MATH 101B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA

MATH 101B: ALGEBRA IIPART A: HOMOLOGICAL ALGEBRA

These are notes for our first unit on the algebraic side of homologicalalgebra. While this is the last topic (Chap XX) in the book, it makessense to do this first so that grad students will be more familiar withthe ideas when they are applied to algebraic topology (in 121b). Atthe same time, it is not my intention to cover the same material twice.The topics are

Contents

1. Additive categories 12. Abelian categories 22.1. some definitions 22.2. definition of abelian category 32.3. examples 43. Projective and injective objects 54. Injective modules 74.1. dual module 74.2. constructing injective modules 84.3. proof of lemmas 94.4. Examples 125. Divisible groups 156. Injective envelope 177. Projective resolutions 207.1. Definitions 207.2. Modules of a PID 217.3. Chain complexes 237.4. Homotopy uniqueness of projective resolutions 267.5. Derived functors 297.6. Left derived functors 34

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MATH 101B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 1

1. Additive categories

On the first day I talked about additive categories.

Definition 1.1. An additive category is a category C for which everyhom set HomC(X, Y ) is an additive group and

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

(2) The category has finite direct sums.

I should have gone over the precise definition of direct sum:

Definition 1.2. The direct sum ⊕ni=1Ai is an object X together withmorphisms ji : Ai → X, pi : X → Ai so that

(1) pi ◦ ji = id : Ai → Ai(2) pi ◦ jj = 0 if i 6= j.(3)

∑ji ◦ pi = idX : X → X.

Theorem 1.3. ⊕Ai is both the product and coproduct of the Ai

Proof. Suppose that fi : Y → Ai are morphisms. Then there is amorphism

f =∑

ji ◦ fi : Y → ⊕Aiwhich has the property that pi ◦ f = pijifi = fi. Conversely, given anymorphism g : Y → ⊕Ai satisfying pi ◦ g = fi for all i, then we have:

f =∑

jifi =∑

jipig = idX ◦ g = g

So, f is unique and ⊕Ai is the product of the Ai. By an analogousargument, it is also the coproduct. �

2 MATH 101B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA

The converse is also true:

Proposition 1.4. Suppose that X =∏Ai =

∐Ai and the composition

of the inclusion ji : Ai → X with pj : X → Aj is

pj ◦ ji = δij : Ai → Aj

I.e., it is the identity on Ai for i = j and it is zero for i 6= j. Then∑ji ◦ pi = idX

Proof. Let f =∑ji ◦ pi : X → X. Then

pj ◦ f =∑i

pj ◦ ji ◦ pi =∑i

δijpi = pj

So, f = idX by the universal property of a product. �

In class I pointed out the sum of no objects is the zero object 0 whichis both initial and terminal. Also, I asked you to prove the following.

Problem. Show that a morphism f : A → B is zero if and only if itfactors through the zero object.

2. Abelian categories

2.1. some definitions. First I explained the abstract definition of ker-nel, monomorphism, cokernel and epimorphism.

Definition 2.1. A morphism f : A → B in an additive category C isa monomorphism if, for any object X and any morphism g : X → A,f ◦ g = 0 if and only if g = 0. (g acts like an “element” of A. It goesto zero in B iff it is zero.)

Another way to say this is that f : A→ B is a monomorphism andwrite 0→ A→ B if

0→ HomC(X,A)f]−→ HomC(X,B)

is exact, i.e., f] is a monomorphism of abelian groups. The lower sharpmeans composition on the left or “post-composition.” The lower sharpis order preserving:

(f ◦ g)] = f] ◦ g]Epimorphisms are defined analogously: f : B → C is an epimor-

phism if for any object Y we get a monomorphism of abelian groups:

0→ HomC(C, Y )f]−→ HomC(B, Y )


An abelian category is an additive category which has kernels andcokernels satisfying all the properties that one would expect which canbe stated categorically. First, I explained the categorical definition ofkernel and cokernel.

Definition 2.2. The kernel of a morphism f : A→ B is an object Kwith a morphism j : 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 h : X → K so that g = j ◦ h.

Since this is a universal property, the kernel is unique if it exists.

Theorem 2.3. A is the kernel of f : B → C if and only if

0→ HomC(X,A)→ HomC(X,B)→ HomC(X,C)

is exact for any object X. In particular, j : ker f → B is a monomor-phism.

If you replace A with 0 in this theorem you get the following state-ment.

Corollary 2.4. A morphism is a monomorphism if and only if 0 is itskernel.

Cokernel is defined analogously and satisfies the following theoremwhich can be used as the definition.

Theorem 2.5. The cokernel of f : A → B is an object C with amorphism B → C so that

0→ HomC(C, Y )→ HomC(B, Y )→ HomC(A, Y )

is exact for any object Y .

Again, letting C = 0 we get the statement:

Corollary 2.6. A morphism is an epimorphism if and only if 0 is itscokernel.

These two theorems can be summarized by the following statement.

Corollary 2.7. For any additive category C, HomC is left exact in eachcoordinate.

2.2. definition of abelian category.

Definition 2.8. 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 epimorphism A � I and a monomorphism I ↪→ B.(5) A morphism f : A → B is an isomorphism if and only if it is

both mono and epi.

Proposition 2.9. The last condition follows from the first four condi-tions.

Proof. First of all, isomorphisms are always both mono and epi. Thedefinition of an isomorphism is that it has an inverse g : B → A sothat f ◦ g = idB and g ◦ f = idA. The second condition implies that fis mono since

(g ◦ f)] = g] ◦ f] = id] = id

which implies that f] is mono and f is mono. Similarly, f ◦ g = idBimplies that f is epi.

Conversely, suppose that f : A→ B is both mono and epi. Then, by(2), it is the kernel of its cokernel which is B → 0. So, by left exactnessof Hom we get:

0→ HomC(B,A)→ HomC(B,B)→ HomC(B, 0)

In other words, f] : HomC(B,A) ∼= HomC(B,B). So, there is a uniqueelement g : B → A so that f ◦ g = idB. Similarly, by (3), there is aunique h : B → A so that h ◦ f = idA. If we can show that g = h thenit will be the inverse of f making f invertible and thus an isomorphism.But this is easy:

h = h ◦ idB = h ◦ f ◦ g = idA ◦ g = g

�

2.3. examples. The following are abelian categories:

(1) The category of abelian groups and homomorphisms.(2) The category of finite abelian groups. This is an abelian cat-

egory since any homomorphism of finite abelian groups has afinite kernel and cokernel and a finite direct sum of finite abeliangroups is also finite.

(3) R-mod= the category of all left R-modules and homomorphisms(4) R-Mod= the category of finitely generated (f.g.) left R-modules

is an abelian category assuming that R is left Noetherian (allsubmodules of f.g. left R-modules are f.g.)

