## On the Smoothness of the

## Quot Functor

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

**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 **

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 xiin matrices X1, . . . , Xm.

We further introduce the concepts of lifts and formal smoothness for functors, and define the
Quotn_{F /A/k}functor acting on the category of k-algebras, taking some k-algebra B to the set of
quotients 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 Hilb4_{k[x,y,z]/k} and Quot2L2_{k[x,y]/k[x,y]/k}

### Contents

1 Introduction 3 1.1 Background . . . 3 1.2 Overview . . . 4 1.3 Acknowledgments . . . 5 2 Preliminaries 6 2.1 Categories . . . 6 2.1.1 Yoneda’s Lemma . . . 6 2.1.2 Representable Functors . . . 8 3 Free Quotients 10 3.1 Quotients in Rank One . . . 103.1.1 One Variable . . . 11

3.1.2 Several Variables . . . 12

3.2 Quotients in Rank Two . . . 13

3.2.1 One Variable . . . 14

3.2.2 Two Variables . . . 14

4 Lifts and Formal Smoothness 16 5 The Quot and Hilb Functors 19 5.1 The Quot Functor and Free Quotients . . . 19

5.2 Representing Simple Cases . . . 20

5.3 Smoothness of the Hilb Functor . . . 23

5.4 Smoothness of the Quot Functor . . . 27

1 INTRODUCTION

### 1

### Introduction

The Hilbert scheme H_{A}n of n points on a k-algebra A parametrizes ideals I ⊆ A such that A/I
is locally free of rank n as a k-module. These parametrizing objects are complicated, but one
has 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] is smooth [2]. There is then an open set that has dimension 2n, and Fogarty showed that any point has tangent space of dimension 2n, which implied smoothness. For A = k[x, y, z, w], the Hilbert scheme has an open set of dimension 4n, and for particular choices of I one can compute and 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 even irreducible.

In this work, we will take a different approach using matrix representations. Instead of starting 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 structure

on E is given by pairwise commuting endomorphisms X1, . . . , Xm. Thereafter we ask if these

actions can be infinitesimally lifted, which is an equivalent way of describing smoothness of
the moduli space H_{A}n. By doing so, we find very small examples where the Hilbert and Quot
functors, which are representable by the Hilbert and Quot schemes respectively ([5]), are not
smooth. 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 structure

through a ring homomorphism k[x] → End_{Z}(E), which is uniquely defined by the endomorphism
X it maps the element x to. Consider for example the matrix

X = 0 0 · · · 0 a0 1 0 · · · 0 a1 0 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 matrices}

X1, . . . , 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

1.2 Overview 1 INTRODUCTION onto E. By the first isomorphism theorem for modules, we then have that E = k[x1, . . . , xm]/ ker(ϕ)

is a k[x1, . . . , xm]-quotient of rank n as a free k-module. Generalizing this to arbitrary k-algebras

A, the functor Quotn_{F /A/k} is then defined as taking a k-algebra B to the set of F ⊗kB-module

quotients F ⊗k B E which are locally free of rank n as B-modules, with F some fixed

A-module. In particular, if F = A we get the functor Hilbn_{A/k} as a special case.

The concept of formal smoothness for these functors arise naturally from the corresponding
concept of formal smoothness for a ring homomorphism f : A → B, which is in this sense an
infinitesimal lifting property. We will then say that the functor Quotn_{F /A/k} is smooth if it fulfills
a certain infinitesimal lifting property itself, namely that any element E in Quotn_{F/A/k}(B/N ),
for some N ⊆ B with certain properties, admits a lift to some E in Quotn_{F /A/k}(B), that reduces
to 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 reducing E modulo ε2 indeed gives us back E, and note that such a lift was bound to exist, since the Hilbert scheme on k[x, y] is smooth.

The aim of this text is to expand on this construction of free module structures, with the ultimate goal of finding elements of certain Hilb and Quot functors that do not admit lifts in the preceding sense. With this in mind, we will develop additional ways of interpreting the smoothness condition given, primarily concerning ourselves with previously mentioned matrix representations of module structures. A brief aside will also be given on the representability of the 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 stages of this paper. In particular, the concepts of categories and functors are introduced, followed by an aside on the Yoneda lemma and representable functors.

