Math 131B: Algebra Ii Part A: Homological Algebra

MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA

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

Contents 1. Additive categories 1 2. Kernels and cokernels 2 3. Projective and injective objects 3 3.1. Suﬃciently many projectives 5 3.2. Injective objects 6 4. Injective modules 7 4.1. dual module 7 4.2. Constructing injective modules 7 4.3. proof of lemmas 9 4.4. Examples 11 5. Divisible groups 12 6. Injective envelope 14 7. Projective resolutions 16 7.1. Deﬁnitions 16 7.2. Modules of a PID 17 7.3. Canonical forms for matrices 19 7.4. Chain complexes 20 8. Homotopy invariance of projective resolutions 24 9. Derived functors 27 9.1. Construction of RiF 27 9.2. connecting morphisms 28 9.3. connecting morphism: standard method 31 9.4. Left derived functors 33

0 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 set HomC(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, L Q i.e., the canonical morphism Ai → Ai is an isomorphism. Proof. By induction it suffices to consider sums and products of pairs. We will show A ⊕ B ∼= A × B. Consider the following diagram.

idA (1.1) A / A ; jA pA jA

# f g # A ⊕ B / A × B / A ⊕ B ; ;

jB pB jB idB # B / B 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. Let g : A × B → A ⊕ B be the sum g = jA ◦ pA + jB ◦ pB. Then, it is easy to see that f, g are 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 morphism h : A ⊕ B → A ⊕ B satisfying h ◦ jA = jA and h ◦ jB = jB. Since h = g ◦ f and h = 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 homomorphisms. 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 want to spend time formally defining what is an abelian category. But I will point out the problems 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 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 uniqueg ˜ : X → K so that g = j ◦ g˜ (such ag ˜ 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 0 g ~ j f K / A / B Theorem 2.2. K is the kernel of f : A → B if and only if

j] f] 0 → HomC(X,K) −→ HomC(X,A) −→ 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 maps g 6= h : C → A, f ◦ g 6= f ◦ h, i,e, f] : HomC(X,A) → HomC(X,B) is a monomorphism of groups. Corollary 2.3. A morphism f : A → B is a monomorphism iff ker f = 0. f Notation 2.4. We say that 0 → K → A −→ B is exact if K = ker f. Remark 2.5. A non-example: The category mod-R of f.g. right R-modules does not have kernels if R is not right Noetherian (right ideals are finitely generated). To see this suppose that I is a right ideal which is not finitely generated. Then IR as R-module is not an object in the category. So, the morphism R → R/I does not have a kernel in the category mod-R. Cokernel is defined analogously and satisfies the following theorem which can be used as the definition. Theorem 2.6. The cokernel of f : A → B is an object C with a morphism p : B → C so that p] f ] 0 → HomC(C,Y ) −→ HomC(B,Y ) −→ 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.

f p A / B / C

g 0 ∃!g Y Again, letting C = 0 we get the statement: MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 3 Corollary 2.7. A morphism f is an epimorphism (i.e., f ] is a monomorphism) if and only if coker f = 0.

f Notation 2.8. We say A −→ B → C → 0 is exact if C = coker f. f g We say 0 → A −→ B −→ 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 2 given by multiplication by 2 is an epimorphism: 0 → Z −→ Z → 0 is exact!! 3. Projective and injective objects We will discuss the definition of projective and injective objects in any additive category. And it was easy to show that the category of R-modules has sufficiently many projectives. 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 any epimorphism p : A → B and any morphism g : P → B there exists a morphism g˜ : P → A so that p ◦ g˜ = g. The mapg ˜ is called a lifting of g to A. A ? ∃g˜ p g P / 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 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 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 be P written 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, for every index α, the element g(xα) comes from some element yα ∈ A. I.e., g(xα) = f(yα). Then a liftingg ˜ of g is given by X X (3.1)g ˜( xarα) = yαrα The verification that this is a lifting is straightforward. 4 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA To do the last step in full detail, we use the universal property of free modules: A homomorphism from the free module F generated by a set S to another module M is uniquely determined by a set mapping S → M which is arbitrary, i.e., (3.1) defines the unique homomorphism F → A sending xα to yα. For every R-module M there is a free R-module which maps onto M, namely the free module F generated by symbols [x] for all x ∈ M and with projection map p : F → M given by X X p rα[xα] = rαxα The notation [x] is used to distinguish between the element x ∈ M and the corresponding generator [x] ∈ F . Using the universal property of free modules, the homomorphism p is defined by the equation p[x] = x. For M a finitely generated R-module, take a finite generating set S. Then the free R-module generated by S maps onto M. MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 5 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: f g ϕ = : X ⊕ Y → A ⊕ B h k 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 projectives, i.e., for every R-module M there is a projective R-module P which maps onto M and, 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 enough projectives. 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” instead of “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 ⊕ 0 to 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 is a 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 > ∃s p

idP P / 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 the category P(Z) of f.g. free abelian groups. (2) This morphism does not have as section. 6 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA

By Theorem 3.8, Z is not projective. The objects of P(Z) are, by definition, isomorphic to Zn for some n. If this were projective, by Theorem 3.7, Z would be projective, which is a contraction. Therefore, P(Z) has no projective objects (except 0 which is always projective). In the category of finite abelian groups, the object Zn is not projective since the epimorphism Zn2 → Zn has no section (left as an exercise). But, every finite abelian group is a direct sum of cyclic groups. So, tor-Z has no nonzero projective objects. Getting back to the good cases: Suppose that C is an additive category with the following two properties: (1) C has enough projectives, i.e., for every object X, there is an epimorphism P → X where P is projective. (2) Every morphism in C has a kernel (in C). Then each object X has a projective resolution:

d0 d1 d2 0 ← X ←− P0 ←− P1 ←− P2 ← · · · constructed as follows. P0 is a projective object with an epimorphism d0 : P0 → X. By assumption (2), this has a kernel K0. Let P1 be a projective object with epimorphism P1 → 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 this morphism 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 any monomorphism 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 theorem will tell us that every R-module M has an injective co-resolution which is an exact sequence: 0 → M → Q0 → Q1 → Q2 → · · · where each Qi is injective. MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 7 4. Injective modules I will go over Lang’s proof (XX§4) that every R-module M embeds in an injective module 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: ∧ M ∼ Y ∧ Mα = Mα

Proof. We already saw that the hom functor HomZ(−,X) is left exact for any abelian group X. It is also obviously additive which means that (f + g)] = f ] + g] for all f, 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 M f : Mα → X is given uniquely by its restriction to each summand: fα : Mα → X and the fα can all be nonzero. (So, (fα)α is in the product not the sum.) 4.2. Constructing injective modules. In order to get an injective right R-module we 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 of copies 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, ∧ M ∧ Y ∧ F = ( RR) = RR is projective by (1) proving the Theorem. 8 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA We need one more lemma to prove the main theorem. Then we have to prove the lemmas. 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 module ∧ F . MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 9 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 evaluation ∧ map 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 of ∧ R on HomZ(M ,T ) so that (evxr)(ϕ) = evx(rϕ). This implies that evaluation x 7→ evx is 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 nonzero element 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 k f(kx) = + ∈ / n Z Q Z 1 ∼ This is nonzero on x since 1/n is not an integer: f(x) = n + Z 6= 0 + Z. If C = Z then let f : C → T be given by k f(kx) = + 2 Z 1 Then, f(x) = 2 + 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. Then Q we 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 f α : B → Jα. I.e., f α|A = fα for all α. By deﬁnition of the product there Q exists a unique morphism f : B → Q = Jα so that pα ◦ f = f α for each α. So,