(5) mod-R=the category of all rightR-modules and homomorphisms(6) Mod-R=the category of f.g. right R-modules is abelian if R is

right Noetherian.


The following are examples of additive categories which are notabelian.

(1) Free abelian groups. (This category does not have cokernels)(2) Let R be a non-Noetherian ring, for example a polynomial ring

in infinitely many variables:

R = k[X1, X2, · · · ]

ThenR-Mod, the category of f.g. R-modules is not abelian sinceit does not have kernels. E.g., the kernel of the augmentationmap

R→ k

is infinitely generated.

3. Projective and injective objects

At the end of the second lecture we discussed the definition of injec-tive and projective objects in any additive category. And it was easy toshow that the category of R-modules has sufficiently many projectives.

Definition 3.1. An object P of an additive category C is called projec-tive if for any epimorphism f : A → B and any morphism g : P → Bthere exists a morphism g̃ : P → A so that f ◦ g̃ = g. The map g̃ iscalled a lifting of g to A.

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

Proof. If 0→ A→ B → C → 0 is exact then, by left exactness of Homwe get an exact sequence:

0→ HomC(P,A)→ HomC(P,B)→ HomC(P,C)

By definition, P is projective if and only if the last map is always anepimorphism, 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 elementof F can be written uniquely as

∑rαxα where the coefficients rα ∈ R

are almost all zero (only finitely many are nonzero). Suppose thatg : F → B is a homomorphism. Then, for every index α, the element


g(xα) comes from some element yα ∈ A. I.e., g(xα) = f(yα). Then alifting g̃ of g is given by

g̃(∑

rαxa) =∑

rαyα

The verification that this is a lifting is “straightforward” or I would say“obvious” but it would go like this: The claim is that, first, g̃ : F → Ais a homomorphism of R-modules and, second, it is a lifting: f ◦ g̃ = g.The second statement is easy:

f◦g̃(∑

rαxa) = f(∑

rαyα) =∑

rαf(yα) =∑

rαg(xα) = g(∑

rαxα)

The first claim says that g̃ is additive:

g̃(∑

rαxα +∑

sαxα

)= g̃

(∑(rα + sα)xα

)=∑

(rα + sα)yα = g̃(∑

rαxα) + g̃(∑

sαxα)

and g̃ commutes with the action of R:

g̃(r∑

rαxα

)= g̃

(∑rrαxα

)=∑

rrαyα = r∑

rαxα = rg̃(∑

rαxα

)�

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

p(∑

rα[xα])

=∑

rαxα

The notation [x] is used to distinguish between the element x ∈M andthe corresponding generator [x] ∈ F . The homomorphism p is actuallyjust defined by the equation p[x] = x.

Corollary 3.4. The category of R-modules has sufficiently many pro-jectives, i.e., for every R-module M there is a projective R-modulewhich maps onto M .

This implies that every R-module M has a projective resolution

0←M ← P0 ← P1 ← P2 ← . . .

This is an exact sequence in which every Pi is projective. The projec-tive modules are constructed inductively as follows. First, P0 is anyprojective which maps onto M . This gives an exact sequence:

P0 →M → 0


By induction, we get an exact sequence

Pn → Pn−1 → Pn−2 → · · · → P0 →M → 0

Let Kn be the kernel of dn : Pn → Pn−1. Since there are enoughprojectives, there is a projective module Pn+1 which maps onto Kn.The composition Pn+1 � Kn ↪→ Pn is the map dn+1 : Pn+1 → Pnwhich extends the exact sequence one more step.

Definition 3.5. An object Q of C is injective if, for any monomorphismA→ B, any morphism A→ Q extends to B. I.e., iff

HomC(B,Q)→ HomC(A,Q)→ 0

As before this is equivalent to:

Theorem 3.6. Q is injective if and only if HomC(−, Q) is an exactfunctor.

The difficult theorem we need to prove is the following:

Theorem 3.7. The category of R-modules has sufficiently many injec-tives. I.e., every R-module embeds in an injective R-module.

As in the case of projective modules this theorem will tell us thatevery R-module M has an injective co-resolution which is an exactsequence:

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

4. Injective modules

I will go over Lang’s proof that every R-module M embeds in aninjective module Q. Lang uses the dual of the module.

4.1. dual module.

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

M∧ := HomZ(M,Q/Z)

with right R-action given by

φr(x) = φ(rx)

for all φ ∈M∧, r ∈ R.


Proposition 4.2. Duality is a left exact functor

( )∧ : R-mod→ mod-R

which is additive and takes sums to products:

(⊕Mα)∧ ∼=∏

M∧α

Proof. We already saw that the hom functor HomZ(−, X) is left exactfor any abelian group X. It is also obviously additive which meansthat (f + g)] = f ] + g] for all f, g : N → M . I.e., the duality functorinduces 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 → Xand the fα can all be nonzero. (So, it is the product not the sum.) �

4.2. constructing injective modules. In order to get an injectiveleft R-module we need to start with a right R-module.

Theorem 4.3. Suppose F is a free right R-module. (I.e., F = ⊕RR

is a direct sum of copies of R considered as a right R-module). ThenF∧ is an injective left R-module.

This theorem follows from the following lemma.

Lemma 4.4. (1) A product of injective modules is injective.(2) HomR(M,R∧R) ∼= HomZ(M,Q/Z)(3) Q/Z is an injective Z-module.

Proof of the theorem. Lemma (3) implies that HomZ(−,Q/Z) is an ex-act functor. (2) implies that HomR(−, R∧R) is an exact functor. There-fore, R∧R is an injectiveR-module. Since duality takes sums to products,(1) implies that F∧ is injective for any F with is a sum of RR’s, i.e. Fis a free right R-module. �

We need one more lemma to prove the main theorem. Then we haveto prove the lemmas.

Lemma 4.5. Any left R-module is naturally embedded in its doubledual:

M ⊆M∧∧

Assume this 4th fact for a moment.


Theorem 4.6. Every left R-module M can be embedded in an injectiveleft R-module.

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

F →M∧ → 0

Since duality is left exact we get:

0→M∧∧ → F∧

By the last lemma we have M ⊆ M∧∧ ⊆ F∧. So, M embeds in theinjective module F∧. �

4.3. proof of lemmas. There are four lemmas to prove. Suppose fora moment that T = Q/Z is injective then the other three lemmas areeasy:

Proof of Lemma 4.5. A natural embedding M →M∧∧ is given by theevaluation map ev which sends x ∈ M to evx : M∧ → T which isevaluation at x:

evx(φ) = φ(x)

Evaluation is additive:

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

Evaluation is an R-module homomorphism:

evrx(φ) = φ(rx) = (φr)(x) = evx(φr) = (revx) (φ)

Finally, we need to show that ev is a monomorphism. In other words,for every nonzero element x ∈ M we need to find some additive mapφ : M → T so that evx(φ) = φ(x) 6= 0. To do this take the cyclic groupC 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. If C ∼= Z then letf : C → T be given by

f(kn) =k

2+ Z

Then again, f(x) is nonzero. Since T is Z-injective, f extends to anadditive map φ : M → T . So, evx is nonzero and ev : M → M∧∧ is amonomorphism. �


Proof that products of injectives are injective. Suppose that Jα are in-jective. Then we want to show that Q =

∏Jα is injective. Let

pα : Q → Jα be the projection map. Suppose that f : A → B isa monomorphism and g : A → Q is any morphism. Then we want toextend g to B.

