functions u1, . . . , ur ∈ OX(U) such that, for each closed point q ∈ U :

(i) The maximal ideal Mq of OX,q is generated by u1 − u1(q), . . . , ur − ur(q).(ii) dimZ ≤ r for any irreducible Z ⊂ X containing q.

Proof: We can assume X = Spec(k[X1, . . . , Xn]/I), that p corresponds to (a1, . . . , an)

and that there are fr+1, . . . , fn ∈ I such that the determinant h of the Jacobian matrix

of fr+1, . . . , fn with respect to Xr+1, . . . , Xn is not zero at p. Set U = D(h) and ui = Xi

for i = 1, . . . , r. Let us see that, for any q ∈ U corresponding to (b1, . . . , bn), the classes

X1 − b1, . . . , Xr − bn generate Mq. For simplicity, after a translation, we can assume that

q corresponds to (0, . . . , 0). The assumption h(0, . . . , 0) 6= 0 implies that, after taking a

linear combination of the original polynomials fr+1, . . . , fn ∈ I, we can assume that we

can write fi = Xi + gi,1X1 + . . .+ gi,nXn, with gij ∈ (X1, . . . , Xn) if j = r + 1, . . . , n. We

can thus consider the following congruences modulo I + (X1, . . . , Xr)

M

Xr+1

...Xn

≡ 0

...0

where

M =

1 + gr+1,r+1 gr+1,r+2 . . . gr+1,n

gr+2,r+1 1 + gr+2,r+2 . . . gr+2,n

......

. . ....

gn,r+1 gn,r+2 . . . 1 + gn,n

with gi,j ∈ (Xr+1, . . . , Xn). If f = det(M), it is clear that f is not in I(q). Multipying the

above congruences by the adjoint matrix of M , we get fXi ∈ I + (X1, . . . , Xr). From this

we get the two parts of the statement:

94

–We have f Xi ∈ (X1, . . . , Xr) in k[X1, . . . , Xn]/I, and hence also in OX,q. Since f is

a unit in OX,q, it follows that each Xi is in the ideal (X1, . . . , Xr), hence Mq is generated

by X1, . . . , Xr, which proves (i).

–For any Z we will also have fXi ∈ I(Z) + (X1, . . . , Xr). Since f(0, . . . , 0) 6= 0, then

X1, . . . , Xn are in the ideal of any component of Z ∩ V (X1, . . . , Xr) containing q. Hence,

q is the only irreducible component of Z ∩ V (X1, . . . , Xr) containing q. By Corollary

6.14(ii), dim(Z) ≤ r, proving (ii).

Definition. A local regular ring is a local ring whose maximal ideal is generated by as

many elements as its dimension.

Definition. A smooth point of a scheme X is a point p ∈ X (not necessarily closed) such

that OX,p is a local regular ring. Otherwise, p is said to be a singular point. When p is a

closed smooth point, the set of regular functions u1, . . . , ur as in Proposition 8.4 is called

a local system of parameters at the point p.

Remark 8.5. This is the most we can imitate the notion of dimension that appears

naturally in other contexts: a smooth variety (what is called manifold in other contexts)

of dimension r is a variety having locally around each point a system of r parameters, that

work as a system of coordinates. However, the map around p to an open set of Ark defined

by the local parameters is not an isomorphism in general.

Remark 8.6. In the affine case, the singular locus of a scheme X ⊂ Ank is determined

by the set of points p for which the rank of the Jacobian matrix of a set of generators

f1, . . . , fm of the ideal of X is at most n − r − 1 (see Exercise 8.2). Hence it is a closed

set. Therefore, also the singular locus of a scheme locally of finite type over k is a closed

set. Non-reduced schemes like Spec(k[X,Y ]/(X2)

)have all their points singular, but in

reduced schemes the singular locus is proper. There is standard way to prove it is, assuming

X is irreducible and the characteristic of k is zero: since K(X) has transcendence degree

over k equal to the dimension of r, it is isomorphic to K(X ′) where X ′ is a hypersurface of

Ar+1k , and this clearly has a non-empty set of smooth points. Since Theorem 5.20 implies

that X and X ′ possess isomorphic open sets, the result follows.

Remark 8.7. We have seen in the proof of Theorem 8.1 why the tangent space at a

point is obtained from Mp/M2p. Indeed, when p is a closed point of an affine scheme X

the linear part (after making a translation to the origin) of a regular function f ∈ OX(X)

gives a zero element of Mp/M2p. The reader with some knowledge of plane curves will

know that, when the point is singular, one gets a tangent cone taking the homogeneous

part of smallest degree. For instance, in Example 8.3, the tangent cone at the origin p of

95

X = V (Y 2 −X3) is V (Y 2); in this case, we get that the class of Y 2 in M2p/M

3p is zero. In

general, if Mp is generated by u1, . . . , ur, for each degree d, we consider the map

ψd : k[T1, . . . , Tr]d →Mdp/M

d+1p

sending each homogeneous polynomial of degree d to the class of F (u1, . . . , ur).

Definition. The tangent cone of a scheme X at a closed point p (where X is locally of

finite type over k) is Spec(k[T1, . . . , Tr]/I), where I =⊕

d≥0 ker(ψd).

Remark 8.8. Since I is a homogeneous ideal, then the tangent cone is the affine cone

of Proj(k[T1, . . . , Tr]/I). Observe that k[T1, . . . , Tr]/I is isomorphic to GMp(OX,p) :=⊕

d≥0 Mdp/M

d+1p . Hence, by [AM], Theorem 11.14, the dimension of OX,p is the degree

of a polynomial P ∈ Q[T ] such that, for large d, we have P (d) = dim(OX,p/Mdp). In the

language of the proof of Theorem 2.17, we have (∆P )(d) = dimMdp/M

d+1p . Therefore,

∆P , which has degree dim(OX,p) − 1, is the Hilbert polynomial of Proj(k[T1, . . . , Tr]/I).

dimension is the sum of the values of the Hilbert function of GMp(OX,p) at 0, 1, . . . , d− 1,

it follows that dim(

Proj(k[T1, . . . , Tr]/I))

= dim(OX,p) − 1, and hence the dimension of

the tangent cone of X at p is precisely the codimension of p in X.

Another application of the maps ψd is to find Taylor expansions of regular functions.

Lemma 8.9. Let p ∈ X be a closed point of a scheme of locally finite type over k such

that Mp is generated by u1, . . . , ur. Then, for each f ∈ OX,p and each d, there exist

homogeneous polynomials F0, F1, . . . , Fd ∈ k[T1, . . . , Tr] such that deg(Fi) = i and such

that f −∑di=0 Fi(u1, . . . , ud) ∈Md+1

p .

Proof: We prove it by induction on d. For d = 0, we just take F0 = f(p), since f =

f(p) + (f − f(p)), and f − f(p) ∈Mp.

Assume now d > 0 and that we have F0, . . . , Fd−1 such that f−∑d−1i=0 Fi(u1, . . . , ud) ∈

Mdp. We can thus write

f −d−1∑i=0

Fi(u1, . . . , ud) =∑

i1+...+ir=d

fi1...irui11 . . . uirr

for some fi1...ir ∈ OX,p. Setting Fd =∑i1+...+ir=d fi1...ir (p)T i11 . . . T irr , we will have

f −d∑i=0

Fi(u1, . . . , ud) =∑

i1+...+ir=d

(fi1...ir − fi1...ir (p)

)ui11 . . . uirr ∈Md+1

p ,

which proves the result.

In the case of smooth points, the Taylor series of a regular function is unique, and

this gives the local irreducibility at the point:

96

Theorem 8.10. Let p ∈ X be a smooth closed point of a scheme of locally finite type

over k such that dim(OX,p) = r and Mp is generated by u1, . . . , ur.

(i) For each homogeneous polynomial F ∈ k[T1, . . . , Tr] of degree d, F (u1, . . . , ur) ∈Md+1p

if and only if F = 0; hence, each ψd is injective and, in particular, the tangent cone

of X at p coincides with the tangent space.

(ii) There is a monomorphism OX,p → k[[T1, . . . , Tr]]; in particular, OX,p is a domain and

there is only one irreducible component of X (embedded or not) passing through p.

Proof: Since one implication of (i) is trivial, it is enough to prove that, if F ∈ k[T1, . . . , Tr]

is a non-zero polynomial of degree d, then F (u1, . . . , ur) /∈ Md+1p . After linear change of

coordinates, we can assume that F is monic in the variable Td, so that we can write

F = T dr + F1(T1, . . . , Tr−1)T d−1r + . . .+ Fd−1(T0, . . . , Tr−1)Tr + Fd(T1, . . . , Tr−1)

with each Fi homogeneous of degree i, in particular Fi ∈ (T1, . . . , Tr−1). Therefore, we

can write

F (u1, . . . , ur) = udr + F ′,

with F ′ ∈ (u1, . . . , ur−1). Assume now, for contradiction, that F (u1, . . . , ur) ∈Md+1p . We

can thus write

F (u1, . . . , ur) =∑

i1+...+ir=d

fi1...irui11 . . . uirr

with fi1...ir ∈Mp. In particular, we can write

F (u1, . . . , ur) = f0...0dudr + F ′′,

with F ′′ ∈ (u1, . . . , ur−1). Putting together the two expressions for F (u1, . . . , ur), we find

(1− f0...0d)udr = F ′′ − F ′ ∈ (u1, . . . , ur−1).

Since 1− f0...0d is not in Mp (because f0...0d is), it follows that 1− f0...0d is a unit, hence

udr ∈ (u1, . . . , ur−1). This implies that, in a neighborhood U ⊂ X of p, the point p is

defined by the regular functions u1, . . . , ur−1, whis is absurd by Theorem 6.16(iii).

For (ii), an induction argument on d and part (ii) prove that the polynomials Fd of

Lemma 8.9 are unique, so that the assignment f 7→∑d≥0 Fd is the wanted homomorphism

OX,p → k[[T1, . . . , Tr]]. Its kernel is⋂d≥0 M

dp, which is zero by [AM], Corollary 10.18.

By definition, the smoothness of a closed point p in X means that p is locally defined

(as a scheme) by as many equations as it codimension in X. Observe that, if X ⊂ Ank ,

we also implicitly proved in the proof of Proposition 8.4 that also the component of X

containing p is also defined by as many equations as its codimension in Ank . We make

explicit in the following result.

97

Lemma 8.11. Let p ∈ X = Spec(k[X1, . . . , Xn]/I) be a smooth closed point and assume

dim(OX,p) = r. Then, there exist n− r local germs fr+1, . . . , fn ∈ OAnk,p that generate the

local ideal of X at p.

Proof: We repeat the notation and choice of coordinates of the proof of Proposition

8.4. Let I ′ ⊂ k[X1, . . . , Xn] be the ideal generated by the polynomials fr+1, . . . , fnwhose homogeneous parts of smallest degree are, respectively, Xr+1, . . . , Xn. Let X ′ =

Spec(k[X1, . . . , Xn]/I ′) be the corresponding scheme. Since I ′ ⊂ I, there is a closed em-

bedding X ⊂ X ′. Clearly, TpX ⊂ TpX′ ⊂ V (Xr+1, . . . , Xn), which implies that we have

a chain of equalities. In particular, X and X ′ must share the only component Z passing

through p (Theorem 8.10(ii)), which necessarily has dimension r, and locally the ideal of

that component is defined by fr+1, . . . , fn.

The same holds if we replace the affine space with another general scheme that is

smooth at the point:

Theorem 8.12. Let Y ⊂ X be a closed embedding of schemes of finite type over k. If

p ∈ Y is a smooth closed point for Y and X, then the ideal of Y in OX,p is generated by

dim(OX,p)− dim(OY,p) equations.

Proof: By Theorem 8.10(ii), we can assume X,Y irreducible, and we can also assume

that they are affine, i.e. subschemes of some ANk . We can choose coordinates such that

p = (0, . . . , 0), TpY = V (Xm+1, . . . , XN ) and TpX = V (Xn+1, . . . , XN ). Applying Lemma

8.11 to X and Y , we can pick fn+1, . . . fN ∈ I(X) defining locally X, and fm+1, . . . fn ∈I(Y ) that, together with fn+1, . . . fN , define Y . Thus the classes of fm+1, . . . fn ∈ I(Y )

locally generate the ideal of Y inside X.

Remark 8.13. In the case in which Y ⊂ X is irreducible of codimension one, the above

result remains true even if Y is not smooth at p. The basic idea to prove so (see [Sh],

Chapter II, §3, Theorem 1) is that the inclusion of OX,p in a ring of formal series (Theorem

8.10(ii)) implies that OX,p is also a unique factorization domain, and then it is possible to

repeat the proof of Proposition 6.6.

We can now deal with the smoothness of points that are not necessarily closed. From

the very definition, that p is smooth in X means that, in some non-empty open U ⊂ X,

the irreducible set p is defined by as many equation as codim(p, X). Let us see that

the precise meaning of the sentence “the generic point of an irreducible set Y ⊂ X is

smooth” is what one could expect (which implies that, in order to study the smoothness

of a scheme X, it is enough to study the smoothness of its closed points):

98

Proposition 8.14. Let X be a scheme of finite type over k and let p ∈ X be the generic

point of an the irreducible subset Y ⊂ X. Then p is smooth in X if and only if there is an

open set V ⊂ Y such that any closed point of Y is smooth in X.

Proof: If p is smooth, i.e. OX,p is a regular ring, by definition there exists an open set U

meeting Y in which IX∩U (Y ∩ U) is generated by dim(OX,p) equations. Hence, if q is a

smooth closed point of Y in the open set V of smooth points of Y ∩U , the maximal ideal

of OY,q = OX,q/I(Y ) is generated by dim(OY,q) elements. Therefore, the maximal ideal

of OX,q, for q ∈ V is generated by dim(OX,p) + dim(OY,q) = dim(OX,q) equations, so that

q is smooth in X.

Reciprocally, assume that a general element of Y is smooth in X. Hence a general

element of Y is smooth both in X and Y . Then Theorem 8.12 implies that the generic

point of Y is smooth in X.

Remark 8.15. Consider an irreducible scheme X of locally finite type over k, and let

K(X) be its field of rational functions. In the case in which we have a smooth p ∈ X

that is the generic point of Y ⊂ X of codimension one, then OX,p is a local regular ring

of dimension one, i.e. a discrete valuation ring. If f is a generator of Mp, then any

non-zero g ∈ OX,p belongs to some Mdp but not to Md+1

p (by [AM], Corollary 10.18).

This means that we can write g = ufd, with u /∈ Mp, hence u is a unit. Therefore, any

non-zero element of the quotient field K(X) can be written as g = ufd, with u a unit in

OX,p and d an integrar, that could be negative. Writing d = νY (g), we have a valuation

νY : K(X) \ 0 → Z, which makes OX,p a discrete valuation ring. The geometrical

meaning of νY is that it measures (when positive) the vanishing order of f at Y or (when

negative) the order of the pole at Y of f .

Proposition 8.16. Let X be an irreducible scheme of finite type over k whose singular

locus has dimension at most dim(X) − 2. Then, for any non-zero rational function f ∈K(X), there exists a finite number of irreducible subvarieties Y ⊂ X of codimension one

such that νY (f) 6= 0.

Proof: The assumption about the singular locus says that any irreducible Y ⊂ X of

codimension one has an open set of smooth points of X. Hence, by Proposition 8.14, the

generic point of Y is smooth, so that the valuation νY makes sense.

Consider a non-empty affine open set U ⊂ X in which f is a regular function. Since X\U ⊆/ X, then dim(X \U) < dim(X), and hence there are at most finitely many irreducible

Y ⊂ X of codimension one not meeting U (by Proposition 4.15, X \ U decomposes as a

finite union of irreducible components). On the other hand, let U ∩V (f) = Y1∪ . . .∪Ym be

the decomposition into irreducible components. By Corollary 6.14, all Yi has codimension

99

one, and for all of them νYi(f) > 0. On the other hand, for any other Y meeting U , the

rational function f is a unit in OX,Y , because the class of (U ∩D(f), 1f ) is an inverse, and

hence νY (f) = 0.

We study now what happens in the projective case.

Remark 8.17. We try to find first an expression for the embedded tangent space, which

will be the Zariski closure of the embedded tangent of the restriction to each D(Xi). We

will make the computations for i = 0. Fix a point p = (a0 : . . . : an) ∈ Pnk with a0 6= 0 and

any homogeneous polynomial F ∈ k[X0, . . . , Xn] of degree d such that F (a0, . . . , an) = 0.

Set f(X1, . . . , Xn) := F (1, X1, . . . , Xn). Then the equation in D(X0) = Ank

∂f

∂X1(a1

a0, . . . ,

ana0

)(X1 −a1

a0) + . . .+

∂f

∂Xn(a1

a0, . . . ,

ana0

)(Xn −ana0

) =

=∂F

∂X1(1,

a1

a0, . . . ,

ana0

)(X1 −a1

a0) + . . .+

∂F

∂Xn(1,

a1

a0, . . . ,

ana0

)(Xn −ana0

) =

=1

ad−10

( ∂F∂X1

(a0, . . . , an)(X1 −a1

a0) + . . .+

∂F

∂Xn(a0, . . . , an)(Xn −

ana0

))

has a homogenization (after removing the factor 1

ad−10

)

(− a1

a0

∂F

∂X1(p)− . . .− an

a0

∂F

∂Xn(p))X0 +

∂F

∂X1(p)X1 + . . .+

∂F

∂Xn(p)Xn =

∂F

∂X0(a)X0 +

∂F

∂X1(p)X1 + . . .+

∂F

∂Xn(p)Xn

where we made the substitution Xi = ai in the Euler identity

∂F

∂X0X0 +

∂F

∂X1X1 + . . .+

∂F

∂XnXn = dF

and use F (a) = 0. Therefore, we get a symmetric expression, showing in particular that it

does not depend on the affine space to which we restrict. We thus have that, given a homo-

geneous ideal I ⊂ k[X0 . . . , Xn] the embedded tangent space of X = Proj(k[X0 . . . , Xn]/I)

at the point p = (a0 : . . . : an) is given by

TpX =⋂F∈I

V (∂F

∂X0(a)X0 + . . .+

∂F

∂Xn(p)Xn)

and, as in Exercise 8.2, it is enough to make the intersection when F varies only in a set

of (homogeneous) generators of I.

100

Exercise 8.18. Use the above expression to prove that, if X ⊂ Pnk is a smooth closed

subscheme of dimension r, then the Gauss map X → G(r, n) that sends each closed point

p to its embedded tangent space in Pnk is a morphism.

In Exercise 8.18, if X is a hypersurface, we get a map in Pnk∗, whose image, in the case

n = 2 or when X is a quadric, is what is called the dual curve or the dual quadric, which

are still a curve or a quadric in the dual space. But if X has arbitrary dimension, then

its image is a subvariety that is more difficult to deal with. In order to keep working in

Pnk∗, the idea is to work not with tangent spaces, but with tangent hyperplanes, i.e. with

hyperplanes containing tangent spaces. We make the idea more precise in the following

example.

Example 8.19. Let X ⊂ Pnk be a closed irreducible reduced subscheme of dimension r.

Then Xsm = X \Sing(X) is a non-empty open subset of X. Consider the incidence variety

Xsm × Pnk∗ ⊃ Ism = (p,H) | TpX ⊂ H

(prove as an exercise that Ism is a closed subset). Since the fiber at any p ∈ Xsm under the

first projection is a linear subspace of dimension n− r− 1 in Pnk∗, it follows from Theorem

6.21 that Ism is irreducible of dimension n− 1, hence the same holds for its Zariski closure

I ⊂ X × Pnk∗. Then its image X∗ ⊂ Pnk

∗ under the second projection is an irreducible

closed subscheme,

Definition. The subscheme X∗ ⊂ Pnk∗ constructed in the above example is called the dual

variety of X.

Remark 8.20. One would expect that a general hyperplane is tangent at most at a finite

number of points, and this would give that X∗ is defined by one equation. This would

imply that one could describe X by just that equation, since the self-duality (X∗)∗ = X

(which is proved(∗) showing that I is also the incidence variety defining the dual of X∗)

implies that we can recover X from X∗. However this is not true as the very self-duality

shows. Indeed, if you start with X ⊂ Pnk that is not a hypersurface, then the dual of X∗

is not a hypersurface. One could think that this strange behavior is due to the fact that

X∗ is in general very singular, but even the dual of a smooth scheme is not necessarily

a hypersurface. For example, a result of L. Ein shows that, when X is smooth, then

dim(X∗) ≥ dim(X) and that equality holds exactly in four cases:

(i) X is a hypersurface.

(ii) X = G(1, 4) ⊂ P9k.

(iii) X = P1k × Pr−1

k ⊂ P2r−1.

(∗) This is true only in characteristic zero.

