MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRApeople.brandeis.edu/~igusa/Math131b/Math131b_notesA.pdf · 2019-02-27 · MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 3 Corollary

MATH 131B: ALGEBRA IIPART A: HOMOLOGICAL ALGEBRA

These are notes for our first unit on the algebraic side of homological algebra. Whilethis is the last topic (Chap XX) in the book, it makes sense to do this first so thatgrad students will be more familiar with the ideas when they are applied to algebraictopology (in 151b). At the same time, it is not my intention to cover the same materialtwice. The topics are

Contents

1. Additive categories 12. Kernels and cokernels 23. Projective and injective objects 33.1. Sufficiently many projectives 53.2. Injective objects 64. Injective modules 74.1. dual module 74.2. Constructing injective modules 74.3. proof of lemmas 94.4. Examples 115. Divisible groups 126. Injective envelope 147. Projective resolutions 167.1. Definitions 167.2. Modules of a PID 177.3. Canonical forms for matrices 197.4. Chain complexes 208. Homotopy invariance of projective resolutions 249. Derived functors 279.1. Construction of RiF 279.2. connecting morphisms 289.3. connecting morphism: standard method 319.4. Left derived functors 33

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MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 1

1. Additive categories

From Lang III§3: Abelian categories we take only the first two axioms: AB1, AB2.

Definition 1.1. An additive category is a category C for which every hom setHomC(X, Y ) is an additive group and

AB1a composition is biadditive, i.e., (f1 + f2) ◦ g = f1 ◦ g + f2 ◦ g and f ◦ (g1 + g2) =f ◦ g1 + f ◦ g2.

AB1b C has a zero object 0 having the property that HomC(X, 0) = 0 = HomC(0, X)for all objects X.

AB2 C has finite products and coproducts (sums).

Theorem 1.2. In an additive category, finite products and finite sums are isomorphic,i.e., the canonical morphism

⊕Ai →

∏Ai is an isomorphism.

Proof. By induction it suffices to consider sums and products of pairs. We will showA⊕B ∼= A×B. Consider the following diagram.

(1.1) AidA //

jA

##

AjA

##A⊕B f // A×B

pA

;;

pB ##

g // A⊕B

BidB //

jB

;;

BjB

;;

where f : A⊕B → A×B is the unique morphism so that pA ◦f ◦ jA = idA, pB ◦f ◦ jB =idB and the other two compositions are zero: pB ◦ f ◦ jA = 0 = pA ◦ f ◦ jB. Letg : A× B → A⊕ B be the sum g = jA ◦ pA + jB ◦ pB. Then, it is easy to see that f, gare inverse morphisms:

(1) g ◦ f ◦ jA = jA ◦ pA ◦ f ◦ jA + jB ◦ pB ◦ f ◦ jA = jA ◦ idA + 0 = jA and,similarly, g ◦f ◦ jB = jB. By definition of coproduct, there is a unique morphismh : A⊕ B → A⊕ B satisfying h ◦ jA = jA and h ◦ jB = jB. Since h = g ◦ f andh = idA⊕B are two solutions, we have g ◦ f = idA⊕B.

(2) pA ◦ f ◦ g = pA ◦ f ◦ jA ◦ pA + pA ◦ f ◦ jB ◦ pB = pA and, similarly, pB ◦ f ◦ g = pB.So, f ◦ g is the identity on A×B.

Therefore, A⊕B and A×B are canonically isomorphic in any additive category. �

Example 1.3. Some examples of additive categories:

(1) The category Mod-R of all right R-modules.(2) The category mod-R of finitely generated right R-modules and R-linear homo-

morphisms. E.g., mod-Z is the category of finitely generated abelian groups.(3) The category tor-Z of finite abelian groups.(4) The category P(Z) of finitely generated free abelian groups (groups isomorphic

to Zn for some finite n).

The first and third categories are “abelian” but the other two are not. I don’t wantto spend time formally defining what is an abelian category. But I will point out theproblems with the second and fourth categories and how to fix them.

2 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA

2. Kernels and cokernels

In a general additive category, a morphism might not have a kernel or cokernel.However, if they do exist, they are unique.

Definition 2.1. The kernel of a morphism f : A→ B is an object K with a morphismj : K → A so that

(1) f ◦ j = 0 : K → B(2) For any other object X and morphism g : X → A so that f ◦ g = 0 there exists

a unique g̃ : X → K so that g = j ◦ g̃ (such a g̃ is called a lifting of g to K).

Since this is a universal property, the kernel is unique (up to isomorphism) if it exists.

X

g��

∃!g̃

~~

0

K

j // Af // B

Theorem 2.2. K is the kernel of f : A→ B if and only if

0→ HomC(X,K)j]−→ HomC(X,A)

f]−→ HomC(X,B)

is exact for any object X. Here f] is left composition with f .

Recall that a morphism f : A → B is a monomorphism if, for any two mapsg 6= h : C → A, f ◦ g 6= f ◦ h, i,e, f] : HomC(X,A)→ HomC(X,B) is a monomorphismof groups.

Corollary 2.3. A morphism f : A→ B is a monomorphism iff ker f = 0.

Notation 2.4. We say that 0→ K → Af−→ B is exact if K = ker f .

Remark 2.5. A non-example: The category mod-R of f.g. right R-modules does nothave kernels if R is not right Noetherian (right ideals are finitely generated). To seethis suppose that I is a right ideal which is not finitely generated. Then IR as R-moduleis not an object in the category. So, the morphism R → R/I does not have a kernel inthe category mod-R.

Cokernel is defined analogously and satisfies the following theorem which can be usedas the definition.

Theorem 2.6. The cokernel of f : A→ B is an object C with a morphism p : B → Cso that

0→ HomC(C, Y )p]−→ HomC(B, Y )

f]−→ HomC(A, Y )

is exact for any object Y where f ] is right composition with f .

The standard diagram defining coker f is the following.

A

0 ��

f // Bp //

g��

C

∃!g��Y

Again, letting C = 0 we get the statement:


Corollary 2.7. A morphism f is an epimorphism (i.e., f ] is a monomorphism) ifand only if coker f = 0.

Notation 2.8. We say Af−→ B → C → 0 is exact if C = coker f .

We say 0→ Af−→ B

g−→ C → 0 is exact if A = ker g and C = coker f .

These two theorems can be summarized by the following statement.

Corollary 2.9. For any additive category C, HomC is left exact in each coordinate.

Exercise 2.10. (1) Show that mod-R has cokernels.(2) In the category of f.g. free abelian groups, show that the morphism f : Z → Z

given by multiplication by 2 is an epimorphism: 0→ Z 2−→ Z→ 0 is exact!!

3. Projective and injective objects

We will discuss the definition of projective and injective objects in any additive cat-egory. And it was easy to show that the category of R-modules has sufficiently manyprojectives. But the category of free R-modules does not.

Definition 3.1. An object P of an additive category C is called projective if for anyepimorphism p : A → B and any morphism g : P → B there exists a morphismg̃ : P → A so that p ◦ g̃ = g. The map g̃ is called a lifting of g to A.

A

p��

Pg //

∃g̃??

B

Theorem 3.2. P is projective if and only if HomC(P,−) is an exact functor.

Proof. If 0→ A→ B → C → 0 is exact then, by left exactness of Hom we get an exactsequence:

0→ HomC(P,A)→ HomC(P,B)→ HomC(P,C)

By definition, P is projective if and only if the last map is always an epimorphism, i.e.,iff we get a short exact sequence

0→ HomC(P,A)→ HomC(P,B)→ HomC(P,C)→ 0

�

Theorem 3.3. Any free R-module is projective.

Proof. Suppose that F is free with generators xα. Then every element of F can bewritten uniquely as

∑xαrα where the coefficients rα ∈ R are almost all zero (only

finitely many are nonzero). Suppose that g : F → B is a homomorphism. Then, forevery index α, the element g(xα) comes from some element yα ∈ A. I.e., g(xα) = f(yα).Then a lifting g̃ of g is given by

(3.1) g̃(∑

xarα) =∑

yαrα

The verification that this is a lifting is straightforward. �


To do the last step in full detail, we use the universal property of free modules: Ahomomorphism from the free module F generated by a set S to another module M isuniquely determined by a set mapping S →M which is arbitrary, i.e., (3.1) defines theunique homomorphism F → A sending xα to yα.

For every R-module M there is a free R-module which maps onto M , namely the freemodule F generated by symbols [x] for all x ∈ M and with projection map p : F → Mgiven by

p(∑

rα[xα])

=∑

rαxα

The notation [x] is used to distinguish between the element x ∈M and the correspondinggenerator [x] ∈ F . Using the universal property of free modules, the homomorphism pis defined by the equation p[x] = x.

For M a finitely generated R-module, take a finite generating set S. Then the freeR-module generated by S maps onto M .


Recall that in an additive category finite coproducts are isomorphic to finite products.They are called finite sums. But the “sum” is defined to be the coproduct.

We use matrix notation for morphisms between direct sums. For example:

ϕ =

[f gh k

]: X ⊕ Y → A⊕B

is the unique morphism satisfying

(1) pA ◦ ϕ ◦ jX = f : X → A(2) pA ◦ ϕ ◦ jY = g : Y → A(3) pB ◦ ϕ ◦ jX = h : X → B(4) pB ◦ ϕ ◦ jY = k : Y → B

3.1. Sufficiently many projectives.

Theorem 3.4. The categories Mod-R and mod-R have sufficiently many projec-tives, i.e., for every R-module M there is a projective R-module P which maps onto Mand, if M is finitely generated, we can also choose P to be f.g.

Remark 3.5. The other two categories in our list: tor-Z and P(Z) do not have enoughprojectives. In fact they have no nonzero projective objects! I added a proof below.

Proposition 3.6. The categories tor-Z and P(Z) have no nonzero projective objects.

The proof uses two facts which are so important that I call them “Theorems” insteadof “Lemmas”.

Theorem 3.7. Any direct summand of a projective object is projective.