Since each Jα is injective each composition pα ◦ g : A→ Jα extendsto a morphism gα : B → Jα. I.e., gα ◦f = pα ◦g for all α. By definitionof the product there exists a unique morphism g : B → Q =

∏Jα so

that pα ◦ g = gα for each α. So,

pα ◦ g ◦ f = gα ◦ f = pα ◦ g : A→ Jα

The uniquely induced map A →∏Jα is g ◦ f = g. Therefore, g is an

extension of g to B as required. �


Finally, we need to prove that

HomR(M,R∧R) ∼= HomZ(M,Q/Z)

To do this we will give a 1-1 correspondence and show that it (thecorrespondence) is additive.

If f HomR(M,R∧R) then f is a homomorphism f : M → R∧R whichmeans that for each x ∈M we get a homomorphism f(x) : R→ Q/Z.In particular we can evaluate 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,R∧R)→ HomZ(M,Q/Z)

We need to know that this is additive. I used “know” instead of“show” since this is one of those steps that you should normally skip.However, you need to know what it is that you are skipping. The factthat we need to know is that

φ(f + g) = φ(f) + φ(g)

This is an easy calculation which follows from the way that f + gis defined, namely, addition of function is defined “pointwise” whichmeans 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)

Finally we need to show that φ is a bijection. To do this we findthe inverse φ−1 = ψ. For any homomorphism g : M → Q/Z letψ(g) : M → R∧R be given by

ψ(g)(x)(r) = g(rx)

Since this is additive in all three variables, ψ is additive and ψ(g) isadditive. We also need to check that ψ(g) is a homomorphisms of leftR-modules, i.e., that ψ(g)(rx) = rψ(g)(x). This is an easy calculation:

ψ(g)(rx)(s) = g(s(rx)) = g((sr)x)

[rψ(g)(x)](s) = [ψ(g)(x)](sr) = g((sr)x)

The verification that ψ is the inverse of φ is also straightforward:For all f ∈ HomR(M,R∧R) 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(1x) = g(x)


So, φ(ψ(g)) = g.I will do the last lemma (injectivity of Q/Z) tomorrow.

4.4. Examples. I prepared 3 examples but I only got to two of themin class.

4.4.1. polynomial ring. LetR = Z[t], the integer polynomial ring in onegenerator. This is a commutative Noetherian ring. It has dimension 2since a maximal tower of prime ideal is given by

0 ⊂ (t) ⊂ (t, 2)

These ideals are prime since the quotient of R by these ideals are do-mains (i.e., have no zero divisors):

R/0 = Z[t], R/(t) = Z

are domains and

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

Proposition 4.7. A Z[t] module M is the same as an abelian grouptogether with an endomorphism M → M given by the action of t. Ahomomorphism of Z[t]-modules f : M → N is an additive homomor-phism which commutes with the action of t.

Proof. I will use the fact that the structure of an R-module on anadditive group M is the same as a homomorphism of rings φ : R →End(M). When R = Z[t], this homomorphism is given by its value ont since φ(f(t)) = f(φ(t)). For example, if f(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. �

What do the injective R-modules look like? We know that Q = R∧Ris injective. 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, · · · . There-fore, an element f ∈ Q, f : Z[t] → Q/Z is given uniquely by thesequence

f(1), f(t), f(t2), f(t3), · · · ∈ Q/ZMultiplication by t shifts this sequence to the left since tf(ti)−f(tit) =f(ti+1). This proves the following.


Theorem 4.8. The injective module Q = Z[t]∧ is isomorphic to theadditive group of all sequences (a0, a1, a2, · · · ) of elements ai ∈ Q/Zwith the action of t given by shifting to the left and dropping the firstcoordinate. I.e.,

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

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

4.4.2. fields. Suppose that R = k is a field. Then I claim that allk-modules are both projective and injective.

First note that a k-module is the same as a vector space over thefield k. Since every vector space has a basis, all k-modules are free.Therefore, all k-modules are projective. Then I went through a roundabout argument to show that all k-modules are injective and I onlymanaged to show that finitely generated k-modules are injective. (Moreon this later.)

Finally, I started over and used the following theorem.

Theorem 4.9. Suppose that R is any ring. Then the following areequivalent (tfae).

(1) All left R-modules are projective.(2) All left R-modules are injective.(3) Every short exact sequence of R-modules splits.

First I recalled the definition of a splitting of a short exact exactsequence.

Proposition 4.10. Given a short exact sequence of left R-modules

(4.1) 0→ Af−→ B

g−→ C → 0

Tfae.

(1) B = f(A)⊕D for some submodule D ⊆ B.(2) f has a retraction, i.e., a morphism r : B → A s.t. r ◦ f = idA.(3) g has a section, i.e., a morphism s : C → B s.t. g ◦ s = idC.

Proof. This is a standard fact that most people know very well. Forexample, (1) ⇒ (2) because a retraction r is given by projection tothe first coordinate followed by the inverse of the isomorphism f :A → f(A). (2) ⇒ (1) by letting D = ker r. [You need to verify thatB = f(A)⊕D which is in two steps: D∩f(A) = 0 and D+f(A) = B.For example, ∀x ∈ B, x = fr(x) + (x− fr(x)) ∈ f(A) +D.] �


Proof of Theorem . (1) ⇒ (3): In the short exact sequence (4.1), C isprojective (since all modules are assumed projective). Therefore, theidentity map C → C lifts to B and the sequence splits.

(3) ⇒ (1): Since any epimorphism g : B → C has a section s, any

morphism f : X → C has a lifting f̃ = s ◦ f : X → B.The equivalence (2) ⇐⇒ (3) is similar. �


5. Divisible groups

Z-modules are the same as abelian groups. And we will see thatinjective Z-modules are the same as divisible groups.

Definition 5.1. An abelian group D is called divisible if for any x ∈ Dand any positive integer n there exists y ∈ D so that ny = x. (We saythat x is divisible by n.)

For example, Q is divisible. 0 is divisible. A finite groups is divisibleif and only if it is 0.

Proposition 5.2. Any quotient of a divisible group is divisible.

