On the Smoothness of the

Quot Functor

S E B A S T I A N W I E G A N D T

Master of Science Thesis Stockholm, Sweden 2013

On the Smoothness of the Quot Functor

S E B A S T I A N W I E G A N D T

Master’s Thesis in Mathematics (30 ECTS credits) Master Programme in Mathematics (120 credits) Royal Institute of Technology year 2013 Supervisor at KTH was Roy Skjelnes Examiner was Roy Skjelnes

TRITA-MAT-E 2013:29 ISRN-KTH/MAT/E--13/29--SE Royal Institute of Technology School of Engineering Sciences KTH SCI SE-100 44 Stockholm, Sweden URL: www.kth.se/sci

Abstract

For a commutative ring k, we consider free k-modules E, endowing them with k[x1, . . . , xm]-module structures through a ring homomorphism k[x1, . . . , xm] → EndZ(E). These struc-tures are then inspected by encoding the actions of the unknowns xi in matrices X1, . . . , Xm.We further introduce the concepts of lifts and formal smoothness for functors, and define theQuotnF/A/k functor acting on the category of k-algebras, taking some k-algebra B to the set ofquotients of the form (F ⊗kB)/N , which are locally free as B-modules. Lastly, we find con-crete examples of modules showing that the functors Hilb4k[x,y,z]/k and Quot

2⊕2 k[x,y]/k[x,y]/kare not formally smooth.

Contents

1 Introduction 31.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Preliminaries 62.1 Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.1.1 Yoneda’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.1.2 Representable Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3 Free Quotients 103.1 Quotients in Rank One . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3.1.1 One Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.1.2 Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.2 Quotients in Rank Two . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.2.1 One Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.2.2 Two Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

4 Lifts and Formal Smoothness 16

5 The Quot and Hilb Functors 195.1 The Quot Functor and Free Quotients . . . . . . . . . . . . . . . . . . . . . . . . 195.2 Representing Simple Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205.3 Smoothness of the Hilb Functor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235.4 Smoothness of the Quot Functor . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

References 31

1 INTRODUCTION

1 Introduction

The Hilbert scheme HnA of n points on a k-algebra A parametrizes ideals I ⊆ A such that A/Iis locally free of rank n as a k-module. These parametrizing objects are complicated, but onehas that the tangent space of H at the point corresponding to I is given by Hom(I/I2, A/I).This fact has often been used to extract information about the Hilbert scheme.

Fogarty, for instance, used this fact to show that the Hilbert scheme with A = k[x, y] issmooth [2]. There is then an open set that has dimension 2n, and Fogarty showed that anypoint has tangent space of dimension 2n, which implied smoothness. For A = k[x, y, z, w], theHilbert scheme has an open set of dimension 4n, and for particular choices of I one can computeand find a tangent space of different dimension. This is for example the case with n = 8 and

I = (x2, xy, y2, z2, zw,w2, xz − yw),

which has tangent space of dimension 25 [3, p. 42]. So, here the Hilbert scheme is not evenirreducible.

In this work, we will take a different approach using matrix representations. Instead ofstarting with an ideal I ⊆ k[x1, . . . , xm] such that the quotient ring is free of rank n as a k-module, we start with an arbitrary free k-module E. Then an k[x1, . . . , xm]-module structureon E is given by pairwise commuting endomorphisms X1, . . . , Xm. Thereafter we ask if theseactions can be infinitesimally lifted, which is an equivalent way of describing smoothness ofthe moduli space HnA. By doing so, we find very small examples where the Hilbert and Quotfunctors, which are representable by the Hilbert and Quot schemes respectively ([5]), are notsmooth. In fact, these are the smallest that can exist.

1.1 Background

Given a free k-module E with basis {e1, . . . , en}, we can endow E with a k[x]-module structurethrough a ring homomorphism k[x]→ EndZ(E), which is uniquely defined by the endomorphismX it maps the element x to. Consider for example the matrix

X =

0 0 · · · 0 a01 0 · · · 0 a10 1 · · · 0 a2...

.... . .

......

0 0 · · · 1 an−1

,encoding the action of x on E. This in fact gives us E as the quotient

k[x]/(xn − (a0 + a1x+ . . .+ an−1xn−1)),

by considering the action of x on {1, x, . . . , xn−1}. In fact, by introducing m commuting matricesX1, . . . , Xm, we can endow E with a k[x1, . . . , xm]-module structure in the same way.

Now, given we have a k[x1, . . . , xm]-module structure on E, suppose we have a k[x1, . . . , xm]-module homomorphism

ϕ : k[x1, . . . , xm] � E

3

1.2 Overview 1 INTRODUCTION

onto E. By the first isomorphism theorem for modules, we then have thatE = k[x1, . . . , xm]/ ker(ϕ)is a k[x1, . . . , xm]-quotient of rank n as a free k-module. Generalizing this to arbitrary k-algebrasA, the functor QuotnF/A/k is then defined as taking a k-algebra B to the set of F ⊗k B-modulequotients F ⊗k B � E which are locally free of rank n as B-modules, with F some fixedA-module. In particular, if F = A we get the functor HilbnA/k as a special case.

The concept of formal smoothness for these functors arise naturally from the correspondingconcept of formal smoothness for a ring homomorphism f : A → B, which is in this sense aninfinitesimal lifting property. We will then say that the functor QuotnF/A/k is smooth if it fulfills

a certain infinitesimal lifting property itself, namely that any element E in QuotnF/A/k(B/N),

for some N ⊆ B with certain properties, admits a lift to some E in QuotnF/A/k(B), that reducesto E modulo N .

Example 1.1. As a descriptive example of this formal smoothness criterion, let D2 = k[ε]/(ε2)

and consider the free D2-module

E = D2[x, y]/(x2 − εy, y2 − εx, xy) ∈ Hilb3k[x,y]/k(D2).

With D3 = k[ε]/(ε3), this admits a lift to

E = D3[x, y]/(x2 − εy, y2 − εx, xy − ε2) ∈ Hilb3k[x,y]/k(D3),

and we can identify B = k[ε]/(ε3) and N = (ε2) in the above discussion. We see that reducingE modulo ε2 indeed gives us back E, and note that such a lift was bound to exist, since theHilbert scheme on k[x, y] is smooth.

The aim of this text is to expand on this construction of free module structures, with theultimate goal of finding elements of certain Hilb and Quot functors that do not admit lifts inthe preceding sense. With this in mind, we will develop additional ways of interpreting thesmoothness condition given, primarily concerning ourselves with previously mentioned matrixrepresentations of module structures. A brief aside will also be given on the representability ofthe Hilb and Quot functors in some simple cases.

1.2 Overview

In Section 2, we introduce some category theoretic language, which we will use in the later stagesof this paper. In particular, the concepts of categories and functors are introduced, followed byan aside on the Yoneda lemma and representable functors.

In Section 3, we consider free k-modules E, and endow them with k[x1, . . . , xm]-modulestructures through a ring homomorphism k[x1, . . . , xm] → EndZ(E), taking xi to an endomor-phism Xi : E → E. Given a basis {e1, . . . , en} for E, these can be considered as n× n matricesacting on E, encoding the actions of the unknown variables. For example the actions of x andy on k[x, y]/(x2, y2, xy) can be encoded in the matrices

X =

0 0 01 0 00 0 0

, Y =0 0 00 0 0

1 0 0

,4

1 INTRODUCTION 1.3 Acknowledgments

by considering the actions of x and y on the elements 1, x and y. Further, given r elementse1, . . . , er ∈ E, we induce a k[x1, . . . , xm]-module homomorphism

ϕ :

r⊕k[x1, . . . , xm]→ E

by mapping (f1, . . . , fr) to f1(X1, . . . , Xm)e1 + . . .+fr(X1, . . . , Xm)er. If this mapping is surjec-tive, we get E as a quotient (

⊕r k[x1, . . . , xm])/ ker(ϕ), which is free of rank n as a k-module.In Section 4, we introduce the concepts of lifts and formal smoothness, in this sense an

infinitesimal lifting property concerning the ability of lifting certain B/N -modules for somek-algebra B and nilpotent N ⊆ B to the larger algebra B.