In Section 3, we consider free k-modules E, and endow them with k[x1, . . . , xm]-module

structures 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 matrices

acting on E, encoding the actions of the unknown variables. For example the actions of x and y on k[x, y]/(x2, y2, xy) can be encoded in the matrices

1 INTRODUCTION 1.3 Acknowledgments by considering the actions of x and y on the elements 1, x and y. Further, given r elements e1, . . . , er∈ E, we induce a k[x1, . . . , xm]-module homomorphism

ϕ :

r

M

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 (Lr

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 some k-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 into
a category theoretic language. We will introduce the Quot functor Quotn_{F /A/k}: k -alg → Set
for a commutative ring k, a k-algebra A and an A-module F. This functor maps a k-algebra
algebra B to quotients of the form (F ⊗kB)/N , which are locally free as B-modules. We will

inspect 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 Quot functor, using concrete examples of non-liftable elements.

1.3 Acknowledgments

2 PRELIMINARIES

### 2

### Preliminaries

In this section, we will introduce some basic category theoretical concepts, and explore their relation 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 move on to our specific objects of interest, namely k-algebras. However, the general formulation will come 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 denoted hom(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 a morphism 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 -alg of 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 A to B, denoted Homk -alg(A, B), consisting of the set of k-linear maps f between vector spaces

A 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. Its objects 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 introintro-duce the following concept of functors:

Definition 2.2. Let C and D be two categories. Then a (covariant) functor F : C → D is a mapping 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 ) in hom(D), such that F (1X) = 1F (X) for any X ∈ ob(C), and for any two morphisms

f : 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 a couple of more concepts.

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) in

hom(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-functor for 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 category k -alg. This functor will allow us to find a correspondence between objects of our category and its set of morphisms.

Definition 2.4. Given a k-algebra A, the hom-functor hA: k -alg → Set is the functor

Hom(A, −), taking a k-algebra B to the set of k-algebra homomorphisms Homk -alg(A, B), and

a homomorphism f : B → B0 to the morphism f ◦ −, denoting composition.

We are now ready to state the theorem, which actually allows us to transform the study of the category k -alg to the study of the category of functors F : k -alg → Set, which might be easier, 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 hAto

F and the elements of F (A).

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

hA(A)
hA(f )
- h_{A}(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

2.1 Categories 2 PRELIMINARIES for any B. Therefore, ΦB is uniquely determined by the element ΦA(idA) ∈ F (A), and the

mapping 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 morphism

f : 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 )
- h_{A}(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. But this 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) is

a bijection, proving the theorem.

In particular, this tells us that when F = hB for some k-algebra B, we have a one-to-one

correspondence 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? In

particular, 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 hAfor some k-algebra A. Further, a representation of F is a pair (A, Φ) such that

Φ : hA → F is a natural isomorphism. Lastly, the element ξ = ΦA(idA) is called the universal

element of the representation.

Now, in light of the Yoneda lemma, we know that natural transformations Φ : hA→ F is in

one-to-one correspondence with elements of F (A), and that ξ = ΦA(idA) corresponds to a given

Φ. Further, given any element ξ ∈ F (A), we can define a natural transformation Φ : hA→ F by

ΦX(f ) = (F (f ))(ξ)

2 PRELIMINARIES 2.1 Categories universal elements are in one-to-one correspondence with representations of the functor F . In other 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, allowing us 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 . Then the 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 any commutative 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 us that there is a unique ring homomorphism bf : S−1A → X such that bf ◦ i = g. Therefore, Φ is a natural isomorphism, and the functor F is indeed representable by S−1A, with the universal

3 FREE QUOTIENTS

### 3

### Free Quotients

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

k[x1, x2, . . . , xm]) /I, and will specifically take an in-depth look

at some cases with the rank 1 ≤ r ≤ 2. We start by proving the following proposition, showing how 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 to having 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, and
having a ring homomorphism ϕ : k → End_{Z}(E), where (ϕ(r))(e) := r · e. We can extend ϕ
to ϕ : A → Ende Z(E) by letting ϕ(r) = ϕ(r) for all r ∈ k and then lettinge ϕ(xe i) = Xi for
some commuting endomorphisms Xi (since xixj = xjxi implies Xi ◦ Xj = Xj ◦ Xi). Lastly,

defining ϕ on k and each variable separately defines it on all of k[x_{e} 1, . . . , xm], since it is a ring

homomorphism.

Corollary 3.1. With A and E as in the proposition, m endomorphisms Xi: E → E and an

element e ∈ E determines an A-module homomorphism A → E, given by the composition ϕ : A → End(E)−−→ E.eve

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

Proof. We begin by noting that HomA(A, E) is isomorphic to E as A-modules, since if ϕ : A → E

is 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 and

elements e = ϕe(1) in E. The map taking ϕe to e is further easily seen to be an A-module

homomorphism, 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 → End_{Z}(E) is our A-module structure on E.
We therefore see that the composition

ϕ : A → End(E)_{−−→ E}eve

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 free

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 by mapping x to some endomorphism X : E → E, which, given we make sure the map

k[x] → End(E)−−→ Eeve

is surjective, gives us that E = k[x]/I for some ideal I ⊆ k[x]. We will inspect these structures in 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 a basis {1, x, . . . , xn−1}.

Proof. Since k is a field, k[x] is a principal ideal domain, and so I = (F (x)) for some (without loss of generality, monic) F ∈ k[x]. Further, I contains some polynomial G of least degree, say m, for which F (x) | G(x). Therefore, deg F ≤ deg G, and so deg F = 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 deg F ≤ deg H, 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 previously
mentioned, 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 a0 1 0 · · · 0 a1 0 1 · · · 0 a2 .. . ... . .. ... ... 0 0 · · · 1 an−1 .

This induces a surjective map k[x] → End(E) −−−→ E, since bev(1) _{0}+ b1x + . . . + bn−1xn−1 maps

to itself in E. Further, xn7→ a0+ a1x + . . . + an−1xn−1, showing that F (x) = xn− (a0+ a1x +

. . . + an−1xn−1) = 0 in E.

3.1 Quotients in Rank One 3 FREE QUOTIENTS which after expanding along the last column comes out to

(−1)n−1 n−1 X 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 is given by k[x1, x2, . . . , xm]/I for some ideal I defining relations so that the quotient is of some

rank n.

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

endomorphisms X = 0 a0 c0 1 a1 c1 0 a2 c2 , Y = 0 c0 b0 0 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 are

satisfied, 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 that

E = k[x, y]/I for some ideal I ⊆ k[x, y]. I is then given by

I = (x2− (a_{0}+ 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 remainder of 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 with k = k0[ε]/(ε2) for some commutative ring k0. Let E be a free k-module with basis {e1, e2, e3, e4},

and consider the matrices

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 0 0 0 ε2 0 0 0 0 0 0 0 0 0 ,

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

composition

k[x, y, z] → Endk(E)−−→ Eev1

is surjective, since each endomorphism maps e1to a different basis element among {e1, e2, e3, e4}.

Therefore, these endomorphisms endow E with a k[x, y, z]-module structure, and further tells us that E = k[x, y, z]/I for some I ⊆ k[x, y, z]. The induced actions of x, y and z then give us that

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 homomorphism A → End(E) still endows E with an A-module structure, we now want to find A-module homomorphisms F → E. Therefore, we generalize the corollary to the following (while the proof remains essentially the same):

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

M

End(E)−→ E,ev

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

Proof. We note that EndA(F, E) =LrE, 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 ith place (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 → End_{Z}(E) is the A-module structure on E. This finally shows that ϕe is indeed

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 homomorphism
k[x] ⊕ k[x] → E. Then a ring homomorphism k[x] → End_{Z}(E) endows E with a k[x]-module
structure as previously. The same machinery as in the previous subsection can therefore be
used in higher ranks. Specifically, if

2 M k[x] → 2 M End(E)−−→ Eeve

is a surjective k[x]-module homomorphism, then E is given by a quotient (k[x] ⊕ k[x])/I. We will 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 b1 a2 b2

.

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

k[x] ⊕ k[x] →

2

M

Endk(E) → E

is surjective, we have that E = (k[x] ⊕ k[x])/N for some submodule N ⊆ k[x] ⊕ k[x]. Choosing to evaluate in e1 and e2, the map will be surjective if X itself is. Considering the action of x

on 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 the same manner as above. The fully general situation is a bit more tedious to write down, so we will 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 on

E be described by the matrices

X =1 0 ε 1 , Y =0 ε 0 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 ε
0 ε2
, Y X =ε
2 _{ε}
0 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

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)).

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 then later extend this to formally smooth transformations between functors. Formal smoothness in this sense is essentially an infinitesimal lifting property, in that a mapping into a small neighborhood 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 be
formally smooth if for every k-algebra homomorphism A → A0, every nilpotent ideal N ⊆ A0
such that N2 = (0) and every A-algebra homomorphismg : B → A_{b} 0/N , there exists an A-algebra
homomorphism (called a lift ) g : B → A0 so that the diagram

B bg - A0/N A f 6 - A0 6 g -is commutative.

Example 4.1. Consider the diagram

B = k[x, y]/(xy) pb - k[ε]/(ε2) = A0/N A = k 6 - k[ε]/(ε3) = A0 6 p

-having identified A, B and A0 in the above definition, with N = (ε2) begin a nilpotent ideal in
k[ε]/(ε3), and the map p : B → A_{b} 0/N taking both x and y to ε. We note that since p(xy) =_{b}
ε2= 0 in k[ε]/(ε2), this is a k-algebra homomorphism. Due to the singularity at the origin, we
would not expect f : A → B to be smooth, and we will see that this is indeed the case.

Suppose that the lift p : B → A0 ofp exists. Then, since the above diagram should commute,_{b}
we must have p(x) = ε + aε2 and p(y) = ε + bε2 for some a, b ∈ k. However, then we have that
p(xy) = p(x)p(y) = (ε + aε2)(ε + bε2) = ε2 6= 0 in k[ε]/(ε3_{), and so no such homomorphism p}

exists. 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, suppose

we have k-algebra homomorphisms f : A → B and g : B → C. Then the functor B 7→ hB will

take the morphism g to the precomposition by it, g ◦ f , and so this functor will invert the arrow f to the transformation f∗: hB → hA.

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 nilpotent
ideal N ⊆ B such that N2 _{= (0) and every transformation b}_{ψ : h}

B/N → F there exists a lift

ψ : hB → F so that the diagram

F
b
ψ
h_{B/N}
G
ϕ
?
h_{B}
?
ψ

is commutative. Further, we call a functor F smooth if for every transformation bψ : hB/N → F

there exists a lift ψ : hB→ F so that

F b ψ 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 formal smoothness, but rather look into what a lift of quotients entails. Suppose E0 is a free A/N -module of rank n for some commutative ring A and nilpotent ideal N as above. We then say that E0 can be lifted to a free A-module E if the projection of E onto A/N is E0 and E is free of rank n.

Further, identifying a quotient of a polynomial ring A/N [x1, . . . , xm] with the actions of its

variables as we have done previously, a lift can also be found by finding commutative matrices with elements in A whose projections onto A/N give us the original matrices. We illustrate this in an example below. This is not yet an argument about functorial smoothness, but rather about free quotients, as noted. However, in the next section, this machinery will immensely help us show the non-smoothness of certain functors that are to be introduced.

4 LIFTS AND FORMAL SMOOTHNESS The products of these are

XY =
0 0 0
0 0 0
0 0 ε2
, Y X =
0 0 0
0 ε2 _{0}
0 0 0
,

showing that they commute over A/N , but that the obvious (trivial) lift is not commutative over 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 ε2 1 0 0 0 ε 0 , Y = 0 ε2 0 0 0 ε 1 0 0 , we have that XY = Y X = ε2 0 0 0 ε2 0 0 0 ε2 ,

rendering a commutative representation. Further, this lifts E to A[x, y]/(x2−εy, y2_{−εx, xy−ε}2_{),}

where

ε2x = (xy)x = x2y = εy2 = ε2x,

voiding the previous issue. ♦

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 abstract this 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 draw several parallels to the previous sections.