Proof. Suppose D is divisible and K is a subgroup. Then any elementof the quotient D/K has the form x+K where x ∈ D. This is divisibleby any positive n since, if ny = x then

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

Therefore D/K is divisible. �

Theorem 5.3. The following are equivalent (tfae) for any abeliangroup 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. It is easy to see that the first two conditions are equivalent.Suppose that x ∈ D and n ≥ 0. Then, A = nZ is a subgroup of thecyclic group B = Z and f : nZ → D can be given by sending thegenerator n to x. The homomorphism f : nZ→ D can be extended toZ if and only if D is divisible. Thus (2) implies (1) and (1) implies (2)in the case B = Z. The argument for any cyclic group is the same.

It follows from the definition of injectivity that (3) ⇒ (2). So, weneed to show that (1) and (2) imply (3).

So, suppose that D is divisible. Then we will use Zorn’s lemma toprove that it is injective. Suppose that A is a submodule of B andf : A → D is a homomorphism. Then we want to extend f to allof B. To use Zorn’s lemma we take the set of all pairs (C, g) whereA ⊆ C ⊆ B and g is an extension of f (i.e., f = g|A). This set ispartially ordered in an obvious way: (C, g) < (C ′, g′) if C ⊆ C ′ andg = g′|C. It also satisfies the hypothesis of Zorn’s lemma. Namely,any totally ordered subset (Cα, gα) has an upper bound: (∪Cα,∪gα).Zorn’s lemma tells us that this set has a maximal element, say, (M, g).We just need to show that M = B. We show this by contradiction.


If M 6= B then there is at least one element x ∈ B which is notin M . Let Zx = {kx | k ∈ Z} be the subgroup of B generated by x.Then M +Zx is strictly bigger than M . So, if we can find an extensiong : M+Zx→ D of g then we have a contradiction proving the theorem.There are two cases.

Case 1. M ∩Zx = 0. In this case, let g = (g, 0). I.e. g(a, kx) = g(a).Case 2. M ∩ Zx = nZx. (n is the smallest positive integer so that

nx ∈ M .) Since D is divisible, there is an element y ∈ D so thatny = g(nx). Let g : M + Zx→ D be defined by g(a+kx) = g(a) + ky.This is well defined by the following lemma since, for any a = knx

g(a) = g(knx) = kg(nx) = kny

�

Lemma 5.4. Suppose that A,B are submodules of an R-module C andf : A→ X, g : B → X are homomorphisms of R-modules which agreeon A∩B. Then we get a well-defined homomorphism f+g : A+B → Xby the formula

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

Proof. Well-defined means that, if the input is written in two differentways, the output is 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. �


6. Injective envelope

There is one other very important fact about injective modules whichwas not covered in class for lack of time and which is also not coveredin the book. This is the fact that every R-module M embeds in aminimal injective module which is called the injective envelope of M .This is from Jacobson’s Basic Algebra II.

Definition 6.1. An embedding A ↪→ B is called essential if everynonzero submodule of 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 containsa/b, then it contains 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 maxi-mal element by Zorn’s lemma. Then A ↪→ B/X must be essential,otherwise there would be a disjoint submodule of the form Y/X andX ⊂ Y,A∩Y = 0 contradicting the maximality of Y . For (2), C existsby 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 aretraction r : M → Q giving a splitting of the sequence. Conversely,suppose that every sequence as above splits. Then for any monomor-phism i : A ↪→ B and any morphism f : A → Q we can form thepushout M in the following diagram

A B

Q M

-i

?

f

?

f ′

-j

As you worked out in your homework, these morphisms form an exactsequence:

A(fi)−−→ Q⊕B (j,−f ′)−−−−→M → 0


Since i is a monomorphism by assumption, A is the kernel of (j,−f ′).Therefore (again using your homework) A is the pull-back in the abovediagram. This implies that j is a monomorphism. [Any morphismg : X → Q which goes to zero in M , i.e., so that j ◦ g = 0, will give amorphism (g, 0) : X → Q ⊕ B which goes to zero in M and thereforelifts uniquely to h : X → A so that(

f

i

)◦ h =

(f ◦ hi ◦ h

)=

(g

0

)But i is a monomorphism. So, i ◦ h = 0 implies h = 0 which in turnimplies that f ◦ h = g = 0. So, j is a monomorphism.]

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 retractionr : M → Q. (r ◦ j = idQ) Then it is easy to see that r ◦ f ′ : B → Q isthe 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 embeddingQ ↪→M is an isomorphism.

Proof. (⇒) Suppose Q is injective and Q ↪→ M is essential. Then theidentity map Q→ Q extends to a retraction r : M → Q whose kernelis disjoint from Q and therefore must be zero making M ∼= Q.

(⇐) Now suppose that every essential embedding of Q is an isomor-phism. We want to show that Q is injective. By the previous lemmait suffices to show that every short exact sequence

0→ Qj−→M → N → 0

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

Theorem 6.6. For any R-module M there exists an essential em-bedding M ↪→ Q with Q injective. Furthermore, Q is unique up toisomorphism under M .

Proof. We know that there is an embedding M ↪→ Q0 where Q0 isinjective. By Lemma 6.3 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 anembedding on Q, ker f ∩ Q = 0. This forces ker f = 0 since Q ↪→ Nis essential. So, f : N → Q0 is a monomorphism. This contradicts themaximality 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, supposeM ↪→ Q′ is another essential embedding of M into an injective Q′.Then the inclusion M ↪→ Q′ extends to a map g : Q→ Q′ which mustbe a monomorphism since its kernel is disjoint from M . Also, g mustbe onto since g(Q) is injective making the inclusion g(Q) ↪→ Q′ splitwhich contradicting the assumption that M ↪→ Q′ is essential unlessg(Q) = Q′. �


7. Projective resolutions

We talked for a week about projective resolutions.

(1) Definitions(2) Modules over a PID(3) Chain complexes, maps and homotopies(4) Homotopy uniqueness of projective resolutions(5) Examples

7.1. Definitions. Suppose that M is an R-module (or, more gener-ally, an object of any abelian category with enough projectives) thena projective resolution of M is defined to be a long exact sequence ofthe form

· · · → Pn+1dn+1−−−→ Pn

dn−→ Pn−1 → · · · → P0ε−→M → 0

where Pi are all projective.The (left) projective dimension of M is the smallest integer n ≥ 0 so

that there is 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 thenthe pd(M) =∞.

The (left) global dimension of the ring R written gl dim(R) is themaximum projective dimension of any module.

Example 7.1. (0) R has global dimension 0 if and only if it issemi-simple (e.g., any field).

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

An injective coresolution of a module M is an exact sequence of theform

0→M → Q0 → Q1 → · · ·where all of the Qi are injective. If an abelian category has enoughinjectives then every object has in injective resolution. We went to alot of trouble to show this holds for the category of R-modules.

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

0→M → Q0 → Q1 → · · · → Qn → 0

We will see later that the maximum injective dimension is equal to themaximum projective dimension.


7.2. Modules of a PID. At this point I decided to go through Lang’sproof of the following well-known theorem that I already mentionedseveral times.