101

(iv) X ⊂ P15k is a 10-dimensional spinor variety.

We can use the dual variety to prove the following:

Theorem 8.21 (Bertini). Let X ⊂ Pnk a reduced irreducible subscheme. Then, for each

d > 0 there is a non-empty open set U ⊂ P(k[X0, . . . , Xn]d) in the space of hypersurfaces

of degree d in Pnk such that the singular locus of the intersection of X with a hypersurface

in U is exactly the intersection of the singular locus of X with the hypersurface.

Proof: Considering the Veronese embedding νd,n : Pn → P(nd)−1

k , the intersection of X

with a hypersurface of degree d becomes the intersection of νd,n(X) (which is isomorphic

to X) with a hyperplane of P(nd)−1

k , so that it is enough to prove the result for d = 1.

If X is defined by a homogeneous ideal I ⊂ k[X0, . . . , Xn], then X ∩ V (H) (where

H is a linear form) is defined by I + (H). Hence, for any closed point p ∈ X ∩ V (H),

Tp(X ∩ V (H)

)= TpX ∩ V (H). This implies that, if p is singular in X, it is also singular

in X ∩ V (H) and that, if p is smooth in X, then it is smooth in X ∩ V (H) if and only if

TpX 6⊂ V (H). Hence, U = Pnk∗ \X∗ is the open set we are looking for.

Remark 8.22. We can iterate the above result and get that, if the dimension of the sin-

gular locus is s, the intersection with more than s general hypersurfaces produces a smooth

subscheme. In particular, the intersection of X with a linear subspace of codimension r

produces a smooth scheme of dimension zero. This is necessarily a scheme of points, each

with multiplicity one, i.e. defined by its maximal ideal. Hence we get exactly as many

different points as the degree of X. This yields a geometric definition of the degree.

102

9. Vector bundles

We have seen in the previous section that, at least locally at smooth points, subva-

rieties are defined by the right number of equations, an we wonder whether we can glue

all of them. For example, we know that subvarieties of codimension one in Ank and Pnk are

defined by one global equation. In the case of Ank , the equation is a regular funcion, and in

the case of Pnk , the equation is a homogeneous polynomial, which is not a regular function,

but can be regarded as an object obtained by glueing regular functions. We will see that

this will be the case in general in codimension one, and that the right way of interpreting

the object obtained by glueing local equations is a section of a line bundle. In the case of

general codimension r, one could expect that the r local equations glue to form a section

of a vector bundle of rank r, but we will see that this is not the case.

Let us start studying a couple of cases in Pnk , one of codimension one, and another of

codimension two:

Example 9.1. Consider X = Proj(k[X0, . . . , Xn]/(F )

), with F ∈ k[X0, . . . , Xn] a non-

zero homogeneous polynomial of degree d > 0. In any D(Xi), the restriction of X is

defined by the regular function FXd

i

. Hence, to pass from the local equation on D(Xi) to

the local equation on D(Xj) we need to multiply byXd

i

Xdj

, which is a regular function on

D(Xi) ∩ D(Xj), having no zeros. Moreover, given i, j, k, passing from D(Xi) to D(Xk),

which is given my multiplication byXd

i

Xdk

, is the same as passing first from D(Xi) to D(Xj)

(multiplying byXd

i

Xdj

) and then passing from D(Xj) to D(Xk) (multiplying byXd

j

Xdk

).

Example 9.2. Let X ⊂ P2k be the subset consisting of the points (1 : 0 : 0), (0 : 1 : 0)

and (0 : 0 : 1). At each open set Ui = D(Xi), X ∩ Ui consists of one point, so its affine

ideal is generated by exactly two elements. More precisely, the ideals of X ∩ U0, X ∩ U1

and X ∩ U2 are respectively generated by x1

x0, x2

x0; x0

x1, x2

x1; and x0

x2, x1

x2. For any λ ∈ k \ 0

(the reader will understand soon why we do not take just λ = 1), it is possible to relate

the two first sets of generators by the expression x1

x0

x2

x0

=

(x1

x0)2 0

(1− λ)x1x2

x20

λx1

x0

x0

x1

x2

x1

and the matrix in the expression has entries regular in U0 ∩ U1 and is invertible in that

open set. Similarly, there is a relation x0

x1

x2

x1

=

λ′ x2

x1(1− λ′)x0x2

x21

0 (x2

x1)2

x0

x2

x1

x2

103

Putting together the last two relations we get x1

x0

x2

x0

=

λ′ x1x2

x20

(1− λ′)x2

x0

λ′(1− λ)(x2

x0)2 (λλ′ − λ′ + 1)

x22

x0x1

x0

x2

x1

x2

Thus if we want to avoid divisions by zero (and get a matrix similar to the previous ones),

we need λλ′ − λ′ + 1 = 0, i.e. λ′ = 11−λ . Hence, if λ 6= 0, 1, we can glue together the

equations.

The precise definition is the following (in which the reader can replace the category

of algebraic varieties with any category of topological spaces):

Definition. A vector bundle of rank r (or line bundle if r = 1) over a scheme X over k

is a k-scheme F equipped with a k morphism π : F → X so that there exists a covering

X = ∪i∈IUi by (Zariski) open subsets such that:

(i) For each i ∈ I there is a k isomorphism ψi : π−1(Ui) → Ui ×k Ark satisfying that the

composition π ψ−1i : Ui ×k Ark → Ui is the first projection.

(ii) For each i, j ∈ I there is an (r× r)-matrix Aij (called transition matrix, or transition

function if r = 1) whose entries are regular functions in Ui ∩ Uj satisfying that the

composition

ϕij := ψj ψ−1i|Ui∩Uj

: (Ui ∩ Uj)×k Ark → π−1(Ui ∩ Uj)→ (Ui ∩ Uj)×k Ark

is determined by the first projection (Ui ∩ Uj) ×k Ark → Ui ∩ Uj and the morphism

(Ui ∩ Uj) ×k Ark → Ark is given by the homomorphism of k-algebras k[T1, . . . , Tr] →O(Ui ∩ Uj)[T1, . . . , Tr] defined byT1

...Tr

7→ Aij

T1...Tr

.

A section of the vector bundle F is a k-morphism s : X → F such that π s = idX .

Remark 9.3. We make some precisions about the definition:

–The morphisms ϕij , when interpreted as maps defined for closed points can be written

as ϕij : (Ui ∩ Uj) × kr → (Ui ∩ Uj) × kr defined by ϕij(x, v) = (x,Aij(x)v). With this

point of view, condition (i) is saying that, for any x ∈ X the set π−1(x) (called the fiber

of the vector bundle at the point x), is bijective to kr, and that locally the fibers are glued

to produce a trivial product Uij × kr. We will in fact write this instead of Ui ×k Ark, since

condition (ii) is just saying that the fibers of F have to be regarded as vector spaces rather

104

than affine spaces, since the way of identifying the fibers from Ui to Uj is by means of a

linear transformation. Observe the det(Aij) can not vanish at any point of Ui ∩ Uj .–If only one open set is needed, i.e. F = X ×k kr and π is the first projection, we say

that F is a trivial bundle of rank r. In this case, a section of F is the same as giving r

regular functions on X. In the general case, giving a section of F is equivalent to give, for

any i, r regular functions fi1, . . . , fir such that, for each i, j we have relations fj1|Ui∩Uj

...fjr |Ui∩Uj

= Aij

fi1|Ui∩Uj

...fir |Ui∩Uj

.

This is saying that the ideals (fi1, . . . , fir) ⊂ OX(Ui) and (fj1, . . . , fjr) ⊂ OX(Uj) coincide

on OX(Ui∩Uj). Therefore, a section s determines a sheaf of ideals, i.e. a closed embedding

Y ⊂ X, that we will call the zero locus of the section, and will be denoted by (s)0. If X is

irreducible of dimension n, all the components of Y have dimension at least n− r.–It is clear that any subpartition of the covering still satisfies conditions (i) and (ii),

so we will always work, when necessary, with sufficiently fine partitions.

Proposition 9.4. Let X be a k-scheme with an open covering X = ∪i∈IUi. Given a set

of matrices Aiji,j∈I whose entries are regular functions on Ui ∩ Uj and satisfying:

(i) Aii is the identity matrix Ir;

(ii) AjkAij = Aik;

there is a vector bundle over X whose matrices are the given matrices.

Proof: Just use Example 4.10 to glue the schemes Ui ×k Ark using the matrices Aij as in

the definition of vector bundle (observe that (i) and (ii) imply that the matrices Aij are

invertible).

Example 9.5. Proposition 9.4 allows to actually construct vector bundles from Examples

9.1 and 9.2. In particular, Example 9.1 yields a line bundle Ld over Pn with transition

functionsxdi

xdj

, in which we can now even allow d to be negative. When d > 0, any homoge-

neous polynomial F ∈ k[X0, . . . , Xn] of degree d defines a section of Ld whose zero locus

is the scheme determined by the ideal (F ) (even if F has multiple components). Similary,

Example 9.2 provides, for any λ 6= 0, 1, a vector bundle Fλ over P2k having a section whose

zero locus consists of the three coordinate points.

Exercise 9.6. Construct a rank-two vector bundle over the smooth quadric X =

V (X0X3+X1X4+X22 ) ⊂ P4 with a section whose zero locus is the line L = V (X0, X1, X2).

This vector bundle is called the spinor bundle over X.

105

In fact the notion of “vector bundle determined by transition matrices” does not make

any sense unless we have the notion of isomorphism (“determined by” always means “up

to isomorphism”). The precise definition is:

Definition. A morphism of vector bundles is a regular map ϕ : F → F ′ such that π = π′ϕ(i.e. it sends fibers Fx to fibers F ′x) and the induced map ϕx : Fx → F ′x is linear. If any

ϕx is an isomorphism we will say that ϕ is an isomorphism.

Remark 9.7. The local description of a morphism of bundles is as follows. We first take

a common partition X =⋃i Ui for which F and F ′ trivialize. Since giving a morphism

Ui× kr → Ui× kr′

is equivalent to give an (r′× r)-matrix Ai with entries in OX(Ui), then

a morphims F → F ′ is equivalent to give a collection of matrices Ai such that, for each

i, j, the restriction to Ui ∩ Uj satisfies A′ijAi = AjAij .

Remark 9.8. It is possible to define in the natural way the notion of dual vector bundle,

direct sum or tensor product of vector bundles, wedge or symmetric power of a vector

bundle... We leave the reader the task of checking how the transition matrices can be

constructed from the transition matrices of the original vector bundles. We concentrate in

just few examples.

Exercise 9.9. Show that we have the following isomorphisms of line bundles over Pnk(following the notation of Example 9.5):

(i) L0 is the trivial line bundle.

(ii) Ld ⊗ Le ∼= Le+d.

(iii) L∗d∼= L−d.

(iv) SymmeLd ∼= Lde.

(v) If n = 2,∧2

Fλ ∼= L3.

Example 9.10. For any smooth variety of dimension n, it is possible to define its

cotangent bundle ΩX (its dual vector bundle TX is called the tangent bundle). For this,

for each closed point p, we consider a local system of parameters u1, . . . , un on some open

neighborhood U ⊂ X of p. Then ΩX |U ∼= U × kn, in which we identify a section of U × kn

(i.e. regular functions f1, . . . , fn) with the formal differential expression f1du1+. . .+fndun,

and the transition matrices are defined using the standard properties of differentials. For

example, to define the cotangent bundle of Pnk , we take the standard covering with Ui =

D(Xi) for i = 0, . . . , n. A differencial form on Ui as a formal expression of the form

ω = f0d(X0

Xi) + . . .+ fi−1d(

Xi−1

Xi) + fi+1d(

Xi+1

Xi) + . . .+ fnd(

Xn

Xi)

106

where f0, . . . , fi−1, fi+1, . . . , fn ∈ k[X0

Xi, . . . , Xi−1

Xi, Xi+1

Xi, . . . , Xn

Xi]. To pass from Ui to Uj we

write, for any k 6= i:

d(Xk

Xi) = d

( (Xk

Xj)

(Xi

Xj)

)=Xj

Xid(Xk

Xj)− XkXj

X2i

d(Xi

Xj)

so that we get

ω =

Xj

Xi

(f0d(

X0

Xj)+. . .+(−X0

Xif0−. . .−

Xi−1

Xifi−1−

Xi+1

Xifi+1−. . .−

Xn

Xifn)d(

Xi

Xj)+. . .+fnd(

Xn

Xj)

).

In other words, the transition matrix from Ui to Uj is given by

Aij =Xj

Xi

1 0 . . . 0 . . . 00 1 . . . 0 . . . 0...

......

...−X0

Xi−X1

Xi. . . −Xj

Xi. . . −Xn

Xi

......

......

0 0 . . . 0 . . . 1

.

In particular,∧n

ΩPnk

∼= L−n−1. For example, in the case n = 2, we have

A10 =

−X20

X21−X0X2

X21

0 X0

X1

, A21 =

X1

X20

−X0X1

X22

−X21

X22

, A20 =

0 X0

X2

−X20

X22−X0X1

X22

.

Exercise 9.11. Show that, if F is a vector bundle determined by transition matrices

Aij , then the dual vector bundle F ∗ is a vector bundle determined by transition matrices

(Atij)−1

(where Atij denotes the transposed matrix of Aij). In particular, the dual vector

bundle F ∗λ of the vector bundle Fλ obtained from Example 9.5 has transition matrices:

A′10 =

x20

x21

λ−1λ

x0x2

x21

0 1λx0

x1

, A′21 =

(1− λ)x1

x20

λx0x1

x22

x21

x22

, A′20 =

0 λ−1λ

x0

x2

x20

x22

1λx0x1

x22

.

Use that to show that the matrices

A0 =

(1− λ 0

0 1

), A1 =

(λ− 1 0

0 λ

), A2 =

(−1 00 −λ

)define an isomorphism between ΩP2 and the vector bundle F ∗λ defined in Example 9.5

(this shows that all the bundles Fλ are isomorphic to the tangent bundle TP2 , in particular

isomorphic to each other).

107

Example 9.12. Assume that we have a closed embedding Y ⊂ X of irreducible smooth

schemes of local finite type over k. By Theorem 8.12, we can find an open covering

X =⋃i Ui such that on each Ui the closed subscheme Y ∩ Ui is defined by as the zero

locus of regular functions f1, . . . , fr ∈ OX(Ui), where r = codim(Y,X). The restriction of

differentials, clearly define a surjective morphism ΩX |Y → ΩY , and we call is kernel the

conormal bundle N∗Y/X of Y in X. It has rank r and, on each Y ∩ Ui, it is generated by

df1, . . . , dfr. Its dual is called the normal bundle of Y in X.

The following result is saying that the above local equations fi glue together to produce

a vector bundle on X if and only if the normal bundle can be extended to be the wanted

vector bundle on X.

Theorem 9.13. Let F be a vector bundle of rank r on a smooth irreducible scheme X

of locally finite type over k. Assume that F has a section whose zero locus is a smooth

irreducible subscheme Y ⊂ X of codimension r. Then, NY/X = F|Y .

Proof: Let X =⋃I Ui be a partition in open sets on which F trivializes, let Aij the cor-

responding transition matrices of F and let fi1, . . . , fir ∈ OX(Ui) be the regular functions

defining the given section of F . We thus have a relation on Ui ∩ Uj fj1...fjr

= Aij

fi1...fir

.

Taking differentias, and restricting to Y (where all the fil vanish) we get dfj1...

dfjr

= Aij

dfi1...dfir

(where the bar over the matrix indicates that all its entries are resticted to Y ). Hence, if

a section of the conormal bundle is written on Y ∩ Ui as g1dfi1 + . . . + grdfir, it will be

written on Ui ∩ Uj as

( g1 . . . gr )

dfi1...dfir

= ( g1 . . . gr )

dfj1...

dfjr

A−1ij

and therefore the transition matrices of N∗Y/X are (Atij)−1. As a consequence, the transition

matrices of NY/X are Aij , which proves the result.

108

Example 9.14. As an application, we get a restriction for a subvariety Y ⊂ X to be the

zero locus of a section of a vector bundle F . Indeed, assume Y,X are smooth irreducible

of respective dimensions m,n. From the exact sequence

0→ N∗Y/X → ΩX |Y → ΩY → 0

it follows∧m

ΩY ∼=∧n−m

NY/X ⊗∧n

ΩX |Y . Therefore, by Theorem 9.13, the line bundle∧mΩY must be the restriction of the line bundle

∧n−mF ⊗

∧nΩX (this is known as the

adjunction formula). For example, consider the twisted cubic Y ⊂ P3k which is isomorphic

to P1k, hence ΩY ∼= L−2. Also, the restriction to the vector bundle L′d over P3

k to Y is

isomorphic to L3d. Therefore, ΩY is not the restriction of any line bundle of the type L′dover P3

k, but we will see that all the line bundles over Pnk are of that type. Therefore,

the twisted cubic cannot be obtained as the zero locus of a section of a rank-two vector

bundle over P3k. It is worth to remark that a construction of Serre and Hartshorne shows

that the extendability of ΩY to a line bundle over X is essentially sufficient (the only extra

condition is a cohomology vanishing that always holds, for example, when X is a projective

space of dimension at least three) in the case in which Y has codimension two.

Example 9.15. Let us consider U ⊂ Pnk×kn+1 to be the subset of pairs (p, v) such that v

is a vector in the vector line of kn+1 defining the point p ∈ Pnk and let π : U → Pnk be the first

projection. For each i = 0, . . . , n, the fiber of any p = (a0 : . . . : an) ∈ D(Xi) is the linear

space of multiples of the vector (a0ai , . . . ,anai

) (which does not depend on the representative

of p). Hence we have an isomorphism ψ−1i : D(Xi) × k → π−1(D(Xi)) determined by(

(a0 : . . . : an), λ)7→((a0 : . . . : an), λ(a0ai , . . . ,

anai

)). Hence, the corresponding composition

maps ϕij : D(Xi)∩D(Xj)× k → π−1(D(Xi)∩D(Xj)

)→ D(Xi)∩D(Xj)× k are defined

by ((a0 : . . . : an), λ

)7→((a0 : . . . : an), λ(

a0

ai, . . . ,

anai

))

=

=((a0 : . . . : an),

ajaiλ(a0

aj, . . . ,

anaj

))7→((a0 : . . . : an),

ajaiλ)

which proves that U is isomorphic to L−1.

Exercise 9.16. Generalize the above example and show that, for any Grassmannian

G(k, n), the subset U ⊂ G(k, n)×kn+1 consisting of the pairs (Λ, v) such that v is a vector

in the vector subspace of kn+1 defining the Λ ⊂ Pnk is a vector bundle of rank k + 1 when

considering the first projection π : U → G(k, n). It is called the tautological or universal

vector bundle.

Example 9.17. Dualizing the above inclusion we have an epimorphism of vector bundles

G(k, n) × (kn+1)∗ → U∗, in which now the fiber of U∗ at Λ is the vector space of linear

109

forms on the (k+1)-dimensional linear space ~Λ ⊂ kn+1 definining Λ. Taking the d-th sym-

metric power of the epimorphism we get another epimorphism G(k, n)×k[X0, . . . , Xm]d →SymmdU∗, whose fiber at a subspace Λ is the restriction of a homogeneous form of degree

d fro kn+1 to ~Λ. Hence, a homogeneous form F ∈ k[X0, . . . , Xm]d produces a section of

SymmdU∗ whose zero locus is the set of linear subspaces Λ of dimension k contained in

the hypersurface V (F ) ⊂ Pnk . For example, given a cubic surface V (F ) ⊂ P3k, the cubic

form F produces a section of the vector bundle Symm3U∗ over G(1, 3), whose zero locus

is the set of lines contained in V (F ). Since Symm3U∗ has rank four, and this is also the

dimension of G(1, 3), one expects to get a zero dimensional zero locus, i.e. a finite number

of lines contained in the cubic surface.

It is clear that the set of sections of a vector bundle has always a structure of vector

space. In the case of SymmdU∗, we have seen that all the elements of k[X0, . . . , Xm]d can

be regarded as sections of the vector bundle. In fact, there are no more, as we will prove

now in the particular case k = 0 (in which SymmdU∗ is nothing but Ld):

Proposition 9.18. There is a natural identification of the set of sections of Ld over Pnkwith the vector space of homogeneous polynomials of degree d in k[X0, . . . , Xn].

Proof: A section of Ld is defined by by giving, at each D(Xi), a polynomial fi ∈k[X0

Xi, . . . , Xn

Xi], or equivalently fi = Fi

Xaii

, where each Fi ∈ k[X0, . . . , Xn] (i = 0, . . . , n)

is a homogeneous polynomial of degree ai (we can clearly take ai ≥ d). Using the

transition functions of Ld, we have that, for i 6= j, it should hold fj = fi(Xi

Xj)d, i.e.