Proof. Suppose that P = X ⊕ Y is projective. Then we show that X is projective:Given any epimorphism f : A → B and morphism g : X → B, we get the morphism[g, 0] : X ⊕ Y → B. Since X ⊕B is projective, there is a lifting h : X ⊕ Y → A of g⊕ 0to A. Then h ◦ jX : X → Y is a lifting of g to A since f ◦ h ◦ jX = [g, 0] ◦ jX = g. �

Theorem 3.8. Any epimorphism p : X → P where P is projective splits, i.e., there isa morphism s : P → X, called a section, so that p ◦ s = idX .

Proof. Since P is projective and p : X → P is an epimorphism, idP : P → P lifts to X.

X

p��

PidP //

∃s>>

P

I.e., p ◦ s = idP as claimed. �

Proof. In class we went through the proof that

(1) The morphism f : Z→ Z given by multiplication by 2 is an epimorphism in thecategory P(Z) of f.g. free abelian groups.

(2) This morphism does not have as section.


By Theorem 3.8, Z is not projective. The objects of P(Z) are, by definition, isomorphicto Zn for some n. If this were projective, by Theorem 3.7, Z would be projective, whichis a contraction. Therefore, P(Z) has no projective objects (except 0 which is alwaysprojective).

In the category of finite abelian groups, the object Zn is not projective since theepimorphism

Zn2 → Znhas no section (left as an exercise). But, every finite abelian group is a direct sum ofcyclic groups. So, tor-Z has no nonzero projective objects. �

Getting back to the good cases: Suppose that C is an additive category with thefollowing two properties:

(1) C has enough projectives, i.e., for every objectX, there is an epimorphism P → Xwhere P is projective.

(2) Every morphism in C has a kernel (in C).Then each object X has a projective resolution:

0← Xd0←− P0

d1←− P1d2←− P2 ← · · ·

constructed as follows. P0 is a projective object with an epimorphism d0 : P0 → X. Byassumption (2), this has a kernel K0. Let P1 be a projective object with epimorphismP1 → K0 = ker d0. (In the diagram d1 : P1 → P0 is the composition P1 � K0 � P0.)By induction on n we have an epimorphism Pn � Kn−1. Let Kn be the kernel of thismorphism and let Pn+1 be a projective object with an epimorphism Pn+1 � Kn+1.

Warning! In a general additive category, Kn−1 may not be the cokernel of Kn � Pn.

Theorem 3.9. In the category Mod-R of all R-modules and, for R right Noetherian,the category mod-R of f.g. right R-modules, every object has a projective resolution.

3.2. Injective objects.

Definition 3.10. An object Q of any additive category C is injective if, for anymonomorphism A→ B, any morphism A→ Q extends to B. I.e., iff

HomC(B,Q)→ HomC(A,Q)→ 0

is exact.

As before this is equivalent to:

Theorem 3.11. Q is injective if and only if HomC(−, Q) is an exact functor.

The difficult theorem we need to prove is the following:

Theorem 3.12. The category Mod-R has sufficiently many injectives. I.e., every R-module embeds in an injective R-module (usually not finitely generated).

As in the case of projective modules, since every morphism has a cokernel, this the-orem will tell us that every R-module M has an injective co-resolution which is anexact sequence:

0→M → Q0 → Q1 → Q2 → · · ·where each Qi is injective.


4. Injective modules

I will go over Lang’s proof (XX§4) that every R-module M embeds in an injectivemodule Q. Lang uses the dual M∧ of a module. (Another dual M∨ is discussed in III§6.)

4.1. dual module.

Definition 4.1. The dual of a right R-module M is defined to be the left R-module

M∧ := HomZ(M,Q/Z)

with left R-action given by(rϕ)(x) = ϕ(xr)

for all ϕ ∈M∧, r ∈ R.

M∧ is, in general, not finitely generated.

Proposition 4.2. Duality is a left exact functor

( )∧ : Mod-R→ R-Mod

which is additive and takes sums to products:(⊕Mα

)∧ ∼= ∏M∧α

Proof. We already saw that the hom functor HomZ(−, X) is left exact for any abeliangroup X. It is also obviously additive which means that (f + g)] = f ] + g] for allf, g : N →M . I.e., the duality functor induces a homomorphism (of abelian groups):

HomR(N,M)→ HomZ(M∧, N∧)

Duality also takes sums to products since a homomorphism

f :⊕

Mα → X

is given uniquely by its restriction to each summand: fα : Mα → X and the fα can allbe nonzero. (So, (fα)α is in the product not the sum.) �

4.2. Constructing injective modules. In order to get an injective right R-modulewe need to start with a left R-module.

Theorem 4.3. Suppose F is a free left R-module. (I.e., F = ⊕ RR is a direct sum ofcopies of R considered as a left R-module). Then F∧ is an injective right R-module.

This theorem follows from the following lemma.

Lemma 4.4. (1) A product of injective modules is injective.(2) HomR(M, RR

∧) ∼= HomZ(M,Q/Z) (isomorphism of left R-modules)(3) Q/Z is an injective Z-module.

Proof of the theorem. Lemma (3) implies that HomZ(−,Q/Z) is an exact functor. (2)implies that HomR(M, RR

∧) is an exact functor. Therefore, RR∧ is an injective R-

module. Since duality takes sums to products,

F∧ = (⊕

RR)∧ =∏

RR∧

is projective by (1) proving the Theorem. �


We need one more lemma to prove the main theorem. Then we have to prove thelemmas.

Lemma 4.5. Any right R-module is naturally embedded in its double dual:

M ⊆M∧∧

Assume this 4th fact for a moment.

Theorem 4.6. Every right R-module M can be embedded in an injective right R-module.

Proof. Let F be a free left R-module which maps onto M∧:

F →M∧ → 0

Since duality is left exact we get:

0→M∧∧ → F∧

By Lemma 4.5, we have M ⊆ M∧∧ ⊆ F∧. So, M embeds in the injective moduleF∧. �


4.3. proof of lemmas. There are four lemmas to prove:

(1) A product of injective modules is injective.(2) HomR(M, RR

∧) ∼= HomZ(M,Q/Z) (isomorphism of left R-modules)(3) Q/Z is an injective Z-module.(4) M ⊆M∧∧.

Assuming (3) T = Q/Z is injective, the other three lemmas are easy:

Proof of (4) Lemma 4.5. A natural embedding M → M∧∧ is given by the evaluationmap ev which sends x ∈M to evx : M∧ → T which is evaluation at x:

evx(ϕ) = ϕ(x).

This is easily seen to be additive:

evx+y(ϕ) = ϕ(x+ y) = ϕ(x) + ϕ(y) = evx(ϕ) + evy(ϕ) = (evx + evy) (ϕ).

The right action of R on M gives a left action of R on M∧ which gives a right action ofR on HomZ(M∧, T ) so that (evxr)(ϕ) = evx(rϕ). This implies that evaluation x 7→ evxis a right R-module homomorphism:

evxr(ϕ) = ϕ(xr) = (rϕ)(x) = evx(rϕ) = (evxr) (ϕ).

Finally, we need to show that ev is a monomorphism. In other words, for every nonzeroelement x ∈M we need to find some additive map ϕ : M → T so that evx(ϕ) = ϕ(x) 6= 0.To do this take the cyclic group C generated by x

C = {kx : k ∈ Z}This is either Z or Z/n. In the second case let f : C → T be given by

f(kx) =k

n+ Z ∈ Q/Z

This is nonzero on x since 1/n is not an integer: f(x) = 1n

+ Z 6= 0 + Z. If C ∼= Z thenlet f : C → T be given by

f(kx) =k

2+ Z

Then, f(x) = 12

+ Z is nonzero in Q/Z. Since T is Z-injective, f extends to an additive

map f : M → T . So, evx is nonzero since evx(f) = f(x) = f(x) 6= 0. So, ev : M →M∧∧

is a monomorphism. �

Proof of (1): products of injectives are injective. Suppose that Jα are injective. Thenwe want to show that Q =

∏Jα is injective. Let pα : Q → Jα be the projection map.

Suppose that A ⊂ B and f : A→ Q is any morphism. Then we want to extend f to B.Since each Jα is injective each composition fα = pα ◦ f : A → Jα extends to a

morphism fα : B → Jα. I.e., fα|A = fα for all α. By definition of the product thereexists a unique morphism f : B → Q =

∏Jα so that pα ◦ f = fα for each α. So,

pα ◦ f |A = fα|A = fα : A→ Jα

The uniquely induced map A →∏Jα is f . Therefore, f is an extension of f to B as

required. �


Proof of (2):HomR(M, RR

∧) ∼= HomZ(M,Q/Z).

To prove this we will give a 1-1 correspondence and show that it (the correspondence)is additive. We also show that this map is an isomorphism of left R-modules.

If f ∈ HomR(M, RR∧) then f : M → RR

∧, an R-linear map, which means that foreach x ∈ M we get f(x) ∈ RR

∧ = HomZ(R,Q/Z), i.e., f(x) : R → Q/Z. Furthermore,for r ∈ R, x ∈ M , we have f(xr) = f(x)r or f(xr)(s) = f(x)(rs). In particular we canevaluate this at 1 ∈ R. This gives ϕ(f) : M → Q/Z by the formula

ϕ(f)(x) := f(x)(1)

This defines a mapping

ϕ : HomR(M, RR∧)→ HomZ(M,Q/Z)

We need to show that this is additive, i.e.,

ϕ(f + g) = ϕ(f) + ϕ(g)

This is an easy calculation which follows from the way that f + g is defined, namely,addition of function is defined “pointwise” which means that (f + g)(x) = f(x) + g(x)by definition. So, ∀x ∈M ,

ϕ(f + g)(x) = (f + g)(x)(1) = [f(x) + g(x)](1) = f(x)(1) + g(x)(1)

= ϕ(f)(x) + ϕ(g)(x) = [ϕ(f) + ϕ(g)](x).

We would also like to know that ϕ is left R-linear, i.e., we want to show ϕ(rf) = rϕ(f).This follows from the fact that 1r = r = r1!

ϕ(rf)(x) = [rf(x)](1) = f(x)(1r)

[rϕ(f)](x) = ϕ(f)(xr) = f(xr)(1) = f(x)(r1)

Finally we need to show that ϕ is a bijection. To do this we find the inverse ϕ−1 = ψ.For any additive g : M → Q/Z let ψ(g) : M → RR

∧ be given by ψ(g)(x) : R→ Q/Z:

[ψ(g)(x)](r) := g(xr)

Since this is additive in all three variables, ψ is additive and ψ(g) is additive. We alsoneed to check that ψ(g) is a homomorphisms of right R-modules, i.e., that ψ(g)(xr) =ψ(g)(x)r. This is an easy calculation:

[ψ(g)(xr)](s) = g((xr)s) = g(xrs)

[ψ(g)(x)r](s) = [ψ(g)(x)](rs) = g(x(rs))

The verification that ψ is the inverse of ϕ is also straightforward: For all f ∈HomR(M, RR

∧) we have

ψ(ϕ(f))(x)(r) = ϕ(f)(rx) = f(rx)(1) = [rf(x)](1) = f(x)(1r) = f(x)(r)

So, ψ(ϕ(f)) = f . Similarly, for all g ∈ HomZ(M,Q/Z) we have:

ϕ(ψ(g))(x) = [ψ(g)(x)](1) = g(x1) = g(x)

So, ϕ(ψ(g)) = g. �

I will do the remaining Lemma (3) (injectivity of Q/Z) next time.


4.4. Examples. First, a trivial example:

4.4.1. fields. Suppose that R = k is a field.Since all short exact sequences of vector spaces split, we have:

Theorem 4.7. In both categories mod-k and Mod-k all objects are both projective andinjective.

4.4.2. polynomial ring. Let R = Z[t], the integer polynomial ring in one generator. Thisis a commutative Noetherian ring. It has dimension 2 since a maximal tower of primeideal is given by

0 ⊂ (t) ⊂ (t, 2)

These ideals are prime since the quotient of R by these ideals are domains (i.e., have nozero divisors):

R/0 = Z[t], R/(t) = Zare domains and

R/(t, 2) = Z/(2) = Z/2Zis a field, making (2, t) into a maximal ideal.

Proposition 4.8. A Z[t]-module M is the same as an abelian group together with anendomorphism M → M given by the action of t. A homomorphism of Z[t]-modulesf : M → N is an additive homomorphism which commutes with the action of t.

Proof. I will use the fact that the structure of an R-module on an additive group Mis the same as a homomorphism of rings ϕ : R → End(M). When R = Z[t], thishomomorphism is given by its value on t since ϕ(f(t)) = f(ϕ(t)). For example, iff(t) = 2t2 + 3 then

ϕ(f(t)) = ϕ(2t2 + 3) = 2ϕ(t) ◦ ϕ(t) + 3idM = f(ϕ(t))

Therefore, ϕ is determined by ϕ(t) ∈ EndZ(M) which is arbitrary. �

For R = Z[t], what do the injective R-modules look like? We know that Q = R∧R isinjective. What does that look like?

Q = HomZ(Z[t],Q/Z)

But Z[t] is a free abelian group on the generators 1, t, t2, t3, · · · . Therefore, an elementf ∈ Q, f : Z[t]→ Q/Z is given uniquely by the sequence

f(1), f(t), f(t2), f(t3), · · · ∈ Q/ZMultiplication by t shifts this sequence to the left since (ft)(ti)− f(tti) = f(ti+1). Thisproves the following.

Theorem 4.9. The injective Z[t]-module Q = Z[t]∧ is isomorphic to the additive groupof all sequences (a0, a1, a2, · · · ) of elements ai ∈ Q/Z with the action of t given by shiftingto the left and dropping the first coordinate. I.e.,

t(a0, a1, a2, · · · ) = (a1, a2, · · · ).

The word “isomorphism” is correct here because these are not the same set.


5. Divisible groups

Z-modules are the same as additive groups (abelian groups with group operation +).And we will see that injective Z-modules are the same as divisible groups.

Definition 5.1. An additive group D is called divisible if for any x ∈ D and anypositive integer n there exists y ∈ D so that ny = x. (We say that x is divisible by n.)Equivalently, multiplication by any n > 0 gives a surjective map D � D.

Example 5.2. Q is divisible. Another interesting example is:∏p Zp⊕p Zp

the product of all cyclic groups of prime order Zp modulo their direct sum. This isdivisible since, for any n, we have ∏

p Zp⊕p Zp

∼=∏

p>n Zp⊕p>n Zp

and multiplication by n is an isomorphism Zp ∼= Zp for all primes p > n.

Proposition 5.3. Any quotient of a divisible group is divisible.

Proof. Suppose D is divisible and K is a subgroup. Then any element of the quotientD/K has the form x + K where x ∈ D. This is divisible by any positive n since, ifny = x then

n(y +K) = ny +K = x+K

Therefore D/K is divisible. �

Example 5.4. Q/Z is divisible.

Theorem 5.5. Tfae (the following are equivalent) for any additive group D:

(1) D is divisible.(2) If A is a subgroup of a cyclic group B then any homomorphism A→ D extends

to B.(3) D is an injective Z-module.

Proof. (2) ⇒ (1): Take B = Z, A = nZ. For any x ∈ D, let f : A → D be given byf(an) = ax. So, f(n) = x. Extend this to f : Z → D be an extension of f to Z. Thenny = f(n) = f(n) = x. So, D is divisible.

(1) ⇒ (2): We may assume A 6= 0. If A is a nontrivial subgroup of a cyclic groupB then the generator a0 of A is a multiple of the generator b0 of B: a0 = nb0. Forany f : A → D, let x = f(a0). Since D is divisible, there is y ∈ D so that ny = x.Then there is a unique homomorphism f : B → D sending b0 to y and f |A = f sincef(a0) = f(nb0) = ny = x.

(3)⇒ (2) follows from the definition of injective module.(2) ⇒ (3). Given that D satisfies (2), we will use Zorn’s Lemma to prove that D is

injective. Suppose that X is a submodule of Y and f : X → D is a homomorphism.Then we want to extend f to all of Y . To use Zorn’s Lemma we take the set of all pairs


(C, g) where X ⊆ C ⊆ Y and g is an extension of f (i.e., f = g|X). This set is partiallyordered in an obvious way: (C, g) < (C ′, g′) if C ⊆ C ′ and g = g′|C. It also satisfies thehypothesis of Zorn’s Lemma. Namely, any totally ordered subset (Cα, gα) has an upperbound: (∪Cα,∪gα). Zorn’s Lemma tells us that this set has a maximal element, say,(M, g). We just need to show that M = Y . We show this by contradiction.

If M 6= Y then there is at least one element y ∈ Y which is not in M . Let B be thecyclic subgroup of Y generated by y. Let A = B ∩M . Then, by (2), the restrictionh = g|A : A → D extends to a homomorphism h : B → D. Then (by Lemma below)g ⊕ h : M ⊕B → D induces a homomorphism

g + h : M +B → D

which is equal to g on M and h on B. Then (M +B, g + h) > (M, g) contradicting themaximality of (M, g). Therefore, M = Y and g : Y → D is an extension of f : X → D.So, D is an injective Z-module. �

Lemma 5.6. Suppose that A,B are submodules of an R-module C and f : A → X,g : B → X are homomorphisms of R-modules which agree on A ∩ B. Then we get awell-defined homomorphism f + g : A+B → X by the formula

(f + g)(a+ b) = f(a) + g(b).

Proof. Well-defined means that, if the input is written in two different ways, the outputis still the same. So suppose that a+ b = a′ + b′. Then a− a′ = b′ − b ∈ A ∩B. So,

f(a− a′) = f(a)− f(a′) = g(b′ − b) = g(b′)− g(b)

by assumption. Rearranging the terms, we get f(a)+g(b) = f(a′)+g(b′) as desired. �

This lemma also follows from left-exactness of Hom: We have a short exact sequenceof R-modules:

0→ A ∩B( jA−jB

)−−−→ A⊕B pA+pB−−−−→ A+B → 0.

Since HomR(A⊕B,X) ∼= HomR(A,X)⊕ HomR(B,X) this gives a left exact sequence:

0→ HomR(A+B,X)→ HomR(A,X)⊕ HomR(B,X)→ HomR(A ∩B,X).

Exactness of this sequence is equivalent to the Lemma.


6. Injective envelope

There is one very important fact about injective modules which is not covered inLang’s book. This is the fact that every R-module M embeds in a minimal injectivemodule which is called the injective envelope of M . This is from Jacobson’s BasicAlgebra II.

Definition 6.1. An embedding A ↪→ B is called essential if every nonzero submoduleof B meets A. I.e., C ⊆ B,C 6= 0⇒ A ∩ C 6= 0.

For example, Z ↪→ Q is essential because, if a subgroup of Q contains a/b, then itcontains a ∈ Z. Also, every isomorphism is essential.

Exercise 6.2. Show that the composition of essential maps is essential.

Lemma 6.3. Suppose A ⊆ B. Then

(1) ∃X ⊆ B s.t. A ∩X = 0 and A ↪→ B/X is essential.(2) ∃C ⊆ B maximal so that A ⊆ C is essential.

Proof. For (1) the set of all X ⊆ B s.t. A ∩ X = 0 has a maximal element by Zorn’sLemma. Then A ↪→ B/X must be essential, otherwise there would be a disjoint sub-module of the form Y/X and X ⊂ Y,A∩ Y = 0 contradicting the maximality of Y . For(2), C exists by Zorn’s Lemma. �

Lemma 6.4. Q is injective iff every short exact sequence

0→ Q→M → N → 0

splits.

Proof. If Q is injective then the identity map Q→ Q extends to a retraction r : M → Qgiving a splitting of the sequence. Conversely, suppose that every sequence as abovesplits. Then for any monomorphism i : A ↪→ B and any morphism f : A → Q we canform the pushout M in the following diagram

Ai //

f ��

Bf ′��

Qj // M

These morphisms form an exact sequence:

A(fi)−−→ Q⊕B (j,−f ′)−−−−→M → 0

This implies that j is a monomorphism. [If j(x) = 0 then (j,−f ′)(x, 0) = 0. So, thereis an a ∈ A so that (x, 0) = (f(a), i(a)). But i(a) = 0 implies a = 0. So, x = f(a) = 0.]