Theorem 7.2. Suppose that R is a PID and E is a free R-module.Then every submodule of E is free.

Proof. (This proof is given on page 880 as an example of Zorn’s lemma.)Suppose that E is free with basis I and let F be a arbitrary submoduleof E. Then we consider the set P of all pairs (J, w) where J ⊆ I and wis a basis for FJ := F ∩EJ where EJ is the submodule of E generatedby J . In other words, FJ is the set of all elements of F which are linearcombinations of elements of the subset J of the given basis of E.

For example, suppose that I = {i, j, k} and J = {i, j}. If F ⊂ E isthe submodule given by

F = {(x, y, z) |x+ y + z = 0}then FJ = {(x,−x, 0)}.

The set P = {(J, w)} is partially ordered in the usual way: (J, w) ≤(J ′, w′) if J ⊆ J ′ and w ⊆ w′. To apply Zorn’s lemma we need to checkthat every tower has an upper bound. So, suppose that {(Jα, wα)} isa tower. Then the upper bound is given in the usual way by

(J, w) = (∪αJα,∪αwα)

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

(1) w is linearly independent.(2) w spans FJ , i.e., w ⊂ FJ and every element of FJ is a linear

combination of elements of w.

The first point is easy since any linear dependence among elements ofw involves only a finite number of elements of w which must all belongto some wα (if x1, · · · , xn ∈ w then each xi is contained in some wαi .Let α be the largest αi. Then wα contains all the xi. Since wα is abasis for FJα , these elements are linearly independent.

The second point is also easy. wα ⊂ FJα ⊆ FJ . So, the unionw = ∪wα is contained in FJ . Any element x ∈ FJ has only a finitenumber of nonzero coordinates which all lie in some Jα. So x ∈ FJαwhich is spanned by wα.

This verifies the hypothesis of Zorn’s lemma. Therefore, the conclu-sion holds and our poset P has a maximal element (J, w). We claimthat J = I. This would mean that FJ = FI = F and w would be abasis for F and we would be done.


To prove that J = I suppose J is strictly smaller. Then there existsan element k of I which is not in J . Let J ′ = J∪{k}. Then we want tofind a basis w′ for FJ ′ so that (J, w) < (J ′, w′), i.e., the basis w′ needsto contain w. In that case we get a contradiction 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 ofR which appear as k-coordinates of elements of FJ ′ . This is the imageof the k-coordinate projection map

FJ ′ ⊆ EJ ′pk−→ R

So, A is an ideal in R. So, A = Rs for some s ∈ R. Let x ∈ FJ ′ so thatpk(x) = s. Then I claim that w′ = w ∪ {x} is a basis for FJ ′ . First,w′ is clearly linearly independent since any linear combination whichinvolves x will have nonzero k-coordinate so cannot be zero. And anylinear combination not involving x cannot be zero since w is linearlyindependent. Finally, w′ spans FJ ′ . Given any z ∈ FJ ′ we must havepk(z) ∈ A = Rs. So pk(z) = rs and pk(z − rx) = 0. This implies thatz−rx ∈ FJ which is spanned by w. So z is rx plus a linear combinationof elements of w. So, w′ spans FJ ′ and we are done. �

Since Z is a PID we get the following.

Corollary 7.3. Every subgroup of a free abelian group is free.


7.3. Chain complexes. At this point I decided to review the basicdefinitions for chain complexes, chain maps and chain homotopies.

Suppose that A is an abelian category. Then a chain complex overA is an infinite sequence of objects and morphisms (called boundarymaps):

· · · → 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 wherethe superscripts are to keep track of which chain complex the boundarymaps dn are in. These morphisms form a big commuting diagram inthe shape of a ladder.

7.3.1. category of chain complexes.

Proposition 7.4. If A is abelian let C∗(A) be the category of chaincomplexes over A and chain maps. Then C∗(A) is also abelian.

I didn’t give a detailed proof but I pointed out how direct sums,kernels and cokernels are constructed. First the direct sum: C∗ ⊕ D∗is the chain complex with objects Cn ⊕Dn and boundary maps

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

Then the kernel of a chain map f∗ : C∗ → D∗ is defined to be the chaincomplex with nth term 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, we get this induced map.) The cokernel complex is givensimilarly by

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

where dn is the morphism induced by dDn .A cochain complex over an abelian categoryA is a sequence of objects

and morphisms (called coboundary maps)

C0 d0−→ C1 d1−→ · · ·so that dn+1 ◦ dn = 0. A morphism of cochain complexes C∗ → D∗

is a sequence of morphisms fn : Cn → Dn which form a commutingladder diagram. It is convenient to use the fact that this is the same asa chain complex over the opposite category Aop which is also abelian.


If the category of cochain complexes over A is denoted C∗(A) then thisduality can be written as

C∗(A)op ∼= C(Aop)

7.3.2. homology. The homology of a chain complex C∗ is defined to bethe sequence of objects:

Hn(C∗) :=ker dn

im dn+1

In theory these are defined only up to isomorphism. So, they are nottrue functors. However, in practice, they can almost always be explic-itly constructed. The construction does not have to be particularlyelegant or simple. But it avoids set theoretic headaches since the Ax-iom of Choice does not apply to categories: We are not allowed to“choose” an object Hn(C∗) for every chain complex C∗ only for a setof chain complexes.

Homology is a functor in the sense that any chain map f∗ : C∗ → D∗induces a morphism in homology Hn(f) : Hn(C∗) → Hn(D∗). Thisis because commutativity of the “ladder” implies that ker dCn maps toker dDn and im dCn+1 maps to im dDn+1. This functor is additive in thesense that Hn(f∗ + g∗) = Hn(f∗) +Hn(g∗). In other words, Hn gives ahomomorphism

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

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

7.3.3. chain homotopy. Two chain maps f∗, g∗ : C∗ → D∗ are calledchain homotopic if there 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. We call h∗ a homotopy from f∗ to g∗ andwe write

h∗ : f∗ ' g∗

Theorem 7.5. Homotopic chain maps induce the same map in homol-ogy.

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∗ andtherefore to zero in H∗(D∗) and h∗ ◦ dC∗ is zero on ker dC∗ . �

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

Corollary 7.6. Any chain homotopy equivalence induces an isomor-phism in homology.

Proof. Theorem 7.5 implies that Hn(f∗) ◦Hn(g∗) = Hn(idD) = idHn(D)

and similarly the other way. So, Hn(f∗) is an isomorphism with inverseHn(g∗). �


7.4. Homotopy uniqueness of projective resolutions. Here I provedthat the projective resolution of any R-module (or any object of anabelian category with enough projectives) is unique up to chain ho-motopy. I used diagrams and the (equivalent) equation. First I wrotedown the understood standard interpretation of a diagram.