FjXai−di = FiX

aj−dj . This means that (reducing denominators in fi) we can take each ai

equal to d and then Fi is the same polynomial F for each i = 0, . . . , n.

Definition. Given a vector bundle F over X and a vector space V of sections of F , the

evaluation map is the morphism evV : X × V → F defined by evV (s, p) = s(p). If evV is

surjective, we say that the sections of V generate the vector bundle.

Theorem 9.19. Let F be a vector bundle of rank r over an irreducible scheme X of finite

type over k. Assume that a space V of sections generates F . Then the zero locus of a

general section in V is either the empty set or a closed subscheme of codimension r.

Proof: We consider the incidence variety I ⊂ X ×k P(V ) consisting on the pairs (p, [s])

such that p is in the zero locus of s. Our assumption about the fact that V generates F

means that, for each closed point p ∈ X, the fiber of p under the projection I → X is a

linear subspace of codimension r in P(V ). Therefore, by Theorem 6.21, I is irreducible and

dim(I) = dim(X) + dim(V )− 1− r. Now we have two possibilities. One could have that

the second projection I → P(V ) is not surjective, so that a general section of V (i.e. in

110

the complement of the image) has empty zero locus. If instead the projection I → P(V ) is

surjective, by Theorem 6.20, there exists a non-empty set U ⊂ P(V ) such that, if [s] ∈ U ,

the fiber of [s] (which is the zero locus of s), has dimension equal to dim(X)−r, as wanted.

Remark 9.20. In fact, much more can be said in case X is also smooth. In that case,

since the fibers of I → X are smooth, it can be proved that then I is also smooth and that

this implies that the general fiber of X → P(V ) is smooth. Hence we get a generalization

of Bertini’s Theorem.

Example 9.21. Given a regular map ϕ : X → Y and a rank-r vector bundle G → Y

over Y , show that the second projection of the fibered product G ×Y X defines a rank-r

vector bundle over X (called pull-back of G via ϕ and denoted by ϕ∗G). If G is determined

by an open covering Y =⋃i Vi and transition matrices Aij , then ϕ∗G is determined by

X =⋃i ϕ−1(Vi) and transition matrices ϕ Aij , obtained by composing each entry of Aij

with ϕ. The vector bundle ϕ∗G can be regarded as the bundle over X whose fiber at each

p ∈ X is the fiber Gϕ(p) of G at ϕ(p).

Theorem 9.22. If a line bundle L is generated by a space of sections V , there is a

morphism ϕV : X → P(V )∗ such that the pull-back of the epimorphism P(V )∗ × V → U∗

is the evaluation map evV : X × V → L.

Proof: As a map of points, we associate to each closed point p ∈ X the class of the linear

map V → Lp defined by s 7→ s(p) (composing with any isomorphism Lp ∼= k we get a linear

form, and the change of the isomorphism produces proportional linear forms). Hence, for

each p ∈ X, the line in V ∗ determining ϕ(p) is L∗p, identified with the space of linear forms

V → k that are zero for each s ∈ V for which s(p) = 0. With this point of view, the

inclusion L∗ ⊂ X×V ∗ dual to the evaluation map is the pull-back to X of the tautological

inclusion U ⊂ P(V )∗ × V ∗, as wanted.

Remark 9.23. In coordinates, if s0, . . . , sN is a basis of V , then the coordinates of each

ϕV (p) can be written as (s0(p) : . . . : sN (p)) (which makes sense again by taking any

isomorphism Lp ∼= k). Since locally a section is given by a regular function, the map ϕV

is locally defined by regular functions, hence it is indeed a morphism. As an example, if V

is the subspace V ⊂ k[T0, T1]4 (regarded as the space of sections of L1 over P1k produces

the map P1k → P3

k defined by (t0 : t1) 7→ (t40 : t30t1 : t0t31 : t41), which maps P1

k to the curve

of Exercise 2.18. The reader is invited to replace in the above result the line bundle with

a vector bundle of arbitrary rank r, and show that a vector space V of sections generating

111

the bundle produces now a map to the Grassmannian of linear subspaces of dimension

r − 1 in P(V )∗.

We will end this section trying to understand better the projective space associated

with the vector sections of a line bundle, since this is the ambient space to which we can

map schemes. This will be the same as studying their zero loci, and in this way we will

get the notion of divisor.

Example 9.24. Let X be an irreducible reduced scheme of finite type over k whose

singular locus has codimension at least two, and let s be a non-zero section of L. Since

X is of finite type, we can take a finite partition X =⋃i Ui for which L trivializes,

with transition functions fij, a section s is given by a collection of regular functions

fi ∈ OX(Ui) such that

fifij = fj (∗)

on Ui∩Uj . In particular, the functions fi, fij can be regarded as rational funcions onX, and

hence apply the valuations νY defined in Remark 8.15. Observe that, for any irreducible

hypersurface Y on X meeting both Ui and Uj (hence Y meets Ui ∩ Uj because Y is

irreducible), equality (∗) implies νY (fi) = νY (fj) (and they are obviously nonnegative). We

can thus associate to the section s a natural number νY (s) for each irreducible hypersurface

Y ⊂ X (we leave the reader the task of checking that this definition does not depend on

the representation of L by a covering and transition matrices).

Definition. A (Weil) divisor on an irreducible scheme X of dimension n is a formal

combination n1Y1 + . . .+ nrYr, where Y1, . . . , Yr are irreducible subvarieties of dimension

n− 1 of Y and n1, . . . , nr are non-zero integers. The support of a divisor D = n1Y1 + . . .+

nrYr is supp(D) = Y1∪ . . .∪Yr The set of all divisors of X is denoted by Div(X), and it is

an abelian group with the natural sum (Div(X) can be thus defined as the abelian group

generated by the set of irreducible subvarieties of dimension n−1 of X). The divisor having

all its coefficients equal to zero is called the empty or zero divisor, and is the zero element

of the group. An effective divisor is a divisor all of whose coefficients are nonnegative. If

s is a non-zero section of a line bundle over X, we can define the zero locus divisor of the

section as

(s)0 =∑Y⊂X

νY (s)Y.

As we have just used, Proposition 8.16 indicates that we can associate a divisor to

any non-zero rational. In some cases, it can be the zero divisor. For example, the class of

X in Spec(k[X,Y ]/(XY − 1)

)defines the zero divisor. However, on a proper scheme we

cannot find that situation:

112

Lemma 9.25. If X is a smooth irreducible proper scheme over k, the map div : K(X) \0 → Div(X) defined by div(f) =

∑Y νY (f)Y is a homomorphism whose kernel is the

set of nonzero constant elements of k.

Proof: It is clear that div is a homomorphism (it is well defined by Proposition 8.16). If we

assume div(f) to be zero, this is equivalent to νY (f) = 0 for any irreducible hypersurface

Y ⊂ X. For any point x ∈ X, let U be an affine open set of X containing x. Then f is

in the quotient field of OX(U). Since νY (f) = 0 for any irreducible hypersurface Y ⊂ X

meeting U , it means that f is in the localization of OX(U) at any prime ideal of length

one, so that f is actually in OX(U). Therefore f is regular at x. Since this is true for

any x ∈ X, it follows that f is a regular function on the whole X. The properness and

irreducibility of X imply that f is a constant in k, which completes the proof.

Definition. A principal divisor is a divisor of the form div(f). We will denote with

Princ(X) the set of all principal divisors, i.e. the image of div, which is a subgroup

of Div(X). Two divisors D and D′ are linearly equivalent (and it will be indicated by

D ∼ D′) if they are congruent modulo Princ(X). In other words, D ∼ D′ if and only if

there exists a nonzero rational function f such that D′ −D = div(f).

This linear equivalence will give the precise relationship between line bundles and

divisors:

Theorem 9.26. Let X be a smooth irreducible proper scheme over k. Then:

(i) For any effective divisor D, there exists a line bundle LD and a section of it whose

zero locus is D

(ii) If D is the zero locus of a section of a line bundle L and D′ is the zero locus of a

section of a line bundle L′, the divisors D and D′ are linearly equivalent if and only

if L and L′ are isomorphic.

(iii) Two sections of the same line bundle have the same zero locus divisors if and only if

they are proportional.

Proof: To prove (i), write D = n1Y1 + . . . + nrYr. Since X is smooth, we know from

Remark 8.13 that, for each closed point p ∈ X, there is an open neighborhood such that

the ideal of each Yi is generated by a regular function gi (that could be 1 if Yi does not

meet the open set). Therefore, the regular function f = gn11 . . . gnr

r defines D on that open

set. We can thus cover X by open sets X =⋃i Ui such that on each Ui, the divisor is the

zero locus of a regular function fi ∈ OX(Ui). Since νY (fjfi

) = 0 for any hypersurface Y

meeting Ui and Uj , it follows as in the proof of Lemma 9.25 that eachfj |Ui∩Uj

fi|Ui∩Uj

is a regular

113

function on Ui ∩ Uj . From this, it is clear that the quotientsfj |Ui∩Uj

fi|Ui∩Uj

are the transition

functions of a line bundle LD, and the regular functions fi determine a section of LD whose

zero locus is D.

For (ii), take an open covering X =⋃i Ui that trivializes both L and L′, with re-

spective transition functions fij and f ′ij . If the section of L giving D is defined by fi,we have fj |Ui∩Uj

= fijfi|Ui∩Ujand, if the section of L′ giving D′ is defined by f ′i, we

also have f ′j |Ui∩Uj= f ′ijf

′i |Ui∩Uj

. Then L and L′ are isomorphic if and only if for each i

there are regular functions gi ∈ OX(Ui) with no zeros such that f ′ijgi|Ui∩Uj= gj |Ui∩Uj

fij .

Putting everything together, this is equivalent tofi|Ui∩Uj

f ′i |Ui∩Uj

gi|Ui∩Uj= gj |Ui∩Uj

fj |Ui∩Uj

f ′j |Ui∩Uj

, i.e.

there exist a rational function f ∈ K(X) such that f|Ui= figi

f ′i

. Since νY (gi) = 0 for each

Y meeting Ui, the above equality is equivalent to D′ + div(f) = D, as wanted.

Part (iii) is as part (ii), but now we have div(f) = 0, which implies, by Lemma 9.25,

that f is a constant and, since all gi can be taken to be 1, we get that the two sections are

proportional, with proportionality equal to f .

Remark 9.27. Parts (i) and (ii) of Theorem 9.26 provide a map from the set of effective

divisors to the Picard group Pic(X) of isomorphism classes of line bundles, and that this

maps factorizes, as an injection through the set of linear equivalence classes. And part

(iii) means that, for each line bundle in the image of the first map, its counterimage is

the projectivization of the vector space of sections of the line bundle. Observe that we

could extend all the above to a homomorphism Div(X)→ Pic(X) factorizing through an

injection Div(X)/Princ(X) if we allow the fi to be rational functions instead of regular

functions. This leads to the following notion:

Definition. A Cartier divisor on an irreducible scheme of finite type over k is a collection

of rational functions fi associated with an open covering X =⋃i Ui such that, for each

i, j, the quotientfjfi

is a regular function on Ui ∩ Uj .

Obviously, when X is proper and smooth, the two notions of divisor coincide. It can

also be proved that, in that case, the monomorphism Div(X)/Princ(X) → Pic(X) is in

fact an isomorphism.

Remark 9.28. After Theorem 9.26, to study linear subspaces V of the space of sections

of a line bundle L, is the same as studying subspaces of effective divisors. More precisely,

given a divisor D corresponding to L, there is an isomorphism between the space of sections

of L and the vector L(D) := f ∈ K(X) | D + div(f) ≥ 0 ∪ 0, and the projectivization

of the space of sections of L is isomorphic to |D| := D′ ≥ 0 | D′ ∼ D.

114

Definition. A linear system of divisors is a linear subspace of some |D|, i.e. the projec-

tivization of a linear subspace V ⊂ L(D). A base-point free linear system is a linear system

Λ such that, for each closed point p ∈ X there is D′ ∈ Λ such that p /∈ Supp(D′).In this language, Theorem 9.22 is saying that a a base-point free linear system Λ yields

a morphism X → Λ∗.

115

10. Sheaves of modules

We have seen that sections of line bundles can be used to give morphisms to a pro-

jective space. The first natural question is to know how many sections a line bundle (or

a vector bundle) possesses. Before approaching that problem, we will need to study the

structure of the sections of a vector bundle.

Remark 10.1. We already observed that the space of sections of a vector bundle has a

structure of vector space. But we can do more than just multiplying a section s : X → F

by a constant: we can allow the constant to vary in each fiber by a regular function f , i.e.

we can define a new section fs by (fs)(p) = f(p)s(p). Of course this is nothing new if the

scheme is proper, since the only regular functions are the constants. But even in this case,

we can consider sections on an open set, and they form a module over the ring of regular

functions on the open set. Moreover, in open sets on which the vector bundle trivializes,

the module is free. All these considerations lead to the following:

Definition. Given a topological space X and a sheaf of rings OX , a sheaf of modules over

OX isa sheaf F such that, for each open set U ⊂ X, F(U) is an OX(U)-module, and the

restriction map F(U) → F(V ) are homomorphisms of OX(U)-modules (observe that the

restriction map OX(U)→ OX(V ) provides F(V ) a structure of OX(U)-module. A locally

free sheaf is a sheaf of modules F such that there exists an open covering X =⋃i Ui

satisfying that each F|Uiis free, i.e. isomorphic to a finite direct sum (OX)|Ui

⊕ . . . ⊕(OX)|Ui

.

We omit the definitions of morphism of sheaves, stalks,..., which are the same as for

rings. It is also natural to define direct sum of sheaves. Other notions, like tensor product

or dual will be studied in detail later.

In the same way as in Theorem 3.23 we produced a sheaf of rings on Spec(A) from

the ring A, we can now get a sheaf of modules from an A-module M :

Theorem 10.2. Let A be a ring, let M be an A-module and set X = Spec(A):

(i) The assignment D(f) 7→Mf defines a a sheaf FM of OX modules (see Remark 3.14).

(ii) For each p ∈ Spec(A), the map Mp → (FM )p that assigns to each mf ∈Mp the germ

class of (D(f), mf ) is an isomorphism.

Proof: It is proved as Theorem 3.23.

We have the similar result for graded rings and modules (as for localization in rings,

the brackets notation indicates that we take quotients mF with F,m homogeneous of the

same degree):

116

Theorem 10.3. Let S be a graded ring, let M be a graded S- module and set X =

Proj(S). Then:

(i) The assignment D(F ) 7→M(F ) defines a sheaf of OX -modules FM on Proj(S).

(ii) For each p ∈ X, there is an isomorphism M(p) → (FM )p sending each mF ∈ M(p) to

the germ of (D(F ), mF ).

(iii) For each F ∈ S homogeneous of positive degree, the restriction of FM to D(F ) is

isomorphic to the sheaf FM(F )defined on Spec(S(F )) (see Proposition 4.5).

Proof: Parts (i) and (ii) are the analogue to parts (iii) and (iv) of Theorem 3.24, while

part (iii) is proved as Proposition 4.5.

Definition. A coherent sheaf is a sheaf F of OX -modules such that there is an open

covering X =⋃i Ui satisfying that each Ui is isomorphic to some Spec(Ai) and F|Ui

takes

the form FMi, where Mi is a finitely generated Ai-module.

Remark 10.4. As in Theorem 3.3, glueing small pieces but with some extra difficulty, it

can be proved (see [Ha], II, Proposition 5.4), that any coherent sheaf on an noetherian affine

scheme Spec(A) takes the form FM for some finitely generated A-module M (precisely,

M = F(X)). The situation for projective schemes is more complicated, and we will discuss

it later.

Example 10.5. From Theorem 10.3(iii), if M is a finitely generated graded module over

a graded ring S, then FM is a coherent sheaf on Proj(S). Also, any locally free sheaf is

coherent, since the free sheaf OX⊕ . . .⊕OX on X = Spec(A) is the sheaf FM correspoding

to the free module M = A⊕ . . .⊕A. In this last case we can get a nice characterization:

Theorem 10.6. Let F be a coherent sheaf over an irreducible reduced noetherian scheme

X over k. Then the following are equivalent:

(i) F is locally free.

(ii) F is the sheaf of sections of a vector bundle over X of rank r.

(iii) Each Fp is a free module over OX,p of rank r.

(iv) For each p ∈ X, the dimension of Fp/MpFp as a vector space over k(p) is r.

Proof: We will prove the implications cyclically.

(i) ⇒ (ii): Let X =⋃i Ui be an affine open covering such that each F|Ui

is a free

sheaf, associated to the module OX(Ui)⊕ri . In particular, F(V ) = OX(V )⊕ri for each

V ⊂ Ui. Thus, for any two open subsets Ui, Uj , the irreducibility of X implies Ui∩Uj 6= ∅,so that the equality OX(Ui∩Uj)⊕ri = F(Ui∩Uj) = OX(Ui∩Uj)⊕rj implies ri = rj . Let us

117

write r for that common rank. We take, for each i, a basis mi1, . . . ,mir of F(Ui). Hence,

we have two basis mi1|Ui∩Uj, . . . ,mir |Ui∩Uj

and mj1|Ui∩Uj, . . . ,mjr |Ui∩Uj

of F(Ui ∩ Uj).There is hence an invertible (r × r)-matrix Aij with entries in OX(Ui ∩ Uj) such thatmj1|Ui∩Uj

...mjr |Ui∩Uj

= Aij

mi1|Ui∩Uj

...mir |Ui∩Uj

.

Since a section fi1mi1 + . . .+ firmir of F(U) is identified with (fi1, . . . , fir), we thus have fj1|Ui∩Uj

...fjr |Ui∩Uj

= (Atij)−1

fi1|Ui∩Uj

...fir |Ui∩Uj

.

This shows that the matrices (Atij)−1 are the transition matrices of a rank r vector bundle

over X whose sheaf of sections is F .

(ii)⇒ (iii): This is trivial.

(iii)⇒ (iv): It is also trivial.

(iv) ⇒ (i): For each p ∈ X, let U = Spec(A) be an open neighborhood of p such

that F|U = FM for some finitely generated A-module M . Now take any prime ideal

p ⊂ A. Assume that the classes of m1, . . . ,mr form a basis of Mp/pMp (it is clear that

we can remove denominators, since their classes are non-zero constants in Ap/pAp). Let

m′1, . . . ,m′s be a set of generators of M as an A-module. Their classes in Mp modulo

pMp can be written as a linear combination of the classes of m1, . . . ,mr. Hence, modulo

(m1, . . . ,mr), we have the following congruences in Mp:

m′1 ≡g11

f11m′1 + . . .+

g1s

f1sm′s

...

m′s ≡gs1fs1

m′1 + . . .+gssfss

m′s

with gij ∈ p and fij /∈ p. Now the determinant of the matrix

1− gs1fs1

. . . − g1sf1s...

. . ....

− gs1fs1. . . 1− gss

fss

can be written as f1

f2, with f1, f2 /∈ p, and it follows that f1

f2m′i ∈ (m1, . . . ,mr) for i =

1, . . . , s. We can thus write each of them as f1f2m′i =

g′i1f ′i1m1 + . . . +

g′irf ′irmr, with f ′ij /∈ p.

Taking f = f1

∏i,j f

′ij , we get that m1, . . . ,mr generate Mf . Replacing U with D(f),

118

we can assume that M is generated by m1, . . . ,mr. Assume we have a linear relation

f1m1 + . . . + frmr = 0. Localizing at each prime p′ ⊂ A, since the classes of m1, . . . ,mr

modulo p′Mp′ generate a vector space of dimension r, they are independent, hence each

fi is in the maximal ideal p′Ap′1 , i.e. fi ∈ p′. Therefore each fi is in the intersection of

all the prime ideals of A, which is zero, since A is an integral domain. Hence M is a free

module of rank r, which implies that F is locally free.

Remark 10.7. If you know the notion of projective module, recall that an A-module M

is projective if and only if each Mp is a projective Ap-module, and this is in turn equivalent

to each Mp is free. Hence, Theorem 10.6 says that the sheaf FM on Spec(A) (where A is a

reduced finitely generated k-algebra) is locally free if and only if M is a projective module.

Remark 10.8. Observe that, in the last implication, we in fact proved that, given

any coherent sheaf F , if Fp/MpFp has dimension r, then there is an open neighborhood

U of p such that Fq/MpFq has dimension at most r for each p ∈ U . Hence, the sets

Xr := p ∈ X | dim(Fp/MpFp) ≥ r are closed sets for each r. If p is the generic point of

X and r = dim(Fp/MpFp), then the fact p ∈ Xr implies X = Xr and U := Xr \Xr+1 6= ∅.Hence r = mindim(Fp/MpFp) | p ∈ X and it is achieved in a whole open set of points.