Since j is a monomorphism there is a short exact sequence

0→ Qj−→M → coker j → 0

We are assuming that all such sequences split. So, there is a retraction r : M → Q(r ◦ j = idQ). It follows that r ◦ f ′ : B → Q is the desired extension of f : A→ Q:

r ◦ f ′ ◦ i = r ◦ j ◦ f = idQ ◦ f = f


So, Q is injective. �

Lemma 6.5. Q is injective if and only if every essential embedding Q ↪→ C is anisomorphism.

Proof. (⇒) Suppose Q is injective and Q ↪→ C is essential. Then the identity mapQ→ Q extends to a retraction r : C → Q whose kernel is disjoint from Q and thereforemust be zero making C ∼= Q.

(⇐) Now suppose that every essential embedding of Q is an isomorphism. We wantto show that Q is injective. By the previous lemma it suffices to show that every shortexact sequence

0→ Qj−→M → N → 0

splits. By Lemma 6.3 there is a submodule X ⊆M so that jQ∩X = 0 and Q ↪→M/Xis essential. Then, by assumption, this map must be an isomorphism. So, M ∼= Q⊕Xand the sequence splits proving that Q is injective. �

Theorem 6.6. For any R-module M there exists an essential embedding M ↪→ Q withQ injective. Furthermore, Q is unique up to isomorphism under M .

Proof. We know that there is an embedding M ↪→ Q0 where Q0 is injective. By Lemma6.3 we can find Q maximal with M ↪→ Q ↪→ Q0 so that M ↪→ Q is essential.

Claim: Q is injective.If not, there exists an essential Q ↪→ N . Since Q0 is injective, there exists f : N → Q0

extending the embedding Q ↪→ Q0. Since f is an embedding on Q, ker f ∩Q = 0. Thisforces ker f = 0 since Q ↪→ N is essential. So, f : N → Q0 is a monomorphism. Thiscontradicts the maximality of Q since the image of N is an essential extension of M inQ0 which is larger than Q.

It remains to show that Q is unique up to isomorphism. So, suppose M ↪→ Q′ isanother essential embedding of M into an injective Q′. Then the inclusion M ↪→ Q′

extends to a map g : Q→ Q′ which must be a monomorphism since its kernel is disjointfrom M . Also, g must be onto since g(Q) is injective making the inclusion g(Q) ↪→ Q′

split which contradicting the assumption that M ↪→ Q′ is essential unless g(Q) = Q′. �


7. Projective resolutions

Today we talk about projective resolutions.

(1) Definitions(2) Modules over a PID(3) Canonical forms(4) Chain complexes, maps and homotopies

7.1. Definitions. Suppose that M is a (right) R-module (or, more generally, an ob-ject of an additive category A with kernels and enough projectives) then a projectiveresolution of M is defined to be a long exact sequence of the form

· · · → Pn+1dn+1−−−→ Pn

dn−→ Pn−1 → · · · → P0ε−→M → 0

where Pi are all projective. Exact means in general that each Pi maps epimorphicallyto the kernel of the previous map. In the homework students showed that, for anyprojective object P , we get a long exact sequence of abelian groups:

· · · A(P, Pn+1)→ A(P, Pn)→ A(P, Pn−1)→ · · · → A(P, P0)→ A(P,M)→ 0

where we use the abbreviation A(X, Y ) = HomA(X, Y ).One application of this long exact sequence is the “uniqueness” of projective resolu-

tions: Given any other projective resolution · · · → P ′1 → P ′0 → M , there is a sequenceof morphisms fi : P ′i → Pi making the following diagram commute:

· · · // P ′2

f2��

d′2 // P ′1

f1��

d′1 // P ′0

f0��

ε′ // M

=

��· · · // P2

d2 // P1d1 // P0

ε // M

The morphism f0 : P ′0 → P0 exists since ε : P0 →M is an epimorphism. The morphismf0 ◦ d′1 : P ′1 → P0 goes to the kernel of ε : P0 → M . Since P1 maps epimorphically ontoker ε, we get a lifting f1 : P ′1 → P1, and so on. (f0◦d′1 ∈ A(P ′1, P0) maps to 0 ∈ A(P ′1,M)and therefore lifts to A(P ′1, P1).)

The (right) projective dimension of M is the smallest integer n ≥ 0 so that thereis a projective resolution of the form

0→ Pn → Pn−1 → · · · → P0ε−→M → 0

We write pd(M) = n. If there is no finite projective resolution then the pd(M) =∞.If pd(M) = 0 we have, in a general additive category, an epimorphism: ε : P0 � M

with kernel 0. This does not necessarily imply that ε is an isomorphism. For example,in P(Z) we saw that multiplication by 2 is an epimorphism Z→ Z with kernel zero. Weneed the following additional assumption on A:

(∗) Every epimorphism is the cokernel of its kernel.

The (right) global dimension of the ring R written gl dim(R) is the maximumprojective dimension of any right R-module.


Example 7.1. (0) R has global dimension 0 if and only if it is semi-simple (e.g.,any field). We will study this in detail in Part C.

(1) Any principal ideal domain (PID) has global dimension ≤ 1 since every submod-ule of a free module is free and every module (over any ring) is (isomorphic to)a quotient of a free module.

An injective coresolution of a module M is an exact sequence of the form

0→M → Q0 → Q1 → · · ·

where all of the Qi are injective. If an additive category has cokernels and enoughinjectives then every object has in injective resolution. We went to a lot of trouble toshow that this holds for the category of R-modules for any ring R: We showed thatevery module M embeds in an injective module Q0 and there is a canonical choice ofQ0 called the injective envelope of M . The next term Q1 is the injective envelope of (orany injective module which contains) the cokernel of M ↪→ Q0 and so on.

The injective dimension id(M) is the smallest integer n so that there is an injectivecoresolution of the form

0→M → Q0 → Q1 → · · · → Qn → 0

We will see later that the maximum injective dimension is equal to the maximum pro-jective dimension.

Exercise 7.2. Show that the shortest injective coresolution of M is given by takinginjective envelopes at each stage.

7.2. Modules of a PID. Here is Lang’s proof of the following well-known theorem thatI already mentioned several times. We skipped this in class.

Theorem 7.3. Suppose that R is a PID and E is a free R-module. Then every sub-module of E is free.

Proof. (Lang, page 880) Let E be a free R-module with basis I and let F ⊆ E be aarbitrary submodule. Notation: for any J ⊆ I, let EJ be the free submodule of Egenerated by the subset J of I. Let FJ = F ∩ EJ . Then F = FI .

Step 1: Define a poset:

P = {(J, w) : J ⊆ I and FJ is free with basis w }

The partial ordering is (J, w) ≤ (J ′, w′) if J ⊆ J ′ and w ⊆ w′.For example, suppose that E = R3 with basis I = {e1, e2, e3} and J = {e1, e3}. If

F ⊂ E is the submodule given by

F = {(x, y, z) : x+ y + z = 0}

then FJ = {(x, 0,−x)} is freely generated by w = e1 − e3. So, (J, e1 − e3) ∈ P .Step 2: Show that each tower {(Jα, wα)} in P has an upper bound.There is an obvious upper bound given in the usual way by

(J, w) = (∪αJα,∪αwα)


This clearly has the property that (Jα, wα) ≤ (J, w) for all α. We need to verify that(J, w) is an element of the poset P . Certainly, J = ∪Jα is a subset of I. So, it remainsto check that

(1) w is linearly independent over R, i.e., if a finite sum∑wiri = 0 then all ri = 0.

(2) w spans FJ , i.e., w ⊂ FJ and every element of FJ is an R-linear combination ofelements of w.

(1) is easy: Any R-linear dependence among elements of w involves only a finitenumber of elements of w which must all belong to some wα: If x1, · · · , xn ∈ w then eachxi is contained in some wαi . Let α be the largest αi. Then wα contains all the xi. Sincewα is a basis for FJα , these elements are linearly independent.

(2) is also easy: wα ⊂ FJα ⊆ FJ . So, the union w = ∪wα is contained in FJ . Anyelement x ∈ FJ has only a finite number of nonzero coordinates which all lie in some Jα.So x ∈ FJα which is spanned by wα.

Therefore, by Zorn’s Lemma, P has a maximal element (J∞, w∞). It remains to:Step 3: Show J∞ = I and thus w∞ is a free basis for F = FI .To prove that J∞ = I suppose J∞ is strictly smaller. Then there exists an element

k of I which is not in J∞. Let J ′ = J∞ ∪ {k}. Then we want to find a basis w′ for FJ ′so that (J∞, w∞) < (J ′, w′), i.e., the basis w′ needs to contain w∞. This would give acontradiction to the maximality of (J∞, w∞). There are two cases.

Case 1. FJ ′ = FJ∞ . In that case take w′ = w∞ and we are done.Case 2. FJ ′ 6= FJ∞ . This means that there is at least one element of FJ ′ whose k-

coordinate is nonzero. Let A be the set of all elements of R which appear as k-coordinatesof elements of FJ ′ . This is the image of the k-coordinate projection map

FJ ′ ⊆ EJ ′pk−→ R.

So, A is an ideal in R. Since R is a PID, A = 〈a0〉 = a0R. Let x ∈ FJ ′ so that pk(x) = a0.Then I claim that

w′ = w∞ ∪ {x}is a basis for FJ ′ .

(a) w′ is R-linearly independent: Any R-linear combination which involves x will havenonzero k-coordinate so cannot be zero. And any R-linear combination not involving xcannot be zero since w∞ is R-linearly independent.

(b) w′ spans FJ ′ : Given any z ∈ FJ ′ we must have pk(z) ∈ A = a0R. So pk(z) = a0rand pk(z − xr) = 0. This implies that z − xr ∈ FJ which is spanned by w. So z is xrplus an R-linear combination of elements of w∞. So, w′ spans FJ ′ and we are done. �

Example 7.4. (1) Z is a PID.(2) K[t] is a PID for any field K.(3) K[s, t] is not a PID.(4) Z[t] is not a PID.

Corollary 7.5. Every subgroup of a free abelian group is free.

Exercise 7.6. Show that, if R is a PID, every quotient module of an injective moduleis injective. (We already know this for R = Z. The proof for any PID is similar.)


7.3. Canonical forms for matrices. Students asked to go over canonical forms, suchas the Jordan canonical form of a square matrix. This uses the following theorem.

Theorem 7.7. Every finitely generated module over a PID is a direct sum of cyclicmodules.

Suppose that T is an n × n matrix with entries in a field K. We view T as anendomorphism

T : Kn → Kn

of Kn. This makes M = Kn into a K[t] module by letting t act by T (tv := Tv). SinceK[t] is a PID, M is a direct sum of cyclic modules: M = A⊕B ⊕ · · · .

The first reduction is that any direct sum decomposition of the K[t] module M givesa block decomposition of T . If M = A⊕B then

STS−1 =

[TA 00 TB

]where TA, TB are the matrices of the action of T on A,B and S is the change of basismatrix using a new basis whose first basis vectors lie in A and other basis vectors lie inB.

The next reduction is: Assume M is a cyclic K[t]-module. Then M = K[t]/ 〈f(t)〉.We can assume f(t) is monic:

f(t) = tn + b1tn−1 + b2t

n−2 + · · ·+ bn

We took n = 4 for simplicity of notation. Then, after changing basis, I claimed that thematrix of T is (rational canonical form):

STS−1 =

0 0 0 −b41 0 0 −b30 1 0 −b20 0 1 −b1

This is the matrix of T with respect to the following basis for M = K[t]/ 〈f(t)〉:

1, t, t2, t3

Multiplication by t sends 1 to t, t to t2, t2 to t3. This gives the first three columns ofthe matrix. The last column comes from the action of t on the last vector:

t4 = −b4 − b3t− b2t2 − t1t3.

In case K = C, every polynomial is a product of linear polynomials:

f(t) =∏

(t− ai)mi

The first step is to isolate each factor (t− ai)m1 using the following.

Theorem 7.8 (Chinese remainder theorem). If f(t) = g(t)h(t) where g(t), h(t) arerelatively prime then

K[t]/ 〈f(t)〉 ∼= K[t]/ 〈g(t)〉 ⊕K[t]/ 〈h(t)〉


This reduces us to the case when M = C[t]/ 〈(t− a)m〉. For example, when m = 4,we get the following Jordan canonical form

STS−1 =

a 1 0 00 a 1 00 0 a 10 0 0 a

We did the case m = 2. Then M = C2 and (t − a)2 = 0 on M . Let V = (t − a)M .This is one dimensional. Take e1 ∈ V . Then (t − a)e1 = 0 means Te1 = ae1. Takev ∈ M, v /∈ V . Then (t − a)v = be1 for some b 6= 0 in C. Take e2 = v/b. Then thematrix of T with respect to e1, e2 is

STS−1 =

[a 10 a

]7.4. Chain complexes. We give basic definitions for chain complexes, chain maps andchain homotopies over an additive category and add hypotheses as needed.

Suppose that A is an additive category. Then a chain complex over A is an infinitesequence of objects and morphisms (called boundary maps):

· · · → Cndn−→ Cn−1

dn−1−−−→ · · · → C1d1−→ C0.

so that the composition of any two arrows is zero:

dn−1 ◦ dn = 0.

The chain complex is denoted either C∗ or (C∗, d∗).Given two chain complexes C∗, D∗ a chain map f∗ : C∗ → D∗ is a sequence of

morphisms fn : Cn → Dn so that dDn ◦ fn = fn−1 ◦ dCn where the superscripts are to keeptrack of which chain complex the boundary maps dn are in. These morphisms form abig commuting diagram in the shape of a ladder:

· · · // Cn

fn��

dCn // Cn−1

fn−1

��

dCn−1 // · · ·dC2 // C1

f1��

dC1 // C0

f0��

· · · // DndDn

// Dn−1dDn−1

// · · ·dD2

// D1dD1

// D0

7.4.1. category of chain complexes.

Proposition 7.9. Let C∗(A) be the category of chain complexes and chain maps overan additive category A . Then C∗(A) is an additive category. If A has kernels, resp.cokernels then so does C∗(A).

Proof. The direct sum: C∗ ⊕ D∗ is the chain complex with objects Cn ⊕ Dn andboundary maps

dC⊕Dn = dCn ⊕ dDn : Cn ⊕Dn → Cn+1 ⊕Dn+1


The kernel of a chain map f∗ : C∗ → D∗ is defined to be the chain complex with nthterm ker fn and boundary map

d′n : ker fn → ker fn−1

induced by the morphism dCn : Cn → Cn−1. (Since fn−1 ◦ dCn = dDn ◦ fn = 0 on ker fn, weget this induced map.) The cokernel complex is given similarly by

· · · → coker fndn−→ coker fn−1 → · · ·

where dn is the morphism induced by dDn . �

7.4.2. Homotopy.

Definition 7.10. Given two chain maps f∗, g∗ : C∗ → D∗, a chain homotopy h∗ :f∗ ' g∗ is a sequence of morphisms hn : Cn → Dn+1 so that

dDn+1 ◦ hn + hn−1 ◦ dCn = gn − fn(This is what you add to f∗ to get g∗.)

Next time we will discuss homology, which requires more conditions on our additivecategory, and the theorem that homotopic chain maps induce the same map in homology.


7.4.3. Homology. Suppose that A is an additive category with kernels and cokernels forall f : X → Y and so that each f factors uniquely as a composition of

X � I ↪→ Y

where I = coker(ker f ↪→ X) = ker(Y � coker f). Then I is called the image of f .The homology of a chain complex C∗ is defined to be the sequence of objects:

Hn(C∗) :=ker dn

im dn+1

.

This is only well defined up to isomorphism. There is a set-theoretic difficulty: wecannot in general choose a representative. We ignore this.

Homology is a functor in the sense that any chain map f∗ : C∗ → D∗ induces amorphism in homology Hn(f∗) : Hn(C∗) → Hn(D∗). This is because commutativity ofthe “ladder” implies that fn : Cn → Dn maps ker dCn to ker dDn and im dCn+1 to im dDn+1.

im dCn+1

fn��

// ker dCn

fn��

// Hn(C∗)

Hn(f∗)

��im dDn+1

// ker dDn // Hn(D∗)//

Exercise 7.11. Find necessary and sufficient conditions on f∗ so that Hn(f∗) = 0.

This functor is additive in the sense that Hn(f∗ + g∗) = Hn(f∗) + Hn(g∗). In otherwords, Hn gives a homomorphism

Hn : HomC∗(A)(C∗, D∗)→ HomA(Hn(C∗), Hn(D∗))

Additivity follows from the shape of the diagram in a way that I will explain later.

7.4.4. abelian category. We have now seen each condition (or its dual) in the definitionof an abelian category. Thus, we have seen that we need all of these conditions. (Wealso need enough projective and/or injective objects!)

Definition 7.12. An abelian category is an additive category C so that

(1) Every morphism has a kernel and a cokernel.(2) Every monomorphism is the kernel of its cokernel.(3) Every epimorphism is the cokernel of its kernel.(4) Every morphism f : A → B can be factored as the composition of an epimor-

phism A� I and a monomorphism I ↪→ B.

Remark 7.13. An abelian category might not have enough projective and/or injectiveobjects. For example, the category of finite abelian groups is an abelian category withoutany nonzero projective or injective objects.

Proposition 7.14. A morphism f : A → B in an abelian category is an isomorphismif and only if it is both mono and epi.

Proof. If f : A→ B is a monomorphism, its kernel is 0. If it is an epimorphism, it mustbe the cokernel of 0 → A which is A

=−→ A. So, f is an isomorphism. The converse isobvious. �


7.4.5. chain homotopy. Recall that a chain homotopy h∗ : f∗ ' g∗ between chainmaps f∗, g∗ : C∗ → D∗ is a sequence of morphisms

hn : Cn → Dn+1

so thatdDn+1 ◦ hn + hn−1 ◦ dCn = gn − fn

for all n ≥ 0 where h−1 = 0.

Cn+1

dn+1

��

// Dn+1

dn+1

��Cn

dn

��

gn−fn //

hn

88

Dn

dn

��Cn−1 //

hn−1

88

Dn−1

Theorem 7.15. Homotopic chain maps induce the same map in homology.

Proof. This follows from the fact that Hn is additive:

Hn(g∗)−Hn(f∗) = Hn(g∗ − f∗) = Hn(dD∗ ◦ h∗ + h∗ ◦ dC∗ )

= Hn(dD∗ ◦ h∗) +Hn(h∗ ◦ dC∗ )

But both of these are zero since dD∗ ◦ h∗ maps to the image of dD∗ and therefore to zeroin H∗(D∗) and h∗ ◦ dC∗ is zero on ker dC∗ . �

7.4.6. homotopy equivalence. Two chain complexes C∗, D∗ are called (chain) homo-topy equivalent and we write C∗ ' D∗ if there exist chain maps f∗ : C∗ → D∗ andg∗ : D∗ → C∗ so that f∗ ◦ g∗ ' idD and g∗ ◦ f∗ ' idC . The chain maps f∗, g∗ are called(chain) homotopy equivalences and we write f∗ : C∗ ' D∗.

Corollary 7.16. Any chain homotopy equivalence induces an isomorphism in homology.

Proof. Theorem 7.15 implies that Hn(f∗) ◦ Hn(g∗) = Hn(idD) = idHn(D) and similarlythe other way. So, Hn(f∗) is an isomorphism with inverse Hn(g∗). �


8. Homotopy invariance of projective resolutions

Here I proved that the projective resolution of any R-module (or any object of anabelian category with enough projectives) is unique up to chain homotopy. I useddiagrams and the (equivalent) equation. First I wrote down the understood standardinterpretation of a diagram.

Lemma 8.1. Given that the solid arrows (in diagram below) form a commuting diagram,there exists a dotted arrow as indicates making the entire diagram commute. [This isthe understood meaning of this kind of diagram.] The dotted arrow is not necessarilyuniquely determined (it is labeled ∃ and not ∃!) The assumptions are that P is projectiveand ker dn = im dn+1.

Cn+1

dn+1

��P

∃f̃==

f //

0 !!

Cn

dn��

Cn−1

Proof. By definition of kernel, f lifts uniquely to ker dn. But dn+1 : Cn+1 � im dn+1 =ker dn is onto by assumption. So, by definition of P being projective, f lifts to Cn+1. �

Lemma 8.2. With standard wording as above. The additional assumptions are that theright hand column is exact (i.e., im dCn+1 = ker dCn ), Pn+1 is projective and the left handcolumn is a chain complex (i.e. dPn ◦ dPn+1 = 0).

Pn+1

∃fn+1 //

dPn+1

��

Cn+1

dCn+1

��Pn

fn //

dPn��

Cn

dCn��

Pn−1fn−1 // Cn−1

Proof. The assumptions implies dCn ◦ (fn ◦ dPn+1) : Pn+1 → Cn−1 is 0. By the previouslemma this implies that fn ◦ dPn+1 : Pn+1 → Cn lifts to Cn+1. �

Lemma 8.3. Let f∗ : P∗ → C∗ be as in the above diagram. Suppose that there is amorphism hn−1 : Pn−1 → Cn so that

fn−1 ◦ dPn = dCn ◦ fn = dn ◦ hn−1 ◦ dPn(fd = df = dhd). Then there exists a morphism hn : Pn → Cn+1 so that (abbreviating):

(1) fn = dhn + hn−1d(2) dhnd = fnd = dfn+1

And by induction we get hn+1, hn+2, · · · satisfying (1), (2).


Proof. (1) implies (2). But the existence of hn satisfying (1): dhn = hn−1d− fn followsfrom the previous lemma and the calculation:

d(hn−1d− fn) = dhd− df = 0

by assumption. �

Pn+1

fn+1 //

dPn+1

��

Cn+1

dCn+1

��dnhn−1dn = dnfn ⇒ ∃hn : Pn

∃hn::

fn //

dPn

��

Cn

dCn

��Pn−1

fn−1 //

hn−1

::

Cn−1

Theorem 8.4. Suppose P∗ → M → 0 is a projective chain complex (augmented) overM and C∗ → N → 0 is a resolution of N (i.e., an exact sequence). Suppose f : M → Nis any morphism. Then

(1) There exists a chain map f∗ : P∗ → C∗ over f . I.e., the following diagramcommutes:

P∗f∗ //

ε��

C∗

ε��

Mf // N

(2) f∗ is unique up to chain homotopy (where h−1 : M → C0 is 0).

Proof. (1) Since ε : C0 → N is an epimorphism and P0 is projective, the map f ◦ ε :P0 → N lifts to a map f0 : P0 → C0. The rest is by induction using Lemma 8.2.

(2) Lemma 8.3 implies the existence of a homotopy: Given two chain maps f∗, g∗ :P∗ → C∗ covering the same morphism f : M → N , their difference covers 0 : M → N .We get the following diagram.

P1g1−f1 //

d

��

C1

d

��P0

∃h0::

g0−f0 //

ε

��

C0

ε

��M

0 //

h−1=0

::

N

�

This gives us the statement that we really want:


Corollary 8.5. In any abelian category with enough projectives, any object A has aprojective resolution P∗ → A. Furthermore, any two projective resolutions of A arehomotopy equivalent.

Proof. If there are two projective resolutions P∗, P′∗ then the first part of the theorem

above tells us that there are chain maps f∗ : P∗ → P ′∗ and g∗ : P ′∗ → P∗ which cover theidentity map on A. Since g∗ ◦ f∗ and the identity map are both chain maps P∗ → P∗over the identity of A, the second part of the theorem tells us that

f∗ ◦ g∗ ' idP∗

and similarly g∗ ◦ f∗ ' idP ′∗ . So, P∗ ' P ′∗. �

The dual argument gives us the following. [In general you should state the dualtheorem but not prove it.]

Theorem 8.6. In any abelian category with enough injectives, any object B has aninjective coresolution. Furthermore, any two injective coresolutions of B are homotopyequivalent.

Following this rule, I should also give the statement of the dual of the previoustheorem:

Theorem 8.7. Suppose 0 → M → Q∗ is an injective cochain complex under M and0 → N → C∗ is a coresolution of N (i.e., a long exact sequence). Suppose f : N → Mis any morphism. Then

(1) There exists a cochain map f∗ : C∗ → Q∗ under f . I.e., the following diagramcommutes:

C∗f∗ // Q∗

N

OO

f // M

OO

(2) f∗ is unique up to chain homotopy starting with h0 : C0 →M equal to 0.

Note: The subscript indicates covariant functor: (fg)∗ = f∗g∗.


9. Derived functors

A functor F : A → B between abelian categories A,B is called left exact if it is an

additive functor which takes every exact sequence 0→ Xj−→ Y

p−→ Z → 0 in A to a leftexact sequence

0→ FXFj−→ FY

Fp−→ FZ

in B. For example, F (X) = HomA(A,X) is a left exact functor

HomA(A0,−) : A →Mod-Z.We would like a formula for the cokernel of Fp : FY → FZ.

The “right derived functors” R1F,R2F, · · · are the answer to this problem. Thissequence of functors complete the left exact sequence to a canonical long exact sequence:

0→ FX → FY → FZ → R1FX → R1FY → R1FZ → R2FX → R2FY → · · ·The formula for RiF also gives R0F = F .

Contravariant left exact functors also give right derived functors. For example F =HomA(−, B0) is such a functor. These are functors which take each short exact sequence0→ X → Y → Z → 0 to left exact sequences

0→ FZ → FY → FX

and there right derived functors RiF extend this to a long exact sequence

0→ FZ → FY → FX → R1FZ → R1FY → R1FX → R2FZ → · · ·Analogously, covariant and contravariant right exact functors (such as −⊗R B) give

left derived functors (extending the sequence to the left):

· · · → L2FZ → L1FX → L1FY → L1FZ → FX → FY → FZ → 0

In this section we construct right derived functors and verify their properties.

9.1. Construction of RiF . Assume the category A is abelian with enough injectives.Then each object X of A has an injective coresolution X → Q∗ unique up to chainhomotopy equivalence.

Definition 9.1. The right derived functors RiF of F are defined to be the ithcohomology of the cochain complex F (Q∗):

0→ F (Q0)→ F (Q1)→ F (Q2)→ · · ·

In the case F = HomR(A,−), the right derived functors are the Ext functors:

ExtiR(A,X) := RiF (X) = H i(HomR(A,Q∗))

Note that the derived functors are only well-defined up to isomorphism. If there isanother choice of injective coresolutions Q′∗ then Q∗ ' Q′∗ which implies that F (Q∗) 'F (Q′∗) which implies that

H i(F (Q∗)) ∼= H i(F (Q′∗))

By definition of F (Q∗) we take only the injective objects. The term F (X) is deliber-ately excluded. But F is left exact by assumption. So we have an exact sequence

0→ F (X)→ F (Q0)→ F (Q1)


Thus

Theorem 9.2. The zero-th derived functor R0F is canonically isomorphic to F . Inparticular,

Ext0R(A,X) ∼= HomR(A,X)

Example 9.3. Compute Ext1Z(Z3,Z2). Take an injective coresolution of Z2:

0→ Z2 → Q/2Z j−→ Q/Z→ 0

This gives a left exact sequence

0→ Hom(Z3,Z2)→ Hom(Z3,Q/2Z)j]−→ Hom(Z3,Q/Z)

Then, by definition, Ext1Z(Z3,Z2) is the cokernel of j] (which is a monomorphism sinceHom(Z3,Z2) = 0).

Claim: This map is an isomorphism and, therefore, Ext1(Z3,Z2) = 0.Proof: A homomorphism Z/3 → Q/Z is given by its value on the generator 1 + 3Z

of Z/3Z. This must be a coset a/b + Z so that 3a/b ∈ Z. In other words b = 3 anda = 0, 1 or 2. Similarly, a homomorphism Z/3→ Q/2Z sends the generator of Z/3 to acoset a/b+ 2Z so that 3a/b ∈ 2Z. So, b = 3 and a = 0, 2 or 4. So, both of there groupshave exactly three elements. We know j] is a monomorphism. So, it is a bijection.

9.2. connecting morphisms. One of the basic properties of the derived functors isthat they fit into a long exact sequence where some maps are induced from the givenmaps α, β in the short exact sequence and every third map is a mysterious morphismcalled the “connecting map”. The main technical detail is to construct this map. Thereare several ways to do this. We can “chase the diagram” (the traditional method) or wecan “rotate” the exact sequence (a conceptual method which is more difficult but makessome sense)

Theorem 9.4. Given any short exact sequence

0→ Aα−→ B

β−→ C → 0

there is a sequence of morphisms (the connecting morphisms)

δn : RnF (C)→ Rn+1F (A)

making the following sequence exact:

0→ F (A)→ F (B)→ F (C)δ0−→ R1F (A)→ R1F (B)→

R1F (C)δ1−→ R2F (A)→ R2F (B)→ R2F (C)

δ2−→ R3F (A)→ · · ·Furthermore, δn is natural in the sense that, given any commuting diagram with exactrows:

0 // A

f��

α // B

g��

β // C

h��

// 0

0 // A′α′ // B′

β′ // C ′ // 0


we get a commuting square:

RnF (C)

h∗��

δn // Rn+1F (A)

f∗��

RnF (C ′)δn // Rn+1F (A′)

Here is one construction of these δ operators. First I needed the following lemmas.

Lemma 9.5. If Q is injective then RnF (Q) = 0 for all n ≥ 1.

Proof. The injective coresolution of Q has only one term:

0→ Q→ Q→ 0

So, R∗F (Q) is the cohomology of the exact sequence with only one term:

0→ F (Q)→ 0

and this is in degree 0. �

Lemma 9.6. If 0 → A → Q → K → 0 is a short exact sequence where Q is injective,then we have an exact sequence

0→ F (A)→ F (Q)→ F (K)π−→ R1F (A)→ 0

andRnF (K) ∼= Rn+1F (A)

for all n ≥ 1.

Proof. We can use Q = Q0 as the beginning of an injective coresolution of A:

0→ Aj−→ Q0

j0−→ Q1j1−→ Q2

j2−→ Q3 → · · ·Since coker j ∼= im j0 ∼= ker j1 ∼= K, we can break this up into two exact sequences:

0→ Aj−→ Q0 → K → 0

0→ K → Q1j1−→ Q2

j2−→ Q3 → · · ·The second exact sequence shows that the injective coresolution of K is the same asthat for A but shifted to the left with the first term deleted. So,

RnF (K) ∼= Rn+1F (A)

for all n ≥ 1.When n = 0 we have, by left exactness of F , the following exact sequence:

0→ F (K)→ F (Q1)(j1)∗−−→ F (Q2)

In other words, F (K) ∼= ker(j1)∗. The image of (j0)∗ : F (Q0) → F (Q1) lands inF (K) = ker(j1)∗. The cokernel is by definition the first cohomology of the cochaincomplex F (Q∗) which is equal to R1F (A). So, we get the exact sequence

F (Q0)→ F (K)π−→ R1F (A)→ 0

We already know that the kernel of F (Q0)→ F (K) is F (A) so this proves the lemma. �


The construction of the delta operator proceeds as follows. Start with any short exactsequence 0→ A→ B → C → 0. Then choose an injective coresolution of A:

0→ Aj−→ Q0

j0−→ Q1j1−→ Q2

j2−→ Q3 → · · ·Let K = ker j1 = im j0 = coker j. Since Q0 is injective, the map A → Q0 extends to Band cokernels map to cokernels giving a commuting diagram:

(9.1) 0 // A

idA��

α // B

f��

β // C

g

��

// 0

0 // Aj // Q0

p // K // 0

The map g : C → K induces a map g∗ : RnF (C) → RnF (K) and the connectingmorphism δn for n ≥ 1 can be defined to be the composition:

δn : RnF (C)g∗−→ RnF (K) ∼= Rn+1(A).

When n = 0, the map on the right is an epimorphism, not an isomorphism:

δ0 : F (C)g∗−→ F (K)

π−→ R1(A).

We will skip the following proof that δ∗ is independent of the choice of g : C → K.This follows from the fact that, for any other choice g′, the difference g−g′ lifts toQ0 sincef −f ′ : B → Q0 is zero on A and therefore factors through C. So, g∗−g′∗ = RnF (g−g′)factors through RnF (Q0) = 0 so g∗ = g′∗.

0 // A

0

��

α // B

f−f ′

��

β // C

g−g′

��

//

h

{{

0

0 // Aj // Q0

p // K // 0

h ◦ β = f − f ′ ⇒ p ◦ h = g − g′

Rn−1F (C) //

��

RnF (A)

0

��

α // RnF (B)

(f−f ′)∗��

β // RnF (C)

g∗−g′∗=0

��

//

h∗

ww

0

Rn−1F (K)∼= if n≥2 // // RnF (A)

j // RnF (Q0) = 0p // RnF (K) // 0

In the case n = 0, g∗ − g′∗ = F (g − g′) still factors through F (Q0), which is not zeroin general. But this is OK since F (Q0) goes to the kernel of π : F (K) → R1F (A). So,π∗(g∗ − g′∗) = 0 making δ0 = π∗ ◦ g∗ = π∗ ◦ g′∗.


9.3. connecting morphism: standard method. We will go over the standard methodof constructing the connecting morphism δ∗, show that it gives a long exact sequence(Theorem 9.4) and also show that, up to sign, it is the same morphism given by “rotat-ing” the short exact sequence. To make this comparison between different δ∗’s we needthe formula for δ∗ in each case.

Recall the rotation construction: Given a short exact sequence

0→ Aα−→ B

β−→ C → 0

we take an injective coresolution of A:

0→ Aj−→ Q0

j0−→ Q1j1−→ Q2

j2−→ · · ·Let K = im j0 = ker j1. This gives a short exact sequence A → Q0 → K and we get amorphism of short exact sequences (9.1).

(9.2) A

idA

��

α // B

f

��

β // // C

g

��

j′

''Q′0 j′0

''g0

��

Aj // Q0

p // //

j0 ++

K

''

Q′1

g1

��

Q1 j1

''Q2

The morphism g induces the connecting morphism:

δi : RiFCRiF (gi)−−−−→ RiFK ∼= Ri+1RA

where gi : Q′i → Qi+1 is the morphism of injective coresolutions induced by g : C → K.The standard construction uses two lemmas. The first lemma we will not prove since

this is usually done in a topology class.

Lemma 9.7 (Long exact sequence). Let A∗f∗−→ B∗

g∗−→ C∗ be a short exact sequence ofchain complexes or cochain complexes (or complexes indexed by all integers). I.e., foreach n, An → Bn → Cn is a short exact sequence. Then we get a long exact sequence inhomology:

· · · → Hn(A∗)Hn(f∗)−−−−→ Hn(B∗)

Hn(g∗)−−−−→ Hn(C∗)δ∗−→ Hn−1(A∗)→ · · ·

where f∗, g∗ are induced from the chain maps A∗ → B∗ and B∗ → C∗ and δ∗ is given by“chasing the diagram”, i.e., δn[z] = [x], z ∈ ker dn : Cn → Cn−1 is given by lifting z toz̃ ∈ Bn and letting x ∈ An−1 be the unique element which maps to dn(z̃) ∈ Bn−1:

0 // An

dn��

fn // ∃z̃ ∈ Bn

dn��

gn // z ∈ Cndn��

// 0

0 // x ∈ An−1fn−1// dn(z̃) ∈ Bn−1

gn−1 // 0 ∈ Cn−1 // 0


You need to verify that δ∗ is well-defined and that, together with the induced mapsHn(f∗), Hn(g∗), gives the long exact sequence in homology.

Corollary 9.8. Given a short exact sequence of chain or cochain complexes, if two ofthem are exact, so is the third. �

Lemma 9.9 (Horseshoe Lemma). Given injective coresolutions A→ Q∗ of A and C →Q′∗ of C, there is an injective coresolution of B of the following form:

B −→ Q0 ⊕Q′0 −→ Q1 ⊕Q′1 −→ Q2 ⊕Q′2 −→ · · ·

and the following commuting diagram with exact rows and columns:

0 // A

j

��

α // B

fj′ ◦ β

��

β // C

j′

��

// 0

0 // Q0

j0

��

10

// Q0 ⊕Q′0j0 h0

0 j′0

��

[0,1]// Q′0

j′0

��

// 0

0 // Q1

j1

��

10

// Q1 ⊕Q′1j1 h1

0 j′1

��

[0,1]// Q′1

j′1

��

// 0

0 // Q2

��

10

// Q2 ⊕Q′2

��

[0,1]// Q′2

��

// 0

Where f : B → Q0 and hi = (−1)i+1gi are given in (9.2).

Proof. The rows are clearly exact seqences. Columns 1 and 3 are given to be exact.Corollary 9.8 implies that Column 2 is exact as long as it is a complex. In other words,[

j0 h00 j′0

][f

j′ ◦ β

]=

[00

]=

[j0 ◦ f + h0 ◦ j′ ◦ β

j′0 ◦ j′ ◦ β

]Looking at diagram (9.2) we see that j0 ◦ f = g0 ◦ j′ ◦ β. So, h0 = −g0 makes this work.(Also j′0 ◦ j′ = 0 is given.) And[

j1 h10 j′1

][j0 h00 j′0

]=

[0 00 0

]=

[j1 ◦ j0 j1 ◦ h0 + h1 ◦ j′0

0 j′1 ◦ j′0

]By diagram (9.2) we see that j1◦g0 = g1◦j′0. So, h0 = −g0, h1 = g1 makes this work. �


Proof of Theorem 9.4. When we apply the functor F to the three coresolutions Q∗ →Q∗ ⊕Q′∗ → Q′∗, using Lemma 9.7 we get the long exact sequence:

0→ FA→ FB → FCδ0−→ R1FA→ R1FB → R1FC

δ1−→ R2FA→ · · ·

By the formula hi = (−1)i+1gi, the connecting morphism δi is the same, up to sign, asthe one given by rotating the exact sequence. �

9.4. Left derived functors. There are two cases when we use projective resolutionsinstead of injective coresolutions:

(1) When the functor is left exact but contravariant, e.g., F = HomR(−, B).(2) When the functor is right exact and covariant, e.g., F = −⊗R B.

In both cases we take a projective resolution P∗ → A→ 0 and define the left derivedfunctors to be LnF (A) = Hn(F (P∗)) in the first case and LnF (A) = Hn(F (P∗)) in thesecond case.

(Whether the contravariant functor ExtiR(−, B) is a right or left derived functor de-pends on whether it is viewed as a left-exact functor (Mod-R)op →Mod-Z, as we do, oras a right-exact functor Mod-R→ (Mod-Z)op as in [Lang].)

Definition 9.10. The left derived functors of F (A) = A ⊗R B are called LnF (A) =TorRn (A,B)

9.4.1. review of tensor product. Following Lang, I will take tensor products only overcommutative rings. The advantage is that A ⊗R B will be an R-module. The tensorproduct is defined by a universal condition.

Definition 9.11. Suppose that A,B are modules over a commutative ring R. Then amap

g : A×B → C

from the Cartesian product A × B to a third R-module C is called R-bilinear if it isan R-homomorphism in each variable. I.e., for each a ∈ A, the mapping b 7→ g(a, b) isa homomorphism B → C and similarly g(−, b) ∈ HomR(A,C) for all a ∈ A. A⊗R B isdefined to be the R-module which is the target of the universal R-bilinear map

f : A×B → A⊗B

When I say that f is universal I mean that for any other R-bilinear map g : A×B → Cthere is a unique R-homomorphism h : A⊗B → C so that g = h ◦ f .

The universal property tells us that A⊗R B is unique if it exists. To prove existencewe need to construct it. But this is easy. You just take A⊗RB to be the free R-modulegenerated by all symbols a⊗b where a ∈ A, b ∈ B modulo the relations that are required,namely:

(1) (ra)⊗ b = r(a⊗ b)(2) (a+ a′)⊗ b = a⊗ b+ a′ ⊗ b(3) a⊗ rb = r(a⊗ b)(4) a⊗ (b+ b′) = a⊗ b+ a⊗ b′


I pointed out that the universal property can be expressed as an isomorphism

HomR(A⊗B,C) ∼= BiLinR(A×B,C)

And the definition of R-bilinear can be expressed as the isomorphisms

BiLinR(A×B,C) ∼= HomR(A,HomR(B,C)) ∼= HomR(B,HomR(A,C))

So, we conclude that

HomR(A⊗B,C) ∼= HomR(A,HomR(B,C))

This is a special case of:

HomR(F (A), C) ∼= HomR(A,G(C))

with F = ⊗B and G = HomR(B, ). When we have this kind of isomorphism, F iscalled the left adjoint and G is called the right adjoint and F,G are called adjointfunctors.

Lemma 9.12. Any left adjoint functor is right exact. In particular, tensor product isright exact. Also, any right adjoint functor is left exact.

Proof. In the first case, suppose that F is a left adjoint functor and

(9.3) 0→ Aα−→ A′

β−→ A′′ → 0

is a short exact sequence. Then for any C, the left exactness of HomR(−, G(C)) givesan exact sequence

0→ HomR(A′′, G(C))→ HomR(A′, G(C))→ HomR(A,G(C))

By adjunction, this is equivalent to an exact sequence

0→ HomR(F (A′′), C))→ HomR(F (A′), C))→ HomR(F (A), C))

The exactness of this sequence for all C is equivalent to the exactness of the followingsequence by definition of cokerF (α):

F (A)F (α)−−→ F (A′)→ F (A′′)→ 0

The left exactness of G is analogous. (Also, the proof uses the left exactness of Hom sothe second case is dumb.) �


9.4.2. Properties of tensor product. Recall that R is a commutative ring with unity. So,left and right R-modules are the same thing. The first property is “obvious”.

Lemma 9.13. A⊗R B ∼= B ⊗R A.

The second property holds for any additive functor.

Lemma 9.14. Let F be any additive functor. Then

F (A1 ⊕ A2) ∼= F (A1)⊕ F (A2).

In particular we have (A1 ⊕ A2)⊗R B ∼= (A1 ⊗R B)⊕ (A2 ⊗R B) andA⊗R (B1 ⊕B2) ∼= A⊗R B1 ⊕ A⊗R B2.

Proof. There are morphisms ji : Ai → A1⊕A2 and pi : A1⊕A2 → Ai so that pi◦ji = idAi ,pk ◦ ji = 0 for k 6= i and

∑ji ◦ pi = idA1⊕A2 . Since F is additive, F (ji), F (pi) satisfy the

same formulas. So, F (A1 ⊕ A2) = F (A1)⊕ F (A2). �

Another important lemma is the following.

Lemma 9.15. R⊗R B is isomorphic to B as R-modules.

Proof. Let f : R × B → B be the map f(r, b) = rb. This is R-bilinear when R iscommutative. So, we get an induced homomorphism

f : R⊗B → B

sending r⊗ b to rb. The inverse is given by g(b) = 1⊗ b. We don’t need to show that gis a homomorphism, just that it is a set map inverse to f . Then f is an isomorphism.

Verification: Clearly fg(b) = 1 · b = b. So, f ◦ g = idB.Conversely, take any element

∑ri ⊗ bi ∈ R⊗B. Then

gf(∑

ri ⊗ bi)

=∑

gf(ri ⊗ bi) =∑

1⊗ ribi =∑

ri ⊗ bi

So, g ◦ f is the identity on R⊗B. So, R⊗B ∼= B. �

9.4.3. computations. With these lemmas, I did some computations. Suppose that R = Zand A = Z/n. Then a projective resolution of A is given by

0→ Z ·n−→ Z→ Z/n→ 0

Since this sequence is exact it gives the following right exact sequence for any abeliangroup B:

Z⊗B ·n−→ Z⊗B → Z/nZ⊗B → 0

Using the lemma that R⊗R B ∼= B this becomes:

B·n−→ B → Z/nZ⊗B → 0

So, we conclude thatZ/nZ⊗B ∼= B/nB

More generally, if A is any finitely generated abelian group then

A ∼= Zr ⊕ Z/t1 ⊕ Z/t2 ⊕ · · · ⊕ Z/tn


and, since tensor product distributes over direct sum we get:

A⊗Z B = Br ⊕B/t1B ⊕B/t2B ⊕ · · · ⊕B/jnBThe derived functor TorZ1 (Z/n,B) is by definition the kernel of the map

Z⊗B ·n−→ Z⊗BSince Z⊗B = B this is just the map B → B given by multiplication by n. So,

TorZ1 (Z/n,B) = {b ∈ B : nb = 0}It is the subgroup of B consisting of all elements whose order divides n. It is the“n-torsion” subgroup of B. Maybe that is why it is called Tor.

9.4.4. extension of scalars. Suppose that R is a subring of S (e.g., Z ⊂ R). A homo-morphism of free R-modules

Rn f−→ Rm

is given by a matrix M(f) = (aij) as follows. If the basis elements of Rn are ej and thebasis elements of Rm are ei then

f(ej) =m∑i=1

eiaij

for some aij ∈ R. These numbers determine f since, for an arbitrary element x =∑ejxj ∈ Rn we have

f(x) = f

(∑j

ejxj

)=∑i,j

eiaijxj

This can be written in matrix form:

f

x1x2· · ·xn

=

∑a1jxj∑a2jxj· · ·∑anjxj

= (aij)

x1x2· · ·xn

When you tensor with S you get Rn ⊗R S = (R⊗R S)n = Sn

Rn ⊗R S = Snf⊗idS−−−→ Rm ⊗R S = Sm

The claim is that M(f⊗idS) = M(f). So, f∗ = f⊗idS is obtained from f by “extendingscalars” to S. If you have an integer matrix, you just take the same matrix and considerit as a real matrix.


9.4.5. two definitions of Ext. The last topic in Homological Algebra is the proof thatthe two definitions of ExtnR(A,B) that we now had are equivalent. Similarly, the twodefinitions of TorRn (A,B) given by taking projective resolution of A or B are equivalent.

Theorem 9.16. If P∗ → A is a projective resolution of A and B → Q∗ is an injectiveresolution of B then

Hn(Hom(P∗, B)) ∼= Hn(Hom(A,Q∗))

So, either formula gives ExtnR(A,B).

Proof. The theorem is true in the case when n = 0 because both sides are isomorphic toHomR(A,B). So, suppose n ≥ 1. I will give the proof in the case n = 2.

The idea is to give a description of ExtnR(A,B) which uses both P∗ and Q∗. Take P∗,Q∗ fixed and consider all possibilities for the following commuting diagram.

(9.4) P3

f3��

d3 // P2

f2��

d2 // P1

f1��

d1 // P0

f0��

d0 // A

f��

// 0

��0 // B // Q0 j0

// Q1 j1// Q2 j2

// Q3

Claim: The group of homotopy classes of chain maps f∗ : (P∗ → A) → (B → Q∗)shifted in degree by n (as in the above diagram when n = 2) is isomorphic to bothHn(Hom(P∗, B)) and Hn(Hom(A,Q∗)).

Proof: The first step is to show that f determines the fi, i ≥ 1 uniquely up tohomotopy. This follows from Theorem 8.4 since P∗ is a projective resolution of A and0→ B → Q0 → Q1 → · · · → Qn is exact.

But the choice of f up to homotopy is:

[f ] ∈ ker((jn)] : HomR(A,Qn)→ HomR(A,Qn+1))

im((jn−1)] : HomR(A,Qn−1)→ HomR(A,Qn))= Hn(Hom(A,Q∗))

Therefore, the group of homotopy classes of chain maps f∗ : (P∗ → A)→ (B → Q∗)shifted in degree by n is isomorphic to Hn(Hom(A,Q∗)).

Similarly, the morphism fn : Pn → B determines the maps fn−1 : Pn−1 → Q0, etcdown to f : A → Qn uniquely up to homotopy since the Qi are injective. And thehomotopy class of fn is given by

[fn] ∈ker(d]n+1 : HomR(Pn, B)→ HomR(Pn+1, B))

im(d]n : HomR(Pn−1, B)→ HomR(Pn, B))= Hn(Hom(P∗, B)).

So, the group of homotopy classes of chain maps f∗ : (P∗ → A)→ (B → Q∗) shiftedin degree by n is isomorphic to Hn(Hom(P∗, B)). Therefore,

Hn(Hom(A,Q∗)) ∼= Hn(Hom(P∗, B))

proving the theorem. �

This proof comes from the following interpretation of the higher Ext groups.Given R-modules A,B consider the set of all exact sequences of length n+ 2:

0→ B → E1 → E2 → · · · → En → A→ 0


Call these n-fold extension of B by A. Define two such sequences to be equivalent ifthere is a chain map between them which is the identity on both A and B:

0 // B

=

��

// E1

��

// E2

��

// · · · // En

��

// A

=

��

// 0

0 // B // E ′1 // E ′2 // · · · // E ′n // A // 0

Theorem 9.17. The set of equivalence classes of n-fold extension of B by A is inbijection with the set ExtnR(A,B).

I won’t go through the details. Just to say that, given any n-fold extension as above,the corresponding homotopy class [g] ∈ Hn(Hom(P∗, B)) is given as follows. Since P∗ isa projective resolution of A, there is, by Theorem 8.4, a unique homotopy class of chainmaps from P∗ to B → E1 → · · · → En covering the identity on A. On level n this givesthe map g : Pn → B.

Similarly, since Q∗ is an injective coresolution of B, there is a unique homotopy classof cochain complexes: E1 → · · · → En → A → 0 to Q∗ under the identity on B. Atlevel n this gives a map f : A→ Qn representing [f ] ∈ Hn(Hom(A,Q∗)).

The difficulty with this construction is that one cannot easily see how to add twon-fold extensions. However, there is one advantage. There is a clear notion of iterationof extension given a map:

ExtnR(A,B)⊗ ExtmR (B,C)→ Extn+mR (A,C)

And this notion is clearly associative and makes⊕

Extn(A,A) into an associative ring!

Exercise 9.18. Show that the identity in this ring is idA ∈ HomR(A,A) = Ext0(A,A).

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MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRApeople.brandeis.edu/~igusa/Math131b/Math131b_notesA.pdf · 2019-02-27 · MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 3 Corollary