Lemma 7.7. Given that the solid arrows (in below) form a commuting

diagram, there exists a dotted arrow (the arrow labeled ∃f̃ is supposedto be dotted) as indicates making the entire diagram commute. [Thisis the understood meaning of this kind of diagram.] The dotted arrowis not necessarily uniquely determined (it is labeled ∃ and not ∃!) Theassumptions are that P is projective and g : A→ B is onto.

Cn+1

P Cn

Cn−1

?

dn+1

��∃f̃

-f

@@@@R

0?

dn

Proof. By definition of kernel, f lifts uniquely to ker dn. But Cn+1 mapsonto ker dn. So, by definition of P being projective, f lifts to Cn+1. �

Lemma 7.8. With standard wording as above. The additional assump-tions are that the right hand column is exact (i.e., im dCn+1 = ker dCn ),Pn+1 is projective and the left hand column is a chain complex (i.e.dPn ◦ dPn+1 = 0).

Pn+1 Cn+1

Pn Cn

Pn−1 Cn−1

-∃fn+1

?

dPn+1

?

dCn+1

-fn

?

dPn

?

dCn

-fn−1

Proof. The assumptions implies dCn ◦ (fn ◦ dPn+1) = 0. By the previouslemma this implies that fn ◦ dPn+1 lifts to Cn+1. �


Theorem 7.9. Suppose P∗ → M → 0 is a projective chain complex(augmented) over M and C∗ → N → 0 is a resolution of N (i.e., anexact sequence). Suppose f : M → N is any morphism. Then

(1) There exists a chain map f∗ : P∗ → C∗ over f . I.e., the follow-ing diagram commutes:

P∗ C∗

M N

-f∗

?

ε

?

ε

-f

(2) f∗ is unique up to chain homotopy.

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 byinduction using lemma we just proved.

(2) To prove the existence of the homotopy, I first restated Lemma7.7 as an equation. It says that for any homomorphism f : P → Cn sothat dn ◦ f = 0, there exists a homomorphism f̃ : P → Cn+1 so thatdn+1 ◦ f̃ = f .

We want to show that f∗ is unique up to homotopy. So, supposef∗, g∗ are two chain maps over f : M → N . Then we want to showthat there exists a sequence of morphisms hn : Pn → Cn+1 so that

dCn+1 ◦ hn + hn−1 ◦ dPn = gn − fnWe set h−1 = 0 by definition. So, for n = 0 we get:

d1 ◦ h0 = g0 − f0

First, h0 exists because ε(g0− f0) = (f − f)ε = 0. If h0, · · · , hn−1 existsatisfying the above equation then in particular we have:

(7.1) dn ◦ hn−1 + hn−2 ◦ dn−1 = gn−1 − fn−1

We want to show that hn exists satisfying the equation

dCn+1 ◦ hn = gn − fn − hn−1 ◦ dPnThe right hand side is the f of Lemma 7.7 and the map that we want(hn) is the f̃ in Lemma 7.7. So, all we need to do is show that dCn ◦f = 0.

dn(gn − fn − hn−1 ◦ dn) = dn ◦ gn − dn ◦ fn − dn ◦ hn−1 ◦ dn= gn−1 ◦ dn − fn−1 ◦ dn − dn ◦ hn−1 ◦ dn

Factoring out the dn and using Equation (7.1) we get:

= (gn−1 − fn−1 − dn ◦ hn−1)dn = hn−2 ◦ dn−1 ◦ dn = 0

This is 0 since dn−1 ◦ dn = 0. Thus hn exists and f∗ ' g∗. �


This gives us the statement that we really want:

Corollary 7.10. In any abelian category with enough projectives, anyobject A has a projective resolution P∗ → A. Furthermore, any twoprojective resolutions of A are homotopy 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 the identity map on A. Since g∗ ◦ f∗ andthe 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 shouldstate the dual theorem but not prove it.]

Theorem 7.11. In any abelian category with enough injectives, anyobject B has an injective coresolution. Furthermore, any two injectivecoresolutions of B are homotopy equivalent.

Following this rule, I should also give the statement of the dual ofthe previous theorem:

Theorem 7.12. Suppose 0→M → Q∗ is an injective cochain complexunder M and 0 → N → C∗ is a coresolution of N (i.e., a long exactsequence). Suppose f : N →M is any morphism. Then

(1) There exists a cochain map f ∗ : C∗ → Q∗ under f . I.e., thefollowing diagram commutes:

C∗ Q∗

N M

-f∗

6

-f

6

(2) f ∗ is unique up to chain homotopy.


7.5. Derived functors.

Definition 7.13. Suppose that A, cB are abelian categories where Ahas enough injectives and F : A → B is a left exact (additive) functor.Then the right derived functors RiF are defined as follows. For anyobject B of A choose an injective coresolution B → Q∗ and let RiF (B)be the ith cohomology 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 Extfunctors:

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

Note that the derived functors are only well-defined up to isomor-phism. If there is another choice of injective coresolutions Q′∗ thenQ∗ ' Q′∗ which implies that F (Q∗) ' F (Q′∗) which implies that

H i(F (Q∗)) ∼= H i(F (Q′∗))

I pointed out later that, for R-modules, there is a canonical minimalinjective coresolution for any module.

By definition of F (Q∗) we take only the injective objects. The termF (B) is deliberately excluded. But F is left exact by assumption. Sowe have an exact sequence

0→ F (B)→ F (Q0)→ F (Q1)

Thus

Theorem 7.14. The zero-th derived functor R0F is canonically iso-morphic to F . In particular,

Ext0R(A,B) ∼= HomR(A,B)

At this point we tried to do an example: Compute Ext1Z(Z/3,Z/2).

We took an injective coresolution of Z/2

0→ Z/2Z→ Q/2Z j−→ Q/Z→ 0

We used that fact that any quotient of a divisible group is divisible.We took Hom(Z/3,−) into the injective part:

j∗ : Hom(Z/3,Q/2Z)→ Hom(Z/3,Q/Z)

Then I claimed that this map is an isomorphism. Here is a simple-minded proof. A homomorphism Z/3 → Q/Z is given by its valueon the generator 1 + 3Z of Z/3Z. This must be a coset a/b + Z sothat 3a/b ∈ Z. In other words b = 3 and a = 0, 1 or 2. Similarly,a homomorphism Z/3 → Q/2Z sends the generator of Z/3 to a coset


a/b + 2Z so that 3a/b ∈ 2Z. So, b = 3 and a = 0, 2 or 4. So, bothof there groups have exactly three elements and a simple calculationshows that j∗ is a bijection.

7.5.1. delta operator. On of the basic properties of the derived functorsis that they fit into an a long exact sequence.

Theorem 7.15. Given any short exact sequence

0→ Aα−→ B

β−→ C → 0

there is a sequence of homomorphisms

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 di-agram with exact rows:

0 A B C 0

0 A′ B′ C 0

- -α

?

f

-β

?

g

-

?

h

- -α′

-β′

-

we get a commuting square:

RnF (C) Rn+1F (A)

RnF (C ′) Rn+1F (A′)

-δn

?h∗

?f∗

-δn

I gave the following construction of these δ operators. First I neededthe following lemmas, the first being obvious.