Lastly, we will in Section 5 revisit many of the examples given in Section 3, translated intoa category theoretic language. We will introduce the Quot functor QuotnF/A/k : k-alg → Setfor a commutative ring k, a k-algebra A and an A-module F. This functor maps a k-algebraalgebra B to quotients of the form (F ⊗k B)/N , which are locally free as B-modules. We willinspect some specific cases, followed by a discussion on the formal smoothness of this functor.This discussion concludes with showing the non-smoothness of two specific cases of the Quotfunctor, using concrete examples of non-liftable elements.

1.3 Acknowledgments

I would like to give my deepest thanks to my supervisor Roy Skjelnes for his invaluable supportand our many rewarding discussions.

5

2 PRELIMINARIES

2 Preliminaries

In this section, we will introduce some basic category theoretical concepts, and explore theirrelation to k-algebras for commutative rings k.

2.1 Categories

We will begin with the fairly abstract general definition of a category, and then swiftly moveon to our specific objects of interest, namely k-algebras. However, the general formulation willcome in handy for some of the statements of the theory.

Definition 2.1. A category C consists of a class of objects and a class of morphisms (arrows)between objects. Further, letting the class of morphisms from object a to object b be denotedhom(a, b), there is a binary operation

hom(a, b)× hom(b, c)→ hom(a, c)

for any three objects a, b, c called composition of morphisms, written g◦f for morphisms f : a→b, g : b → c. Further, morphism composition is associative, and for every object x there is amorphism 1x : x→ x such that for any morphism f : a→ b, we have 1b ◦ f = f = f ◦ 1a.

To get a sense of this rather abstract definition, let us specifically look at the category k-algof k-algebras for a commutative ring k. Here, the objects are k-algebras, while the morphisms(or arrows) f : A → B for two k-algebras A and B are the k-algebra homomorphisms from Ato B, denoted Homk-alg(A,B), consisting of the set of k-linear maps f between vector spacesA and B such that f(xy) = f(x)f(y) for all x, y ∈ A and f(1) = 1.

Another important and simple category is that of sets, which we will denote by Set. Itsobjects are, as one would assume, sets, and its morphisms are functions between sets.

Now that categories can be viewed as an extension of the concept of sets, we want to intro-duce a concept analogous to functions, defined on general categories. Therefore, we introducethe following concept of functors:

Definition 2.2. Let C and D be two categories. Then a (covariant) functor F : C → D is amapping such that

• for any object X ∈ ob(C), F (X) is an element of ob(D), and

• for any morphism f : X → Y in hom(C) we get a morphism F (f) : F (X) → F (Y ) inhom(D), such that F (1X) = 1F (X) for any X ∈ ob(C), and for any two morphismsf : X → Y and g : Y → Z in hom(C) we have F (g ◦ f) = F (g) ◦ F (f).

2.1.1 Yoneda’s Lemma

There is one result of category theory that will be of special interest and importance to us,called the Yoneda lemma. In order to state the result, however, we will need to introduce acouple of more concepts.

Having formulated the definition of a functor, we would now like to have some way oftransforming a functor F : C → D into some other functor G : C → D in a natural way. Indeed,the below definition gives us just that.

6

2 PRELIMINARIES 2.1 Categories

Definition 2.3. Given two functors F and G between categories C and D, a natural transfor-mation Φ from F to G associates to every object X ∈ ob(C) a morphism ΦX : F (X)→ G(X) inhom(D), such that for every morphism f : X → Y in hom(C), the following diagram commutes:

F (X)F (f)- F (Y )

◦

G(X)

ΦX

?

G(f)- G(Y ).

ΦY

?

Further, if ΦX is an isomorphism for every X ∈ C, then Φ is called a natural isomorphism,and F and G are said to be naturally isomorphic.

Now, the last step before formulating the Yoneda lemma is introducing the hom-functorfor small enough categories (in the sense that hom(a, b) form sets for all a, b in the category).However, in order to not get bogged down in details, we will restrict the definition to the categoryk-alg. This functor will allow us to find a correspondence between objects of our category andits set of morphisms.

Definition 2.4. Given a k-algebra A, the hom-functor hA : k-alg → Set is the functorHom(A,−), taking a k-algebra B to the set of k-algebra homomorphisms Homk-alg(A,B), anda homomorphism f : B → B′ to the morphism f ◦ −, denoting composition.

We are now ready to state the theorem, which actually allows us to transform the study ofthe category k-alg to the study of the category of functors F : k-alg → Set, which might beeasier, considering Set is a very well understood category.

Theorem 2.1 (Yoneda’s lemma). Let F be a functor from k-alg to Set. Then, for any k-algebra A, there is a one-to-one correspondence between the natural transformations from hA toF and the elements of F (A).

Proof. Consider a natural transformation Φ: hA → F . Then, the diagram

hA(A)hA(f)- hA(B)

F (A)

ΦA

?

F (f)- F (B)

ΦB

?

commutes for every k-algebra B and every k-algebra homomorphism f : A → B. Specifically,we have that

ΦB((hA(f))(idA)) = ΦB(f ◦ idA) = ΦB(f) = (F (f))(ΦA(idA))

7

2.1 Categories 2 PRELIMINARIES

for any B. Therefore, ΦB is uniquely determined by the element ΦA(idA) ∈ F (A), and themapping Φ 7→ ΦA(idA) is an injection Nat(hA, F )→ F (A).

For any x ∈ F (A), consider now the induced map ΨB : f 7→ (F (f))(x), for any morphismf : A→ B. If this defines a natural transformation Ψ: hA → F , then we specifically have

ΨA(idA) = (F (idA))(x) = idF (A)(x) = x,

showing that Φ 7→ ΦA(idA) is in fact also surjective, establishing a bijection between Nat(hA, F )and F (A). For this to be the case, we want the diagram

hA(B)hA(f)- hA(C)

F (B)

ΨB

?

F (f)- F (C)

ΨC

?

to commute for any k-algebras B and C, and for any k-algebra homomorphism f : B → C. Butthis is indeed the case, since for any morphism g : A→ B, we have

ΨC((hA(f))(g)) = ΨC(f ◦ g) = (F (f ◦ g))(x) = (F (f) ◦ F (g))(x) = (F (f))(ΨB(g)),

showing that the diagram commutes. Thus, we have shown that the mapping Φ 7→ ΦA(idA) isa bijection, proving the theorem.

In particular, this tells us that when F = hB for some k-algebra B, we have a one-to-onecorrespondence between Homk-alg(B,A) and the natural transformations from hA to hB.

2.1.2 Representable Functors

We may now ask ourselves, what can be said about a general functor F in relation to hA? Inparticular, in the case that F is naturally isomorphic to hA, we give the following definition.

Definition 2.5. A functor F : k-alg → Set is said to be representable if it is naturally iso-morphic to hA for some k-algebra A. Further, a representation of F is a pair (A,Φ) such thatΦ: hA → F is a natural isomorphism. Lastly, the element ξ = ΦA(idA) is called the universalelement of the representation.

Now, in light of the Yoneda lemma, we know that natural transformations Φ: hA → F is inone-to-one correspondence with elements of F (A), and that ξ = ΦA(idA) corresponds to a givenΦ. Further, given any element ξ ∈ F (A), we can define a natural transformation Φ: hA → F by

ΦX(f) = (F (f))(ξ)

for f ∈ Hom(A,X). In order for (A,Φ) to be a representation of F , we need Φ to be an isomor-phism, which is clearly dependent on ξ. Therefore, we see that by definition, ξ is a universalelement if and only if the natural transformation induced by it is an isomorphism. Thus, the

8

2 PRELIMINARIES 2.1 Categories

universal elements are in one-to-one correspondence with representations of the functor F . Inother words, we can identify any representation of F using its corresponding universal element.

All of the previous is true for the category of commutative rings, CRing, as well, allowingus to end the section by giving this fairly elementary example of a representable functor:

Example 2.1. Let A be a commutative ring and S ⊆ A be a multiplicative subset of A.Consider the functor F : CRing→ Set, defined by

F (B) = {ϕ : A→ B | ϕ a ring homomorphism such that ϕ(s) ∈ B∗ for all s ∈ S}

for any commutative ring B, and F (f) = f ◦ − for any ring homomorphism f : X → Y . Thenthe localization of A by S, S−1A, is a member of CRing, and i : A→ S−1A, taking a to (a, 1)is a member of F (S−1A).