You can thus regard a coherent sheaf as a generalization of vector bundle, in which you

allow special fibers to be vector spaces of higher dimension.

Definition. The rank of a sheaf F on an irreducible scheme X is dim(Fp/MpFp), where

p is the generic point of X. The support of the sheaf F is the closed set Supp(F) = p ∈X | dim(Fp/MpFp) > 0.

We can pass now to the main example to understand coherent sheaves on projective

schemes:

Example 10.9. Given a graded module M , we can construct, for each integer d, a new

one M(d), defined as the same set, with the same additive group structure, but with the

difference that the part of degree m of M(d) is Md+e. In the particular case M = S, we

can thus construct the module S(d), which is no longer a ring (except for d = 0). It is

easy to check that M(d) = M ⊗S S(d). If X = Proj(S), we will write OX(d) for the sheaf

FS(d). We thus have that FM(d) = FM ⊗OXOX(d). In general, we will write F(d) to

denote F ⊗OX(d).

To understand better this construction, let us do in detail the easiest example:

Proposition 10.10. The sheaf OPnk(d) is the sheaf of sections of the line bundle Ld over

Pnk .

119

Proof: Write S = k[X0, . . . , Xn] and take the standard covering Pnk =⋃D(Xi). By

Theorem 10.3(ii), the restriction(OPn

k(d))|D(Xi)

is the sheaf associated to S(d)(Xi) on

Spec(S(Xi)

). Now we observe that the elements of S(d)(Xi) take the form F

Xmi

, where

F has degree m in S(d), i.e. it is a homogeneous polynomial of degree m + d. We can

thus write FXm

i= F

Xm+di

Xdi , and F

Xm+di

∈ S(Xi). This shows that S(d)(Xi) is a free module

generated by Xdi . Reproducing the proof of the first implication in Theorem 10.6, we get

that OPnk(d) is the sheaf of sections of the line bundle with transition functions

Xdi

Xdj

, i.e.

Ld.

We thus found a way to recover the graded ring S = k[X0, . . . , Xn], since Proposition

9.18 indicates that Sd is just the space of sections of Ld, i.e.(OPn

k(d))(Pnk ). In general,

one can wonder whether, given a graded ring S, it is still true that Sd is(OX(d)

)(X).

However, this is not true:

Example 10.11. Consider (see Exercise 2.18) the graded ring S = k[X0, X1, X2, X3]/I,

where I = (X0X3−X1X2, X31 −X2

0X2, X32 −X1X

23 , X

21X3−X0X

22 ), and let X = Proj(S).

Recall that, regarded as a projective set, X consisted of the set of points of the form

(t40 : t30t1 : t0t31 : t41). In other words, S ⊂

⊕d≥0 k[T0, T1]4d, and the inclusion is strict

since the monomial T 20 T

21 is missing. However, we can find this monomial as a section

of OX(1). Indeed, we can cover X = D(X0) ∪ D(X3), and the expression T 20 T

21 can be

written asX2

1

X0, which is an element of

(OX(1)

)(D(X0)

)and as

X22

X3, which is an element

of(OX(1)

)(D(X3)

). Both sections coincide in D(X0X3) because

X21 X3

X0X3=

X0X22

X0X3. In fact,

it can be proved that(OX(d)

)(X) = k[T0, T1]4d. In general,

⊕d≥0

(OX(d)

)(X) is the

integral closure of S inside its quotient field. Observe, for example, that α =X2

1

X0satisfies

the integral relation α2 = X1X2.

Definition. Given a graded ring S and X = Proj(S), we define

S(OX) =⊕d≥0

(OX(d)

)(X),

which is a graded ring. Given a coherent sheaf F over X, we define

M(F) =⊕d≥0

(F(d)

)(X),

which is a graded S(OX)-module.

Remark 10.12. Let S be a graded ring of the form A[X0, . . . , Xn]/I and let M be a

graded S-module. We clearly have, for each degree d, a natural homomorphism Md →

120

(FM (d)

)(X), hence a homomorphism of graded S-modules M → M(FM ). A section of

FM (d) consists of giving for each i = 0, . . . , n a quotient mi

Xsi

, with mi ∈ M homogeneous

of degree d + s, and with the compatibility conditionXs

jmi

XsiXs

j=

Xsimj

XsiXs

jin M(XiXj). This

means that there is a relation in M

(XiXj)s+a(Xs

jmi −Xsimj) = 0.

Hence, an element of M(FM )(X0) will consist of a collection of quotients mi

Xd0X

si

and we can

writemi

Xd0X

si

=(X0Xi)

s+aXs0mi

(X0Xi)s+aXs0X

d0X

si

=(X0Xi)

s+aXsim0

(X0Xi)s+aXs0X

d0X

si

=m0

Xs+d0

.

This gives an isomorphism between M(X0) and M(FM )(X0), and the same holds when

replacing X0 with any other Xi. This shows that M and M(FM ) define the same sheaf.

In general, the situation is similar, and we state it without proof:

Theorem 10.13. Let S be a graded ring of the form A[X0, . . . , Xn]/I and let F be a

coherent sheaf over X = Proj(S). Then:

(i) F = FM(F).

(ii) The A-module F(X) is finitely generated.

(iii) F = FM if and only if Md =(F(d)

)(X) for d >> 0.

Proof: Part (i) is not so difficult, and its proof can be found in [Ha]§II Proposition 5.15.

Part (ii) is the hard part ([Ha]§II Theorem 5.19), and part (iii) can be done using the

techniques of the proof of (ii) ([Ha]§II Exercise 5.9).

Remark 10.14. The above result is much stronger than what it could look at a first

glance. For example, since any non-zero coherent sheaf F comes from a non-zero graded

module M , for d >> 0 we have that Md is not zero and, by part (iii), it is also the space

of sections of F(d). Hence, F(d) has non-zero sections. This is part of what is known

as Serre’s Theorem B (used in fact for the proof of Theorem 10.13), which essentially

says that any F(d) is in some sense “positive” if d >> 0 (see also Remark 11.5). As

an application, consider any line bundle L over a projective scheme X ⊂ Pnk . It must

correspond to some locally free sheaf L of rank one. As we just said, some L(d) must

have non-zero sections, i.e. the line bundle L⊗ (Ld)|X has a non-zero section, whose zero

locus is an effective divisor D. Since (Ld)|X has also some non-zero section, writing D′

for its zero locus divisor, we get that L corresponds to the divisor D − D′. Therefore,

the injection Div(X)/Princ(X) → Pic(X) of Remark 9.27 is in fact an isomorphism of

groups. For example, since any hypersurface in Pnk is a section of some Ld, it follows that

any line bundle on Pnk is isomorphic to some Ld. Hence, Pic(Pnk ) is isomorphic to Z.

121

Since the class of a divisor on Pnk is given by an integer d and the sheaf of sections of

the corresponding line bundle is OPnk(d), we will copy this notation in general:

Definition. Given a divisor D on X, we will write OX(D) for the sheaf of sections of

the line bundle associated to D. As we have seen, OX(D) does not depend on the linear

equivalence class of D, and any locally free sheaf of rank one (also called invertible sheaf)

takes the form OX(D) for some class of divisors D. Typically one also writes K for the

class of a canonical divisor, i.e. the class corresponding to the canonical line bundle∧n

ΩX ;

its sheaf of sections will be thus denoted with OX(K) or ωX .

Remark 10.15. Theorem 10.13(iii) reminds what happened for ideals and their satura-

tion. In fact, the theory of sheaves of ideals is a particular case of the theory of sheaves

of modules. Indeed, an ideal over a ring A is nothing but a submodule of A, so that a

sheaf of ideals in X is nothing but a subsheaf of modules of OX . In this context, Theo-

rem 10.13 is just saying that a closed embedding in Pnk is the same as a saturated ideal

of k[X0, . . . , Xn]. For a general graded module M over k[X0, . . . , Xn], one can consider

M(FM ) as the “saturation” of M , although the difference now is that we cannot compare

M and M(FM ), since they do not live in a common ambient space.

Proposition 10.16. Let D ⊂ X be an effective divisor of a smooth irreducible subvariety.

Then the sheaf of ideals of D is the OX(−D).

Proof: The divisor D is the zero locus of a section a line bundle determined by a covering

X =⋃i Ui and transition functions fij ∈ OX(Ui ∩Uj). The section whose zero locus is D

is given by a collection of regular functions fi ∈ OX(Ui) and, for each i, the ideal IY,X(Ui)

is (freely) generated by fi. Therefore, IY,X is locally free of rank one and, repeating the

proof of Theorem 10.6, it is the sheaf of sections of a line bundle with transition functionsfi|Ui∩Uj

fj |Ui∩Uj

, which are equal to f−1ij , proving the result.

Remark 10.17. In general, if Y ⊂ X has pure codimension r and is smooth, then

we know that each IY,X(Ui) is generated by r elements, but this is never free if r > 1.

However, in the open set U = X \ Y , the restriction of IY,X is the trivial line bundle,

so that IY,X has rank one. It can be seen that instead, for any p ∈ Y , stalk of IY,Xat p is still generated by r elements, and modulo Mp(IY,X)p they form a basis of the

corresponding vector space over k(p). The precise algebraic statement is that the tensor

product IY,X ⊗OXOY is isomorphic to IY,X/I2

Y,X , which can be identified with the sheaf

of sections of the cotangent bundle N∗Y/X .

Remark 10.18. Coming back to the case of codimension one D ⊂ X, observe that we

have an inclusion ID,X ⊂ OX corresponding to the morphism of line bundles L∗D → X×k.

122

However, the morphism of line bundles is not injective, since it is the zero map in the fibers

of the points of D. This means that the natural notion of being a monomorphism of vector

bundles does not coincide (it is in fact stronger) with the notion of being monomorphism

of sheaves. This last notion should be (as we have seen in the proof of Proposition 3.11)

either that the maps in open sets are injective or, equivalently, that the corresponding

germ maps among stalks are injective. However, if we replace “injective” with “surjective”

both notions of epimorphism of sheaves are not equivalent. We thus have to choose one of

the two possible definitions.

Definition. An exact sequence of sheaves is a sequence of morphisms of sheaves of OX -

modules 0 → F ′ ϕ−→F ψ−→F ′′ → 0 such that, for each p ∈ X, the sequence of homomor-

phisms of OX,p-modules 0→ F ′pϕp−→Fp

ψp−→F ′′p → 0 is exact.

The following result shows that taking sections preserves the exactness except at the

right-hand side:

Proposition 10.19. If 0 → F ′ ϕ−→F ψ−→F ′′ → 0 is an exact sequence of OX -modules,

then for any open subset U ⊂ X the corresponding sequence

0→ F ′(U)ϕ(U)−→F(U)

ψ(U)−→F ′′(U)

is exact.

Proof: We start proving the injectivity of ϕ(U). If (ϕ(U))(s′) = 0, this means that((ϕ(U))(s′)

)p

= 0 for all p ∈ U , i.e. ϕp(s′p) = 0. Since ϕp is injective for any p, it follows

s′p = 0 for all x ∈ U , so that s′ = 0.

Similarly, one proves ψ(U) ϕ(U) = 0, since for any s′ ∈ F ′(U) and p ∈ U it holds(ψ(U)(ϕ(U)(s′))

)p

= (ψx(ϕp(s′p)) = 0.

Finally, let s ∈ F(U) such that (ψ(U))(s) = 0. This means that, for any p ∈ U , we

have ψp(sp) = 0. Therefore, sp is in the image of ϕp, which means that we can find an

open subset U ′ ⊂ U containing x and a section s′ ∈ F ′(U ′) such that (ϕ(U ′))(s′) = s|U ′ .

In other words, we can find an open covering U =⋃Ui and sections s′i ∈ F ′(Ui) such that

(ϕ(Ui))(s′i) = s|Ui

. The point now is that we would like to glue the sections s′i to build

up a section s′ in the whole U (and necessarily its image by ϕ(U) will be s, finishing the

proof). To do so, we would need these sections to agree on the intersections Ui ∩ Uj . But

this is because

(ϕ(Ui ∩ Uj))(s′i|Ui∩Uj) = (ϕ(Ui))(s

′i)|Ui∩Uj) = s|Ui∩Uj

and, analogously, (ϕ(Ui ∩ Uj))(s′j |Ui∩Uj) = s|Ui∩Uj

. Now, the injectivity of ϕ(Ui ∩ Uj)proved at the beginning concludes s′i|Ui∩Uj

= s′j |Ui∩Uj, completes the proof.

123

Let us try to understand what happens with the surjectivity in a couple of examples:

Example 10.20. Let us consider a projective subscheme X ⊂ Pnk defined by a grades

ideal I ⊂ S = k[X0, . . . , Xn]. The exact sequence 0→ IX,Pnk→ OPn

k→ i∗OX → 0 (where

i is the inclusion) comes by taking the sheaves corresponding to the graded modules of the

exact sequence 0 → I → S → S(X) → 0 (an sequence of modules is exact if and only if

it is exact when localizing at any prime ideal). In fact, taking any shift with any d we get

0→ I(d)→ S(d)→(S(X)

)(d)→ 0 we get an exact sequence

0→ IX,Pnk(d)→ OPn

k(d)→ i∗OX(d)→ 0.

Taking global sections, we get in the middle the space Sd of homogeneous polynomials of

degree d (see Proposition 10.10) and the sections of IX,Pnk(d) are, by Proposition 10.19,

those polynomials vanishing on X, i.e. the space of sections is precisely Id. However,

the space of sections of i∗OX(d), which is the space of sections of OX(d) on X, it is not

necessarily S(X)d, as we have seen in Example 10.11.

Example 10.21. We consider now an example which is a priori a little bit different. If

S = k[X0, . . . , Xn], we can consider the exact sequence

0→M → S(−1)⊕(n+1) → S → S/M→ 0

in which the middle map is given by (F0, . . . , Fn) 7→ F0X0 + . . .+ FnXn and M is defined

as a kernel. Observe that S/M has only part of degree zero, so it defines the same sheaf

as the zero module, i.e. it defines the zero sheaf. We thus arrive to an exact sequence

0→ FM → OPnk(−1)⊕(n+1) → OPn

k→ 0

and, taking global sections, the last map is S(−1)⊕(n+1)0 → S0, i.e. 0→ k, which obviously

is not surjective. We will complete the example analyzing the sheaf FM . We first prove

that it is locally free(∗) of rank n. Indeed, for any i = 0, . . . , n, it is easy to check that

M(Xi) is freely generated by the n elements

ek,i := (0, . . . , 0,−xkx2i

, 0, . . . , 0,1

xi, 0, . . . , 0)

for k = 0, . . . , i−1, i+1, . . . , n. It is a simple exercise to show that the (matricial) relations

among the elements ek,i and el,j are the same as the ones among the elements d(xk

xi) and

d( xl

xj), which shows that FM is the sheaf of sections of the cotangent bundle of Pnk . The

above exact sequence of sheaves on Pnk is called the Euler exact sequence.

(∗) We will prove it directly, although it is not difficult to prove, using Theorem 10.6,

that the kernel of a surjective morphisms of locally free sheaves is locally free.

124

11. Cohomology and duality

Applying standard techniques of cohomology (existence of enough injectives,...), one

could easily deduce from Proposition sectionsleftexact, that we can complete the exact

sequence there by plugging at the right suitable cohomology spaces. We will do it instead

by constructing more explicitly this cohomology.

Remark 11.1. Let us try to imitate the proof of Proposition 10.19 to find out why the

surjectivity of ψ(U) fails. Take thus a section s′′ ∈ F ′′(U). The surjectivity of ψ implies

the surjectivity of any ψx, so that we can find an open covering U =⋃i Ui and sections

si ∈ F(Ui) such that (ψ(Ui))(si) = s′′|Ui. Where is now the obstruction to glue the sections

si? For each intersection Ui ∩ Uj we can still prove

(ψ(Ui ∩ Uj))(si|Ui∩Uj) = s′′|Ui∩Uj

= (ψ(Ui ∩ Uj))(sj |Ui∩Uj).

But now ψ(Ui ∩ Uj) is not injective. The most we can say, using the exactness we proved

in Proposition 10.19 is that we can find sections s′ij ∈ F ′(Ui ∩ Uj) such that

(ϕ(Ui ∩ Uj))(s′ij) = (sj − si)|Ui∩Uj.

Observe that we still can say more about these new s′ij using the injectivity of ϕ. Indeed

in the intersection of three open sets Ui ∩ Uj ∩ Uk we can write

0 = (sk − sj)|Ui∩Uj∩Uk− (sk − si)|Ui∩Uj∩Uk

+ (sj − si)|Ui∩Uj∩Uk=

= (ϕ(Ui ∩ Uj ∩ Uk))(s′jk|Ui∩Uj∩Uk− s′ik|Ui∩Uj∩Uk

+ s′ij |Ui∩Uj∩Uk)

so that we conclude

s′jk|Ui∩Uj∩Uk− s′ik|Ui∩Uj∩Uk

+ s′ij |Ui∩Uj∩Uk= 0.

Inspired in the above remark, we can give the following definitions (Remark 11.1 will

correspond to the case p = 1; observe that somehow the indices were ordered in the sense

that i precedes j, which in turn precedes k, thus explaining the minus sign in the middle

term of the last expression)):

Definition. Given a sheaf of OX -modules F and an ordered set of open subsets U =

Uii∈I (i.e. I is a well-ordered set of indices) such that X =⋃i∈I Ui, we write for each

p ≥ 0

Cp(X,F ,U) = Πi0<...<ipF(Ui0 ∩ . . . ∩ Uip).

125

We moreover define homomorphisms dp : Cp(X,F ,U) → Cp+1(X,F ,U) by sending the

element having the element si0...ip ∈ F as its (i0, . . . , ip)-coordinate to the element whose

(i0, . . . , ip+1)-coordinate is

p+1∑j=0

(−1)jsi0...j...ip+1 |Ui0∩...∩Uip+1

.

One easily verifies dp dp−1 = 0, so that we call the p-th Cech cohomology of the sheaf Fassociated to the covering U to the quotient

Hp(X,F ,U) = ker(dp)/ Im(dp−1).

Remark 11.2. Notice that, for any covering U , the definition of sheaf implies that

the map F(X) → H0(X,F ,U) sending a section s ∈ F(U) to the element whose i-th

coordinate is s|Uiis an isomorphism (observe that there is no d−1 so there is no need to

quotient in the definition of H0(X,F ,U)).

One natural question that arises is: what is the covering one should choose in order

to have the right cohomology completing the exact sequence in Proposition 10.19 in order

to get a long exact sequence of cohomology. Remark 11.1 suggests that one needs in fact

to find a “sufficiently fine covering”. The precise way of doing so is the following:

Definition. Consider the partial ordering to the set of open coverings of a noetherian

separated scheme(∗) X by setting that U ′ ≤ U if there is a map σ : I ′ → I respecting

the ordering such that for each i′ ∈ I ′ the open set U ′i′ is contained in Uσ(i′) (i.e. if the

partition U ′ is finer than partition U). If U ′ ≤ U one can naturally define a homomorphism

Hp(X,F ,U)→ Hp(X,F ,U ′). Then the p-th cohomology of the sheaf F is the direct limit

Hp(X,F) = lim→Hp(X,F ,U).

By Remark 11.2, H0(X,F) = F(X) for any sheaf. It is now a tedious but straightfor-

ward calculation to show that we can complete the exact sequence of Proposition 10.19:

Theorem 11.3. Given an exact sequence of coherent sheaves

0→ F ′ → F → F ′′ → 0

(∗) The same definition could be done for an arbitrary topological space, but then the

cohomology defined in this way does not coincide with the one obtained from standard

homological methods; see Chapter II.4 of [Ha] for details.

126

there exists a long exact sequence

0→ H0(X,F ′)→ H0(X,F)→ H0(X,F ′′)→ H1(X,F ′)→ H1(X,F)→ H1(X,F ′′)→ . . .

Proof: We will only complete the exactness at H0(X,F ′′), the rest following essentially the

same techniques (and definition of the maps). So, following the notation of Remark 11.1,

we define the map δ1 : H0(X,F ′′)→ H1(X,F ′) by sending any s′′ ∈ H0(X,F ′′) = F ′′(X)

to the class defined by s′ij in H1(X,F ′,U), where U is the covering X =⋃i Ui (in which

we fix some ordering), and s′ij ∈ F ′(Ui ∩ Uj) satisfy

(ϕ(Ui ∩ Uj))(s′ij) = (sj − si)|Ui∩Uj,

where (ψ(Ui))(si) = s′′|Ui. Since we had s′jk|Ui∩Uj∩Uk

− s′ik|Ui∩Uj∩Uk+ s′ij |Ui∩Uj∩Uk

= 0,

this implies that they define an element in ker(d′1). Moreover, a different choice of sections

ti ∈ F ′(Ui) such that (ψ(Ui))(ti) = s′′|Uiis equivalent to si − ti ∈ kerψ(Ui), so that

si − ti =(ϕ(Ui)

)(t′i) for some t′i ∈ F ′(Ui). These new ti would lead to different sections

t′ij ∈ F ‘(Ui ∩ Uj) such that

(ϕ(Ui ∩ Uj))(t′ij) = (tj − ti)|Ui∩Uj= (sj − si)|Ui∩Uj

−(ϕ(Ui ∩ Uj)

)((t′j − t′i)|Ui∩Uj

).

Since ϕ(Ui∩Uj) is injective, this is equivalent to s′ij = t′ij + (t′j− t′i)|Ui∩Uj, i.e. the families

s′ij and t′ij differ in an element of Im(d′0). This implies that δ1 is well defined.

Repeating these computations, δ1(s′′) = 0 if and only if the class of s′ij is zero, i.e.

we can write s′ij = (t′j− t′i)|Ui∩Ujfor some family t′i. But this is in turn equivalent to the

fact that the family of sections si −ϕ(Ui)(t′i) ∈ F(Ui) glue together in the intersections of

U , i.e. there is a section s ∈ F(X) such that s|Ui= si − ϕ(Ui)(t

′i). By the definition of

si, this is equivalent to say that(ψ(X)

)(s) and s′′ coincide when restricting to any Ui, i.e.

s′′ =(ψ(X)

)(s), i.e. it is in the image of ψ(X). This completes the proof of the exactness

at H0(X,F ′′).

Remark 11.4. When X = Spec(A) is an affine scheme we know (see Remark 10.4) that

there is an equivalence between the category of coherent sheaves over X and the category

of finitely generated A-modules. Therefore the map ψ(X) in Proposition 10.19 is also

surjective. As a standard consequence in cohomology theory, sheaves over affine schemes

do not possess cohomology, in the sense that Hp(X,F) = 0 for any p > 0 and any coherent

sheaf F . Using again standard techniques in cohomology, one can deduce that, if U is a

covering by affine open subsets of a scheme X, then Hp(X,F ,U) = Hp(X,F) for any

p ≥ 0 (see [Ha], §III, Theorem 4.5).

127

Remark 11.5. When X = Proj(S), what we have now is that we have an equivalence

among the category of coherent sheaves over X and the equivalence classes of graded mod-

ules over S in which two modules are equivalent if and only if they coincide for sufficiently

high degree. Recall also that a module representating of a coherent sheaf F is given by⊕d

(F(d)

)(X). In particular, an exact sequence of coherent sheaves 0 → F ′ → F →

F ′′ → 0 is exact if and only if 0 →(F ′(d))(X) →

(F(d)

)(X) →

(F ′′(d)

)(X) → 0 is

exact for d >> 0. This has as a conseguence that, given a coherent sheaf F over X, then

Hp(X,F(d)) = 0 for d >> 0 and any p > 0.

Example 11.6. For p = 1, our definition coincides with the standard definition of

cohomology (see [Ha], §III, Exercise 4.4). It is then a nice exercise to consider a proper

irreducible scheme X and consider the sheaves of multiplicative groups O∗X (of regular

functions with no zeros), M∗X of non-zero rational functions (in fact this is the constant

sheaf whose group of sections at any non-empty open set is K(X)∗) and their quotient

M∗X/O∗X (defined as the sheaf associated to the presheaf U 7→ M∗(U)O∗

X(U) ). It is then a nice

exercise to check that, taking cohomology in exact sequence for 0 → O∗X → M∗X →M∗X/O∗X → 0, one gets 1 → k∗ → K(X)∗

div−→Div(X) → Pic(X) (see Lemma 9.25 and

Remark 9.27).

The following result is true for any scheme, but we can give an easy proof for projective

schemes:

Proposition 11.7. Let X be a projective scheme of dimension n. Then, for any coherent

sheaf F over X, one has Hp(X,F) = 0 if p > n.

Proof: By Remark 2.19, we can find n+ 1 hypersurfaces V (F0), . . . , V (Fn) in the ambient

projective space ofX such thatX∩V (F0)∩. . .∩V (Fn) = ∅. Hence U = D(F0), . . . , D(Fn)is an affine open covering of X. By Remark 11.4, we have Hp(X,F) = Hp(X,F ,U) for

any p ≥ 0. On the other hand, since U has only n+ 1 elements, it follows Cp(X,F ,U) = 0

if p > n, so that also Hp(X,F ,U) = 0 if p > n, as wanted.

To see how it works, let us compute cohomology in a concrete case:

Example 11.8. Let us compute by hand H1(P2k,OP2

k(l)) for any l. We take the affine

covering U = D(X0), D(X1), D(X2). The elements of C1(P2k,OPn

k(l),U) take the form

(F01

(X0X1)a,

F02

(X0X2)a,

F12

(X1X2)a),

where F01, F02, F03 are homogeneous polynomials of degree l + 2a. Such an element is in

ker(d1) if and only if

F12Xa0 − F02X

a1 + F01X

a2 = 0.

128

Thus, for example F01Xa2 is in the primary ideal (Xa

0 , Xa1 ). Since Xa

2 is not in the radical

of (Xa0 , X

a1 ), it follows that we can write

F01 = G1Xa0 −G0X

a1

for G0, G1 homogeneous polynomials of degree l+ a. Substituting in the above expression

we get

Xa0 (F12 +G1X

a2 ) = Xa

1 (F02 +G0Xa2 )

from where we conclude the existence of G2 such that

F12 = G2Xa1 −G1X

a2

F02 = G2Xa0 −G0X

a2

Hence

(F01

(X0X1)a,

F02

(X0X2)a,

F12

(X1X2)a) = d0(

G0

Xa0

,G1

Xa1

,G2

Xa2

).

This proves ker(d1) = Im(d0), so that H1(P2k,OP2

k(l)) = 0.

Remark 11.9. In generalOPnk(l) has no intermediate cohomology, i.e. Hi(Pnk ,OPn

k(l)) = 0

for i = 1, . . . , n−1 and any l ∈ Z. The reason (if we want to imitate the above proof) is that,

fixing a, there is an exact sequence with arrows∏

0≤i1<...<ir≤n S →∏

0≤i1<...<ir−1≤n S

sending the u-ple (Fi1...ir ) to (Fi1...ir−1) in which

Fi1...ir−1=

r∑j=1

(−1)jFi1...j...irXaj .

In fact, this exact sequence holds more generally by replacing Xa0 , . . . , X

an with what is

called a regular sequence F0, . . . , Fn, i.e. with the property that no Fr belongs to the

radical of any primary component of (F0, . . . , Fr−1) (this general exact sequence is called

Koszul complex). In fact, this property characterizes these sheaves, in the sense that

Horrocks proved that any locally free sheaf F over Pnk without intermediate cohomology

is necessarily of the form F =⊕

iOPnk(ai) (not having intermediate cohomology means

now Hi(Pnk ,F(l)) = 0 for i = 1, . . . , n − 1 and any l ∈ Z). This result is a generalization

of the fact (proved by Grothendieck) that any locally free sheaf over the line P1k splits as

F =⊕

iOP1k(ai).

The reason why we care about the vanishing of intermediate cohomology and not

also the cohomology of maximal order n is that there is a kind of duality (as it happens

with Poincare duality in Algebraic Topology), and the cohomology of order i is dual to

some cohomology of order n − i. Since the cohomology of order zero does not vanish for

high degree (see Remark 10.14) we do not expect maximal cohomology to vanish. Let us

indicate the first easy case of duality:

129

Proposition 11.10. The following cohomology properties for OPnk(l) hold:

(i) If l > −n− 1 then Hn(Pnk ,OPnk(l)) = 0.

(ii) Hn(Pnk ,OPnk(−n− 1)) has dimension one.

(iii) If l ≤ −n− 1 there is a perfect pairing(*)

H0(Pnk ,OPnk(−l − n− 1))×Hn(Pnk ,OPn

k(l))→ Hn(Pnk ,OPn

k(−n− 1)),

and hence Hn(Pnk ,OPnk(l)) ∼= H0(Pnk ,OPn

k(−l − n− 1))∗.

Proof: We consider U = D(X0), . . . , D(Xn). An element of Cn(Pnk ,OPnk(l),U) is given

by a quotient F(x0...xn)a , where F has degree a(n + 1) + l. The image of dn−1 is the set

of quotients for which F ∈ (xa0 , . . . , xan). If l > −n − 1, then deg(F ) > (a − 1)(n + 1),

which implies that any monomial of F is divisible by at least some xai . This implies

Im(dn−1) = Cn(Pnk ,OPnk(l),U), so that Hn(Pnk ,OPn

k(l)) = 0, which proves (i).

The same reasoning as before shows that, when l = −n − 1 the only monomial not

belonging to Im(dn−1) is xa−10 . . . xa−1

n . Thus a basis of Hn(Pnk ,OPnk(−n− 1)) is given by

the class of 1x0...xn

, which proves (ii).

Since we know H0(Pnk ,OPnk(−l− n− 1)) = S−l−n−1, we define the pairing by sending

(G, [ F(x0...xn)a ]) to [ GF

(x0...xn)a ] (where brackets mean classes modulo dn−1). The vanishing of

[ GF(x0...xn)a ] means that GF has no term in xa−1

0 . . . xa−1n . If [ F

(x0...xn)a ] is no zero, then it has

some monomial of the type xi00 . . . xinn with all ij < a; hence [ GF(x0...xn)a ] 6= 0 when taking

G = xa−1−i00 . . . xa−1−in

n . Analogously, if G 6= 0, it will have some nonzero monomial

xj00 . . . xjnn . If a = maxj0 + 1, . . . , jn + 1, then we have that [Gxa−1−j00 ...xa−1−jn

n

(x0...xn)a ] 6= 0.

In order to state precisely the general duality, we need to introduce more general

derived functors. We do that in the next series of examples.

Example 11.11. We start by being precise with the definition of tensor product of

sheaves. Given two sheaves of OX -modules, F and G, one would be tempted to define

the tensor product as the sheaf F ⊗OXG such that (F ⊗OX

G)(U) = F(U)⊗OX(U) G(U).

However, this is not in general a sheaf, but only a presheaf, and the correct definition is to

take the sheaf associated with that presheaf. For practical purposes, one can define F⊗OXG

(*) A perfect pairing is a bilinear form F : V ×W → k of vector spaces over k such that,

for each v ∈ V \ 0, there is w ∈ W for which F (v, w) 6= 0 and, symmetrically, for each

w ∈W \ 0, there is v ∈ V for which F (v, w) 6= 0. This yields monomorphisms V →W ∗

and W → V ∗. Hence, in finite dimension, we deduce dim(V ) = dim(W ), which implies

that the monomorphisms are necessarily isomorphisms.

130

by its universal property and then check that the above sheaf satisfies that property. The

tensor product of coherent sheaves is a coherent sheaf, since when U = Spec(A) and F|U =

FM ,G|U = FN , with M,N finitely generated A-modules, then (F ⊗OXG)|U = FM⊗AN .

Since the tensor product of modules is right exact, the same holds for sheaves of modules.

Hence one can define de Tor functor to complete the exact sequence on the right. Again,

each Tori(F ,G) are coherent, since they are locally of the form Tori(M,N). These higher

Tor are zero when one of the sheaves is locally flat (for example, if it is locally free). In

particular, the tensor product of an exact sequence with a locally free sheaf remains an

exact sequence.

Example 11.12. Similarly, one can also define a sheafHom(F ,G) by(Hom(F ,G)

)(U) =

Hom(F|U ,G|U ), the OX(U)-module of all morphisms of sheaves from F|U to G|U . When

G = OX we get the dual sheaf F∗. Observe that, if F is locally free, then also F∗ is locally

free, and in fact it is the sheaf of sections of the dual bundle associated with F . As in the

previous example, if F ,G are coherent, so Hom(F ,G) is, locally defined by Hom(M,N).

The functor Hom is left exact, and we get a derived functor Ext, which is called the local

Ext. It is also coherent, locally defined by Exti(M,N). The sheaf Exti(F ,G) will be zero

for p > 0 if F is locally projective, i.e. locally free (see Remark 10.7).

Example 11.13. Maybe you wonder why the above definition of Hom did not consist

of assigning to each open set U the module Hom(F(U),G(U)

)the reason is that, doing

so, there is no way of defining a restriction map when we have V ⊂ U . If we consider

the whole space Hom(F ,G), we still have a left exact functor, and we can construct a new

Ext (called, the global Ext) that completes the exact sequence on the right. We will be

more precise here, since this is precisely the derived functor we will need. Since Hom is

a bifunctor, we have two ways of completing exact sequences. First of all, if we have an

exact sequence of coherent sheaves

0→ F ′ → F → F ′′ → 0,

we get a long exact sequence of modules

0→ Hom(F ′′,G)→ Hom(F ,G)→ Hom(F ′,G)→ Ext1(F ′′,G)→ Ext1(F ,G)→ . . .

and secondly, given another exact sequence of coherent sheaves

0→ G′ → G → G′′ → 0,

we get a long exact sequence of modules

0→ Hom(F ,G′)→ Hom(F ,G)→ Hom(F ,G′′)→ Ext1(F ,G′)→ Ext1(F ,G)→ . . .

Observe that now we cannot guarantee the vanishing of Extp(F ,G) if F is locally free for

p > 0

The relation of this last functor with the cohomology is the following:

131

Proposition 11.14. If F is locally free and G is any coherent sheaf, then Exti(F ,G) =

Hi(X,F∗ ⊗OXG) for all i ≥ 0.

Proof: It is enough to show that the space of morphisms from F to G is isomorphic to

the space of sections of F∗⊗OXG, since this would imply that the functors Extp(F , ) and

Hp(F∗ ⊗ ) are the derived functors completing the same left exact functor. To prove so,

it would suffice to show that the sheaves Hom(F ,G) and F∗ ⊗OXG are isomorphic (and

then take global sections).

First, using the universal property of the tensor product, the natural bilinear mor-

phism F∗ × G → Hom(F,G) produces a morphism F∗ ⊗OXG → Hom(F ,G). And now

this is an isomorphism, because the stalks of F are free.

We can now state the precise duality for the projective space:

Theorem 11.15. Let F be a coherent sheaf over Pnk . Then, for each i = 0, 1, . . . , n, there

is a perfect pairing Hn−i(Pnk ,F)× Exti(F ,OPnk(−n− 1))→ Hn(Pnk ,OPn

k(−n− 1)).

Proof: We prove it by induction on i. The main trick is to construct exact sequences

0→ K0 → P0 → F → 0 (∗)

and

0→ K1 → P1 → K0 → 0 (∗∗)

in which P0 and P1 are direct sums of locally free sheaves of rank one. This is possible

because we can write F = FM , where M =⊕

d F(d). If m1, . . . ,mr are homogeneous

generators of M and we set di = deg(mi), we thus have a graded epimorphism ϕ : N0 :=

S(−d1)⊕ . . .⊕ S(d− r)→M defined by ϕ(F1, . . . , Fr) = F1m1 + . . .+ Frmr. Therefore,

the kernel N ′0 := ker(ϕ) is a finitely generated graded module, and we obtain (*) by

taking P0 = FN0= OPn

k(−d1) ⊕ . . . ⊕ OPn

k(−dr) and K0 = FN ′0 . Observe that the above

construction implies that the map P0(X)→ F(X) is surjective, and hence H1(X,K0) = 0.

Repeating the same trick for the module N ′0 in the role of M , we get (**), and again

H1(X,K1) = 0.

To prove the case i = 0, the sequences (*) and (**) show that there are exact sequences

Hn(Pnk , P1)α−→Hn(Pnk , P0)

β−→Hn(Pnk ,F)→ 0

and

0→ Ext0(F ,OPnk(−n− 1))

γ−→Ext0(P0,OPnk(−n− 1))

δ−→Ext0(P1,OPnk(−n− 1)).

132

On the other hand, Proposition 11.10 tells us that there are perfect pairings

Fj : Hn(Pnk , Pj)× Ext0(Pj ,OPnk(−n− 1))→ Hn(Pnk ,OPn

k(−n− 1))

for j = 0, 1, giving dualities Ext0(Pj ,OPnk(−n − 1)) ∼= Hn(Pnk , Pj)∗, and it is not difficult

to see that, under this duality, δ is the transpose of α. This provides in a natural way a

duality among Hn(Pnk .F) and Ext0(F ,OPnk(−n − 1))∗, and this is the same as giving a

perfect pairing for i = 0.

We consider now the case i = 1. In this case, (*) provides exact sequences (using

Remark 11.9), i.e. that P0 has no intermediate cohomology)

0→ Hn−1(X,F)→ Hn(X,K0)→ Hn(X,P0)

and

Ext0(P0,OPn

k(−n− 1)

)→ Ext0

(K0,OPn

k(−n− 1)

)→ Ext1

(F ,OPn

k(−n− 1)

)→ 0.

In the case i = 0 we already proved that the last arrow the first exact sequence is the dual

of the first arrow of the second exact sequence which proves the duality also in this case.

Assume now 2 ≤ i ≤ n − 1. Now the fact that P0 has no intermediate cohomology

implies that (*) gives isomorphisms

Hn−i(Pnk ,F) ∼= Hn−i+1(Pnk ,K0)

and

Exti(F ,OPnk(−n− 1)) ∼= Exti−1(K0,OPn

k(−n− 1)).

Hence the induction hypothesis applied to the coherent sheaf K0 completes the proof in

these cases.

In the case i = n, we obtain now exact sequences from (*) and (**) (using H1(X,Ki) =

0 and, by induction hypothesis, its dual counterpart Extn−1(Ki,OPn

k(−n− 1)

)= 0)

H0(X,P1)→ H0(X,P0)→ H0(X,F)→ 0

and

0→ Extn(F ,OPnk(−n− 1))→ Extn(P0,OPn

k(−n− 1))→ Extn(P1,OPn

k(−n− 1)).

Using again that the first part of the first exact sequence is the dual of the last part of the

second exact sequence, we complete the duality also in this remaining case.

To generalize the above result to a general variety, we first observe that OPnk(−n−1) is

nothing but the canonical line bundle over Pnk . One of the main ingredients for the duality

we had was the fact that Hn(Pnk , ωPnk) had dimension one. This is a more general fact for

any variety. We discuss a more general fact for complex varieties, which is known as Hodge

theory:

133

Theorem 11.16. Let X be a compact smooth irrreducible complex n-dimensional variety.

For each p, q = 0, . . . , n, let Hp,q(X) = Hq(X,∧p

ΩX). Then:

(i) Hi(X,C) =⊕

p+q=iHp,q(X).

(ii) The class of any subvariety Y ⊂ X of codimension r, regarded as a cycle in H2r(X,C),

lives in Hr,r(C).

(iii) For each p, q, the spaces Hp,q(X) and Hq,p(X) are complex conjugate (in particular

they have the same dimension).

As a consequence, since X is connected and hence H2n(X,C) = C, it follows from (i)

that also Hn(X,ωX) has dimension one (and a generator would correspond to the class of

a point). Then the general duality reads:

Theorem 11.17. Let F be a coherent sheaf over X, a proper irreducible smooth scheme

over k of dimension n. Then, for each i = 0, 1, . . . , n, there is a perfect pairing

Hn−i(X,F)× Exti(F , ωX)→ Hn(X,ωX).

Proof: Although we will not give the prove, at least we give and idea of how to construct

the pairing. As in the projective case, it is enough to have it for i = 0 and then elaborate

a more sophisticated induction argument. In this case, Ext0(F , ωX) = Hom(F , ωX) and

obviously any morphism F → ωX provides a homomorphism Hn(X,F) → Hn(X,ωX).

Of course, the difficult part is to prove that the pairing is perfect. The reader can find a

complete proof of this theorem in [Ha]§III, Theorem 7.6 (in a more general framework).

Remark 11.18. There is a non used but useful way of understanding Serre’s duality. The

first observation is that, as it happens in homological algebra, the elements of Exti(F ,G)

parametrize classes (under certain relation) of extensions

0→ G → F1 → . . .→ Fi → F → 0.

With this point of view, the fact that Hn(X,ωX) has dimension one means that any

extension of length n from ωX to OX that generates the whole space Extn(OX , ωX) =

Hn(X,ωX). For example, if X = Pnk , this extension is given by taking associated sheaves

in the Koszul exact sequence

0→ S(−n− 1)→ S(−n)(n+1n ) → . . .→ S(−1)(

n+11 ) → S → S/M→ 0

corresponding to X0, . . . , Xn (see Remark 11.9). Now the pairing consists of plugging an

extension of length i from ωX → F with another extension of length n− i from F to OX(recalling that Hn−i(F) can be identified with Extn−i(OX ,F)).

Although the reader probably assumed it from the beginning, we can now prove this

important finiteness result:

134

Theorem 11.19. Let F be a coherent sheaf on X = Proj(S), where S is a finitely gener-

ated k-graded algebra. Then, for each i, the vector space Hi(X,F) has finite dimension.

Proof: Since there is a closed embedding i : X → Pnk and i∗F is a coherent sheaf, it is

enough to do the case X = Pnk . In this case, given a coherent sheaf F , by Serre’s duality

it is enough to prove that each Exti(F ,OPn

k(−n− 1)

)is a vector space of finite dimension.

We will prove the result by induction on i. Consider, as in the proof of Theorem 11.15,

the exact sequence

0→ K0 → P0 → F → 0 (∗)

where P0 is a direct sum of locally free sheaves of the type OPnk(a). In particular, we have

an inclusion Ext0(F ,OPn

k(−n − 1)

)⊂ Ext0

(P0,OPn

k(−n − 1)

), and the second space has

finite dimension, what proves the result for i = 0. In general, we have an exact sequence

Exti−1(K0,OPn

k(−n− 1)

)→ Exti

(F ,OPn

k(−n− 1)

)→ Exti

(P0,OPn

k(−n− 1)

).

Now, by induction hypothesis, Exti−1(K0,OPn

k(−n − 1)

)has finite dimension, and also

Exti(P0,OPn

k(−n− 1)

)has finite dimension (in fact it is zero except for i = 0, n), so that

the result follows.

135

12. Riemann-Roch theorem

The goal of this section is to find a method to compute the dimension of the coho-

mology spaces of coherent sheaves over projective schemes, once we have seen that they

are finite-dimensional. Although there is a modern version (due to Grothendieck) of doing

this in general, we will restrict to the classical case of locally free sheaves of rank one on

curves and surfaces.

We start with a motivating example:

Example 12.1. Let L be a line bundle over a scheme X, and let L be its corresponding

sheaf of sections. We know that a subspace V ⊂ L(X) defines a morphism to a projective

space if L is generated by the sections of V , and one can wonder when this map provides an

embedding. First, it is not difficult to guess when the map ϕV : X → P(V )∗ is injective.

The image of the points p, q ∈ X will be different if and only if the set of hyperplanes

passing through ϕ(p) is different from the set of hyperplanes passing through ϕ(q). Or

equivalently, if and only if the set of hyperplanes passing through ϕ(p) and ϕ(q) is a

subspace of codimension two in the set of hyperplanes. In other words, the subspace of

sections of V vanishing at p and q has codimension two in V . This is what happens with

the subspace V ⊂ k[T0, T1]3 generated by T 30 , T0T

21 , T

31 , and this defines an injective map

ϕV : P1k → P2

k

(t0 : t1) 7→ (t30 : t0t21 : t31)

However, its image is the curve V (X0X21 −X3

2 ), which is singular at (1 : 0 : 0), the image

of the point (1 : 0). What happens at that point is that any section in V vanishing at

(1 : 0) is a linear combination of T0T21 , T

31 , hence it vanishes twice at the point. In other

words, the space of sections vanishing twice at the points (i.e. vanishing at (1 : 0) and the

infinitely close point in the only possible tangent direction) has codimension one. In fact,

it can be proved that, in general ϕV is an embedding if and only if, for any subscheme

Z ⊂ X corresponding to two points (different or infinitely close) the space of sections of V

vanishing at Z has codimension two (intuitively, this is saying that that map is injective,

when the two points are different, and that the differential map is injective, when the two

points are infinitely close). In the case in which we take V = L(X), tensoring with L and

talking sections in the exact sequence

0→ IZ,X → OX → i∗OZ →

we identify the space of sections of L⊗ IZ,X as the subspace of sections of L vanishing at

Z. Hence, we need to check that any such subspace has codimension two in L(X). The

general Riemann-Roch theorem intends to be a tool to compute both dimensions. The

136

most interesting case is when X is a curve, since then IZ,X is a locally free sheaf of rank

one (Proposition 10.16), hence we only need to compute the dimension of the space of

sections of such particular type of sheaves. We finish this discussion with an interesting

remark, which is another part of Serre’s Theorem B. Assume that X is projective. Then,

as we observed in Remark 11.5, H1(L(d) ⊗ IZ,X) = 0 for d >> 0. This can be done in a

way that the same d works for all Z corresponding to two points, which implies that L(d)

defines an embedding. As a useful consequence, by Bertini Theorem, L(d) has a section

whose zero locus is a smooth divisor D′, and also OX(d) has a section whose zero locus

is a smooth divisor D′. Therefore we can write L = OX(D′ − D′′), with D′, D′′ smooth

divisors.

Before starting, we fix the following:

Notation. For any coherent sheaf F on a k-scheme X, we will write hi(X,F) to denote

the dimension of Hi(X,F) as a vector space over k.

We start thus with the case of curves. From now on, a curve will be for us an

irreducible smooth proper scheme of dimension one. By Serre’s duality, we have that

h1(X,OX) = h0(X,ωX), and we give this number a name:

Definition. The geometric genus of a smooth irreducible curve X is g = h0(X,ωX).

Theorem 12.2 (Riemann-Roch for curves). If D is a divisor on a curve X, then

h0(X,OX(D))− h1(X,OX(D)) = deg(D) + 1− g.

Proof: Assume first that D is a positive divisor. Hence D can be considered as a subscheme

of X, i.e. from the inclusion i : D → X we get an exact sequence

0→ ID,X → OX → i∗OD → 0.

Since D is a divisor, we know that ID,X ∼= OX(−D) (Proposition 10.16) and, since D is

zero dimensional, i∗OD has nonzero stalks only at a finite number of points. Therefore,

tensoring with OX(D) we get

0→ OX → OX(D)→ i∗OD → 0.

Taking the long exact sequence of cohomology we get

h0(X,OX(D))−h1(X,OX(D)) = h0(X, i∗OD)+h0(X,OX)−h1(X,OX) = deg(D)+1−g.

We take now a general divisor D. and write it as D = D′ −D′′, with D′ and D′′ effective

divisors. Considering now the exact sequence induced by the inclusion i′′ : D′′ ⊂ X

0→ ID′′,X → OX → i′′∗OD′′ → 0

137

and tensoring it with OX(D′) (recalling that ID′′,X ∼= OX(−D′′)) we get

0→ OX(D)→ OX(D′)→ i′′∗OD′′ → 0.

We now take the long exact sequence of cohomology, and using that we already proved the

theorem for effective divisors, we have

h0(X,OX(D))− h1(X,OX(D)) = h0(X,OX(D′))− h1(X,OX(D′))− h0(X, i′′∗OD′′) =

= deg(D′) + 1− g − deg(D′′) = deg(D) + 1− g

which proves the theorem.

Riemann-Roch theorem implies that all divisors coming from the same line bundle L

(i.e. all linearly equivalent divisors) have the same degree. This yields to the following:

Definition. The degree of a line bundle over a curve (or the degree of its corresponding

sheaf of sections) is the degree of any divisor associated with it.

Theorem 12.3. The canonical line bundle of a curve has degree 2g − 2.

Proof: If we apply Riemann-Roch theorem to ωX , then we get h0(X,ωX)− h0(X,OX) =

deg(ωX) + 1− g, i.e. g − 1 = deg(ωX) + 1− g, from where we get the result.

Remark 12.4. Dualizing the above result, the degree of the tangent bundle is 2 − 2g.

When k = C, this agrees with Poincare Theorem: a vector field has as many zeros and

poles (counted with multiplicity) as its topological Euler characteristic which, in the case

of a topological surface of genus g, is 2 − 2g. This means that, in the complex case, the

geometrical genus coincides with the topological genus. We also defined the arithmetic

genus for a projective curve X ⊂ Pnk as 1− pX(0), where PX is the Hilbert polynomial of

X. Observe that, when d >> 0, we have, on one hand that pX(d) = dimS(X)d and, on

the other hand, that S(X)d = h0(X,OX(d)) while h1(X,OX(d)) = 0. Hence, it must be

pX(d) = deg(OX(d)) + 1− g = deg(X)d+ 1− g.

This re-confirms that the leading coefficient is the degree and that the arithmetic genus

coincides with the geometrical genus.

The notion of degree of line bundle over a curve is very important to define an in-

tersection product of divisors on surfaces. In fact, intersection theory is a very difficult

subject in Algebraic Geometry, but the main ideas are easy to explain for surfaces. As

before, surface will mean a smooth irreducible proper scheme of dimension two.

138

Definition. Let D,D′ be two divisors on a surface X. If D =∑i Ci and D′ corresponds

to the line bundle L′, then we define the intersection product D ·D′ =∑i deg(L′|Ci

). This

product is clearly associative, and it is possible to see also that it is commutative and, in

particular, it does not depend on the linear equivalence class of the divisors (this fact also

avoids the technical detail of defining precisely the degree of a line bundle over a singular

curve).

We illustrate this product with a series of examples.

Example 12.5. Assume that we want to count the points of intersection of two plane

curves C,D ⊂ P2k. If deg(D) = d, we can regard D as the zero locus of a section of Ld.

We can restrict both Ld and its section to C, and we thus get a line bundle Ld of degree

ddeg(C), with a section vanishing precisely at the points of C ∩D. By Bezout Theorem,

if the curves have no common components, the number of points of intersection counted

with multiplicity is ddeg(C), which is precisely our definition of C ·D as divisor. The key

observation is that we define this product even if the intersection is not proper. Recalling

that Pic(P2k) ∼= Z, the isomorphism given by the degree, the intersection product can be

defined as the standard product Z× Z→ Z.

Example 12.6. We complicate a little bit the above example by considering X = P1k×kP1

k

instead of the plane P2k. In this case, since we know (see Example 6.17) that any curve is

defined by a bihomogeneous polynomial in k[X0, X1;Y0, Y1]. Therefore, its group of classes

of divisors is isomorphic to Z⊕Z, which is also its Picard group (see Remark 10.14). The

intersection product of a curve of bidegree (d1, d2) with V (X0) ∼= P1k is the degree of Ld2

over P1k, which is d1. Similarly, the intersection of the curve with V (Y0) is d1. This means

that the intersection of the class corresponding with (d1, d2) with the class (1, 0) is d2 while

its intersection with (0, 1) is d1. This means that the intersection product can be described

as (d1, d2) · (e1, e2) = d1e2 +d2e1. In this language, since the cotangent bundle of P1k×k P1

k

is clearly p∗1ωP1k⊕ p∗2ωP1

k(where p1, p2 are the two projection maps to P1

k), it follows that

the class of the canonical line bundle is (−2,−2).

The above examples show how to intersect two curves, and the essence seems to be

to move one of them (in its class of linear equivalence), if necessary, so that eventually the

two curves meet in a finite number of points. However, the general definition allows to

intersect curves that “do not move”.

Example 12.7. Consider X ⊂ P2k × P1

k the set of pairs((x0 : x1 : x2), (y0 : y1)

)such

that y1x0 − y0x1 = 0. Identifying P1k with the pencil of lines through p = (0 : 0 : 1) by

associating to (a0 : a1) the line passing through p and (y0 : y1 : 0), then X can be regarded

as the set of pairs (q, L) such that q ∈ L. Hence, the projection π : X → P2k has as fiber

for each q 6= p the pair (q,< p, q >); while the fiber of p is E := p × P1k. The morphism

139

π is called the blow-up of P2k at p, and can be regarded as a transformation that replaces

the point p with the set of lines (or tangent directions) through it. The curve E is called

the exceptional divisor, and it is defined by the equations V (X0, X1). Let us compute the

self-intersection of E. For that, we need to identify the line bundle LE associated with E

and restrict it to E (which we will identify with P1k). Since E ⊂ D(X2Y0) ∪D(X2Y1), it

is enough to understand the bundle in those two open sets:

–In the open set D(X2Y0), since we can write X1 = Y1

Y0X0, a local equation of E is

given by the regular function X0

X2.

–In the open set D(X2Y1), we have now X0 = Y0

Y1X1, and a local equation of E is X1

X2.

Hence, the transition function of LE from D(X2Y0) to D(X2Y1) is given by X1

X0= Y1

Y0.

Therefore, identifying E with P1k, the line bundle (LE)|E has transition function Y1

Y0, which

implies that it is the line bundle of degree −1. By definition of the intersection product,

we thus get E2 = −1. Of course, the fact that the self-intersection is negative shows that

it is impossible to move E to an equivalent curve intersecting properly E.

Before stating the Riemann-Roch Theorem for surfaces, we will give a definition that

will simplify the writing:

Definition. The Euler-Poincare characteristic of a coherent sheaf F on a scheme X of

dimension n is the alternating sum χ(F) =∑ni=0(−1)ihi(X,F).

Theorem 12.8 (Riemann-Roch for surfaces). Let D be a divisor on a surface X. Then

χ(OX(D)) = χ(OX) + D2−D·K2 , where K is a canonical divisor.

Proof: We prove the result first assuming that D is a smooth curve. Let i : D ⊂ X denote

the inclusion, and consider the standard exact sequence

0→ OX(−D)→ OX → i∗OD → 0

where, as usual we identify ID,X with OX(−D). The difference with the case of curves is

that now, twisting with OX(D), also the last term varies and we get

0→ OX → OX(D)→ i∗OD(D)→ 0

where OD(D) is the sheaf of sections of the restriction to D of LD. Taking the long exact

sequence of cohomology, we get χ(OX(D)) = χ(OX) + χ(OD(D)). Hence, the theorem

will be proved if we show χ(OD(D)) = D2−D·K2 . To show that, we apply Riemann-Roch

Theorem applied to the sheaf OD(D) of D, and using the definition of intersection product

we get

χ(OD(D)) = D2 + 1− g

140

where g is the genus of D. The main point is thus to compute the genus of D. This can be

done from the adjunction formula (Example 9.14) ωD =(ωX ⊗OX(D)

)|D. Indeed, taking

degrees, we get 2g − 2 = (K +D) ·D, from which we can substitute

1− g = −D ·K +D2

2

in the previous formula for χ(OD(D)), and get the wanted result when D is a smooth

curve.

Assume now that D is an arbitrary divisor. We will assume (although the final

theorem is true in a more general context) that X is projective. In this case, we have seen

in Example 12.1 that we can write OX(D) = OX(D′ −D′′), with D′, D′′ smooth curves.

We thus repeat the proof we did for curves, but now we will have an exact sequence

0→ OX(D′ −D′′)→ OX(D′)→ i′′∗OD′′(D′)→ 0.

Taking the long exact sequence in cohomology we will get

χ(OX(D)) = χ(OX(D′))− χ(OD′′(D′)).

As in the previous case we will get

χ(OX(D′)) = χ(OX) +D′2 −D′ ·K

2

and

χ(OD′′(D′)) = D′ ·D′′ + 1− g(D′′) = D′ ·D′′ − D′′ ·K +D′′2

2.

Putting everything together we get

χ(OX(D)) = χ(OX) +D′2 − 2D′ ·D′′ +D′′2 −D′ ·K −D′′ ·K

2= χ(OX) +

D2 −D ·K2

,

as wanted.

As the reader can see, with some intersection theory, a recursion argument on the

dimension, one could get a Riemann-Roch Theorem for divisors on varieties of any dimen-

sion. We will concentrate on getting geometrical consequences from the theorem for curves

and surfaces.

We start with curves, giving a criterion to know when the Riemann-Roch Theorem

gives the dimension of the space of sections of line bundle, and hence a criterion for a line

bundle to give a morphism or an embedding:

141

Proposition 12.9. Let L be a line bundle of degree d over a curve X of genus g and let

L be the corresponding sheaf of sections. Then:

(i) If d ≥ 2g − 1 then h0(X,L) = d− g + 1.

(ii) If d ≥ 2g then L is generated by its global section, hence defines a morphism to Pd−gk .

(iii) If d ≥ 2g + 1 then L defines an embedding into Pd−gk .

Proof: For part (i) we recall, from Serre’s duality, that h1(X,L) = h0(X,ωX ⊗L∗). Since,

from Theorem 12.3, ωX has degree 2g − 2, it follows that ωX ⊗ L∗ has negative degree if

d ≥ 2g − 1, so that it cannot have global sections and h1(X,L) = 0, and the result comes

now form Riemann-Roch Theorem.

Parts (ii) and (iii) follow now from our observations in Example 12.1. For example, L

defines an embedding if, for any p ∈ X, the space of sections of L⊗Ip,X has codimension one

in the space H0(X,L). But, since Ip,X is the sheaf of sections of L∗p, a line bundle of degree

−1, it follows that L ⊗ Ip,X is the sheaf of sections of a line bundle of degree d− 1. Now

part one provides that, when d ≥ 2g, then h0(X,L) = d−g+1 and h0(X,L⊗Ip,X) = d−g,

so that H0(X,L ⊗ Ip,X) has codimension one in H0(X,L), as wanted.

Similarly, for (iii) we need to check that, for any divisor Z ⊂ X of degree two,

H0(X,L⊗ IZ,X) has codimension two in H0(X,L). But this follows now because IZ,X is

the sheaf of sections of a line bundle of degree −2 and our assumption d ≥ 2g + 1 allows

to use (i) again.

Corollary 12.10. A line bundle L over a curve is ample (i.e. a sufficiently high multiple

of it defines an embedding) if and only if deg(L) > 0.

Proof: Assume L⊗d defines an embedding for d >> 0. Then, obviously, deg(L⊗d) > 0.

Since deg(L⊗d) = ddeg(L), it follows deg(L) > 0.

Reciprocally, if deg(L) > 0, a multiple L⊗d, with d ≥ 2g+1, has degree at least 2g+1,

and Proposition 12.9(iii) implies that it defines an embedding. Hence L is ample.

Remark 12.11. Observe that we can now rephrase the proof of Proposition 12.9(i) in

the sense that any locally free sheaf of rank one and degree d ≥ 2g − 1 can be written as

ωX ⊗ L, where L has positive degree, i.e. it is ample. Hence, the proof of Proposition

12.9(i) is saying that H1(X,ωX⊗L) = 0. This is a particular case of the Kodaira vanishing

Theorem that states that on any complex variety (or defined over a field of characteristic

zero), for any ample line bundle L, if L is its sheaf of sections, then Hi(X,ωX ⊗ L) = 0

for any i > 0. However, this is not helpful to decide when a line bundle determines an

embedding or a morphism, since we have noticed that IpX is not locally free unless the

142

point p has codimension one, i.e. if X is a curve. Nevertheless, again in the complex

context, there is another result, the Kodaira Embedding Theorem, that characterizes in

terms of positivity of metrics, when a line bundle is ample.

We start now studying the cases of low genus.

Example 12.12. If X = P1k, we already know ωX = OP1

k(−2), which implies g = 0.

Reciprocally, if g = 0, Proposition 12.9(iii) with d = 1 implies that X is isomorphic to P1k.

In this case, the only line bundle of degree d is Ld, and we already know H0(P1k,OP1

k(d)) =

k[X0, X1]d, which has dimension d+ 1 if d ≥ −1. Moreover, for any d ≥ 1, the line bundle

Ld defines the d-uple Veronese embedding as the rational normal curve in Pdk, while for

d = 0 we get a constant map. We thus recover the situation of Proposition 12.9. Let us

see that having the above dimension for divisors of degree d of characterizes P1k.

Theorem 12.13. Let X be a curve and let L be a line bundle over X of degree d with

sheaf of sections L satisfying h0(X,L) ≥ d+ 1.

(i) If d = 0 , then L is the trivial line bundle.

(ii) If d ≥ 1, then X ∼= P1k and L ∼= Ld.

Proof: The hypothesis of part (i) means that L has a non-zero section, whose zero-locus

divisor is effective and of degree zero, hence necessarily the zero divisor. By Theorem

9.26(ii), it follows that L is the trivial line bundle.

We will prove part (ii) by induction on d. The main idea is to fix any point p ∈ Xand consider the exact sequence

0→ L⊗ Ip,X → L → i∗Op → 0.

Taking global sections, we have

h0(X,L ⊗ Ip,X) ≥ H0(X,L)− h0(X, i∗Op) ≥ (d+ 1)− 1 = (d− 1) + 1.

Since L ⊗ Ip,X is locally free of degree d− 1, this works that the induction process works

as soon as we prove the case d = 1. Hence, if we assume d = 1, we will get from (i) that

L ⊗ Ip,X is isomorphic to OX and all the displayed inequalities become equalities. Hence

L is a line bundle having a two-dimensional space of sections and, for any p ∈ X, the space

of sections vanishing at p has dimension one. Therefore, L defines a map ϕ : X → P1k.

Moreover, for any divisor Z of degree two on X, since L ⊗ IZ,X has degree −1, it has no

sections, i.e. the space of sections of L vanishing at Z has codimension two in H0(X,L).

Therefore, ϕ is an embedding, hence an isomorphism, as wanted. Of course, once we know

that X is isomorphic to P1k, any line bundle of degree d is isomorphic to Ld.

143

Remark 12.14. The above result has a geometrical interpretation. Assume that we

have nondegenerate (i.e. not contained in a hyperplane) curve X ⊂ Pn of degree d ≤n. This means that the map H0(Pnk ,OPn

k(1)) → H0(X,OX(1)) is injective and hence

h0(X,OX(1)) ≥ n+ 1 ≥ d+ 1. Since OX(1) has degree d, Theorem 12.13(ii) implies that

X ∼= P1k and OX(1) corresponds, under this isomorphism, with OP1

k(d). In other words, X

is the rational normal curve of degree d in Pdk.

We pass now to the case of genus one.

Theorem 12.15. Let X be an elliptic curve (i.e. its genus is zero). Then:

(i) ωX = OX .

(ii) For any line bundle L of degree d > 0 on X, if L is its sheaf of sections, h0(X,L) = d.

(iii) Any line bundle over X of degree one is the line bundle associated with a unique point

p ∈ X.

(iv) A non-trivial line bundle L over X is generated by its global sections if and only if it

has degree d ≥ 2.

(v) Any line bundle of degree two defines a double covering X → P1k.

(vi) A line bundle L over X defines an embedding if and only if it has degree d ≥ 3.

(vii) Pic(X) ∼= X × Z.

Proof: Part (i) follows from Theorem 12.13(i), since ωX has degree zero (Theorem 12.3)

and h0(X,ωX) = 1. Part (ii) is a consequence of Proposition 12.9(i).

To prove part (iii), for any line bundle L of degree one, part (ii) implies that, up to

multiplication by a nonzero constant, L has only one nonzero section. Hence, by Theorem

9.26, L = Lp for some p, which is unique.

For part (iv), obviously no line bundle of negative degree has sections and, by Theorem

12.13(i), the trivial bundle is the only line bundle of degree zero having a section (which

clearly generates the bundle), so that we only need to deal with line bundles of positive

degree. That line bundles of degree d ≥ 2 are generated by their global sections follows

from Theorem 12.9(ii). If instead L has degree one, by (ii) and (iii) it has only a one

dimensional space of sections, vanishing at a point p, hence it has p as a base point and

thus it is not generated by its global sections.

To prove (v), if L has degree two, by (ii) and (iv) it defines a map ϕ : X → P1k. Since

the fibers are preimages of hyperplanes of P1k, they are the zero loci of the sections of L,

i.e. divisors of degree two. Hence ϕ is a double covering.

Part (vi) is immediate, since Theorem 12.9(iii) implies that any line bundle of degree

d ≥ 3 gives an embedding, while the other globally generated line bundles are, by (iv), the

144

trivial bundle (which, obviously does not define an embedding) and those of degree two,

and (v) implies that they do not provide an embedding.

We finally go to part (vii). For that, we fix a point o ∈ X and consider Lo the line

bundle of degree one having a section whose zero locus is o. We define ψ : X×Z→ Pic(X)

by ψ(p, d) = Lp ⊗ L⊗(d−1)o (where the tensor product of Lo a negative number of times

means to take the tensor product of L∗o the absolute value number of times). Its inverse

is the map that sends to any line bundle L of degree d the pair (p, d), where p is the zero

locus of any non-zero section of the line bundle of degree one L⊗ L⊗(−d+1)o .

Remark 12.16. The above result allows to fully understand the group structure of any

elliptic curve X, the standard case being a plane cubic curve. The first observation is that

Theorem 12.15(iii) provides a bijection C → Pic1(X), where Pic1(X) of line bundles of

degree one, associating to any point p the line bundle Lp of degree one having a section

whose zero locus is p. Now fixing a point o ∈ X, we have a bijection X → Pic0(X) (where

Pic0(X) is the subgroup of Pic(X) consisting of those line bundles of degree zero) by

assigning to any point p ∈ X the line bundle Lp⊗L∗o. Since Pic0(X) has a group structure,

that bijection provides X with a structure of (commutative) group. The zero element in

Pic0(X) is the trivial line bundle, hence the zero element in X is the fixed element o (as

it happened in the plane cubic, in which one needed to choose the zero element). Assume

now that X is a plane cubic and that o is an in inflection point. This means that there

is a linear form in P2k (which is a section of OX(1)) whose zero locus is the divisor 3o (or

recyprocally, by Theorem 12.15(vi), the line bundle L3o produces an embedding of X as a

plane cubic, and the image of o is an inflection point). In this situation, the sum of three

points p, q, r ∈ X is zero if and only if the line bundle (Lp⊗L∗o)⊗ (Lq ⊗L∗o)⊗ (Lr ⊗L∗o) is

trivial, i.e. if and only if Lp+q+r ∼= L3o. This is equivalent to say that the divisor p+ q+ r

is the zero locus of a section of L3o, which means it is the intersection of X with a line,

i.e. p, q, r are alined.

Remark 12.17. We can generalize Theorem 12.15(vii) to any curve X in the following

way. Considering the degree map and its kernel Pic0(X), we have an exact sequence

0→ Pic0(X)→ Pic(X)→ Z→ 0

that splits. In fact, if we fix o ∈ X, the assignment d 7→ Ldo is a section of the degree map.

Therefore, Pic(X) ∼= Pic0(X)⊕Z. When g = 0, we have Pic0(X) = 0, so that Pic(X) ∼= Z,

while if g = 1, we have that Pic0(X) is isomorphic (in a non-canonical way, depending on

the choice of o ∈ X) to X, and thus Pic(X) ∼= X ×Z. It can be proved that, for arbitrary

g, Pic0(X) has dimension g and, when k = C, it is isomorphic to the quotient of Cg with

a lattice of rank 2g.

145

So far we have seen that all curves of genus zero are isomorphic to P1k and, although it

is no longer true that all elliptic curves are isomorphic to each other (it can be proved that

they form a one-dimensional family), at least all their line bundles have the same behavior

(Theorem 12.15). This is no longer true for curves of genus g ≥ 2; and the difference starts

with the canonical line bundle:

Theorem 12.18. Let X be a curve of genus g ≥ 2. Then the canonical line bundle gives

an embedding if and only if there is no line bundle L of degree two such that its associated

sheaf L satisfies h0(X,L) ≥ 2.

Proof: Recall that giving an effective divisor Z of degree two is equivalent to giving a line

bundle L of degree two such its associated sheaf L satisfies h0(X,L) ≥ 1; also, IZ,X = L∗.Therefore, by Example 12.1, the canonical line bundle gives an embedding if and only if

for any L of degree two with h0(X,L) ≥ 1 we have that H0(X,ωX ⊗L∗) has codimension

two in H0(X,ωX), i.e. h0(X,ωX ⊗L∗) = g− 2 or, using Serre’s duality, h1(X,L) = g− 2.

Since, by Riemann-Roch Theorem we have h0(X,L) − h1(X,L) = 3 − g, this is in turn

equivalent to h0(X,L) = 1, proving the Theorem.

Definition. A hyperelliptic curve is a curve of genus g ≥ 2 for which exists a locally

free sheaf L of rank one and degree two such that h0(X,L) = 2 (it cannot have bigger

dimension by Theorem 12.13(ii)).

Let us see that there exist hyperelliptic curves for any genus:

Example 12.19. Recall from Example 12.6 that a canonical divisor of P1×k P1k has bide-

gree (−2,−2). Hence, by the adjunction formula (and recalling the intersection product

in P1 ×k P1k, the degree of the canonical divisor of a curve X ⊂ P1 ×k P1

k of bidegree (a, b)

is 2g − 2 = 2ab − 2a − 2b, which implies that g has genus g = (a − 1)(b − 1). Therefore,

any curve of bidegree (2, g + 1) has genus g. On the other hand, the second projection

ϕ : X → P1k is a morphism of degree two, and it will be determined by two independent

sections of L := ϕ∗L1 Hence the sheaf of sections of L has degree two and two linearly

independent, which proves that X is hyperelliptic.

Example 12.20. A curve of genus two is always hyperelliptic, since its canonical line

bundle has degree two and two linearly independent sections. On the contrary, a “general”

curve is not hyperelliptic. For g = 3, for example, we can consider a smooth plane quartic

X ⊂ P2k. The adjunction formula shows that ωX = OX(1), hence the canonical line bundle

gives an embedding, which proves that X is not hyperelliptic. Since the degree of the

canonical line bundle is four, it follows that the genus is three. Reciprocally, if a curve

of genus three is not hyperelliptic, its canonical line bundle embeds X as a plane quartic

curve.

146

Remark 12.21. We have now for curves a complete picture of how the canonical line

bundle works for defining maps:

–For a curve of genus zero, the canonical line bundle has negative degree, so no positive

multiple of it could define a map.

–A curve of genus one has trivial canonical line bundle, so any multiple of it defines a

constant map.

–For a curve of genus g ≥ 2, either the canonical line bundle (if the curve is not

hyperelliptic) or twice the canonical line bundle define an embedding.

In general, one defines Kodaira dimension of a scheme X as the dimension of the image

of the rational map defined by a sufficiently big multiple of the canonical line bundle. If

no multiple of the canonical divisor has even a section, then X is said to have Kodaira

dimension equal to −∞. Obviously, the Kodaira dimension is at most the dimension of the

scheme; in case of equality, the scheme is said to be of general type. For curves, curves of

genus zero have Kodaira dimension equal to −∞, elliptic curves have Kodaira dimension

equal to zero, and curves of genus g ≥ 2 are of general type.

Remark 12.22. We analyze now the different invariants that we get for a surface. There

are now different possibilities of defining a genus. One could define the arithmetic genus

pa(X) = χ(OX) − 1, but now this is not the dimension of any cohomology space, but

pa(X) = h2(X,OX) − h1(X,OX). By Serre’s duality, h2(X,OX) = h0(X,ωX), and this

is called the geometric genus, and denoted by pg(X). Observe that now the arithmetic

genus does not coincide with the geometric genus. The classical school (of course with-

out using cohomology and with the Riemann-Roch Theorem known only in the classical

version for curves), called that difference pg(X) − pa(X), the irregularity of the surface

q(X), and noticed that it was equal to h0(X,ΩX). This is because the irregularity is

the dimension of H1(X,OX) = H0,1(X), and (see Theorem 11.16) this is conjugate to

H1,0(X) = H0(X,ΩX). The reader can still wonder if, as it happened in the case of

curves, the topological Euler characteristic χ(X) is related to some of these genera. The

answer is given by Noether’s formula χ(OX) = χ(X)+K2

12 .

147

A. Appendix: Categories and universal properties

In this appendix we will recall the main definitions in category theory, mainly those

coming from universal properties. We will devote the last part to the tensor calculus. We

do not intend to be rigorous, but just to give the flavor, specially for readers who never

saw category theory before.

The main idea of category theory is that in any branch of mathematics we deal with

objects (sets, vector spaces, affine spaces, projective spaces, Euclidean spaces, groups,

rings, modules, topological spaces, differential varieties,...), and morphisms among them

(maps, linear maps, affine maps, projective maps, isometries, group homomorphisms, ring

homomorphisms, module homomorphisms, continuous maps, differential maps,...). In fact,

the morphisms we allow among the objects describe the category we are working on; a

typical example is Rn, which can be regarded as a vector space, affine space or a Euclidean

space depending on whether allowable maps are linear maps, affine maps or isometries.

To speak properly of the category of sets, vector spaces,... we need a precise definition of

category:

Definition. A category C is an algebraic object consisting of a class Obj(C) (whose elements

are called the objects of the category), and, for each two objects X,Y ∈ Obj(C), a set

MorC(X,Y ) (whose elements are called morphisms from X to Y ), with the following

properties (we will just write Mor(X,Y ) when there is no ambiguity for the category):

(i) For each X,Y, Z ∈ Obj(C), there exists a map Mor(X,Y )×Mor(Y,Z)→Mor(X,Z)

(called the composition of morphisms), in which the image of (f, g) will be denoted

by g f .

(ii) For each f ∈ Mor(X,Y ), g ∈ Mor(Y, Z), f ∈ Mor(Z,W ), associativity holds, i.e.

h (g f) = (h g) f .

(iii) Each Mor(X,X) contains a morphism idX with the property that, for any f ∈Mor(X,Y ), we have f idX = f = idY f . It is a simple exercise to prove that

idX is unique, and it is called the identity morphism.

We will give the main notions in category theory, giving the most standards examples

of them, starting with the category of sets. In many cases, we will need some extra

structure, for example that the objects have the structure of abelian groups (as it happens

also for the category of rings, fields, vector spaces over a fixed field, modules over a fixed

ring,...). More precisely, we will need to work with the so-called abelian categories, which

we will define below.

Example A.1. Besides the “natural” categories we mentioned, we can have strange

categories in which the morphisms among objects are not necessarily maps. For example,

148

let I be a partially ordered set. Then we can define a category CI , in which the objects

are the elements of I, and, for any i, j ∈ I then Mor(i, j) = ∅ if i 6≤ j, while Mor(i, j)

consists of only one element if i ≤ j. If you want to avoid ambiguities, you can assign

Mor(i, j) = ∅.

Despite of the above example, we will think always of morphisms as maps, and we will

denote them with arrows, i.e. we will write f : X → Y for f ∈Mor(X,Y ). However, it is

not possible to define a priori notions like injective or surjective morphism. The categorical

definition is:

Definition. A monomorphism in a category is a morphism f ∈Mor(X,Y ) such that, for

each g, h ∈Mor(W,X) satisfying f g = f h, it follows g = h.

Definition. An epimorphism in a category is a morphism f ∈ Mor(X,Y ) such that, for

each g, h ∈Mor(Y,Z) satisfying g f = h f , it follows g = h.

Unfortunately, being at the same time monomorphism and epimorphism (such a mor-

phism is called bimorphism) is not equivalent to be an isomorphism, whose precise definition

is:

Definition. An isomorphism in a category is a morphism f ∈Mor(X,Y ) such that there

exists g ∈ Mor(Y,X) satisfying f g = idY and g f = idX . Such g is unique and it is

called inverse morphism and denoted f−1.

Remark A.2. Observe that we have a nice symmetry among the notions of monomor-

phism and epimorphism. This is due to the fact that, given a category C, we can define

its opposite category Cop in which the objects are the same, but we “reverse the arrows”,

i.e. MorCop(X,Y ) = MorC(Y,X) (we obviously need to reverse the composition order of

morphisms). With this language, an epimoprhism in C is a monomorphism in Cop and

viceversa. We will refer to this fact saying that monomorphism and epimorphism are dual

notions.

Definition. A constant morphism in a category is a morphism c : X → Y such that, for

any f, g ∈Mor(Y, Z), it follows c f = c g.

Its dual notion is not so intuitive, unless we think of the zero map (i.e. in each object

there is a canonical zero element and the constant map is zero):

Definition. A coconstant morphism in a category is a morphism c : X → Y such that, for

any f, g ∈Mor(Z,X), it follows f c = g c.

Definition. A zero morphism is a morphism that is simultaneously constant and cocon-

stant. A category with zero morphisms is a category such that, for each pair of objects,

149

there exists a morphism 0XY satisfying the following properties (which imply the collection

of morphisms 0XY to be unique):

(i) For each f ∈Mor(Y,Z), then f 0X = 0XZ .

(ii) For each f ∈Mor(Z, Y ), then 0XY f = 0XY .

If you enjoy philosophical questions, you could wonder whether there exists a category

consisting of all categories. As you can imagine, this is the same kind of dilemma and leads

to the same type of contradictions as in the discussion about the set of all sets. Of course we

will not treat this here, but at least we have to define what the right notion of morphisms

among categories should be:

Definition. A functor F among to categories C and D is a map F : Obj(C) → Obj(D)

together with a collection of maps F : MorC(X,Y )→MorD(F (X), F (Y ) when X,Y vary

in Obj(C), such that:

(i) For each X ∈ Obj(C), F (idX) = idF (X).

(ii) For each X,Y, Z ∈ Obj(C) and f ∈ Mor(X,Y ), g ∈ Mor(Y, Z) it follows F (g f) =

F (g) F (f).

You can define composition of functors, and hence isomorphism of categories.

Example A.3. There are lots of examples of functors, most of them you already met:

1) One easy example is the so-called forgetful functor, which consists in “forgetting”

part of some structure. For example, you can associate to each ring A its underlying

additive group, so that you get a functor from the category of rings to the category of

abelian groups. Observe that, in this example, all the maps of the functor are bijections,

but the functor is not an isomorphism, since the inverse is not even defined (each group

structure cannot be enlarged to a ring structure).

2) If you assign to any topological space its fundamental group, then you have a

functor from the category of topological spaces to the category of groups.

3) Let A be a commutative ring with unit, and let ModA denote the category of A-

modules. Fixing an A-modulo M , we have a functor Hom(M, ) :ModA →ModA defined

in the natural way. However, the other natural choice Hom( ,M) : ModA → ModA is

not a functor, since it reverses arrows, in the sense that a morphism M1 → M2 induces

a morphism Hom(M2,M) → Hom(M1,M). The way of solving this is to consider the

opposite category in either the source or the target category.

Definition. A contravariant functor F : C → D is a functor F : Cop → D. Usually, a

functor as defined first is called covariant functor and the word “functor” can mean either

covariant or contravariant.

We pass to the important notion of limit, for which we first give an illustrative example:

150

Example A.4. The ring of formal series A[[X]] (where A is in general a ring, but the

reader can think it is a field) can be considered as a kind of limit, when n tends to infinity,

of the space of polynomials of degree at most n. Of course, that space, if defined like this,

is not a ring, so that we consider instead the quotient ring An := A[X]/(Xn+1), which is

in natural bijection with the set of polynomials of degree at most n. Observe that, for each

m ≤ n, we have a natural homomorphism ρnm : An → Am. These homomorphisms satisfy

ρnl = ρml ρnm when l ≤ m ≤ n. Then, A[[X]] can be characterized (up to isomorphism)

as the ring that “is at the left side” of the chain

. . .→ An → An−1 → . . .→ A1 → A0

but “it is at the right” of any other ring A′ “at the left” of the chain. In a more precise

way, A[[X]] satisfies the universal property:

(*) For each n ∈ N. there is a homomorphism ρn : A[[X]] → An such that, if m ≤ n it

holds ρm = ρnm ρn and, for any other ring A′ with homomorphisms ρ′n : A′ → Ansuch that, if m ≤ n it holds ρ′m = ρnm ρ′n, there exists a unique ϕ : A′ → A[[X]]

such that ρ′n = ϕ ρn for any n ∈ N.

The general definition is the following:

Definition. Let I be a set with a preorder ≤ (i.e. ≤ is a reflexive and transitive binary

relation on I). An inverse system of objects in a category is a collection of objects Xii∈Iparametrized by I (most authors ask I to be a directed set, i.e. for each i, j ∈ I, there

exists k ∈ I such that i ≤ k and j ≤ k) satisfying that, any time that j ≤ i, there is a

morphism ρij : Xi → Xj such that

(i) ρii = idXi .

(ii) If k ≤ j ≤ i, then ρik = ρjk ρij .

The inverse limit (also called projective limit) of the system Xii∈I is an object X (usually

denoted as X = lim←−Xi) endowed with morphisms ρi : X → Xi such that, for each j ≤ i,

it holds ρj = ρij ρi, and satisfying the following property: for any other object X ′ and

morphisms ρ′i : X → X ′i such that, for each j ≤ i, it holds ρ′j = ρ′ij ρi, there exists a

unique morphism ϕ : X ′ → X such that, for each i ∈ I, it holds ρ′i = ρi ϕ.

Observe that the definition of inverse system is nothing but a contravariant functor

from CI (see Example A.1) to C.

Any object defined through a universal property is unique up to isomorphism (an

isomorphism in a category is a morphism having an inverse that is also a morphism). We

do it in detail in the case of inverse limits, and leave it as an exercise for the rest of the

notions we will introduce.

151

Proposition A.5. If it exists, the inverse limit is unique up to isomorphism.

Proof: Assume that X (with morphisms ρi : X → Xi) and X ′ (with morphisms ρ′i : X ′ →Xi) are inverse limits of the inverse system Xii∈I . Applying to the morphisms ρ′i the

universal property for X, we get a morphism ϕ : X ′ → X such that ρ′i = ρi ϕ for each

i ∈ I. Reciprocally, applying to the morphisms ρi the universal property for X ′, we get a

morphism ϕ′ : X → X ′ such that ρi = ρ′i ϕ′ for each i ∈ I.

The main trick now is that, applying to the morphisms ρi : X → Xi the universal

property for X, there exists a unique morphism ν : X → X such that ρi = ρi ν for each

i ∈ I. Obviously ν = idX satisfies that property and, by construction, also ν = ϕ ϕ′

does, hence ϕ ϕ′ = idX . Symmetrically, replacing X with X ′ we also get ϕ′ ϕ = idX′ ,

which proves that ϕ is an isomorphism.

Example A.6. We introduce now the main examples of inverse limits:

(i) Consider the ideal M = (X1, . . . , Xn) in the polynomial ring A[X1, . . . , Xn]. For

each i ∈ N, let Ai = A[X1, . . . , Xn]/Mi+1. Then the inverse limit of the system Aii∈Nis the ring of formal series A[[X1, . . . , Xn]].

(ii) When the preorder in I is = (i.e. any element is only related with itself), then the

inverse limit of any system Xii∈I is called the product of the Xi’s, and it is denoted by

Πi∈IXi. In the category of sets, the usual cartesian product is the categorical product, and

the morphisms ϕi are just the projections. In fact, in any reasonable category, the product

is also the cartesian product together with the projection maps. The point is to give the

right structure to the cartesian product. For vector spaces, groups, rings,... that structure

is clear (just defining the operation coordinate by coordinate). In the case of topological

spaces there is a notion of product topology in a cartesian product, and that is exactly

what produces the projection maps to be continue and satisfy the universal property for

the categorical product.

(iii) In many categories, the inverse limit of a system Xii∈I can be constructed as

the subset of the product Πi∈IXi consisting of the tuples (xii∈I) such that ρij(xi) = xj

if j ≤ i.

(iv) When the system consists of three objects X,Y, S in which the only relations are

arrows f : X → S and Y → S, then the limit is called pullback or fibered product and it is

usually denoted as X ×S Y . In many categories, it is the subset of the product consisting

of pairs (x, y) such that f(x) = g(y). When f and g are inclusions of objects, then the

pullback can be identified with the intersection.

(v) In a category with zero morphisms, the kernel of a morphism f ∈ Mor(X,Y ) is

defined to be the pullback of f and 0XY .

152

(vi) When the system is empty, its product is called terminal object (or final object).

In other words, a category has a terminal object O if for any object X there is a unique

element in Mor(X,O). In the category of sets, a terminal object is any set with only one

element, while in what we call abelian categories, the terminal object is the zero vector

space, group, ring,... depending on the category we are dealing with.

There is a dual notion of inverse limit. We will illustrate it with an example concerning

again Taylor series, but this time convergent.

Example A.7. For each open neighborhood U of 0 ∈ C, consider the ring O(U) of

holomorphic functions on U . When V ⊂ U we have a restriction map ρUV : O(U) →O(V ) that is a homomorphism of rings and satisfying that, when W ⊂ V ⊂ U , it holds

ρUW = ρVW ρUV . When we want to study the behavior of functions in a sufficiently

small neighborhood of 0, we arrive to a new notion of limit. In this case, we can order the

set U of open neighborhoods of 0 with the preorder U ≤ V defined as V ⊂ U (the reason of

reversing the order is that we want an open neighborhood to be “bigger” when it is closer

to 0, i.e. when it is smaller in size). With this order, U becomes a directed set (given open

sets U, V ∈ U , its intersection U ∩ V is bigger than both). Now, the local ring CZ of

convergent series in a neighborhood of 0 is characterized by the following universal property

(saying now that this ring “is at the right side” of all the rings O(U), and “it is at the left

side” of any other such ring): There are homomorphisms ρU : O(U)→ Cz (sending to

each function its Taylor expansion at 0) such that, if V ⊂ U , we have ρU = ρV ρUV ; and,

for any other ring A′ with homomorphisms ρ′U : O(U)→ A′ satisfying ρ′U = ρ′V ρUV when

V ⊂ U , then there exist a unique homomorphism ϕ : CZ → A′ such that ρ′U = ϕ ρUfor each U ∈ U .

Definition. A direct system of objects in a category is a covariant functor CI → C, i.e.

a collection of objects Xii∈I parametrized by a set I with a preorder ≤ (again, most

authors ask I to be a directed set) such that, any time that i ≤ j, there is a morphism ρijsuch that

(i) ρii = idXi .

(ii) If i ≤ j ≤ k, then ρik = ρjk ρij .

The direct limit (also called inductive limit) of the system is an object X (usually denoted

as X = lim−→Xi) endowed with morphisms ρi : Xi → X such that, for each i ≤ j, it holds

ρi = ρj ρij , and satisfying the following property: for any other object X ′ and morphisms

ρ′i : Xi → X ′ such that, for each i ≤ j, it holds ρ′i = ρ′j ρij , there exists a unique morphism

ϕ : X → X ′ such that, for each i ∈ I, it holds ρ′i = ϕ ρi.

As in Proposition A.5, one proves that, if it exists, the direct limit is unique up to

isomorphism.

153

Example A.8. This is a list of representative examples of direct limit, which can be

considered as the dual ones to the list in Example A.6:

(i) If X is a complex analytic manifold of dimension n and x ∈ X is a point, let Ube the set of open neighborhoods of X. The preorder U ≤ V if V ⊂ U provides U with

a structure of directed set. If O(U) denotes the set of holomorphic functions U → C.

Then, as in Example A.7, the direct limit of the system O(U)U∈U is isomorphic to

CZ1, . . . , Zn (just take local coordinates around x and associate to any holomorphic

function its Taylor expansion at the point).

(ii) When the preorder ≤ on I reduces to =, then the direct limit is called coproduct

and denoted by∐i∈I Xi. In the category of sets, it coincides with the disjoint union. In

categories like vector spaces, groups, rings, modules,... (and in general in what we will call

abelian categories), the coproduct is usually called direct sum, and denoted by⊕

i∈I Xi;

in this case, it can be regarded as the subset of Πi∈IXi of those elements whose all but

a finite number of coordinates are zero (or 1, in the case of multiplicative groups), and

the maps ρi : Xi →⊕

i∈I Xi ared defined by sending each x ∈ Xi to the tuple in which

all coordinates are zero except the one in the i-th place in which we put x. Hence, if I is

finite, the direct sum and the product coincide.

(iii) In many categories, the direct limit can be constructed as the quotient of the

coproduct by the equivalence relation generated by identifying ρi(x) with ρj(ρij(x)) if

j ≤ i (here ρi : Xi →∐i∈I Xi represents the structure maps of the coproduct).

(iv) When the system consists of three objects X,Y, S in which the only relations are

arrows f : S → X and S → Y , then the limit is called pushout or fibered coproduct and it

is usually denoted as X∐S Y . In many categories, it is the quotient of the coproduct by

the relation generated by identifying f(s) with g(s) for any s ∈ S.

(v) In a category with zero morphisms, the cokernel of a morphism f ∈ Mor(X,Y )

is the pushout of f and 0XY .

(vi) If the system is empty, its coproduct is called initial object (or coterminal object).

In other words, an initial object is an object O such that, for any object in the category,

there exists a unique morphism in Mor(O,X). In the category of sets, the initial object

is the empty set, while in an abelian category the initial object is again the zero object.

Definition. A zero object in a category is an object that is both terminal and initial.

Observe that, if a zero object 0 exists, the composition of the morphisms X → 0 and

0→ Y is a zero morphism.

Exercise A.9. Show that, in any category with zero morphisms, there exists a morphism∐i∈I Xi → Πi∈IXi such that pi ji = idXi for each i ∈ I.

We can finally give the precise definition of abelian category:

154

Definition. An abelian category is a category with zero object, pullbacks (in particular

kernels) and pushouts (in particular cokernels) and such that any monomorphism is a

kernel and any epimorphism is a cokernel. Observe that we can define the image of a

morphism as the kernel of the map to the cokernel, so that we can define the notion of

exact sequence.

There are much other objects in concrete categories that can be characterized by

universal properties. We illustrate them in the following examples.

Example A.10. Let A be a commutative ring with unity. and let S ⊂ A be a multiplica-

tive set (i.e. s1, s2 ∈ S implies s1s2 ∈ S) such that 0 /∈ S, 1 ∈ S. The localization of A at

S (denoted as S−1A) is “the smallest ring” containing A in which the elements of S are

units. The precise definition is that S−1A is a ring with a homomorphism i : A → S−1A

such that the image of any element of S is a unit and, for any other ring A′ with a ho-

momorphism j : A → A′ satisfying that the elements of S map to units, there exists a

unique homomorphism ϕ : S−1A → A′ such that j = ϕ i. The standard construction is

to consider the set A× S modulo the equivalence relation

(a, s) ∼= (a′, s′) if and only if there exists s′′ ∈ S such that (as′ − a′s)s′′ = 0

(observe that, if one allows 0 ∈ S in the definition, automatically S−1A = 0). The

equivalence class of (a, s) is denoted by as , and the ring structure of S−1A is given by

a

s+a′

s′=as′ + a′s

ss′

a

s· a′

s′=aa′

ss′.

The homomorphism i : A→ S−1A is defined by i(a) = a1 , and it is not necessarily injective.

Indeed its kernel consists of all those a ∈ A for which there exists s ∈ S such that as = 0.

For example, if A is an integral domain, i is injective. The main examples of localization

are:

(i) If p is a prime ideal, its complement S := A \ p is closed under multiplication, and

S−1A is denoted by Ap and called the localization of A at p. If A is an integral domain,

the localization at the prime ideal 0 is a field called the quotient field of A.

(ii) If f ∈ A is a non-nilpotent element, then we can take S = fm | m ∈ N. In that

case, S−1A is denoted by Af . An element afm is zero if and only if there exists n ∈ N such

that afn = 0. Observe that, if A is reduced (i.e. its only nilpotent element is 0), then this

is equivalent to af = 0.

Exercise A.11. Prove that the inverse image by i : A → S−1A defines a bijection

between the set of ideals in S−1A and the set of ideals of A not meeting S, and that this

155

bijection restricts to a bijection of prime ideals. In particular, the set of ideals of Ap is

in bijection with the set of ideals of A contained in p (hence Ap is a local ring, i.e. it has

a unique maximal ideal or, equivalent, the set of non-units is an ideal, namely the only

maximal ideal), and the set of ideals of Af is in bijection with the set of ideals of A not

containing f in its radical.

We present now the main notions of homological algebra. We express them in the

category of modules, although they are valid in a more general context, as we will explain

later.

Example A.12. Consider an exact sequence of A-modules

0→M ′ →M →M ′′ → 0.

Then, for any other A-module N , there are exact sequences

0→ Hom(N,M ′)→ Hom(N,M)→ Hom(N,M ′′)

and

0→ Hom(M ′′, N)→ Hom(M,N)→ Hom(M ′, N).

If A were a vector space, so that we are dealing with vector spaces, both exact sequences

could be completed with a zero at the right. However, this is not true for a general ring

A, although we can complete the first exact sequence if N is a free module. This leads to

the following:

Definition. An object in an abelian category A is projective if, for any epimorphism

M → M ′′, any morphism N → M ′′ lifts to a morphism N → M (in particular, free

modules are projective). Similarly, N is injective if, for any monomorphism M ′ →M , any

morphism M ′ → N lifts to a morphism M → N . We say that the category has enough

injectives (resp. has enough projectives) if, for any M ∈ Obj(A), there is a monomorphism

M → N (resp. an epimorphism N →M) where N is an injective (resp. projective) object.

Definition. A (covariant) functor F : A → B among abelian categories is a left-exact

functor (resp right-exact functor) if, for any exact sequence on A

0→M ′ →M →M ′′ → 0

the corresponding image by F

0→ F (M ′)→ F (M)→ F (M ′′)

is exact (resp.

F (M ′)→ F (M)→ F (M ′′)→ 0

is exact. We can extend these definitions to contravariant functors in the natural way.

The main result is the following:

156

Theorem A.13. If an abelian category has enough injectives then, for any (covariant) left-

exact functor F , there exist functors RiF unique up to isomorphism such that, R0F = F

and, for any exact sequence 0→M ′ →M →M ′′ → 0, there exists a long exact sequence

0→ R0F (M ′)→ R0F (M)→ R0F (M ′′)→ R1F (M ′)→ R1F (M)→ R1F (M ′′)→ R2F (M ′)→ . . .

Proof: We sketch the construction, which can be found in any text on homological algebra

(see for example [HS], Theorem IV-6.1). If C has enough injectives, any object M admits

a resolution

0→M → I0 → I1 → I2 → . . .

(with each Ii injective), so that, applying the functor F , we get a complex

F (I0)→ F (I1)→ F (I2)→ . . .

and we define RiF (M) to be the i-th homology of the complex.

Remark A.14. Taking the opposite category in the source and/or target in the above

theorem, we get equivalent statements:

1) If an abelian category has enough projectives then, for any right-exact covariant

functor F , there exist functors LiF unique up to isomorphism such that, L0F = F and,

for any exact sequence 0→M ′ →M →M ′′ → 0, there exists a long exact sequence

. . .→ L2F (M ′′)→ L1F (M ′)→ L1F (M)→ L1F (M ′′)→ L0F (M ′)→ L0F (M)→ L0F (M ′′)→ 0

2) If an abelian category has enough injectives (resp. enough projectives) then, for

any contravariant left-exact functor F , there exist functors RiF unique up to isomorphism

such that, R0F = F and, for any exact sequence 0→M ′ →M →M ′′ → 0, there exists a

long exact sequence

0→ R0F (M ′′)→ R0F (M)→ R0F (M ′)→ R1F (M ′′)→ R1F (M)→ R1F (M ′)→ R2F (M ′′)→ . . .

3) If an abelian category has enough projectives then, for any contravariant right-

exact functor F , there exist functors LiF unique up to isomorphism such that L0F = F

and, for any exact sequence 0→M ′ →M →M ′′ → 0, there exists a long exact sequence

. . .→ L2F (M ′)→ L1F (M ′′)→ L1F (M)→ L1F (M ′)→ L0F (M ′′)→ L0F (M)→ L0F (M ′)→ 0

Definition. The above functor RiF (resp. LiF ) is called the i-th right (resp. left) derived

functor of F .

157

Example A.15. We apply now Theorem A.13 to the functor Hom(N, ), which we

have seen in Example A.12 to be left-exact (we skip the proof of the existence of enough

injectives). Thus there are functors Exti(N, ) (coinciding with Hom(N, ) for i = 0) such

that, for any exact sequence as in the example, there are long exact sequences

0 → Hom(N,M ′)→ Hom(N,M)→ Hom(N,M ′′)→

→ Ext1(N,M ′)→ Ext1(N,M)→ Ext1(N,M ′′)→

→ Ext2(N,M ′)→ Ext2(N,M)→ Ext2(N,M ′′)→ . . .

The striking point is that we can also apply part (ii) of Remark A.13 to the functor

Hom( , N), (in this case it is an easy exercise to show that ModA has enough projec-

tives). Hence we get other derived functors Exti( , N), and it turns out that Exti(M,N)

and Exti(M,N) are naturally isomorphic for all M,N, i, so that we have also long exact

sequences0 → Hom(M ′′, N)→ Hom(M,N)→ Hom(M ′, N)→

→ Ext1(M ′′, N)→ Ext1(M,N)→ Ext1(M ′, N)→

→ Ext2(M ′′, N)→ Ext2(M,N)→ Ext2(M ′, N)→ . . .

One has that a module N is projective if and only if Exti(N,M) = 0 for all i > 0 and all

modules M , while N is injective if and only if Exti(M,N) = 0 for all i > 0 and all modules

M .

We will devote the rest of this appendix to tensor calculus, whose objects are defined

by universal properties. We start with the first notion:

Definition. The tensor product of two A-modules M,N is the A-module M ⊗A N deter-

mined by the universal property that there exists a bilinear map ϕ : M ×N → M ⊗A Nsuch that, for any other A-module P and any other bilinear map ϕ′ : M ×N → P , there

exists a unique homomorphism ψ : M ⊗AN → P such that ϕ = ψ ϕ′. One usually writes

m⊗ n := ϕ(m,n).

The standard construction (an often use directly as definition) of tensor product is

the following:

Exercise A.16. Let F be the free module generated by the Cartesian product M ×N ,

and let R be the submodule of F generated by the expressions of the form

(a1m1 + a2m2, n)− a1(m1, n)− a2(m2, n)

(m, a1n1 + a2n2)− a1(m.n1)− a2(m,n2)

158

with m,m1,m2 ∈M , n, n1, n2 ∈ N and a1, a2 ∈ A. Prove that F/R is the tensor product

of M and N .

However, there are many cases in which the tensor product has an ad-hoc expression:

Example A.17. Consider the bilinear form of vector spaces over k:

ϕ : k[X1, . . . , Xn]× k[Y1, . . . , Ym]→ k[X1, . . . , XnY1, . . . , Ym]

defined by ϕ(f, g) = fg. Since a basis of the vector space of polynomials is given

by the set of all monomials, the bilinear form ϕ is univoquely determined by the im-

age of all pairs (Xi11 . . . Xin

n , Yj11 . . . Y jmm ). Hence, given any other bilinear form ϕ′ :

k[X1, . . . , Xn] × k[Y1, . . . , Ym] → W , the map ψ : k[X1, . . . , XnY1, . . . , Ym] → W deter-

mined by ψ(Xi11 . . . Xin

n Yj11 . . . Y jmm ) = ϕ′(Xi1

1 . . . Xinn , Y

j11 . . . Y jmm ) is the unique homo-

morphism such that ϕ = ψ ϕ′. Hence k[X1, . . . , Xn] ⊗k k[Y1, . . . , Ym] can be identified

with k[X1, . . . , XnY1, . . . , Ym]. In general, the tensor product of a vector space with basis

uii∈I and a vector space with basis vjj∈J is a vector space with basis ui⊗vji∈I,j∈J .

Example A.18. Let U, V be vector spaces over a field k and let Homfin(U, V ) the vector

space of linear maps of finite rank. Consider the bilinear map ϕ : U∗ × V → Hom(U, V )

defined by(ϕ(f, v)

)(u) = f(u)v. If fii∈I is a basis of V and vjj∈J is a basis of V ,

then ϕ is determined by assigning the image of the pairs (fi, vj), which form a basis of

Homfin(U, V ). Hence, any other bilinear map ϕ′ : U∗ × V → W induces a unique linear

map ψ : Homfin(U, V )→W mapping ϕ(fi, vj) to ϕ′(ψ(fi, vj)). Hence the tensor product

U∗⊗k V is naturally isomorphic to Homfin(U, V ). Of course, this is just Hom(U, V ) when

U or V has finite dimension.

Remark A.19. Example A.18 yields a way of understanding the tensor product of vector

spaces (if you regard one of them as dual of another one). However, the way a categorist

will understand tensor product is using the notion of adjoint functor. From the universal

definition of tensor product, it follows that Hom(M⊗AN,P ) is naturally isomorphic to the

module of A-bilinear forms M×N → P , which can be identified with Hom(N,Hom(M,P )).

Hence, if we consider the functors F = Hom(M, ) and G := M ⊗A , we will get that, for

any A-modules N,P we have Hom(N,F (P )) = Hom(G(M), P ).

Definition. An adjuntion among two categories C and D is a pair of functors F : C → Dand G : D → C such that, for any X ∈ Obj(C) and any Y ∈ Obj(D), one has a natural

bijection MorC(G(Y ), X)→MorD(Y, F (X)).

159

References

[A] E. Arrondo, Introduction to projective varieties, informal notes like these (in continu-

ous change) available at http://www.mat.ucm.es∼arrondo/projvar.pdf

[AM] M. F. Atiyah, I.G. Macdonald, Introduction to commutative algebra, Addison-Wesley

Publishing Co., 1969.

[BCR] J. Bochnak, M. Coste, M.F. Roy,Geometrie algebrique reelle, Springer, 1987. (English

translation and revision published in 1998).

[CLO] D. Cox, J. Little, D. O’Shea, Ideals, Varieties and Algorithms, Springer-Verlag 1992.

[GH] P. Griffiths, J. Harris, Principles of Algebraic Geometry, a Wiley-Interscience Publi-

cation, 1978.

[H] J. Harris, Algebraic Geometry: a first course, Springer-Verlag 1992.

[Ha] R. Hartshorne, Algebraic Geometry, Springer-Verlag 1977.

[HS] P.J. Hilton, U. Stammbach, A Course in Homological Algebra, Springer 1970.

[Ma] H. Matsumura, Commutative Algebra, Benjamin/Cumming co., 1981.

[Mu] D. Mumford, Algebraic Geometry I: Complex Projective Varieties, Springer 1991

(reprinted ed.).

[R] M. Reid, Undergraduate Algebraic Geometry, London Mathematical Society Student

Texts, 1988.

[Se] E. Sernesi, Topics on families of projective schemes, Queen’s Papers in Pure and

Applied Mathematics, 73, 1986.

[Sh] I. R. Shafarevich, Basic Algebraic Geometry, vol. 1, Springer-Verlag 1994 (rev. and

expanded ed.).

160