Lemma 7.16. If Q is injective then RnF (Q) = 0 for all n ≥ 1.

Lemma 7.17. If 0 → A → Q → K → 0 is a short exact sequencewhere Q is injective, then we have an exact sequence

0→ F (A)→ F (Q)→ F (K)→ R1F (A)→ 0

andRnF (K) ∼= Rn+1F (A)

for all n ≥ 1.


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

0→ Aj−→ Q0

j0−→ Q1j1−→ Q2

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

0→ Aj−→ Q0 → K → 0

0→ K → Q1j1−→ Q2

j2−→ Q3 → · · ·The second exact sequence shows that the injective coresolution of Kis the same as that for A but shifted to the left with the first termdeleted. 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 in F (K) = ker(j1)∗. The cokernel is by definition the firstcohomology of the cochain complex F (Q∗) which is equal to R1F (A).So, we get the exact sequence

F (Q0)→ F (K)→ R1F (A)→ 0

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

My construction of the delta operator proceeded as follows. Startwith any short exact sequence 0→ A→ B → C → 0. Then choose aninjective 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 B and cokernels map to cokernels giving a commutingdiagram:

(7.2)

0 A B C 0

0 A Q0 K 0

- -α

?

idA

-β

?

f

-

?

g

- -j

-p

-

The map g : C → K induces a map g∗ : RnF (C) → RnF (K) and Idefined the connecting homomorphism δn for n ≥ 1 to be the compo-sition:

δn : RnF (C)g∗−→ RnF (K) ∼= Rn+1(A)


I showed that this is independent of the choice of g : C → K since, forany other choice g′, the difference g−g′ lifts to Q0 since f−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′∗. To show independence fromthe choice of Q0 I said that there was a canonical choice for Q0 calledthe injective envelope of A and I promised to write up the proof of that.

What about n = 0? In this case, Lemma 7.17 gives us a 4 term exactsequence:

0→ F (A)→ F (Q0)→ F (K)→ R1F (A)→ 0

So, we can define δ0 : F (C)→ R1F (A) to be the composition

δ0 : F (C)g∗−→ F (K)→ R1F (A)

Again, for any other choice g′ : C → K, the difference g − g′ factorsthrough Q0. This time F (Q0) 6= 0. But that is OK since F (Q0) isin the kernel of the next map F (K) → R1F (A) by the 4 term exactsequence.

7.5.2. Proof of Theorem 7.15. From your homework you might remem-ber that the sequence (7.2) gives a short exact sequence:

0→ B(fβ)−−→ Q0 ⊕ C

(−p,g)−−−→ K → 0

Since RnF (Q0) = 0 the top row in the following sequence is supposedto be exact:

Rn−1F (K) RnF (B) RnF (C) RnF (K)

RnF (A) RnF (B) RnF (C) Rn+1F (A)

- -δn−1

?∼=

-β∗

?=

-g∗

?=

-

?∼=

- -α∗ -β∗ -δn -

In the top sequence RnF (C) occurs in position 3n − 1 and in thebottom sequence it occurs in position 3n. Therefore, exactness of thebottom sequence for all 0 → A → B → C → 0 at position k − 1implies the exactness of the top sequence at position k−1 which impliesthe exactness of the bottom sequence at position k. So, it is exacteverywhere, proving theorem.

We just need to check that the diagram commutes. (Actually itdoesn’t. But that’s OK.) The middle square obviously commutes. Theright hand square commutes by definition of δn. The left square anti-commutes (i.e., going one way is negative the other way). But that isgood enough for the argument to work. This will follow from the waythat α∗ and δn−1 are defined.


The morphism α : A→ B induces a cochain map of injective cores-olutions:

0 A Q0 Q1 Q2 · · ·

0 B QB0 QB

1 QB2

· · ·

- -j

?

α

-

?α0

-

?α1

-

?α2

- -jB

- - -

The cochain map α∗ induces the cochain map α∗ : F (Q∗) → F (QB∗ ).

The induced map in cohomology is α∗ : RnF (A)→ RnF (B) by defini-tion. If the cokernels of j, jB are K,L we get the commuting diagrams

0 A Q0 K 0

0 B QB0 L 0

- -j

?

α

-p

?α0

-

?α

- -jB

-pB

-

0 K Q1 Q2 · · ·

0 L QB1 QB

2· · ·

- -

?

α

-

?α1

-

?α2

- - - -

Just as we hadRnF (K) ∼= Rn+1F (A) we also haveRnF (L) ∼= Rn+1F (B)and the above diagrams show that the maps of injective coresolutionsinduced by α : A → B and α : K → L are the same but shifted.In other words, α∗ : RnF (A) → RnF (B) is the same as the mapα∗ : Rn−1F (K)→ Rn−1F (L).

We need one more commuting diagram:

0 B Q0 ⊕ C K 0

0 B QB0 L 0

- -(fβ)

?=

-(−p,g)

?(α0,−h)

-

?−α

- -jB

-pB

-

Here h : C → QB0 is the morphism needed to make the diagram com-

mute, i.e., the maps α0 ◦ f, jB : B → QB0 are not equal. But they agree

on A. So their difference factors through C. I.e. ∃h : C → QB0 so that

h ◦ β = α0 ◦ f − jB

The coboundary map δn−1 : Rn−1F (K) → RnF (B) is given by thecomposition

δn−1 : Rn−1F (K)−α∗−−→ Rn−1F (L) ∼= RnF (B)

By what I said in the last paragraph, this is the same as−α∗ : RnF (A)→RnF (B) proving the theorem. (The n = 1 case is slightly different.) �


7.6. Left derived functors. There are two cases when we use pro-jective resolutions instead 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 = −⊗RB.

In both cases we take a projective resolution P∗ → A→ 0 and definethe left derived functors to be LnF (A) = Hn(F (P∗)) in the first caseand LnF (A) = Hn(F (P∗)) in the second case.

Definition 7.18. The left derived functors of F (A) = A ⊗R B arecalled LnF (A) = TorRn (A,B)

7.6.1. review of tensor product. Following Lang, I will take tensor prod-ucts only over commutative rings. The advantage is that A⊗RB will bean R-module. The tensor product is defined by a universal condition.

Definition 7.19. Suppose that A,B are modules over a commutativering R. Then a map

g : A×B → C

from the Cartesian product A × B to a third R-module C is calledR-bilinear if it is an R-homomorphism in each variable. I.e., for eacha ∈ A, the mapping b 7→ g(a, b) is a homomorphism B → C andsimilarly g(−, b) ∈ HomR(A,C) for all a ∈ A. A⊗R B is defined to bethe 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-bilinearmap g : A×B → C there is a unique R-homomorphism h : A⊗B → Cso that g = h ◦ f .