Now, consider the natural transformation Φ: Hom(S−1A,−) → F induced by i. For anycommutative ring X and ring homomorphism f : S−1A→ X, we have that

ΦX(f) = (F (f))(i) = f ◦ i.

Suppose now that g : A → X ∈ F (X). The universal property of the localization then tells usthat there is a unique ring homomorphism f̂ : S−1A → X such that f̂ ◦ i = g. Therefore, Φ isa natural isomorphism, and the functor F is indeed representable by S−1A, with the universalelement i : A→ S−1A. ♦

9

3 FREE QUOTIENTS

3 Free Quotients

Let k be a commutative ring. We will in this section introduce and study some properties offree k-modules on the form (

⊕r k[x1, x2, . . . , xm]) /I, and will specifically take an in-depth lookat some cases with the rank 1 ≤ r ≤ 2. We start by proving the following proposition, showinghow we can extend a k-module structure to a k[x1, . . . , xm]-module structure.

Proposition 3.1. Let k be a commutative ring, E be a k-module and A = k[x1, . . . , xm]. Then,an A-module structure on E, compatible with the given k-module structure, is equivalent tohaving m commuting k-linear endomorphisms Xi : E → E (i = 1, . . . ,m).

Proof. The fact that E is a k-module can be expressed as (E,+) being an abelian group, andhaving a ring homomorphism ϕ : k → EndZ(E), where (ϕ(r))(e) := r · e. We can extend ϕto ϕ̃ : A → EndZ(E) by letting ϕ̃(r) = ϕ(r) for all r ∈ k and then letting ϕ̃(xi) = Xi forsome commuting endomorphisms Xi (since xixj = xjxi implies Xi ◦ Xj = Xj ◦ Xi). Lastly,defining ϕ̃ on k and each variable separately defines it on all of k[x1, . . . , xm], since it is a ringhomomorphism.

Corollary 3.1. With A and E as in the proposition, m endomorphisms Xi : E → E and anelement e ∈ E determines an A-module homomorphism A→ E, given by the composition

ϕ : A→ End(E) eve−−→ E.

Further, if ϕ : A → E is surjective, we have that E is a quotient A/I, with I the kernel of themap.

Proof. We begin by noting that HomA(A,E) is isomorphic to E as A-modules, since if ϕ : A→ Eis an A-module homomorphism, it is uniquely determined by ϕ(1), by noting that ϕ(a) = a·ϕ(1)for all a ∈ A. Therefore, we have a one-to-one correspondence between homomorphisms ϕe andelements e = ϕe(1) in E. The map taking ϕe to e is further easily seen to be an A-modulehomomorphism, by the definition of the map.

Now, suppose ϕe(1) = e for some e ∈ E. Then, for arbitrary a ∈ A, we have that ϕe(a) = a·e,which by definition equals (ψ(a))(e), where ψ : A→ EndZ(E) is our A-module structure on E.We therefore see that the composition

ϕ : A→ End(E) eve−−→ E

does indeed give us back ϕe.

3.1 Quotients in Rank One

Letting A = k[x1, . . . , xm], we will now be inspecting k-modules of the form A/I that are freeof rank, say, n. In fact, we will simplify it even further, by looking at just a single variable. Inthis case, the categorization is exceedingly easy to do.

10

3 FREE QUOTIENTS 3.1 Quotients in Rank One

3.1.1 One Variable

Suppose E is a free k-module of rank n. We can endow E with a k[x]-module structure bymapping x to some endomorphism X : E → E, which, given we make sure the map

k[x]→ End(E) eve−−→ E

is surjective, gives us that E = k[x]/I for some ideal I ⊆ k[x]. We will inspect these structuresin more detail in the below example, following a lemma simplifying the discussion.

Lemma 3.1. With k a field, any k-vector space E of dimension n of the form k[x]/I has abasis {1, x, . . . , xn−1}.

Proof. Since k is a field, k[x] is a principal ideal domain, and so I = (F (x)) for some (withoutloss of generality, monic) F ∈ k[x]. Further, I contains some polynomial G of least degree, saym, for which F (x) | G(x). Therefore, degF ≤ degG, and so degF = m.

Suppose now that m < n. Then F (x) = xm− (a0 + a1x+ . . .+ am−1xm−1) = 0 in E, and so{1, x, . . . , xm−1} spans E, which therefore cannot be of dimension n. However, if m = n, then{1, x, . . . , xn−1} spans E by the same reasoning, and further,

H(x) = b0 + b1x+ . . .+ bn−1xn−1 = 0

in E implies that F (x) | H(x), and therefore degF ≤ degH, which is clearly untrue. Therefore,{1, x, . . . , xn−1} is a basis for E.

Example 3.1. Let k be a field, and E a free k-module of rank n. We can, as previouslymentioned, endow E with a k[x]-module structure by introducing a matrix X acting on E,and given consideration to its construction, we have that E = k[x]/I for some ideal I ⊆ k[x].Further, by the lemma we then have that {1, x, . . . , xn−1} is a basis for E.

Consider the endomorphism given by

X =

0 0 · · · 0 a01 0 · · · 0 a10 1 · · · 0 a2...

.... . .

......

0 0 · · · 1 an−1

.

This induces a surjective map k[x] → End(E) ev(1)−−−→ E, since b0 + b1x + . . . + bn−1xn−1 mapsto itself in E. Further, xn 7→ a0 + a1x+ . . .+ an−1xn−1, showing that F (x) = xn − (a0 + a1x+. . .+ an−1x

n−1) = 0 in E.Moreover, the (not necessarily monic) characteristic polynomial of x is given by∣∣∣∣∣∣∣∣∣∣∣

−t 0 · · · 0 a01 −t · · · 0 a10 1 · · · 0 a2...

.... . .

......

0 0 · · · 1 an−1 − t

∣∣∣∣∣∣∣∣∣∣∣,

11

3.1 Quotients in Rank One 3 FREE QUOTIENTS

which after expanding along the last column comes out to

(−1)n−1(n−1∑i=1

(−1)2ibiti + (−1)2n(bn−1 − t) · tn−1)

= (−1)nF (t).

Therefore, the monic characteristic polynomial of x is given by F (t). Then, by Cayley-Hamilton,in fact F (X) = 0 as well. ♦

3.1.2 Several Variables

We will now move forward by looking at free quotients of polynomial rings in several variables,and specifically at some examples in two and three variables. Generally, such a free quotient isgiven by k[x1, x2, . . . , xm]/I for some ideal I defining relations so that the quotient is of somerank n.

Example 3.2. Let E be a free k-module of rank 3, and let {e1, e2, e3} be a basis. Consider theendomorphisms

X =

0 a0 c01 a1 c10 a2 c2

, Y =0 c0 b00 c1 b1

1 c2 b2

on E, with ai, bi, ci some elements in k. If the nine equations given by XY − Y X = 0 aresatisfied, Proposition 3.1 tells us that they endow E with a k[x, y]-module structure. Further,letting e = e1 in Corollary 3.1, the mapping

k[x, y]→ Endk(E)ev(e1)−−−−→ E

is surjective (since a + bx + cy maps to ae1 + be2 + ce3 for any a, b, c ∈ k), showing thatE = k[x, y]/I for some ideal I ⊆ k[x, y]. I is then given by

I = (x2 − (a0 + a1x+ a2y), y2 − (b0 + b1x+ b2y), xy − (c0 + c1x+ c2y)),

taken from the actions of x and y induced by the endomorphisms. ♦

This example illustrates how we will use Proposition 3.1 and Corollary 3.1 in the remainderof this text. Given a free k-module E, we consider a commuting set of endomorphisms on E,and if the composition

k[x1, . . . , xm]→ Endk(E)→ E

is surjective, E is given by a k[x1, . . . , xm]-quotient.

Example 3.3. We will again look at a specific example, this time in three variables, and withk = k′[ε]/(ε2) for some commutative ring k′. Let E be a free k-module with basis {e1, e2, e3, e4},and consider the matrices

X =

0 0 0 01 0 0 00 0 0 00 0 0 0

, Y =

0 0 0 00 0 0 01 0 0 00 0 ε 0

, Z =

0 0 0 00 0 0 ε0 0 0 01 0 0 0

,12

3 FREE QUOTIENTS 3.2 Quotients in Rank Two

along with the 4× 4 identity matrix. We have that XY = Y X = XZ = XZ = Y Z = 0 and

ZY =

0 0 0 00 0 ε2 00 0 0 00 0 0 0

,and so the map into the endomorphism ring is commutative (since ε2 = 0 in k). Further, thecomposition

k[x, y, z]→ Endk(E)ev1−−→ E

is surjective, since each endomorphism maps e1 to a different basis element among {e1, e2, e3, e4}.Therefore, these endomorphisms endow E with a k[x, y, z]-module structure, and further tellsus that E = k[x, y, z]/I for some I ⊆ k[x, y, z]. The induced actions of x, y and z then give usthat

I = (xy, xz, yz, x2, y2 − εz, z2 − εx).

♦

3.2 Quotients in Rank Two

We will now focus our attention to quotients of the form F/I with F = k[x1, . . . , xm] ⊕k[x1, . . . , xm], and more specifically the case with one and two variables.

Before we begin, we will have to alter Corollary 3.1 slightly. While a ring homomorphismA → End(E) still endows E with an A-module structure, we now want to find A-modulehomomorphisms F → E. Therefore, we generalize the corollary to the following (while theproof remains essentially the same):

Corollary 3.2. Let k be a commutative ring and A = k[x1, . . . , xm]. With F =⊕r A and E a

k-module, having m commuting endomorphisms Xi : E → E and r elements ei ∈ E (i = 1, . . . , r)is equivalent to an A-module homomorphism F → E, given by the composition

ϕ : F →r⊕

End(E)ev−→ E,

where the last arrow denotes the map (F1, . . . , Fr) 7→ F1(e1) + . . .+ Fr(er).

Proof. We note that EndA(F,E) =⊕r E, again since an A-module homomorphism F → E is

uniquely determined by where it maps the r elements (0, 0, . . . , 1, 0, . . . , 0), with a 1 in the ithplace (i = 1, . . . , r).

Having chosen e1, . . . , er, the corresponding homomorphism ϕe : F → E is given by

ϕe(a1, . . . , ar) = ϕe((a1, 0, . . . , 0) + . . .+ (0, 0, . . . , ar)) = a1 · e1 + . . .+ ar · er,

which by definition equals(ψ(a1))(e1) + . . .+ (ψ(ar))(er),

where ψ : A → EndZ(E) is the A-module structure on E. This finally shows that ϕe is indeedthe map ϕ as given above.

13

3.2 Quotients in Rank Two 3 FREE QUOTIENTS

3.2.1 One Variable

Suppose E is a free k-module of rank n, and we want to find a k[x]-module homomorphismk[x] ⊕ k[x] → E. Then a ring homomorphism k[x] → EndZ(E) endows E with a k[x]-modulestructure as previously. The same machinery as in the previous subsection can therefore beused in higher ranks. Specifically, if

2⊕k[x]→

2⊕End(E)

eve−−→ E

is a surjective k[x]-module homomorphism, then E is given by a quotient (k[x] ⊕ k[x])/I. Wewill inspect a specific example below.

Example 3.4. Let E be a free k-module of rank 2 with basis {e1, e2}. Consider the endomor-phism on E given by

X =

(a1 b1a2 b2

).

This again endows E with a k[x]-module structure, and if the composition

k[x]⊕ k[x]→2⊕

Endk(E)→ E

is surjective, we have that E = (k[x]⊕ k[x])/N for some submodule N ⊆ k[x]⊕ k[x]. Choosingto evaluate in e1 and e2, the map will be surjective if X itself is. Considering the action of xon E, we get that N = ((x, 0)− (a1, a2), (0, x)− (b1, b2)). ♦

3.2.2 Two Variables

Given a basis for a free module in two variables, we can describe the actions of x and y in thesame manner as above. The fully general situation is a bit more tedious to write down, so wewill restrict the scope and analyze the following example instead.

Example 3.5. Let E be a free A-module with basis {e1, e2}, and let the actions of x and y onE be described by the matrices

X =

(1 0ε 1

), Y =

(0 ε0 0

),

with A = k[ε]/(ε2) for some commutative ring k. Again, for these to give rise to an A[x, y]-module structure, the endomorphisms need to commute. We have that

XY =

(0 ε0 ε2

), Y X =

(ε2 ε0 0

),

and so they do over A, where ε2 = 0. Further, evaluating in e1 and e2, the map k[x, y] →EndA(E)→ E is surjective, since

X(ce1 + (d− εc)e2) = ce1 + de2,

14

3 FREE QUOTIENTS 3.2 Quotients in Rank Two

showing that the image spans all of E. Thus, the structure they generate is a free A-module,given by the quotient

(A[x, y]⊕A[x, y])/((x− 1,−ε), (0, x− 1), (y, 0), (−ε, y)).

♦

The reason for these somewhat obscure examples using the dual numbers will become clearin the next section, where we will define and discuss lifts and formal smoothness of maps.

15

4 LIFTS AND FORMAL SMOOTHNESS

4 Lifts and Formal Smoothness

We will now introduce the concept of formally smooth maps, at first for k-algebras, but thenlater extend this to formally smooth transformations between functors. Formal smoothnessin this sense is essentially an infinitesimal lifting property, in that a mapping into a smallneighborhood can be lifted to a larger space.

Definition 4.1. Let f : A → B be a k-algebra homomorphism. The map f is then said to beformally smooth if for every k-algebra homomorphism A → A′, every nilpotent ideal N ⊆ A′such thatN2 = (0) and every A-algebra homomorphism ĝ : B → A′/N , there exists an A-algebrahomomorphism (called a lift) g : B → A′ so that the diagram

Bĝ - A′/N

A

f

6

- A′

6g

-

is commutative.

Example 4.1. Consider the diagram

B = k[x, y]/(xy)p̂- k[ε]/(ε2) = A′/N

A = k

6

- k[ε]/(ε3) = A′

6p

-

having identified A, B and A′ in the above definition, with N = (ε2) begin a nilpotent ideal ink[ε]/(ε3), and the map p̂ : B → A′/N taking both x and y to ε. We note that since p̂(xy) =ε2 = 0 in k[ε]/(ε2), this is a k-algebra homomorphism. Due to the singularity at the origin, wewould not expect f : A→ B to be smooth, and we will see that this is indeed the case.

Suppose that the lift p : B → A′ of p̂ exists. Then, since the above diagram should commute,we must have p(x) = ε+ aε2 and p(y) = ε+ bε2 for some a, b ∈ k. However, then we have thatp(xy) = p(x)p(y) = (ε + aε2)(ε + bε2) = ε2 6= 0 in k[ε]/(ε3), and so no such homomorphism pexists. Therefore, k → k[x, y]/(xy) is not formally smooth. ♦

Drawing inspiration from this definition, we note that by Yoneda, every k-algebra homomor-phism A → B corresponds to exactly one natural transformation hB → hA. Further, supposewe have k-algebra homomorphisms f : A → B and g : B → C. Then the functor B 7→ hB willtake the morphism g to the precomposition by it, g ◦f , and so this functor will invert the arrowf to the transformation f∗ : hB → hA.

Therefore, the following definition is an intuitive analog of the previous one in the case ofrepresentable functors that also extends to arbitrary functors.

16

4 LIFTS AND FORMAL SMOOTHNESS

Definition 4.2. Let F,G : k-alg→ Set be two functors, and ϕ : F → G be a natural transfor-mation. We say that the map ϕ is formally smooth if for every k-algebra B and every nilpotentideal N ⊆ B such that N2 = (0) and every transformation ψ̂ : hB/N → F there exists a liftψ : hB → F so that the diagram

F �ψ̂

hB/N

G

ϕ

?� hB

?

�

ψ

is commutative. Further, we call a functor F smooth if for every transformation ψ̂ : hB/N → Fthere exists a lift ψ : hB → F so that

F �ψ̂

hB/N

hB?

�

ψ

commutes.

Remark. Observe that all the arrows have been reversed in relation to the definition for smooth-ness of k-algebra homomorphisms.

For now, we will not specifically concern ourselves that much with the concept of formalsmoothness, but rather look into what a lift of quotients entails. Suppose E′ is a free A/N -module of rank n for some commutative ring A and nilpotent ideal N as above. We then saythat E′ can be lifted to a free A-module E if the projection of E onto A/N is E′ and E is freeof rank n.

Further, identifying a quotient of a polynomial ring A/N [x1, . . . , xm] with the actions of itsvariables as we have done previously, a lift can also be found by finding commutative matriceswith elements in A whose projections onto A/N give us the original matrices. We illustratethis in an example below. This is not yet an argument about functorial smoothness, but ratherabout free quotients, as noted. However, in the next section, this machinery will immenselyhelp us show the non-smoothness of certain functors that are to be introduced.

Example 4.2. Let A be the ring k[ε]/(ε3) and N = (ε2). The element E = (A/N)[x, y]/(x2 −εy, y2 − εx, xy) is a free A/N -module with basis {1, x, y}, for which the actions of x and y aregiven by

X =

0 0 01 0 00 ε 0

, Y =0 0 00 0 ε

1 0 0

.17

4 LIFTS AND FORMAL SMOOTHNESS

The products of these are

XY =

0 0 00 0 00 0 ε2

, Y X =0 0 00 ε2 0

0 0 0

,showing that they commute over A/N , but that the obvious (trivial) lift is not commutativeover A. This is also evident from direct computation in E, where we have

0 = (xy)x = x2y = εy2 = ε2x,

meaning that the obvious lift would not be free as an A-module generated by {1, x, y}. However,lifting the matrices to

X =

0 0 ε21 0 00 ε 0

, Y =0 ε2 00 0 ε

1 0 0

,we have that

XY = Y X =

ε2 0 00 ε2 00 0 ε2

,rendering a commutative representation. Further, this lifts E to A[x, y]/(x2−εy, y2−εx, xy−ε2),where

ε2x = (xy)x = x2y = εy2 = ε2x,

voiding the previous issue. ♦

As this example shows, it is often not enough to consider the most obvious lift of a quotient,but we rather have to find some clever way to construct a lift that is actually a free module.In some cases, which we will cover in the next section, a lift is indeed not even possible to find,which will lead us to conclude that certain functors are not formally smooth.

18

5 THE QUOT AND HILB FUNCTORS

5 The Quot and Hilb Functors

Having given an overview of the construction of free quotients, we are now ready to abstractthis concept to a more general setting involving functors acting on the category of k-algebras.However, regardless of this abstraction, we will continue looking at concrete examples, and drawseveral parallels to the previous sections.

Definition 5.1. An A-module M is said to be locally free of rank n if there exist elementsf1, . . . , fp ∈ A such that (f1, . . . , fp) = (1), and further,

Mfi =

n⊕j=1

Afi

for i = 1, . . . , p. In other words, the localization of M at any of the elements fi is free as amodule over the corresponding localization of A.

Remark. Equivalent to the above definition is the condition that Mp is free of rank n as anAp-module for every prime ideal p in A.

Definition 5.2. Let k be a commutative ring, k → A a homomorphism of rings, and F anA-module. Let Quotn(F/A/k) be the set of A-module quotients F � E, where E is locallyfree of rank n as a k-module, and two quotients F � E and F � E′ are equivalent if theirkernels coincide. For any k → k′, let A′ = A ⊗k k′ and F ′ = F ⊗k k′. The Quot functorQuotnF/A/k : k-alg→ Set is then defined as

QuotnF/A/k(k′) = Quotn(F ′/A′/k′)

for any k′.

Remark. If F = A, the Quot functor is also called the Hilbert functor, denoted HilbnA/k. Wewill use this notation in the remainder of this text.

There is a rather straightforward connection between this definition and our previous expo-sition of locally free quotients, as we will see in the next subsection.

5.1 The Quot Functor and Free Quotients

Before jumping in and starting to analyze our previous examples in this new language, we willbridge the gap between these two points of view.

Let k be a commutative ring, A = k[x1, . . . , xm], F =⊕r A and k′ be an arbitrary k-algebra.

Further let E be a free k′-module. We can endow E with an A ⊗k k′-module structure with amapping

A⊗k k′ = k′[x1, . . . , xm]→ End(E),

by fixing m commuting matrices Xi with entries in k′, as seen previously. Then, an A ⊗k k′-

module homomorphism ϕ : F ⊗k k′ → E is given by the composition

F ⊗k k′ →r⊕

End(E)ev−−→ E,

19

5.2 Representing Simple Cases 5 THE QUOT AND HILB FUNCTORS

after picking r elements e1, . . . , er ∈ E to evaluate in. If this map is surjective, we have thatE is given by the quotient (F ⊗k k′)/ ker(ϕ). This is indeed identical to the work we did onfree quotients, with the main difference when working in the general setting of the Quot functorbeing that E is generally only locally free. However, we will continue working with our freeexamples given in the previous section.

For some concrete examples, letting F = A = k[x], we have that any k′-valued elementof QuotnF/A/k is a locally free k

′-module of rank n given by a quotient k′[x]/I for some ideal

I. This, of course, corresponds to the examples given in Section 3.1.1. Further, if A = k[x]and F = A ⊕ A, we have that any k′-valued element is given by a quotient (k′[x] ⊕ k′[x])/N ,corresponding to Section 3.2.1.

5.2 Representing Simple Cases

We will begin by examining the Hilb functor in the simplest of cases, namely A = k[x]. Given anyk-algebra B, we have that k[x]⊗kB ' B[x], and so for any element E of Hilbnk[x]/k(B) we have asurjective map ϕ : B[x] � E, or equivalently an ideal I such that, locally, B[x]/I =

⊕nB. Wewill begin with a proposition that somewhat relaxes the conditions in the case of one variable,after proving the following lemma:

Lemma 5.1. Let A be a ring, and M a finitely generated A-module. We have that if {(x1, 1),. . . , (xn, 1)} generates Mp for every p ∈ Spec(A), then {x1, . . . , xn} generates M .

Proof. Let ϕ :⊕nA→M denote the map taking (a1, . . . , an) to a1x1+. . .+anxn. This induces

an exact sequence

0→ K →n⊕A→M → C → 0,

with K the kernel and C the cokernel of ϕ. This, in turn, induces an exact sequence

0→ Kp →n⊕Ap →Mp → Cp → 0

for every p ∈ Spec(A). Now, if Mp = ((x1, 1), . . . , (xn, 1)), we have that ϕp is surjective, andtherefore that Cp = (0). Coupling this with the fact that an A-module N is trivial if and only ifNp is trivial for every prime p, we see that this implies that C = (0), and so ϕ is surjective.

Using this lemma, we can now show the following proposition:

Proposition 5.1. Any locally free A-module M of the form A[x]/I is free.

Proof. The Nakayama lemma tells us that if {x1, . . . , xn} generate Mp/pM as an Ap-module,then in fact {x1, . . . , xn} generate Mp as an Ap-module. Therefore, since M = A[x]/I is locallyfree as an A-module, we have that

(Mp)/pM =

n⊕Ap/p

is actually a vector space with basis S = {1, x, . . . , xn−1}, and therefore that S generates(A[x]/I)p as an Ap-module for every prime p ⊆ A. Then by the previous lemma, S generatesA[x]/I as an A-module, and it is therefore free.

20

5 THE QUOT AND HILB FUNCTORS 5.2 Representing Simple Cases

The following proposition then somewhat formalizes the argument from Section 3.1.1.

Proposition 5.2. The functor Hilbnk[x]/k is representable by k[t1, . . . , tn]. Further, the universalelement is given by

ξ = k[t1, . . . , tn][x]

/(xn −

n∑i=1

tixi−1

).

Proof. Let E be a free B-module of rank n, given a B[x]-module structure through X : E → Esuch that E = B[x]/I for some I ⊆ B[x]. Representing X as an n × n matrix with elementsin the commutative ring B, we have that, with F the characteristic polynomial of X, F (X) =Xn − (a0I + . . . + an−1Xn−1) = 0 by Cayley-Hamilton. Therefore, with ψ : B[x] → EndZ(E)being the B[x]-module structure on E, we get ψ(F (x)) = 0, and thus F (x) ∈ ker(ψ).

Further, with ϕ : EndZ(E)→ E being the evaluation map granting E its quotient structure,we have that F (x) ∈ ker(ϕ ◦ ψ). Thus, {1, x, . . . , xn−1} is a minimal spanning set of E as aB-module, and hence a basis.

Conversely, if G(x) is a polynomial of degree n, then B[x]/(G(x)) is free of rank n. Wetherefore see that there is a one-to-one correspondence between ideals I such that E = B[x]/Iand polynomials of degree n.

Now, with A = k[t1, . . . , tn], ξ induces the natural transformation Φ: hA → Hilbnk[x]/k givenby ΦB(f) = (Hilb

nk[x]/k(f))(ξ) for any k-algebra B and k-algebra homomorphism f : A → B.

Any such homomorphism is uniquely determined by where it maps t1, . . . , tn, so let f(ti) = bifor some bi ∈ B. Then, we have that

ΦB(f) = B[x]

/(xn −

n∑i=1

bixi−1

),

which by the previous is a bijection for all k-algebras B. Therefore, Hilbnk[x]/k is naturallyisomorphic to hA, with aforementioned universal element.

Having translated this simple example of a quotient in rank one into our new categorytheoretic language, we now want to do the same for some simple case in rank two. Doingthis, we move away from the Hilb functor, and will instead be inspecting the Quot functor.However, as the general Quot functor is fairly difficult to work with, we will start by defining amodification of it, taking into account a specified basis for its quotients.

Definition 5.3. The relative Quot functor, relQuotSF/A/k : k-alg→ Set, is defined as

relQuotSF/A/k(B) ={E ∈ Quot|S|F/A/k(B) | E is free as a B-module with basis S

},

for any k-algebra B.

Letting F = k[x]⊕ k[x], we then have for any E ∈ relQuotSF/A/k with |S| = n, that

E = (F ⊗k B)/N = (B[x]⊕B[x])/N =n⊕B

21

5.2 Representing Simple Cases 5 THE QUOT AND HILB FUNCTORS

for some N ⊆ B[x] ⊕ B[x]. We can thus fix a basis of n elements, chosen (for example)among all elements of the form (xi, xj). Let us assume that, for instance, n = 5, and fixthe basis {(1, 0), (x, 0), (x2, 0), (0, 1), (0, x)}. From the previous proposition, one might thennaively assume that we can therefore represent relQuotSF/A/k with Hom(k[t1, . . . , t5],−). Thiswill however not be the case. We state the following non-general proposition, regarding therelQuot functor taking into account a specified basis. For a more in depth discussion of therelative Quot functor, see [7].

Proposition 5.3. With F = k[x] ⊕ k[x], the functor relQuotSF/k[x]/k, with S = {(1, 0), (x, 0),(x2, 0), (0, 1), (0, x)}, is representable by k[t1, . . . , t10].

Proof. We can write

(B[x]⊕B[x])/N = Be1 + . . .+Be5,

where we identify e1 = (1, 0), . . . , e5 = (0, x). Upon examination, we see that (x3, 0) = b1e1 +

. . .+ b5e5 for some bi ∈ B, and that therefore,

(x3, 0)− [(b1, 0) + (b2x, 0) + (b3x2, 0) + (0, b4) + (0, b5x)] ∈ N.

Similarly, we have that

(0, x2)− [(c1, 0) + (c2x, 0) + (c3x2, 0) + (0, c4) + (0, c5x)] ∈ N

for some ci ∈ B, and soN is in fact determined by 10 coefficients. By a similar argument as in theprevious case, we can therefore construct a natural isomorphism between Hom(k[t1, . . . , t10],−)and relQuotSF/A/k.

As a last introductory example of the functorial definitions, we will revisit Example 3.3,as this will play a major role in subsequent parts of this text. We do this mainly to give atranslation of an earlier example into the language of Quot and Hilb functors, and to show theintimate connection with free quotients.

Example 5.1. Let A = k[x, y, z] and B = k[ε]/(ε2). Then the B-module

E = B[x, y, z]/(xy, xz, yz, x2, y2 − εz, z2 − εx)

is a member of Hilb4A/k(B), as A⊗k B = B[x, y, z] and as we saw in Example 3.3, the endomor-phisms

X =

0 0 0 01 0 0 00 0 0 00 0 0 0

, Y =

0 0 0 00 0 0 01 0 0 00 0 ε 0

, Z =

0 0 0 00 0 0 ε0 0 0 01 0 0 0

are surjective, with the induced map A → End(E) ev(e1)−−−−→ E spanning E. Further, {1, x, y, z}forms a basis of E, and so it is free of rank 4. This again emphasizes the fact that elementsof the Quot and Hilb functors are intimately connected with the previous exposition of freequotients. ♦

22

5 THE QUOT AND HILB FUNCTORS 5.3 Smoothness of the Hilb Functor

5.3 Smoothness of the Hilb Functor

We will now revisit some of our previous examples of quotients in rank one, in a slightly moregeneral setting. We will go through the creation and implications of our matrix representationsof elements of the Hilb functor, and eventually arrive at a discussion of when the functor is andis not smooth.

Previously, we have started with a free k-module E, and given this module a k[x1, . . . , xm]-module structure through a map k[x1, . . . , xm]→ EndZ(E), defining the actions of the variablesxi on E. Conversely, however, suppose we are given a free k-module with a k[x1, . . . , xm]-modulestructure. We can then define a map Φ: k[x1, . . . , xm]→ End(E) by considering the actions ofxi within the given module structure. This way, we can represent any free k-module of the form(⊕r k[x1, . . . , xm])/N with its corresponding matrices. The following definition lets us use this

language in the remainder of the text.

Definition 5.4. Let k be a commutative ring and E a free k-module given a k[x1, . . . , xm]-module structure. We then say that the matrices Xi encoding the action of xi on E representthe k[x1, . . . , xm]-module structure on E.

With this definition, we note that we have already come across a number of representationsof free k-modules, namely our maps k[x1, . . . , xm] → End(E) endowing E with k[x1, . . . , xm]-module structures, which we have shown induce surjective maps k[x1, . . . , xm] � E. We will inthe remainder revisit many of these examples, using our knowledge of their structures.

We begin by summarizing the results from Example 3.1 and Proposition 5.2 in the followingproposition:

Proposition 5.4. Letting k be a commutative ring and B be a k-algebra, elements E ofHilbnk[x]/k(B) have representations given by

X =

0 0 . . . 0 a01 0 . . . 0 a10 1 . . . 0 a2...

.... . .

......

0 0 . . . 1 an−1

,

where E = B[x]/(xn − (a0 + a1x+ . . .+ an−1xn−1)).

Before moving on to analyzing the smoothness of these functors, we will look at one moreexample. Suppose that instead of just one variable, we have three. In this case, a representationconsists of three matrices, each corresponding to the action of one of the variables. Indeed, thisis the case in our next example, where we look at one such representation.

Example 5.2. Let k be a commutative ring andB a k-algebra, and supposeE ∈ Hilb4k[x,y,z]/k(B)has a basis {1, x, y, z} as a B-module. We then claim that E has a representation of the form

X =

0 axx axy axz1 bxx bxy bxz0 cxx cxy cxz0 dxx dxy dxz

, Y =

0 ayx ayy ayz0 byx byy byz1 cyx cyy cyz0 dyx dyy dyz

, Z =

0 azx azy azz0 bzx bzy bzz0 czx czy czz1 dzx dzy dzz

,23

5.3 Smoothness of the Hilb Functor 5 THE QUOT AND HILB FUNCTORS

along with the 4 × 4 identity matrix, where the three above matrices commute pairwise. Thiswill in fact generate a quotient k[axx, . . . , dzz]/I, where the ideal I is given by the 48 equations in36 unknowns stemming from the relations XY = Y X, XZ = ZX and Y Z = ZY . Further, thisquotient will completely represent all elements of the functor Hilb4k[x,y,z]/k with basis {1, x, y, z}.

As noted previously, the fact that these matrices commute immediately forces the columnswith the same mixed indices to be equal. Further, because {1, x, y, z} is a basis for E, we canfind relations

x2 = axx + bxxx+ cxxy + dxxzy2 = ayy + byyx+ cyyy + dyyzz2 = azz + bzzx+ czzy + dzzzxy = axy + bxyx+ cxyy + dxyzxz = axz + bxzx+ cxzy + dxzzyz = ayz + byzx+ cyzy + dyzz,

with coefficients in k, which indeed correspond to the actions of x, y and z on E and thereforealso with the entries in the matrices. These identities are also respected by the matrices (thesteps of which are a bit tedious to write out, but simple to verify). For example, we have thatX2 = axxI + bxxX + cxxY + dxxZ, which is most easily checked by noting that both sides actin the same way on our basis elements. This fact, however, is dependent on the commutativityof the set of endomorphisms. ♦

We can now use these matrix representation of elements of the Hilb functor to check itssmoothness in some cases. We begin by stating a necessary and sufficient condition for themore general Quot functor to be smooth.

Lemma 5.2. Let k be a commutative ring, A a k-algebra and F an A-module. The functorQuotnF/A/k is then smooth if and only if for every k-algebra B and every nilpotent ideal N ⊆ Bsuch that N2 = (0), any element E ∈ QuotnF/A/k(B/N) admits a lift E ∈ Quot

nF/A/k(B), such

that E reduces to E modulo N .

Proof. This is basically just a restatement of the definition. Suppose the diagram

QuotnF/A/k�̂ψ hB/N

hB?

�

ψ

commutes. Then the composition hB/N → hB → Qn equals the map hB/N → Qn; however,hB/N → hB corresponds to the natural map B → B/N , and both mappings into Qn corre-spond to elements of Qn(B/N) and Qn(B), respectively. Therefore, the map hB → Qn thencorresponds to the lift of the given element in Qn(B/N).

Conversely, if a lift E exists, then the corresponding map into Qn(B) makes the diagramcommute.

24

5 THE QUOT AND HILB FUNCTORS 5.3 Smoothness of the Hilb Functor

Consider now an element E = B[x1, x2, . . . , xm]/I in Hilbnk[x1,...,xm]/k

(B) with B = B/N

for some nilpotent N such that N2 = (0) and some k-algebra B, admitting a lift to E inHilbnk[x1,...,xm]/k(B). We then have the diagram

B[x1, . . . , xm] - EndB(E)eve - E

B[x1, . . . , xm]

??- EndB(E)

??eve - E,

??

where the horizontal compositions are surjective, for some e ∈ E, showing that finding a lift forE is equivalent to finding a lift for its matrix representation that is still commutative.

We will start by examining the simplest possible case, where we have only one variable:

Proposition 5.5. The functor Hilbnk[x]/k is formally smooth.

Proof. A representation of any element in Hilbnk[x]/k(A) for any k-algebra A is given by matricesX as shown in Proposition 5.4. The composition A[x]→ EndA(V )→ V is also indeed surjective,since Xie1 = ei+1 for 0 ≤ i < n, and so evaluating in the first column will span all of V .

Further, since powers of X always commute, any lift of a representation {Xi} of an ele-ment E ∈ Hilbnk[x]/k(B/N) to a representation {X

i} of E ∈ Hilbnk[x]/k(B) will be commutative.Therefore, any such element E has a lift, and the functor is smooth.

Remark. Of course, as we saw in Proposition 5.2, Hilbnk[x]/k is representable by k[t1, . . . , tn],the polynomial ring in n variables. This proposition then tells us that this polynomial ring isnon-singular, which is as we would expect.

Essentially, this shows that in the case of one variable, there is nowhere for anything to gowrong so to speak, since any representation will automatically commute. We will see that thisbreaks down when introducing more variables, and that at a certain point this leads the Hilbfunctor to become non-smooth.

However, we will continue with an example over two variables, where one could be ledto believe that the functor is non-smooth, but a clever choice of lift indeed makes the liftedrepresentation commute. In fact, this is a revisit of Example 4.2, and we will therefore omitsome of the details that were given previously.

Example 5.3. Consider Hilb3k[x,y]/k(B/N) with B = k[ε]/(ε3) and N = (ε2). The element

E = (B/N)[x, y]/(x2 − εy, y2 − εx, xy) is a free B/N -module with basis {1, x, y}, and is repre-sented by

X =

0 0 01 0 00 ε 0

, Y =0 0 00 0 ε

1 0 0

,for which we have

XY =

0 0 00 0 00 0 ε2

, Y X =0 0 00 ε2 0

0 0 0

.25

5.3 Smoothness of the Hilb Functor 5 THE QUOT AND HILB FUNCTORS

This representation therefore commutes over B/N , but not obviously over B. However, liftingthe matrices to

X =

0 0 ε21 0 00 ε 0

, Y =0 ε2 00 0 ε

1 0 0

,we have that

XY = Y X =

ε2 0 00 ε2 00 0 ε2

,rendering a commutative representation. ♦

We are now ready to approach our first example of a non-smooth functor. In fact, we havealready come in contact with the quotient used in the proof, namely in Example 3.3.

Proposition 5.6. The functor Hilb4k[x,y,z]/k is not formally smooth.

Proof. Consider again B = k[ε]/(ε3) and N = (ε2). We then have that

(B/N)[x, y, z]/(xy, xz, yz, x2, y2 − εz, z2 − εx)

is represented by

X =

0 0 0 01 0 0 00 0 0 00 0 0 0

, Y =

0 0 0 00 0 0 01 0 0 00 0 ε 0

, Z =

0 0 0 00 0 0 ε0 0 0 01 0 0 0

,for which XY = Y X = XZ = ZX = Y Z = 0 and

ZY =

0 0 0 00 0 ε2 00 0 0 00 0 0 0

,showing that the representation is commutative over B/N , but not over B. If Y and Z arelifts of Y and Z, we have that Y = Y + ε2C and Z = Z + ε2D, with C = (cij) and D = (dij)arbitrary 4 × 4 matrices with elements in k. Then, we have that Y Z − ZY = (Y Z − ZY ) +ε2(CZ + Y D −DY − ZC). Carrying this out, we get

Y Z − ZY = ε2

c14 − d13 0 0 0c24 − d23 0 −1 0

c34 + d11 − d33 d12 d13 d14c44 − c11 − d43 −c12 −c13 −c14

,showing that no matter our choices of C and D, the lift will never commute since the elementin position (2, 3) in the product differs.

Thus, we have shown that there exists a B and a nilpotent ideal N such that an element ofHilb4k[x,y,z]/k(B/N) does not admit a lift to an element in Hilb

4k[x,y,z]/k(B), and that the functor

therefore is not smooth.

26

5 THE QUOT AND HILB FUNCTORS 5.4 Smoothness of the Quot Functor

5.4 Smoothness of the Quot Functor

We would now like to do the same thing when F =⊕r A, and specifically take a closer look at

the case r = 2. Again, finding a mapping A ⊗k B → EndB(E) corresponding to the action ofA⊗k B on E grants us an A⊗k B-module structure on E. We begin by revisiting Example 3.4,again to get a sense of this example set against this new language.

Example 5.4. Let A = k[x] and F = A ⊕ A. Then, if E ∈ Quot2F/A/k(B) for some k-algebraB, we have that

E = (B[x]⊕B[x])/N loc=2⊕B

for some N ⊆ B[x] ⊕ B[x]. Suppose further that {(1, 0), (0, 1)} forms a basis for E as a freeB-module. We then have that (x, 0) = a1(1, 0) + a2(0, 1) and (0, x) = b1(1, 0) + b2(0, 1) forsome ai, bi ∈ B. Considering the action of x on the basis elements, we can again write this as amatrix

X =

(a1 b1a2 b2

),

defining the submodule N = ((x, 0)− (a1, a2), (0, x)− (b1, b2)). ♦

The biggest difference between representations of elements of the Hilbert functor and thoseof elements of the Quot functor is that, in general, representations of elements of Hilb havematrices with very simple first columns, corresponding to the actions of xi on 1. The Quotfunctor, on the other hand, takes into account the module F , and as such there is generallyno multiplicative identity. As such, the first column of the matrices no longer have this simplestructure.

Example 5.5. Let again A = k[x] and F = A ⊕ A. Suppose further that an element E ofQuot4F/A/k is given by

E = (B[x]⊕B[x])/((x2, 0)− (a1e1 + a2e2 + a3e3 + a4e4), (0, x2)− (b1e1 + b2e2 + b3e3 + b4e4)),

where e1 = (1, 0), e2 = (x, 0), e3 = (0, 1) and e4 = (0, x), so that E has basis {(1, 0), (x, 0),(0, 1), (0, x)} as a B-module. The action of x on E can then be described by the matrix

X =

0 a1 0 b11 a2 0 b20 a3 0 b30 a4 1 b4

,giving a distinct appearance of being separated into two vertical components. ♦

Just as in the case with the Hilb functor, there is no concern with commutativity in onevariable. We have the following proposition, mirroring Proposition 5.5.

27

5.4 Smoothness of the Quot Functor 5 THE QUOT AND HILB FUNCTORS

Proposition 5.7. Let A = k[x] and F =⊕r A. Then the functor QuotnF/A/k is formally

smooth.

Proof. Just as in the proof of Proposition 5.5, any matrix encoding X of the action of x on anelement in QuotnF/A/k is commutative (with itself) and can be lifted to a commutative matrix

X. Further, if the determinant of X is nonzero, then the same is true for the lift, showing thatit is also surjective.

Moving on to several variables, it is in fact much easier to find an example of a non-smoothQuot functor than finding a similar example for the Hilb functor. We will begin by giving acomplete description of a simple case in two variables.

Example 5.6. Let A = k[x, y] and F = A ⊕ A. Suppose E = (F ⊗k B)/N is an element ofQuot2F/A/k(B) with basis {e1, e2} for some k-algebra B. We can then describe the actions of xand y on E with the matrices

X =

(a bc d

), Y =

(e fg h

),

where the composite map F ⊗k B →⊕2 EndB(E) → E is surjective, and the two matrices

commute. The fact that the two matrices commute give rise to three equations in the variablesa, . . . , h, as seen by

[X,Y ] = XY − Y X =(

bg − cf af + bh− be− dfce+ dg − ag − ch cf − bg

).

Rearranging, we get the three equationsbg − cf = 0

f(a− d) + b(h− e) = 0g(a− d) + c(h− e) = 0,

showing that as long as the mapping is surjective, bg = cf , a = d and e = h, this induces ak[x, y]-module structure on E. ♦

Note that for a case of this, we can look at Example 3.5, where a = d = 1, e = h = 0 andbg = cf = 0 in the given ring. We will now expand on this example, to get a second case of anon-smooth functor.

Proposition 5.8. With A = k[x, y] and F = A⊕A, the functor Quot2F/A/k is non-smooth.

Proof. Let B = k[ε]/(ε3) and N = (ε2), and consider the endomorphisms

X =

(1 0ε 1

), Y =

(0 ε0 0

)in End(

⊕2B/N). As seen in Example 3.5, these matrices commute, and the evaluation in e1and e2 is surjective onto

⊕2B/N , giving us the quotientE = (B/N [x, y]⊕B/N [x, y])/((x− 1,−ε), (0, x− 1), (y, 0), (−ε, y)).

28

5 THE QUOT AND HILB FUNCTORS 5.4 Smoothness of the Quot Functor

Suppose now that there is a lift of these matrices to X and Y . Then, since the projection ofthe lifts to B/N must be X and Y , they take the form

X = X + ε2(c11 c12c21 c22

)= X + ε2C, Y = Y + ε2

(d11 d12d21 d22

)= Y + ε2D,

whereby we have that XY − Y X = XY − Y X + ε2(XD+CY −DX − Y C). However, we alsohave that

ε2X =

(ε2 0ε3 ε2

)= ε2I

and

ε2Y =

(0 ε3

0 0

)= 0

in B, and so XD −DX = CY − Y C = 0, modulo ε3. Therefore, since

XY − Y X = ε2(

1 00 −1

)6= 0

in B, no lift X and Y will commute, and so the given quotient has no lift. Further, by directcomputation in E, we have that

(0, 0) = x(y, 0) = y(x, 0) = y(1, ε) = ε(0, y) = (ε2, 0),

which is true modulo ε2, but not modulo ε3.

To conclude, we have found concrete examples to show that the functors Hilb4k[x,y,z]/k and

Quot2⊕2 k[x,y]/k[x,y]/k are non-smooth. Using more sophisticated arguments, one can prove thefollowing, more general, theorem.

Theorem 5.1. The functor QuotnF/A/k is smooth if

• A = k[x],

• A = k[x, y] and F = A, or

• A = k[x1, . . . , xm], F = A and n ≤ 3.

Proof sketch. We have proved the first point previously. For a proof of the second point, see[2], Theorem 2.4. Lastly, an outline of the proof of the third point is given by the observationthat a basis for a module of rank 3 is of the form {1, xi, x2i } or {1, xi, xj}, bringing us back toone of the earlier two cases.

Remark. This theorem points to an interesting difference between the Hilbert and Quot schemes.Consider the affine variety C of commuting pairs of 2× 2 matrices (X,Y ). If

X =

(a bc d

), Y =

(e fg h

),

29

5.4 Smoothness of the Quot Functor 5 THE QUOT AND HILB FUNCTORS

we have that C = k[a, b, . . . , h]/I, where

I = (bg − cf, f(a− d) + b(h− e), g(a− d) + c(h− e)),

describes all such pairs. The ideal I is in fact prime, and C is thus irreducible and singular atsome point P .

The affine variety C⊗k k[u1, u2] corresponds to pairs of commuting matrices and an elementu = (u1, u2) ∈ E = k ⊕ k. Any point (X,Y, u) then gives a map k[x, y]→ E of k[x, y]-modules.The condition that this map is surjective is Zariski open, and therefore determines an opensubset U1 ⊂ Spec(C[u1, u2]). One can further check that the set of invertible 2 × 2 matricesGL2 acts freely on U1, and that U1/GL2 gives an open subset of Hilb

2k[x,y]/k. Since Hilb

2k[x,y]/k

is smooth, we have in particular that U1 avoids the singular set, which is determined by thesingularities of C.

The Quot scheme Quot2F/k[x,y]/k with F = k[x, y] ⊕ k[x, y], on the other hand, is singular.In the variety C[u1, u2, v1, v2], points correspond to k[x, y]-module maps

k[x, y]⊕ k[x, y]→ E,

and subjectivity of this map is again a Zariski open condition and determines an open subsetU2 ⊂ Spec(C[u1, u2, v1, v2]). Again, the group GL2 acts freely on U2, and U2/GL2 is an opensubset of Quot2F/k[x,y]/k. Since this scheme is singular, this implies that U2 does not avoid thesingularities of C.

Hence, even though the variety of commuting matrices is the same for both Hilb and Quot,the surjectivity condition on the module homomorphisms determines whether the scheme issmooth or not. In particular, with the matrices given in the proof of Proposition 5.8 we cannotfind a single element e ∈ E corresponding to surjective maps into E.

30

REFERENCES REFERENCES

References

[1] Torsten Ekedahl and Roy Skjelnes. Recovering the good component of the hilbert scheme,2000, arXiv:math/0405073.

[2] John Fogarty. Algebraic families on an algebraic surface. American Journal of Mathematics,90(2):511–521, April 1968.

[3] Robin Hartshorne. Deformation Theory. Graduate Texts in Mathematics. Springer, 2010.

[4] Saunders Mac Lane. Categories for the Working Mathematician. Graduate Texts in Math-ematics. Springer, 2nd edition, 1998.

[5] Nitin Nitsure. Construction of hilbert and quot schemes, 2005, arXiv:math/0504590.

[6] Norbert Roby. Lois polynomes et lois formelles en théorie des modules. Annales scientifiquesde l’É.N.S., 80(3):213–348, 1963.

[7] Gustav Sædén St̊ahl. Gradings of the affine line and its quot scheme. Master’s thesis,Kungliga Tekniska högskolan, 2011.

31

TRITA-MAT-E 2013:29 ISRN-KTH/MAT/E—13/29-SE

www.kth.se

framsia - wiegandtinlägg - wiegandtSchool of Engineering Sciences

rapport - WiegandtIntroductionBackgroundOverviewAcknowledgments

PreliminariesCategoriesYoneda's LemmaRepresentable Functors

Free QuotientsQuotients in Rank OneOne VariableSeveral Variables

Quotients in Rank TwoOne VariableTwo Variables

Lifts and Formal SmoothnessThe Quot and Hilb FunctorsThe Quot Functor and Free QuotientsRepresenting Simple CasesSmoothness of the Hilb FunctorSmoothness of the Quot Functor

References

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