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

Mfi =

n

M

j=1

Afi

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

module over the corresponding localization of A.

Remark. Equivalent to the above definition is the condition that Mp is free of rank n as an

Ap-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 an A-module. Let Quotn(F/A/k) be the set of A-module quotients F E, where E is locally free of rank n as a k-module, and two quotients F E and F E0 are equivalent if their kernels coincide. For any k → k0, let A0 = A ⊗kk0 and F0 = F ⊗k k0. The Quot functor

Quotn_{F /A/k}: k -alg → Set is then defined as

Quotn_{F /A/k}(k0) = Quotn(F0/A0/k0)

for any k0.

Remark. If F = A, the Quot functor is also called the Hilbert functor, denoted Hilbn_{A/k}. We
will 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 will bridge the gap between these two points of view.

Let k be a commutative ring, A = k[x1, . . . , xm], F =LrA and k0be an arbitrary k-algebra.

Further let E be a free k0-module. We can endow E with an A ⊗kk0-module structure with a

mapping

A ⊗kk0 = k0[x1, . . . , xm] → End(E),

by fixing m commuting matrices Xi with entries in k0, as seen previously. Then, an A ⊗kk0

-module homomorphism ϕ : F ⊗kk0 → E is given by the composition

F ⊗kk0 → r

M

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 that

E is given by the quotient (F ⊗kk0)/ ker(ϕ). This is indeed identical to the work we did on

free quotients, with the main difference when working in the general setting of the Quot functor being that E is generally only locally free. However, we will continue working with our free examples given in the previous section.

For some concrete examples, letting F = A = k[x], we have that any k0-valued element
of Quotn_{F /A/k} is a locally free k0-module of rank n given by a quotient k0[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 k0-valued element is given by a quotient (k0[x] ⊕ k0[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 any k-algebra B, we have that k[x] ⊗kB ' B[x], and so for any element E of Hilbnk[x]/k(B) we have a

surjective map ϕ : B[x] E, or equivalently an ideal I such that, locally, B[x]/I =Ln

B. We will 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 : Ln

A → M denote the map taking (a1, . . . , an) to a1x1+. . .+anxn. This induces

an exact sequence

0 → K →

n

M

A → M → C → 0,

with K the kernel and C the cokernel of ϕ. This, in turn, induces an exact sequence 0 → Kp→

n

M

Ap→ Mp→ Cp→ 0

for every p ∈ Spec(A). Now, if Mp = ((x1, 1), . . . , (xn, 1)), we have that ϕp is surjective, and

therefore that Cp = (0). Coupling this with the fact that an A-module N is trivial if and only if

Np 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 locally

free as an A-module, we have that

(Mp)/pM = n

M 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 generates

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 Hilbn_{k[x]/k} is representable by k[t1, . . . , tn]. Further, the universal

element is given by ξ = k[t1, . . . , tn][x] , xn− n X i=1 tixi−1 ! .

Proof. Let E be a free B-module of rank n, given a B[x]-module structure through X : E → E such that E = B[x]/I for some I ⊆ B[x]. Representing X as an n × n matrix with elements in 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 : End_{Z}(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 a
B-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. We therefore see that there is a one-to-one correspondence between ideals I such that E = B[x]/I and polynomials of degree n.

Now, with A = k[t1, . . . , tn], ξ induces the natural transformation Φ : hA→ Hilbn_{k[x]/k} given

by ΦB(f ) = (Hilbnk[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) = bi

for some bi ∈ B. Then, we have that

ΦB(f ) = B[x] , xn− n X i=1 bixi−1 ! ,

which by the previous is a bijection for all k-algebras B. Therefore, Hilbn_{k[x]/k} is naturally
isomorphic to hA, with aforementioned universal element.

Having translated this simple example of a quotient in rank one into our new category theoretic language, we now want to do the same for some simple case in rank two. Doing this, 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 a modification of it, taking into account a specified basis for its quotients.

Definition 5.3. The relative Quot functor, relQuotS_{F /A/k}: k -alg → Set, is defined as

relQuotS_{F /A/k}(B) =
n

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

, for any k-algebra B.

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

E = (F ⊗kB)/N = (B[x] ⊕ B[x])/N = n

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 fix
the basis {(1, 0), (x, 0), (x2_{, 0), (0, 1), (0, x)}. From the previous proposition, one might then}

naively assume that we can therefore represent relQuotS_{F /A/k} with Hom(k[t1, . . . , t5], −). This

will however not be the case. We state the following non-general proposition, regarding the relQuot functor taking into account a specified basis. For a more in depth discussion of the relative Quot functor, see [7].

Proposition 5.3. With F = k[x] ⊕ k[x], the functor relQuotS_{F /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 so N is in fact determined by 10 coefficients. By a similar argument as in the

previous case, we can therefore construct a natural isomorphism between Hom(k[t1, . . . , t10], −)

and relQuotS_{F /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 a translation of an earlier example into the language of Quot and Hilb functors, and to show the intimate 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 Hilb4_{A/k}(B), as A ⊗kB = B[x, y, z] and as we saw in Example 3.3, the

endomor-phisms X = 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 , Y = 0 0 0 0 0 0 0 0 1 0 0 0 0 0 ε 0 , Z = 0 0 0 0 0 0 0 ε 0 0 0 0 1 0 0 0

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

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 more general setting. We will go through the creation and implications of our matrix representations of elements of the Hilb functor, and eventually arrive at a discussion of when the functor is and is 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 variables

xi on E. Conversely, however, suppose we are given a free k-module with a k[x1, . . . , xm]-module

structure. We can then define a map Φ : k[x1, . . . , xm] → End(E) by considering the actions of

xi within the given module structure. This way, we can represent any free k-module of the form

(Lr

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 represent

the k[x1, . . . , xm]-module structure on E.

With this definition, we note that we have already come across a number of representations of 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 in

the 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 following proposition:

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

X = 0 0 . . . 0 a0 1 0 . . . 0 a1 0 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 more example. Suppose that instead of just one variable, we have three. In this case, a representation consists of three matrices, each corresponding to the action of one of the variables. Indeed, this is the case in our next example, where we look at one such representation.

Example 5.2. Let k be a commutative ring and B a k-algebra, and suppose E ∈ Hilb4_{k[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

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. This will in fact generate a quotient k[axx, . . . , dzz]/I, where the ideal I is given by the 48 equations in

36 unknowns stemming from the relations XY = Y X, XZ = ZX and Y Z = ZY . Further, this
quotient will completely represent all elements of the functor Hilb4_{k[x,y,z]/k} with basis {1, x, y, z}.
As noted previously, the fact that these matrices commute immediately forces the columns
with the same mixed indices to be equal. Further, because {1, x, y, z} is a basis for E, we can
find relations
x2 = axx+ bxxx + cxxy + dxxz
y2 = ayy+ byyx + cyyy + dyyz
z2 = azz+ bzzx + czzy + dzzz
xy = axy+ bxyx + cxyy + dxyz
xz = axz+ bxzx + cxzy + dxzz
yz = ayz+ byzx + cyzy + dyzz,

with coefficients in k, which indeed correspond to the actions of x, y and z on E and therefore also with the entries in the matrices. These identities are also respected by the matrices (the steps of which are a bit tedious to write out, but simple to verify). For example, we have that X2 = axxI + bxxX + cxxY + dxxZ, which is most easily checked by noting that both sides act

in the same way on our basis elements. This fact, however, is dependent on the commutativity

of the set of endomorphisms. ♦

We can now use these matrix representation of elements of the Hilb functor to check its smoothness in some cases. We begin by stating a necessary and sufficient condition for the more 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 functor
Quotn_{F /A/k} is then smooth if and only if for every k-algebra B and every nilpotent ideal N ⊆ B
such that N2 _{= (0), any element E ∈ Quot}n

F /A/k(B/N ) admits a lift E ∈ Quot n

F /A/k(B), such

that E reduces to E modulo N .

Proof. This is basically just a restatement of the definition. Suppose the diagram
Quotn_{F /A/k}
b
ψ
h_{B/N}
hB
?
ψ

commutes. Then the composition h_{B/N} → h_{B} → Qn _{equals the map h}

B/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 then

corresponds to the lift of the given element in Qn(B/N ).

5 THE QUOT AND HILB FUNCTORS 5.3 Smoothness of the Hilb Functor
Consider now an element E = B[x1, x2, . . . , xm]/I in Hilbn_{k[x}_{1}_{,...,x}_{m}_{]/k}(B) with B = B/N

for some nilpotent N such that N2 = (0) and some k-algebra B, admitting a lift to E in
Hilbn_{k[x}

1,...,xm]/k(B). We then have the diagram

B[x1, . . . , xm] - End_{B}(E) ev
e
- E
B[x1, . . . , xm]
?
?
- End_{B}(E)
?
? _{ev}
e
- E,
?
?

where the horizontal compositions are surjective, for some e ∈ E, showing that finding a lift for E 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 Hilbn_{k[x]/k} is formally smooth.

Proof. A representation of any element in Hilbn_{k[x]/k}(A) for any k-algebra A is given by matrices
X 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 ∈ Hilbn_{k[x]/k}(B/N ) to a representation {Xi} of E ∈ Hilbn

k[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, Hilbn_{k[x]/k} is representable by k[t1, . . . , tn],

the polynomial ring in n variables. This proposition then tells us that this polynomial ring is non-singular, which is as we would expect.

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

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

Example 5.3. Consider Hilb3_{k[x,y]/k}(B/N ) with B = k[ε]/(ε3_{) and N = (ε}2_{). 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 0 1 0 0 0 ε 0 , Y = 0 0 0 0 0 ε 1 0 0 ,

for which we have

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, lifting the matrices to X = 0 0 ε2 1 0 0 0 ε 0 , Y = 0 ε2 0 0 0 ε 1 0 0 , we have that XY = Y X = ε2 0 0 0 ε2 0 0 0 ε2 ,

rendering a commutative representation. ♦

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

Proposition 5.6. The functor Hilb4_{k[x,y,z]/k} is not formally smooth.
Proof. Consider again B = k[ε]/(ε3_{) and N = (ε}2_{). We then have that}

(B/N )[x, y, z]/(xy, xz, yz, x2, y2− εz, z2− εx) is represented by X = 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 , Y = 0 0 0 0 0 0 0 0 1 0 0 0 0 0 ε 0 , Z = 0 0 0 0 0 0 0 ε 0 0 0 0 1 0 0 0 ,

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

ZY = 0 0 0 0 0 0 ε2 0 0 0 0 0 0 0 0 0 ,

showing that the representation is commutative over B/N , but not over B. If Y and Z are lifts 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 0 c24− d23 0 −1 0 c34+ d11− d33 d12 d13 d14 c44− c11− d43 −c12 −c13 −c14 ,

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

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 =Lr

A, and specifically take a closer look at the case r = 2. Again, finding a mapping A ⊗kB → EndB(E) corresponding to the action of

A ⊗kB on E grants us an A ⊗kB-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 ∈ Quot2_{F /A/k}(B) for some k-algebra
B, we have that

E = (B[x] ⊕ B[x])/N =loc

2

M B

for some N ⊆ B[x] ⊕ B[x]. Suppose further that {(1, 0), (0, 1)} forms a basis for E as a free B-module. We then have that (x, 0) = a1(1, 0) + a2(0, 1) and (0, x) = b1(1, 0) + b2(0, 1) for

some ai, bi∈ B. Considering the action of x on the basis elements, we can again write this as a

matrix

X =a1 b1 a2 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 those of elements of the Quot functor is that, in general, representations of elements of Hilb have matrices with very simple first columns, corresponding to the actions of xi on 1. The Quot

functor, on the other hand, takes into account the module F , and as such there is generally no multiplicative identity. As such, the first column of the matrices no longer have this simple structure.

Example 5.5. Let again A = k[x] and F = A ⊕ A. Suppose further that an element E of
Quot4_{F /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 b1 1 a2 0 b2 0 a3 0 b3 0 a4 1 b4 ,

5.4 Smoothness of the Quot Functor 5 THE QUOT AND HILB FUNCTORS Proposition 5.7. Let A = k[x] and F = Lr

A. Then the functor Quotn_{F /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 an
element in Quotn_{F /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 that
it is also surjective.

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

Example 5.6. Let A = k[x, y] and F = A ⊕ A. Suppose E = (F ⊗kB)/N is an element of

Quot2_{F /A/k}(B) with basis {e1, e2} for some k-algebra B. We can then describe the actions of x

and y on E with the matrices

X =a b c d , Y =e f g h , where the composite map F ⊗k B →

L2

EndB(E) → E is surjective, and the two matrices

commute. The fact that the two matrices commute give rise to three equations in the variables a, . . . , h, as seen by [X, Y ] = XY − Y X = bg − cf af + bh − be − df ce + dg − ag − ch cf − bg . Rearranging, we get the three equations

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

showing that as long as the mapping is surjective, bg = cf , a = d and e = h, this induces a

k[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 and bg = cf = 0 in the given ring. We will now expand on this example, to get a second case of a non-smooth functor.

Proposition 5.8. With A = k[x, y] and F = A ⊕ A, the functor Quot2_{F /A/k} is non-smooth.
Proof. Let B = k[ε]/(ε3) and N = (ε2), and consider the endomorphisms

X =1 0 ε 1 , Y =0 ε 0 0

in End(L2_{B/N ). As seen in Example 3.5, these matrices commute, and the evaluation in e}
1

and e2 is surjective ontoL2B/N , giving us the quotient

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 of the lifts to B/N must be X and Y , they take the form

X = X + ε2c11 c12 c21 c22 = X + ε2C, Y = Y + ε2d11 d12 d21 d22 = Y + ε2D,

whereby we have that XY − Y X = XY − Y X + ε2(XD + CY − DX − Y C). However, we also
have that
ε2X =ε
2 _{0}
ε3 ε2
= ε2I
and
ε2Y =0 ε
3
0 0
= 0

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

XY − Y X = ε21 0

0 −1

6= 0

in B, no lift X and Y will commute, and so the given quotient has no lift. Further, by direct computation 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 ε}3_{.}

To conclude, we have found concrete examples to show that the functors Hilb4_{k[x,y,z]/k} and
Quot2L2_{k[x,y]/k[x,y]/k} are non-smooth. Using more sophisticated arguments, one can prove the

following, more general, theorem.

Theorem 5.1. The functor Quotn_{F/A/k} is smooth if
• A = k[x],

• A = k[x, y] and F = A, or

• A = k[x_{1}, . . . , 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 observation that a basis for a module of rank 3 is of the form {1, xi, x2i} or {1, xi, xj}, bringing us back to

one 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

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 at some point P .

The affine variety C ⊗kk[u1, u2] corresponds to pairs of commuting matrices and an element

u = (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 open subset U1 ⊂ Spec(C[u1, u2]). One can further check that the set of invertible 2 × 2 matrices

GL2 acts freely on U1, and that U1/GL2 gives an open subset of Hilb2k[x,y]/k. Since Hilb2k[x,y]/k

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

singularities of C.

The Quot scheme Quot2_{F /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 subset U2 ⊂ Spec(C[u1, u2, v1, v2]). Again, the group GL2 acts freely on U2, and U2/GL2 is an open

subset of Quot2_{F /k[x,y]/k}. Since this scheme is singular, this implies that U2 does not avoid the

singularities of C.

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 th´eorie des modules. Annales scientifiques

de l’ ´E.N.S., 80(3):213–348, 1963.

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