The universal property tells us that A⊗RB is unique if it exists. Toprove existence we need to construct it. But this easy. You just takeA⊗R B to be the free R-module generated by all symbols a⊗ b wherea ∈ 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 anisomorphism

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 ofisomorphism, F is called the left adjoint and G is called the rightadjoint and F,G are called adjoint functors.

Lemma 7.20. Any left adjoint functor is right exact. In particular,tensor product is right exact. Also, any right adjoint functor is leftexact.

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

(7.3) 0→ Aα−→ A′

β−→ A′′ → 0

is a short exact sequence. Then for any C, the left exactness ofHomR(−, G(C)) gives an 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 exactnessof the following sequence 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 leftexactness of Hom so the second case is dumb.) �

Take the sequence (7.3) and suppose that it splits. I.e., A′ ∼= A⊕A′′and there is a retraction r : A′ → A so that r ◦ α = idA. Then, in theexact sequence

F (A)F (α)−−→ F (A′)→ F (A′′)→ 0

F (α) is a monomorphism since F (r) ◦ F (α) = F (r ◦ α) = idF (A). So,we get a short exact sequence which furthermore splits. This provesthe following.


Lemma 7.21. If F is any right (or left) exact functor then F (A ⊕A′′) ∼= F (A)⊕ F (A′′). In particular,

(A⊕ A′′)⊗R B ∼= (A⊗R B)⊕ (A′′ ⊗R B)

Another important lemma was the following.

Lemma 7.22. R⊗R B is isomorphic to B as R-modules.

Proof. I showed that B satisfies the universal property. Let

f : R×B → B

be the map f(r, b) = rb. This is R-bilinear when R is commutative.Suppose that g : R × B → C is another R-bilinear map. Then wecan define h : B → C by h(b) = g(1, b). This is R-linear since g isR-bilinear. The required diagram commutes since

h ◦ f(r, b) = h(rb) = g(1, rb) = rg(1, b) = g(r, b)

Furthermore, h is unique since it has no choice but to send b to g(1, b).Since B satisfies the universal property, B ∼= R ⊗ B. Also the proofgives the isomorphism. r ⊗ b ∈ R⊗B corresponds to rb ∈ B. �

There was one other lemma that I didn’t prove because it was “ob-vious.”

Lemma 7.23. A⊗B ∼= B ⊗ A

7.6.2. computations. With these lemmas, I did some computations.Suppose that R = Z and A = Z/n. Then a projective resolutionof A is given by

0→ Z ·n−→ Z→ Z/n→ 0

Since this sequence is exact it gives the following right exact sequencefor any abelian group 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 that

Z/nZ⊗B ∼= B/nB

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

A ∼= Zr ⊕ Z/t1 ⊕ Z/t2 ⊕ · · · ⊕ Z/tnand, since tensor product distributes over direct sum we get:

A⊗Z B = Br ⊕B/t1B ⊕B/t2B ⊕ · · · ⊕B/jnB


The 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 multiplicationby n. So,

TorZ1 (Z/n,B) = {b ∈ B |nb = 0}

It is the subgroup of B consisting of all elements whose order dividesn. It is the “n-torsion” subgroup of B. Maybe that is why it is calledTor.

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

Rn f−→ Rm

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

f(ej) =m∑i=1

aijei

for some aij ∈ R. These numbers determine f since, for an arbitraryelement x =

∑xjej ∈ Rn we have

f(x) = f

(∑j

xjej

)=∑i,j

xjaijei =∑i,j

aijxjei

since R is commutative. (Take free right R-modules when R is notcommutative and this will still work.) This can be written in matrixform:

f

x1

x2

· · ·xn

=

∑a1jxj∑a2jxj· · ·∑anjxj

= (aij)

x1

x2

· · ·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 obtainedfrom f by “extending scalars” to S. If you have an integer matrix, youjust take the same matrix and consider it as a real matrix.


7.6.4. two definitions of Ext. The last thing I did was to prove thatthe two definitions of ExtnR(A,B) that we now had were equivalent.

Theorem 7.24. If P∗ → A is a projective resolution of A and B → Q∗is an injective resolution 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 sidesare are isomorphic to HomR(A,B). So, suppose n ≥ 1. I gave theproof in the case n = 2.

I want to construct a homomorphism

Hn(Hom(A,Q∗))→ Hn(Hom(P∗, B))

So, take an element [f ] ∈ Hn(Hom(P∗, B)). The notation means

f ∈ ker((j2)∗ : HomR(A,Q2)→ HomR(A,Q3))

[f ] = f + im((j1)∗ : HomR(A,Q1)→ HomR(A,Q2))

This gives the following diagram

(7.4)

P3 P2 P1 P0 A 0

0 B Q0 Q1 Q2 Q3

-d3

?

f3@@@R

0

-d2

?

f2

-d1

?f1

-d0

?f0

-

?

f@@@R

0

?- - -

j0

-j1

-j2

Since j2 ◦ f = 0, f maps to the kernel K of j2. But P∗ is a projectiveresolution of A and B → Q0 → · · · → Qn−1 → K is a resolution of K.So, we proved that there is a chain map from P∗ to this resolution ofK which is unique up to chain homotopy. This gives the maps f0, f1,etc in the diagram. Note that f2 ◦ d3 = 0 ◦ f3 = 0. So,

f2 ∈ ker((d3)∗ : HomR(P2, B)→ HomR(P3, B))

But f2 is only well defined up to homotopy h : P1 → B. So, we couldget f ′2 = f2 + h ◦ d+ d ◦ h. But the second term must be zero since itgoes through 0 and the first term

h ◦ d2 ∈ im((d2)∗ : HomR(P1, B)→ HomR(P2, B))

This means that

[f2] = f2 + im(d2)∗

is a well defined element of H2(Hom(P∗, B)) and we have a homomor-phism:

HomR(A, ker j2)→ H2(Hom(P∗, B))


But this homomophism is zero on the image of (j1)∗ : Hom(A,Q1) →Hom(A,Q2) because, if f = j1 ◦ g then we can take f0 = g ◦ d0 andf1 = 0 = f2. Therefore, we have a well defined map

H2(Hom(A,Q∗))→ H2(Hom(P∗, B))

which sends [f ] to [f2].This is enough! The reason is that the diagram (7.4) is symmetrical.

We can use the dual argument to define a map

H2(Hom(P∗, B))→ H2(Hom(A,Q∗))

which will send [f2] back to [f ]. So, the two maps are inverse to eachother making them isomorphisms. �

By the way, this gives a symmetrical definition of Extn, namely it isthe group of homotopy classes of chain maps from the chain complexP∗ → A → 0 to the cochain complex 0 → B → Q∗ shifted by n.Elements of ExtnR(A,B) are represented by vertical maps as in Equation(7.4).

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MATH 101B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA