Formal Moduli Spaces for Modules over Planar Noncommutative Quadrics
by
Magne Agnalt Myhren
THESIS for the degree of
MASTER OF EDUCATIONAL SCIENCE
(Master i realfagsutdanning)Faculty of Mathematics and Natural Sciences University of Oslo
May 30, 2011
Contents
Preface 5
1. Introduction 7
2. Hochschild Cohomology 9
3. Deformation Theory 11
3.1. Properties of formal deformations 12
4. Ext 13
4.1. Derivations, Extensions and Ext 13
4.2. The Ext-group defined by resolutions 15
4.3. Ext1A(M, N) and ExtA(M, N) 17
5. Moduli Spaces and Simplifyable Modules 21
5.1. Moduli Spaces 21
5.2. Simplifyable Modules 22
6. Calculating Local Moduli Spaces of Noncommutative Rings 23
6.1. A commutative example 23
6.2. Noncommutative partial differentiation 24
6.3. Example: F=xy 26
6.4. Example: F=x2+y2−λ2 27
6.5. For which pairsM andN of one dimensionalA-modules,
will Ext1A(M, N)̸= 0? 29
References 31
Preface
Given an associative k-algebra A for a field k, we want to give a local description of the moduli space of indecomposable leftA-modules. The tool for this will be deformation theory and homological algebra. The goal of this thesis is to calculate formal moduli spaces of 2- dimensional indecomposable modules generated by two elements with a quadratic relation.
We will also draw some conclusions about simplifyability of modules, via calculations of cycles in the induced extension graph.
I would like to thank my supervisor Arne B. Sletsjøe for his enthusiastic guidance and advice through the process of writing this thesis. I will also thank Fredrik Meyer, Christine Furuseth, Christian Agrell, Johanna Furuholmen and my family for their help and support.
1. Introduction
A general goal of mathematical research is classification of objects [3]. Given a certain class of objects, we want to classify some objects within that class as being more similar in a particular way. This work often reveals hidden structures and relations of the studied objects, which again deepens our understanding of the subject.
When studying modules of rings in modern abstract algebra, it is a natural question to ask whether there is a convenient way of classifying the set of all modules belonging to a ringA.
A way of doing this is to construct a space called themoduli space of A. This space can be thought of as a parameter set, where each point in the space corresponds to an equivalence class of indecomposable A-modules. The space itself often has an interesting geometrical structure, but unfortunately it can be very difficult and sometimes impossible to give a global characterization of it [6]. A usual approach is therefore to use the theory of deformations in a local study as an attempt to obtain information on the moduli space. In many ways, deformation theory is the calculus of moduli spaces describing all of its small perturbations [3].
A way to familiarize with these abstract notions is to exemplify theory by studying one type of ring and do calculations on it. This thesis reflects my attempt at this technique while approaching unfamiliar and abstract theory. The thesis is divided into two parts, one part presenting theory and the other exemplifying it by calculation.
We begin inSection 2by introducing the Hochschild cohomology induced by the Hochschild differential operator. The cup product and the commutator bracket will also be defined. In Section 3 we will present the theory of deformations and deduce some properties of R- deformations. Both Section 2 and 3 are based on the work of Christoff Geiss in [2]. Section 4is a discussion of different ways of defining the Ext-group, and that these definitions produce the same result. This section is a study of theory presented in [7]. We continue inSection 5with a basic introduction to moduli spaces and simplifyable modules based on [5]. Here we will discuss a geometrical condition satisfied when dealing with simplifyable modules. Sec- tion 6contains various examples illustrating calculations of local moduli spaces in the affine noncommutative plane. We will also present the noncommutative partial differentiation [4], and how to use it to produce projective resolutions of free modules. An example involving the geometrical condition for simplifiable modules is displayed, before ending the thesis with a discussion of when there is a non-zero Ext between two modules over a general planar non- commutative quadric. Much of the work done in this thesis is inspired by the private lectures given by Arne B. Sletsjøe [6], in the Spring of 2011.
Notation. Letk be a closed field andAa Noetherian associative unitary k-algebra with underlying vector spacekn and ring multiplicationα∈Homk(A⊗kA, A). Let alsoM be an A-module with module operationµ∈Homk(A⊗kM, M). WhenAis a polynomial ring with two indeterminatesx, yand M =k withµdefined as x7→aand y7→b, we denote M with µasM =k(a, b).
2. Hochschild Cohomology
Given a k-algebra A over a field k with multiplication α ∈ Homk(A⊗kA, A), and an A-bimoduleM with multiplicationµ∈Homk(A⊗kM, M). We will often writeα(a, b) :=ab andµ(a, m) :=amfora, b∈Aandm∈M. Define
A⊗n :=A⊗kA⊗k· · · ⊗kA
| {z }
n
and Cn := Homk(A⊗n, M)
with thecomposition product
◦:Cn×Cm→C(n+m−1) β◦γ=
∑n i=1
(−1)miβ(a0,· · ·, ai−2, γ(ai−1, ai,· · · , ai+m−1), ai+m,· · ·, an+m−1).
The gradedcommutator bracket is defined via this composition product by [β, γ] :=β◦γ−(−1)|β||γ|γ◦β
for homogenous elementsβ and γ. The degree of β ∈Cn is|β|=n−1. An immediate and useful result is the following lemma
Lemma 2.1. The multiplicationαofA is associative if and only ifα◦α= 0.
Proof. α◦αtakes 2 + 2−1 = 3 elements ofA. So leta, b, c∈A, then (α◦α)(a, b, c) =−α(a, α(b, c)) +α(α(a, b), c)
=−a(bc) + (ab)c
which equals 0 if and only ifαis associative.
Definition 2.1. The Hochschild differential δn:Cn→Cn+1 with respect toαis defined by (δnϕ)(a0, . . . , an) =a0ϕ(a1, . . . , an) +
∑n i=1
(−1)iϕ(a0,· · ·, ai−1ai,· · · , an) + (−1)n+1ϕ(a0, . . . , an−1)an.
This expression is rather complicated, but we can simplify it by using the bracket notation.
If we force the last term to have a negative sign, we can combine the first and the last term to beα◦ϕ. The big middle sum then equalsϕ◦α, so combined we get
(δnϕ)(a0, . . . , an) =a0ϕ(a1, . . . , an)−ϕ(a0, . . . , an−1)an
−(−1)|ϕ|
∑n i=1
(−1)iϕ(a0,· · ·, ai−1ai,· · · , an)
=α◦ϕ−(−1)|ϕ|ϕ◦α
= [α, ϕ].
In this thesis we will only considergraded Lie algebras in which the Jacobi identity applies.
We will state this as a lemma, even though it is considered an identity.
Lemma 2.2. The Jacobi identity on graded Lie algebras is as follows [α,[β, γ]] = [[α, β], γ] + (−1)|α||β|[β,[α, γ]]
From the Jacobi identity we immediately get the following result.
Lemma 2.3. Ifαis associative and char(k)̸= 2, thenδi+1δi= 0 for any i∈Z+.
Proof. αis of degree 1, so the Jacobi identity yields δi+1δi(ϕ) =δi+1(δi(ϕ)) = [α,[α, ϕ]]
= [[α, α], ϕ] + (−1)1[α,[α, ϕ]] = [0, ϕ]−[α,[α, ϕ]] =−[α[α, ϕ]]
⇒2[α,[α, ϕ]] = 0
⇒δi+1δi(ϕ) = 0
Sinceδi+1δi = 0,Cn can be characterized as cochains in a complex, denoted theHochschild complex. Set
Zi := Ker(δi) Bi := Im(δi−1),
Define the cohomology of the Hochschild complex is as follows:
Definition 2.2. The Hochschild cohomology is defined as HHi=Zi/
Bi
To further simplify notation we will later make use of thecup product. Let ψ1, ψ2∈Homk(A⊗M, M) anda, b∈A, then the cup product is defined by
(ψ1⌣ ψ2)(a, b) =ψ1
(a, ψ2(b,−)) ,
Given these four elements, the cup product (ψ1⌣ ψ2)(a, b) is a homomorphism fromM to M.
3. Deformation Theory
Let R be in the categoryℓ of punctured Artin rings 1. An Artin ring R is punctured if there is a morphismπ:R→ksuch that the inclusion morphismk ,→Rcomposed withπis the identity onk. A morphismψ∈Mor(ℓ) of punctured rings is ak-algebra mapψ:R→S such that the following diagram commutes:
R ψ //
π
S
π′
k
FixingAand M, the deformation functor DefM :ℓ→sets is defined as DefM(R) ={
MR∈A⊗R-mod|MR⊗πk∼=M andMR is flat overR}/
∼, whereMR∼MR′ if there exists a commutative diagram:
MR
∼= //
(−⊗πk)
MR′
(−⊗π′k)
||zzzzzzzz
M
Examples of rings satisfying these conditions are thetruncated polynomial ringsR=k[t]/(tn) forn≥2. For n= 2 we writeR =k[ϵ]. In both cases the puncture morphism π is simply t7→0 and the identity onM.
Definition 3.1. Let R∈ℓ. An R-deformation ofµ is an element µR∈HomR((A⊗kR)⊗R(M⊗kR),(M⊗kR)) which reduces modulo R toµand is associative, i.e.
(3.1) µR
(a⊗Rµ(b, m))
=µR
(α(a, b)⊗Rm)
fora, b∈A, m∈M
IfR=k[ϵ] we say thatµR is aninfinitesimal deformation. WhenAandM are given, the goal of deformation theory is to classify the set of allR-deformations of theA-moduleM. If we take the inverse limit of a sequence of the Artin ringsk[t]/(tn), we get theformal polyno- mial ringR=k[[t]] yielding aformal deformation. Ris then a pro-Artin ring containing the information of all the truncated rings, and is hence the classification we seek.
We say that an infinitesimal deformation µk[ϵ] of µ is integrable if there exists a formal deformationµk[[t]] of µ which reduces via the projection pt,ϵ :k[[t]]→k[ϵ] toµk[ϵ]. We will come back to integrable deformations and in which circumstances they occur in section 5.
Let R=k[t]/(tn+1),S =k[t]/(tn), MS ∈DefM(S) a deformation and a mapψ:R S.
We say thatψis asmall surjection morphisminℓ. In deformation theory we need an existence of a lifting fromStoR. The following lemma gives us this lifting, however with anobstruction telling us where the lifting doesn’t behave as wanted.
Lemma 3.1. GivenR,S,MS andψ as above, there exists a canonical obstruction o(ψ, MS)∈Ext2A(M, M)
such that o(ψ, MS) = 0if and only if there exists a deformationMR∈DefM(R)lifting MS. If this is the case, the set of deformations in DefM(R)is in a bijective correspondance with thek-vector space Ext1A(M, M).
Proof. See [1, prop. 5.1].
1Not necessarily commutative.
We have not yet discussed the Ext-functor, but we will keep in mind that the obstructions are located in thek-vector space Ext2A(M, M).
3.1. Properties of formal deformations.
As the formal deformations classifies all R-deformations withR=k[t]/(tn), we can study properties of the formal deformations and treat the rest as special cases. LetR = k[[t]] ∼= k⊕kt⊕kt2⊕ · · · andµRbe anR-deformation. Then
µR:(
A⊗kk[[t]])
⊗R
(M⊗kk[[t]])
→(
M⊗kk[[t]]) and byR-linearity we get:
µR(a, m) =am+tψ1(a, m) +t2ψ2(a, m) +· · · (3.2)
for a family ofk-linear mapsψi: Homk(A⊗M, M). This indicates thatµRis determined by {ψi}and µ. Because we know thatµR satisfies condition (3.1), we can use this and (3.2) to deduce some properties of{ψi}. Calculating the left side of (3.1), we get
µR(a, µR(b, m))
=aµR(b, m) +tψ1(a, µR(b, m)) +t2ψ2(a, µR(b, m)) +· · ·
=a(bm) +taψ1(b, m) +t2aψ2(b, m) +t3aψ3(b, m) +· · · +tψ1(a, bm) +t2ψ1(a, ψ1(b, m)) +t3ψ1(a, ψ2(b, m) +· · · +t2ψ2(a, bm) +t3ψ2(a, ψ1(b, m)) +t4ψ2(a, ψ2(b, m)) +· · · , whereas the right side yields
µR(α(a, b), m) =µR(ab, m)
= (ab)m+tψ1(ab, m) +t2ψ2(ab, m) +t3ψ3(ab, m) +· · · .
Sorting the terms by the degree oft, we get a set of equations describing the properties of{ψi}:
• (ab)m=a(bm)
• aψ1(b, m)−ψ1(ab, m) +ψ1(a, bm) = 0
• aψ2(b, m)−ψ2(ab, m) +ψ2(a, bm) =−ψ1(a, ψ1(b, m))
• aψ3(b, m)−ψ3(ab, m) +ψ3(a, bm) =−ψ1(a, ψ2(b, m))−ψ2(a, ψ1(b, m))
... ...
By omittingm, using the Hochshild differential and the previously defined cup product, we can simplify these equations as follows:
• (ab)m=a(bm)
• δ1ψ1(a, b) = 0
• δ1ψ2(a, b) =−(ψ1⌣ ψ1)(a, b)
• δ1ψ3(a, b) =−(ψ1⌣ ψ2)(a, b)−(ψ2⌣ ψ1)(a, b)
... ...
• δ1ψn(a, b) =−∑
i+j=nψi⌣ ψj(a, b)
... ...
This means that we should be able to findµR by successively calculatingψi and using these formulae. However, we must find out whether there are obstructions to doing this in Ext2A or not. Even if we know that these relations exists, no procedure for finding{ψi} is appar- ent. First of all, there are infinitely many terms to calculate, and second it is sometimes very difficult or even impossible to do so. We can simplify the first problem by studying the infinitesimal deformationsR=k[ϵ], and using the lifting property to proceed with studying
R = k[t]/(t3) and so on. In many ways, this procedure resembles Taylor expansion with higher degree of accuracy the higher power oft we calculate. We shall later see that if there are no obstructions in the liftings(
Ext2A(M, M)̸= 0)
, every infinitesimal deformation is inte- grable. This basically means that it suffices to calculate the infinitesimal deformation to get all information about the tangent space ofM.
In the next chapter we will introduce Ext to provide us tools to give more thorough answers to these questions.
4. Ext
In this section we will look at various ways of defining the Ext-group and its relation to the deformation of a module. The Ext-group, ExtA(M, N), was originally defined as the equivalence classes ofextensions of M by N [6].
4.1. Derivations, Extensions and Ext.
Definition 4.1. An extensionξ of M byN forA-modules M andN, is an exact sequence 0 → N → E → M → 0. Two extensions ξ and ξ′ are equivalent if there is a commuting diagram ofA-module homomorphisms:
ξ 0 //N f //E
∼=
g //M //0
ξ′ 0 //N //E′ //M //0 Finally,ξis a split extension ifE=N⊕M as A-modules.
So an element of Ext is the equivalence class of an extension. Ext emerges in many areas of homological algebra, and it can be difficult to understand what this concept really means.
To expand our understanding of Ext and what it really is, we shall in this subsection derive properties of an extension and its equivalence class.
Another way to define Ext is byderivations. The set of derivations ofAis defined as:
Der(
A,Homk(M, N)) :={
D:A→A∈Homk(M, N)|D(ab)(m) =aD(b)(m) +D(a)b(m)} , for elementsa, b∈Aandm∈M.
Letξbe an extension ofM byN withA-module homomorphismsf andg as in the above diagram. As f is injective and g is surjective, it will be convenient to understand E as a direct sumN⊕M with some equivalence relation identifying suitable elements. Becausef is injective andg is surjective, we writef(n) = (n,0) andg(n, m) =mforn∈N andm∈M. We use theA-module properties ofEand theA-module homomorphism properties off and gto explore the structure ofξ:
(1) f(an) =af(n) fora∈A andn∈N, (2) g(a(n, m)) =ag(n, m) form∈M, which implies
(1) (an,0) =a(n,0)∈N⊕M
(2) g(a(n, m)) =g(n′, m′) =m′ andag(n, m) =am
⇒am=m′
By using the above identities we can now calculate a(n, m) =a(
(n,0) + (0, m))
=a(n,0) +a(0, m) = (an,0) + (γ(a, m), am)
=(
an+γ(a, m), am) ,
whereγ(a, m)∈N depending onaand m. Another requirement from the module structure ofEis:
(ab)(n, m) =a(b(n, m))
⇒(abn+γ(ab, m), abm) = (abn+aγ(b, m) +γ(a, bm), abm)
⇒abn+γ(ab, m) =abn+aγ(b, m) +γ(a, bm)
⇒γ(ab, m) =aγ(b, m) +γ(a, bm) By observing thatγ is behaving as a derivation, we define
D:A→Homk(M, N) by lettingD(a)(m) =γ(a, m) and observe that
D(ab)(m) =aD(b)(m) +D(a)b(m)
⇒D∈Der(
A,Homk(M, N))
By investigating the properties of ξ, we found that there is a derivation belonging to ξ.
There are many types of derivations. If we take an homomorphismβ ∈ Homk(M, N), we can construct aninner derivation by the commutator [−, β]. The set of inner derivations is defined as
Inderk(A,Homk(M, N)) :={Dβ∈Der(A)|Dβ= [−, β], β∈Homk(M, N)}. It is easy to check that an inner derivationDβ is in fact a derivation by:
Dβ(ab) = [ab, β] =abβ−βab=abβ+ (aβb−aβb)−βab
=a(bβ−βb) + (aβ−βa)b=a[b, β] + [a, β]b
=aDβ(b) +Dβ(a)b.
In Ext, we know that some extensions are equivalent, so to get a definition for Ext by derivations we must find a type of derivation yielding the same equivalence. Consider again the commuting diagram of two equivalent extensionsξ andξ′:
0 //N f //E
ψ ∼=
g //M //0
0 //N //E′ //M //0 from which we can deduce the following relations:
g(ψ(0, m)) =g(0, m) =m
ψ(0, m) = (β(m), m) for someβ ∈Homk(M, N) ψ(n,0) =ψ(f(n)) =f(n) = (n,0)
⇒ψ(n, m) =ψ(n,0) +ψ(0, m) =ψ(f(n)) +ψ(0, m)
=f(n) +ψ(0, m) = (n,0) +ψ(0, m)
= (n,0) + (β(m), m) = (n+β(m), m).
These identities,D and the homomorphism property ofψcombined yields ψ(a(0, m)) =aψ(0, m)
ψ(
D(a)(m), am)
=a(
β(m), m) (D(a)(m) +β(am), am)
=(
aβ(m) +D′(a)(m), am)
whereD′(a)(m) :=D(a)( β(m))
. This implies that
D(a)(m) +β(am) =aβ(m) +D′(a)(m)
⇒D(a)(m)−D′(a)(m) =aβ(m)−β(am) Using the bracket notation introduced in the previous chapter, we get
⇒D(a)−D′(a) =aβ−βa= [a, β]
⇒D−D′= [−, β] =Dβ∈Inderk(A,Homk(M, N)).
By studying the properties of two equivalent extensions, we found thatξ′ has an inner deriva- tion. We can now give a definition of the Ext-group ofM byN in terms of derivations:
ExtA(M, N)) = Derk(A,Homk(M, N))/
Inderk(A,Homk(M, N)).
Later work in this field has shown that the Ext-group also emerges in other areas, giving rise to other more general defeninitions [6]. In the next subchapter we will have a look at these.
4.2. The Ext-group defined by resolutions.
An algebraic resolution is a way of describing the structure of a module. There are two im- portant types of resolutions we use to define theExt-functor, namely projective and injective resolutions.
Definition 4.2. LetP be anA-module. We say thatP is projective if for every mapP →M and surjectionE→M there exists at least one map P →E such that the following diagram commutes:
P
∃
M oooo E
Definition 4.3. A projective resolution of anA-moduleM is an exact seqence 0oooo M oo P0oo P1 oo P2oo · · ·
where eachPi is a projectiveA-module for everyi∈N.
Applying the functor HomA(−, N) to a projective resolution and we obtain the following chain complex
HomA(P0, N) d
0 //HomA(P1, N) d
1 //HomA(P2, N) //· · ·
Definition 4.4. The Ext-functor of projective resolutions is the cohomology of the previously defined chain complex. This means explicitly
ExtiA(M, N)P =Ker(di)/Im(di−1)fori≥1
A convenient alternative way of viewing projective resolutions is apparent in the following lemma:
Lemma 4.1. An A-module is projective if and only if it is a direct summand of a free A- module.
Proof. See [7, page 33].
As a dual point of view, we will also define Ext via injective resolutions .
Definition 4.5. Let I be anA-module. We say that I is injective if for every mapN →I and injectionN ,→E there exists at least one map E →I such that the following diagram commutes:
I
N //
OO
E
∃
``
Definition 4.6. An injective resolution of an A-moduleN is an exact seqence 0 //N ////I0 //I1 //I2 //...
where eachIi is an injective A-module for every i≥0.
Use the functor HomA(M,−) on an injective resolution and we obtain the following chain complex
HomA(M, I0) d0 //HomA(M, I1) d1 //HomA(M, I2) //· · · This leads us to another definition of Ext:
Definition 4.7. The Ext-functor of injective resolutions is the homology of this chain com- plex. That is
ExtiA(M, N)I =Ker(di)/Im(di−1)fori≥1
Note that the Ext-functor yields, not only one Ext-group as before, but several numbered from 1 to possibly infinitely many depending on the given resolution of M or N. In the examples provided in this thesis we only look at Noetherian rings yielding finite length of their resolutions [7, Prop. 4.1.5]. We shall later see that this way of defining the Ext-functor yields Ext1A(M, N)∼= ExtA(M, N).
The two ways of defining the Ext-functor look very similar, and we shall prove that they in fact produce isomorphic results. Given an extension
ξ: 0→N→E→M →0,
a projective resolution ofM and an injective resolution ofN, consider the following diagram obtained by successively applying the functors HomA(M,−), HomA(P0,−), HomA(P1,−),
· · · on the injective resolution ofN. The result is a commutative diagram:
... ... ...
0 //HomA(M, I2)
OO
s0 //HomA(P0, I2)
OO
s1 //HomA(P1, I2)
OO //
. ..
0 //HomA(M, I1)
d1
OO
r0 //HomA(P0, I1)
j2
OO
r1 //HomA(P1, I1)
OO //HomA(P2, I1)
OO //· · ·
0 //HomA(M, I0)
d0
OO
q0 //HomA(P0, I0)
j1
OO
q1 //HomA(P1, I0)
i1
OO
q2 //HomA(P2, I0)
OO //· · ·
0 //HomA(M, N)
OO //HomA(P0, N)
j0
OO
d0 //HomA(P1, N)
i0
OO
d1 //HomA(P2, N)
k0
OO //· · ·
0
OO
0
OO
0
OO
0
OO
In the diagram above, every row except the bottom one, and every column except the left one, is exact. This is true becauseIi andPj respectively are injective and projective.
Lemma 4.2.Given twoA-modulesM andN and an extension ofM byN, then Ker(d1)/Im(d0)∼= Ker(d1)/Im(d0)implying that Ext1A(M, N)P ∼=Ext1A(M, N)I.
Proof. Considering the above diagram, our goal is to find a well defined morphism Ker(d1)→ Ker(d1) and Im(d0)→Im(d0). Let x∈Ker(d1). Then
(k0◦d1)(x) = 0
⇒(q2◦i0)(x) = 0
⇒i0(x)∈Ker(q2) = Im(q1)
⇒∃y∈HomA(P0, I0) :q1(y) =i0(x) Furthermore we have
(i1◦i0)(x) = 0
⇒j1(y)∈Ker(r1) = Im(r0)
⇒∃z∈HomA(M, I1) :r0(z) =j1(y)
Also (j2◦j1)(y) = 0 which implies that (s0◦d1)(z) = (j2◦r0)(z) = (j2◦j1)(y) = 0. Buts0
is injective, sof1(z) = 0. We’ve now got a well defined morphism Ker(d1)→Ker(d1).
Let nowx∈Im(d0) andd0(y) =x. Because both j0 andi0are injective we write i0(x) =x and j0(y) = y. From the diagram we see that q1(y) = x. Let z be another element with q1(z) =x(Remark: If no suchz exists, q1 is injective and HomA(M, I0) = 0, collapsing the diagram). Nowz−yis an element satisfying
q1(z−y) =q1(z)−q1(y) =x−x= 0
⇒(z−y)∈Ker(q1) = Im(q0).
The mapq0 is injective so again we simplify by denotingq0(z′′) =z−y. From the diagram we see thatj1(y) = (j1◦j0)(y) = 0, which indicates
(r0◦d0)(z′′) =j1(z′′) =j1(z)−0 =j1(z) :=z′.
The fact thatr0 is injective implies f0(z−y) = z′ ∈ Im(d0). Now we have a well defined mapping from Im(d0) to Im(d0). This mapping is obviously an isomorphism so we can now conclude
Ext1A(M, N) = Ker(d1)/Im(d0)∼= Ker(d1)/Im(d0)
Although we only treat Ext1A(M, N) in this lemma, a more abstract approach will prove the same result for ExtiA(M, N) fori≥1 [7, Thm. 2.7.6].
4.3. Ext1A(M, N) and ExtA(M, N).
In this section we will prove that there is a one-to-one correspondence between the equiv- alence classes of extensions ofN byM and elements in Ext1A(M, N) defined by resolutions.
To do this we need a couple of definitions and lemmata.
Definition 4.8. The pushout of two A-module homomorphisms f and g with a common domain Z consists of an A-module P and two A-module homomorphisms i1 : X → P and i2:Y →P for which the following diagram commutes:
Z
g
f //X
i1
Y i2 //P
Moreover the diagram is universal such that if there exists another A-moduleQand A-module homomorphismsj1andj2for which the diagram also commutes, there exist a unique A-module
homomorphismhsuch that the following diagram commutes:
Z
g
f //X
j1
i1
Y i2 //
j2 ++
P
∃!h
Q
The definition of a pushout can also be generalized to other categories. For example, a pushout in set theory is the union of two sets. IfZ =X∩Y andf andg are inclusions, then P=X∪Y with the equivalence relationx∼y ifx∈Z andy∈Z. It is convenient to think of the pushout P as a sum of X and Y divided out by the equivalence relation ∼. For the main result of this section we will need the next lemma:
Lemma 4.3. For a resolution0→K→P0→M →0 ofA-modules whereP0 is projective, and for everyA-moduleN we have the following exact sequence:
0→HomA(M, N)→HomA(P0, N)→HomA(K, N)→Ext1A(M, N)→0
Proof. Consider the following diagram where P → M is a projective resolution of M and every straight sequence is exact:
0
@
@@
@@
@@
@ 0
!!B
BB BB BB
B 0
K2
B
BB BB BB
B K
OO BBBBBBBB
... //P2BBBBBBBB// P1 //
OO
P0
A
AA AA AA
A //M //0
K1
OO !!BBBBBBBB M
?
??
??
??
? 0
OO
0 0
Applying HomA(−, N) to this diagram, we get
0
0 //HomA(M, N) //HomA(P0, N) p //
d0
((Q
QQ QQ QQ QQ QQ
QQ HomA(K, N) ∂ //
q1
Ext1A(M, N) //0
HomA(P1, N)
rmmmmmm6666 mm mm mm
q2
d1
((Q
QQ QQ QQ QQ QQ
Q oo ? _Ker(dOOOO 1)
0 //HomA(K1, N) q3 //HomA(P2, N)
where the upper horizontal sequence arises from the upper right diagonal sequence of the first diagram.
The kernel of d1 maps injectively to HomA(P1, N) and surjectively to Ker(d1)/Im(d0) = Ext1A(M, N).
Letz∈Ext1A(M, N). Thenz∈Ker(d1)⇒d1(z) = 0. Becauseq3 is injective, it follows that q2(z) = 0 implyingz∈Im(q1) and∃x∈HomA(K, N) such thatq1(x) =z.
Becausez∈Ker(d1) = Im(d0) there∃y∈HomA(P0, N) such thatd0(y) =z.
⇒(q1◦p)(y) =z⇒p(y) =x(becauseq1 is injective)
⇒x∈Im(p)
Letr: HomAP1, N)→Ext(M, N) which must be surjective, and let ∂=r◦q1. We are now left to show that∂(x) = 0. Since Ext1A(M, N) = Ker(d1)/Im(d0), we see that
∂(x) =∂(p(y)) =r (
q1( p(y)))
=r(d0(y)) = 0
⇒Im(p) = Ker(∂)
which proves the desired result.
This lemma tells us that Ext1A(M, N) gives us a ”measure” of how Hom(−, N) fails to be exact. Keeping in mind the original notion of the Ext-group, the next lemma is of importance.
Lemma 4.4. Ext1A(M, N) = 0if and only if every extension of M by N is a split extension.
Proof. Let ξ denote an extension sequence 0 → N → E −→q M → 0. By applying the left exact functor HomA(M,−) onξ, we get the following exact sequence:
HomA(M, E)−→q∗ HomA(M, M)−→∂ Ext1(M, N)
Ifξ splits (E =N⊕M), then there exists aσ∈ HomA(M, E) such that q∗(σ) = idM. By exactness it follows that∂(idM) = 0, which means that Ext1(M, N) = 0.
Conversely, if Ext1(M, N) = 0, then q∗ is surjective. This means that there exists a σ ∈ HomA(M, E) such that q∗(σ) = idM. Because the image of N is mapped to 0 by q, Im(σ)∼=M ⇒ξ splits.
ξ: 0 //N //E q //M
|| σ //0
We now define Θ :{ξ} →∂(idM) and restate the lemma as follows:
ξsplits ⇐⇒ Θ(ξ) = 0 in Ext1A(M, N).
Theorem 4.1. There is a one-to-one correspondance between the equivalence classes of ex- tensions of M by N and Ext1A(M, N) via the mapping ofΘ.
Proof. Choose a truncated projective resolution of M; 0 → K −→j P → M → 0. We use lemma (4.3) to produce the following exact sequence
HomA(P, N)→HomA(K, N)−→∂ Ext1(M, N)→0
For an x∈Ext1A(M, N) we can now chooseβx ∈HomA(K, N) withβx 7→x. Let (E, σ, f) be the pushout ofβx and j such that k∈ K 7→ (−βx(k), j(k))∈E, then we can form the following diagram:
0 //K
βx
j //P
σ
//M //0
ξ: 0 //N f //E g //M //0
where g is induced by N −→0 M and the map P → M. We are now left to prove that ξ is exact. Letn∈Ker(f), then:
f(n) = (n,0) = (0,0)
⇒∃y∈K withβx(y) =nandj(y) = 0
⇒j injective givesy= 0 andn=βx(0) = 0
⇒f is injective.
Furthermore, (n, p)∈Ker(g)⇒p=j(y) for somey∈K, which means that (n, p) = (n, p) + (0,0) =(
n, j(y)) +(
βx(y),−(j(y))
=(
n+βx(y),0)
⇒Ker(g)⊆Im(f) Let (n, p)∈Im(f),
⇒∃n′∈N such thatf(n′) = (n, p) = (n′,0)
⇒g(n, p) =g(n′,0) = 0
⇒Im(f)⊆Ker(g)
⇒Im(f) = Ker(g).
Finallyg is surjective because for every m ∈M there is a p∈ P; P 7→ m with g(σ(p)) = idM(m) =m, implying thatξis exact and hence an extension ofM byN. By the naturality of∂we see that Θ(ξ) =xand that Θ is surjective.
Ifβx′ ∈HomA(K, N) is another lift ofx, thenβx−βx′ ∈Ker(∂) so there is ani∈HomA(P, N) with βx′ = βx+ij. If E′ is the pushout of j and βx′, then the map ψ : E → E′ with ψ(n, p) = (n−i(p), p) is an isomorphism. ψis well defined because
ψ(0,0) =ψ(βx(k),−j(k)) = (β(k) +ij(k),−j(k)
= (βx′(k),−j(k)) = (0,0).
It is injective since
(0,0) =ψ(n, p) = (n−i(p), p)
⇒(β′(k),−j(k)) = (n−i(p), p) for somek∈K
⇒p=−j(k)
⇒n−i(p) =n+ij(k) =βx′(k)
⇒n=β′x(k)−ij(k) =βx(k)
⇒(n, p) = (βx(k),−j(k)) = (0,0)
and it is obviously surjective for (n, p) =ψ(n+i(p), p). This shows that there is an equivalence betweenξ and the exact sequence induced byβx′. In fact, this construction gives therefore a set map between Ext1A(M, N) and the equivalence classes of extensions ofM byN.
Conversely, given an extensionξofM byN, the lifting property ofP gives a mapτ:P→E and hence a commutative diagram
0 //K
γ
j //P
τ
//M //0
ξ: 0 //N i //E g //M //0
NowEis the pushout of j andγ. Hence Θ is injective.
5. Moduli Spaces and Simplifyable Modules
The main goal is to find the space containing all indecomposable modulesof A, called the moduli space of A. In this space every point represents a module which either is simple or a non-splitting extension module. By this we mean an A-module E in between two other A-modules N and M in a non-splitting extension, ξ : 0 → N → E → M → 0. The decomposable modules, modules that can be written as the sum of two non-zero submodules, are not of interest because they don’t contain any additional information about the modules ofA. The moduli space is in most cases very hard or even impossible to express globally, but deformation theory is a useful tool to explore the moduli space locally.
5.1. Moduli Spaces.
By choosing an A-module M, we use deformation theory to obtain a local tangent space giving us information about possible directions to move. For example the local moduli space ofA=k⟨x, y⟩/(xy) inM =k(0,0) is two dimensional (Ext1A(M, M)∼=k2), because we can move in the x = 0 and y = 0 direction to find other A-modules2. An important remark here, is that even though the local moduli space aboutM is two dimensional, we cannot infer exactlywhereto go, only the number of dimensions. Because of this, deformation theory will give us a non-zero Ext2A(M, M) meaning that there are obstructions in the deformations of A in M. In this given example, any point not on the two lines x= 0 andy = 0 is not an A-module, and is hence not in the moduli space of A. In the example we just discussed, it can seem obvious what the moduli space is, but that is generally not the case as it is risky to use intuition to conclude anything in the noncommutative plane. Therefore we must take care of not infering too much globally, when computing something locally. We will have a closer look on the calculations involved in this example in section 6.3.
Given ak-algebraA, anA-moduleM and the formal deformation ringR=k[[t]], take the R-deformation with
µR(a, m) =am+tψ1(a, m) +t2ψ2(a, m) +· · ·.
We define S = k[[Ext1A(M, M)∗]] = k[[v1, v2,· · ·, vn]] where {vi}1≤i≤n is a dual basis for Ext1A(M, M). S is a polynomial ring in dim Ext1A(M, M) variables. The local moduli space of first order is then
M′ ={
S →k}∼=kp forp= dim(Ext1A(M, M))
which is the family of all punctuations ofS. This is just our first candidate of the local moduli space, because it depends on there being obstructions in Ext2A(M, M) or not. There are two possibilities. Let us state the first one as a lemma:
Lemma 5.1. If Ext2A(M, M) = 0, every infinitesimal deformation is integrable, and the local moduli space isM′.
Proof. From chapter 3.1 we get from associativity of the R-deformation that δ1(ψ1) = 0 ⇒ ψ1 ∈Der(A, A) and δ1(ψ2) =−(ψ1⌣ ψ1). Using Lemma (2.3) we know that (δ2δ1) = 0 so Im(δ1)⊆Ker(δ2). But Ext2A(M, M) = Ker(δ2)/Im(δ1) = 0, implying that Ker(δ2) = Im(δ1) and that there are no obstructions in the deformation in M. In other words, for every ψi, ψj ∈Im(δ1) i, j∈Z+, we are sure there are no problems for later ψk for k=i+j, and
the infinitesimal deformation is hence integrable.
However if Ext2A(M, M)̸= 0, there exists an elementψi ⌣ ψj ∈Ker(δ2) with
ψi ⌣ ψj ∈/ Im(δ1). This means that for some ψi, ψj, with ψi ⌣ ψj ̸= 0 in Ext2A(M, M), meaningψi⌣ ψj∈/ Im(δ1). This causes trouble, because we then risk choosing some ψi and ψj that satisfies their respective properties, but when combined by the cup product we don’t know ifψk ∈Im(δ1) for ak=i+j. To avoid this problem, we need to find these pairs and
2Remember thatk(0,0) means that the module iskand the module operation isx7→0 andy7→0
divide them out of the moduli space. LetP ={
(ψi, ψj)|ψi ⌣ ψj ̸= 0} and {
w1,· · ·, wm
} be a basis for Ext2A(M, M), then the moduli space ofquadratic order is
M′′=k[[v1,· · ·, vn]]/
(f1,· · · , fm) where fl=∑
i,j
wl(ψi⌣ ψj)vivj (5.1)
In this way we get rid of the basis vectors containing the obstructions. When Ext2A(M, M)̸= 0 we must work out the moduli space of higher and higher order, to get better and better ap- proximation of the local moduli space. It is clear that calculating the local moduli space of high orders will need a lot of tedious calculation, but we will only concentrate on the first and second order.
5.2. Simplifyable Modules.
The notion of a tangent space is intuitively something local, but when working with non- commutative rings there is also possible to study tangent spaces between two modules M andN. This is done by simply calculating Ext1A(M, N) instead of Ext1A(M, M). The answer of this gives us the space containing every extension ofM byN. If this space is zero, every extension ofM byN is split and is hence not contained in the moduli space. If it is non-zero, there exists an indecomposable extension moduleE satisfying 0→N →E → M →0 and we say that there is atangent from M toN.
As previously stated, the moduli space consists of simple modules and non-splitting ex- tensions. In our work of describing the moduli space, it is of interest to find out whether an extension module issimplifyable or not.
Definition 5.1. Let k be an algebraically closed field. AnA-module E satisfying EndA(E)∼=k is simplifyable if there exists a flat familyE →T of A-modules with T a non-singular curve, together with a point0 in T such that the fiberE0 ∼=E, and such that for t ∈T, t̸= 0, the fiberEtis simple.
In other words,Eis simplifyable if it is possible to deform it into a simple module. Of course, every simple module is simplyfyable, so the interesting part is to classify the simplifyable extension modules. The following theorem gives us a necessary criterion for simplifyability.
Theorem 5.1. Let Abe an associativek-algebra and letE be anA module saisfying EndA(E) ∼= k. Let V = Supp(E) = {M1,· · · , Mr} and let QV be the extension graph.
SupposeE is simplifyable. Then there exists a complete cycle inQV.
Proof. [5, Thm. 4.7]
To understand this theorem, we must first clarify some vocabulary. Theextension graph QV is the directed graph with the modules inV as vertices and the Ext1A(Mi, Mj) as arrows for i̸=j. The arrow from Mi to Mj exist if and only if Ext1A(Mi, Mj)̸= 0. A cycle is a path starting and ending in the same vertex, and is said to becompleteif it contains all the vertices ofQV.
If we find a non-zero Ext1A(M, N) and Ext1A(N, M) for twoA-modulesM andN, we know from theorem (5.1) that there is a good chance that the extension moduleE ofM byN is simplifyable. In section 6, we will try to find such complete cycles for various rings in the noncommutative affine plane.
6. Calculating Local Moduli Spaces of Noncommutative Rings We will in the following examples consider noncommutative rings denoted k⟨x, y⟩/(F) where (F) is aquadric.
Definition 6.1. A quadric in the noncommutative affine planeA=k⟨x, y⟩is the zero locus of a quadratic polynomial on the form:
F(x, y) =ax2+bxy+cyx+dy2+ex+f y+g wherea, b, c, d, e, f, g∈k.
By alternating (F), we can explore different kinds of structures evolving in the noncom- mutative case.
6.1. A commutative example.
As an introductary example we will look at the commutative ring k⟨x, y⟩/(xy−yx) :=k[x, y].
LetA=k[x, y] andM =N =k(a, b). First, we need to find a projective resolution ofM, so by using lemma (4.1) we start off with freeA-modules:
0oo M oo µ A A2
oo f oo g X
wheref is defined by (1,0)7→(x−a) and (0,1)7→(y−b). Now we need to findX with a homomorphismgsuch that Im(g) = Ker(f). TryingX=Aand settingg= (u, v)T, we must solve this equation to achieve exactness
f◦g= (x−a, y−b)(u, v)T =u(x−a) +v(y−b) = 0
⇒u=y−b and v=−(x−a).
Note that this answer is easy to spot, becausexy=yx. But when (F) is different, we must potentially solve many linear equations to findf. Also noticing thatgis injective, we get the following projective resolution ofM:
0oo M oo µ Aoo f A2oo g Aoo 0
To calculate Ext1A(M, N) we use the functor HomA(−, N) on this resolution to obtain:
HomA(A, N) dˆ0 //HomA(A2, N) dˆ1 //HomA(A, N) dˆ2 //HomA(0, N)
where ˆd0 =f ◦hfor all h∈HomA(A, N) and ˆd1 =g◦h′ for everyh′ ∈HomA(A2, N), are the induced homomorphisms off and g. In our exampleM =N =k(a, b), so this sequence is isomorphic to
k d
0 //k2 d
1 //k d
2 //0
withd0= (a−a, b−b)T = (0,0)T andd1= (b−b,−a+a) = (0,0). Finally we get Ext1A(M, N) = Ext1A(M, M) = Ker(d1)/Im(d0) =k2/0∼=k2 with basis{u∗, v∗} and
Ext2A(M, N) = Ext2A(M, M) = Ker(d2)/Im(d1) =k/0∼=k.
with basis{w∗}. Because Ext2A(M, M) ̸= 0, the local moduli space of M is notM′ so we must calculate further to approximate it withM′′. We use formula (5.1) to find a polynomial