pα ◦ f|A = f α|A = fα : A → Jα Q The uniquely induced map A → Jα is f. Therefore, f is an extension of f to B as required. 10 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 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 for ∧ each 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 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, 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 also need 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. MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 11 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 and injective. 4.4.2. polynomial ring. Let R = Z[t], the integer polynomial ring in one generator. This is a commutative Noetherian ring. It has dimension 2 since 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 domains (i.e., have no zero divisors): R/0 = Z[t],R/(t) = Z are domains and R/(t, 2) = Z/(2) = Z/2Z is 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 an endomorphism M → M given by the action of t. A homomorphism of Z[t]-modules f : 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 M is the same as a homomorphism of rings ϕ : R → End(M). When R = Z[t], this homomorphism is given by its value on t since ϕ(f(t)) = f(ϕ(t)). For example, if f(t) = 2t2 + 3 then 2 ϕ(f(t)) = ϕ(2t + 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 = RR is 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, ··· . Therefore, an element f ∈ Q, f : Z[t] → Q/Z is given uniquely by the sequence 2 3 f(1), f(t), f(t ), f(t ), · · · ∈ Q/Z Multiplication by t shifts this sequence to the left since (ft)(ti) − f(tti) = f(ti+1). This proves the following. Theorem 4.9. The injective Z[t]-module Q = Z[t]∧ is isomorphic to the additive group of all sequences (a0, a1, a2, ··· ) of elements ai ∈ Q/Z with the action of t given by shifting to 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. 12 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 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 any positive 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: Q p Zp L p Zp the product of all cyclic groups of prime order Zp modulo their direct sum. This is divisible since, for any n, we have Q Q p Zp ∼ p>n Zp L = L p 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 quotient D/K has the form x + K where x ∈ D. This is divisible by any positive n since, if ny = 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 by f(an) = ax. So, f(n) = x. Extend this to f : Z → D be an extension of f to Z. Then ny = 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 group B then the generator a0 of A is a multiple of the generator b0 of B: a0 = nb0. For any 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 since f(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 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 13 (C, g) where X ⊆ C ⊆ Y and g is an extension of f (i.e., f = g|X). This set is partially ordered in an obvious way: (C, g) < (C0, g0) if C ⊆ C0 and g = g0|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 = 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 the cyclic subgroup of Y generated by y. Let A = B ∩ M. Then, by (2), the restriction h = 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 the maximality 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 a well-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 output is still the same. So suppose that a + b = a0 + b0. Then a − a0 = b0 − b ∈ A ∩ B. So, f(a − a0) = f(a) − f(a0) = g(b0 − b) = g(b0) − g(b) 0 0 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 sequence of R-modules: jA (−j ) p +p 0 → A ∩ B −−−→B A ⊕ B −−−−→A B 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. 14 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 6. Injective envelope There is one very important fact about injective modules which is not covered in Lang’s book. This is the fact that every R-module M embeds in a minimal 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 every nonzero 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 contains a/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 maximal element by Zorn’s Lemma. Then A,→ B/X must be essential, otherwise there would be a disjoint submodule 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 → Q giving a splitting of the sequence. Conversely, suppose that every sequence as above splits. Then for any monomorphism i : A,→ B and any morphism f : A → Q we can form the pushout M in the following diagram

i A / B f f 0 j Q / M These morphisms form an exact sequence:

f ( ) (j,−f 0) A −−→i Q ⊕ B −−−−→ M → 0 This implies that j is a monomorphism. [If j(x) = 0 then (j, −f 0)(x, 0) = 0. So, there is 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 j 0 → Q −→ M → coker j → 0 We are assuming that all such sequences split. So, there is a retraction r : M → Q 0 (r ◦ j = idQ). It follows that r ◦ f : B → Q is the desired extension of f : A → Q: 0 r ◦ f ◦ i = r ◦ j ◦ f = idQ ◦ f = f MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 15

So, Q is injective. Lemma 6.5. Q is injective if and only if every essential embedding Q,→ C is an isomorphism. Proof. (⇒) Suppose Q is injective and Q,→ C is essential. Then the identity map Q → Q extends to a retraction r : C → Q whose kernel is disjoint from Q and therefore must be zero making C ∼= Q. (⇐) Now suppose that every essential embedding of Q is an isomorphism. We want to show that Q is injective. By the previous lemma it suﬃces to show that every short exact sequence j 0 → Q −→ M → N → 0 splits. By Lemma 6.3 there is a submodule X ⊆ M so that jQ ∩ X = 0 and Q,→ M/X is essential. Then, by assumption, this map must be an isomorphism. So, M ∼= Q ⊕ X and the sequence splits proving that Q is injective. Theorem 6.6. For any R-module M there exists an essential embedding M,→ Q with Q 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 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 an embedding on Q, ker f ∩ Q = 0. This forces ker f = 0 since Q,→ N is essential. So, f : N → Q0 is a monomorphism. This contradicts the maximality of Q since the image of N is an essential extension of M in Q0 which is larger than Q. It remains to show that Q is unique up to isomorphism. So, suppose M,→ Q0 is another essential embedding of M into an injective Q0. Then the inclusion M,→ Q0 extends to a map g : Q → Q0 which must be a monomorphism since its kernel is disjoint from M. Also, g must be onto since g(Q) is injective making the inclusion g(Q) ,→ Q0 0 0 split which contradicting the assumption that M,→ Q is essential unless g(Q) = Q . 16 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 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 object of an additive category A with kernels and enough projectives) then a projective resolution of M is deﬁned to be a long exact sequence of the form

dn+1 dn · · · → Pn+1 −−−→ Pn −→ Pn−1 → · · · → P0 −→ M → 0 where Pi are all projective. Exact means in general that each Pi maps epimorphically to the kernel of the previous map. In the homework students showed that, for any projective 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- 0 0 tions: Given any other projective resolution · · · → P1 → P0 → M, there is a sequence 0 of morphisms fi : Pi → Pi making the following diagram commute:

d0 d0 0 0 2 0 1 0 ε ··· / P2 / P1 / P0 / M

f2 f1 f0 =

d2 d1 ε ··· / P2 / P1 / P0 / M

0 The morphism f0 : P0 → P0 exists since ε : P0 → M is an epimorphism. The morphism 0 0 f0 ◦ d1 : P1 → P0 goes to the kernel of ε : P0 → M. Since P1 maps epimorphically onto 0 0 0 0 ker ε, we get a lifting f1 : P1 → P1, and so on. (f0 ◦d1 ∈ A(P1,P0) maps to 0 ∈ A(P1,M) 0 and therefore lifts to A(P1,P1).) The (right) projective dimension of M is the smallest integer n ≥ 0 so that there is a projective resolution of the form 0 → Pn → Pn−1 → · · · → P0 −→ 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. We need 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 maximum projective dimension of any right R-module. MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 17 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 submodule 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 enough injectives then every object has in injective resolution. We went to a lot of trouble to show that this holds for the category of R-modules for any ring R: We showed that every module M embeds in an injective module Q0 and there is a canonical choice of Q0 called the injective envelope of M. The next term Q1 is the injective envelope of (or any 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 injective coresolution of the form

0 → M → Q0 → Q1 → · · · → Qn → 0 We will see later that the maximum injective dimension is equal to the maximum projective dimension. Exercise 7.2. Show that the shortest injective coresolution of M is given by taking injective envelopes at each stage. 7.2. Modules of a PID. Here is Lang’s proof of the following well-known theorem that I 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 submodule of E is free. Proof. (Lang, page 880) Let E be a free R-module with basis I and let F ⊆ E be a arbitrary submodule. Notation: for any J ⊆ I, let EJ be the free submodule of E generated 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 0, w0) if J ⊆ J 0 and w ⊆ w0. 3 For example, suppose that E = R 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) 18 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA

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 remains to check that P (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 of elements of w. (1) is easy: Any R-linear dependence among elements of w involves only a ﬁnite number of elements of w which must all belong to 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 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 . Any element 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 0 0 k of I which is not in J∞. Let J = J∞ ∪ {k}. Then we want to ﬁnd a basis w for FJ0 0 0 0 so that (J∞, w∞) < (J , w ), i.e., the basis w needs to contain w∞. This would give a contradiction to the maximality of (J∞, w∞). There are two cases. 0 0 Case 1. FJ = FJ∞ . In that case take w = w∞ and we are done. 0 0 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-coordinates of elements of FJ0 . This is the image of the k-coordinate projection map

pk FJ0 ⊆ EJ0 −→ R.

So, A is an ideal in R. Since R is a PID, A = ha0i = a0R. Let x ∈ FJ0 so that pk(x) = a0. Then I claim that 0 w = w∞ ∪ {x} is a basis for FJ0 . (a) w0 is R-linearly independent: Any R-linear combination which involves x will have nonzero k-coordinate so cannot be zero. And any R-linear combination not involving x cannot be zero since w∞ is R-linearly independent. 0 (b) w spans FJ0 : Given any z ∈ FJ0 we must have pk(z) ∈ A = a0R. So pk(z) = a0r and pk(z − xr) = 0. This implies that z − xr ∈ FJ which is spanned by w. So z is xr 0 0 plus 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 module is injective. (We already know this for R = Z. The proof for any PID is similar.) MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 19 7.3. Canonical forms for matrices. Students asked to go over canonical forms, such as 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 cyclic modules. Suppose that T is an n × n matrix with entries in a field K. We view T as an endomorphism T : Kn → Kn of Kn. This makes M = Kn into a K[t] module by letting t act by T (tv := T v). Since K[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 gives a block decomposition of T . If M = A ⊕ B then T 0 STS−1 = A 0 TB where TA, TB are the matrices of the action of T on A, B and S is the change of basis matrix using a new basis whose first basis vectors lie in A and other basis vectors lie in B. The next reduction is: Assume M is a cyclic K[t]-module. Then M = K[t]/ hf(t)i. We can assume f(t) is monic: n n−1 n−2 f(t) = t + b1t + b2t + ··· + bn We took n = 4 for simplicity of notation. Then, after changing basis, I claimed that the matrix of T is (rational canonical form):   0 0 0 −b4 −1 1 0 0 −b3 STS =   0 1 0 −b2 0 0 1 −b1 This is the matrix of T with respect to the following basis for M = K[t]/ hf(t)i: 1, t, t2, t3 Multiplication by t sends 1 to t, t to t2, t2 to t3. This gives the first three columns of the matrix. The last column comes from the action of t on the last vector: 4 2 3 t = −b4 − b3t − b2t − t1t . In case K = C, every polynomial is a product of linear polynomials:

Y mi f(t) = (t − ai)

m1 The ﬁrst step is to isolate each factor (t − ai) using the following. Theorem 7.8 (Chinese remainder theorem). If f(t) = g(t)h(t) where g(t), h(t) are relatively prime then K[t]/ hf(t)i ∼= K[t]/ hg(t)i ⊕ K[t]/ hh(t)i 20 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA

This reduces us to the case when M = C[t]/ h(t − a)mi. For example, when m = 4, we get the following Jordan canonical form a 1 0 0 −1 0 a 1 0 STS =   0 0 a 1 0 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 T e1 = ae1. Take v ∈ M, v∈ / V . Then (t − a)v = be1 for some b 6= 0 in C. Take e2 = v/b. Then the matrix of T with respect to e1, e2 is a 1 STS−1 = 0 a

7.4. Chain complexes. We give basic deﬁnitions for chain complexes, chain maps and chain 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 inﬁnite sequence of objects and morphisms (called boundary maps):

dn dn−1 d1 · · · → Cn −→ Cn−1 −−−→· · · → C1 −→ 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 D C morphisms fn : Cn → Dn so that dn ◦ fn = fn−1 ◦ dn where the superscripts are to keep track of which chain complex the boundary maps dn are in. These morphisms form a big commuting diagram in the shape of a ladder:

C C C C dn dn−1 d2 d1 ··· / Cn / Cn−1 / ··· / C1 / C0

fn fn−1 f1 f0 ··· / D / D / ··· / D / D n D n−1 D D 1 D 0 dn dn−1 d2 d1

7.4.1. category of chain complexes.

Proposition 7.9. Let C∗(A) be the category of chain complexes and chain maps over an 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 and boundary maps C⊕D C D dn = dn ⊕ dn : Cn ⊕ Dn → Cn+1 ⊕ Dn+1 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 21

The kernel of a chain map f∗ : C∗ → D∗ is deﬁned to be the chain complex with nth term ker fn and boundary map 0 dn : ker fn → ker fn−1 C C D induced by the morphism dn : Cn → Cn−1. (Since fn−1 ◦ dn = dn ◦ fn = 0 on ker fn, we get this induced map.) The cokernel complex is given similarly by

dn · · · → coker fn −→ coker fn−1 → · · · D where dn is the morphism induced by dn . 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 D C dn+1 ◦ hn + hn−1 ◦ dn = gn − fn

(This is what you add to f∗ to get g∗.) Next time we will discuss homology, which requires more conditions on our additive category, and the theorem that homotopic chain maps induce the same map in homology. 22 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 7.4.3. Homology. Suppose that A is an additive category with kernels and cokernels for all 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:

ker dn Hn(C∗) := . im dn+1 This is only well deﬁned up to isomorphism. There is a set-theoretic diﬃculty: we cannot in general choose a representative. We ignore this. 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∗). This is because commutativity of C D C D the “ladder” implies that fn : Cn → Dn maps ker dn to ker dn and im dn+1 to im dn+1. C C im dn+1 / ker dn / Hn(C∗)

fn fn Hn(f∗) D D im dn+1 / ker dn / 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 other words, 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 definition of an abelian category. Thus, we have seen that we need all of these conditions. (We also 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 epimorphism A I and a monomorphism I,→ B. Remark 7.13. An abelian category might not have enough projective and/or injective objects. For example, the category of finite abelian groups is an abelian category without any nonzero projective or injective objects. Proposition 7.14. A morphism f : A → B in an abelian category is an isomorphism if 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 must be the cokernel of 0 → A which is A −→= A. So, f is an isomorphism. The converse is obvious. MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 23

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

hn : Cn → Dn+1 so that D C dn+1 ◦ hn + hn−1 ◦ dn = gn − fn for all n ≥ 0 where h−1 = 0.

Cn+1 / Dn+1 8 hn dn+1 dn+1

gn−fn C / D n 8 n

dn dn hn−1 Cn−1 / Dn−1 Theorem 7.15. Homotopic chain maps induce the same map in homology.

Proof. This follows from the fact that Hn is additive: D C Hn(g∗) − Hn(f∗) = Hn(g∗ − f∗) = Hn(d∗ ◦ h∗ + h∗ ◦ d∗ ) D C = Hn(d∗ ◦ h∗) + Hn(h∗ ◦ d∗ ) D D But both of these are zero since d∗ ◦ h∗ maps to the image of d∗ and therefore to zero C C in H∗(D∗) and h∗ ◦ d∗ is zero on ker d∗ .

7.4.6. homotopy equivalence. Two chain complexes C∗,D∗ are called (chain) homotopy equivalent and we write C∗ ' D∗ if there exist chain maps f∗ : C∗ → D∗ and g∗ : 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 similarly the other way. So, Hn(f∗) is an isomorphism with inverse Hn(g∗). 24 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 8. Homotopy invariance of projective resolutions Here I proved that the projective resolution of any R-module (or any object of an abelian category with enough projectives) is unique up to chain homotopy. I used diagrams and the (equivalent) equation. First I wrote down the understood standard interpretation 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 is the understood meaning of this kind of diagram.] The dotted arrow is not necessarily uniquely determined (it is labeled ∃ and not ∃!) The assumptions are that P is projective and ker dn = im dn+1.

Cn+1 ∃fe = dn+1 f P / Cn

dn 0 ! 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 the C C right hand column is exact (i.e., im dn+1 = ker dn ), Pn+1 is projective and the left hand P P column is a chain complex (i.e. dn ◦ dn+1 = 0).

∃fn+1 Pn+1 / Cn+1

P C dn+1 dn+1

fn Pn / Cn

P C dn dn fn−1 Pn−1 / Cn−1

C P Proof. The assumptions implies dn ◦ (fn ◦ dn+1): Pn+1 → Cn−1 is 0. By the previous P lemma this implies that fn ◦ dn+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 a morphism hn−1 : Pn−1 → Cn so that P C P fn−1 ◦ dn = dn ◦ fn = dn ◦ hn−1 ◦ dn

(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). MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 25

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

d(hn−1d − fn) = dhd − df = 0 by assumption.

fn+1 Pn+1 / Cn+1 : ∃h P n C dn+1 dn+1

fn dnhn−1dn = dnfn ⇒ ∃hn : Pn / Cn : hn−1 P C dn dn

fn−1 Pn−1 / Cn−1

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

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

f∗ P∗ / C∗

f M / N

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

Proof. (1) Since : C0 → N is an epimorphism and P0 is projective, the map f ◦ : 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.

g1−f1 P / C 1 : 1 ∃h0 d d

g0−f0 P / C 0 : 0 h−1=0 ε ε

0 M / N This gives us the statement that we really want: 26 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA Corollary 8.5. In any abelian category with enough projectives, any object A has a projective resolution P∗ → A. Furthermore, any two projective resolutions of A are homotopy equivalent. 0 Proof. If there are two projective resolutions P∗,P∗ then the ﬁrst part of the theorem 0 0 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∗ 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∗ 0 0 and similarly g∗ ◦ f∗ ' idP∗ . So, P∗ ' P∗. The dual argument gives us the following. [In general you should state the dual theorem but not prove it.] Theorem 8.6. In any abelian category with enough injectives, any object B has an injective coresolution. Furthermore, any two injective coresolutions of B are homotopy equivalent. Following this rule, I should also give the statement of the dual of the previous theorem:

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

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

f N / M

(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∗. MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 27 9. Derived functors A functor F : A → B between abelian categories A, B is called left exact if it is an j p additive functor which takes every exact sequence 0 → X −→ Y −→ Z → 0 in A to a left exact sequence F j F p 0 → FX −→ FY −→ 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 F p : FY → FZ. The “right derived functors” R1F,R2F, ··· are the answer to this problem. This sequence 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 sequence 0 → 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 chain homotopy equivalence. Deﬁnition 9.1. The right derived functors RiF of F are deﬁned to 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 Ext functors: i i i ExtR(A, X) := R F (X) = H (HomR(A, Q∗)) Note that the derived functors are only well-deﬁned up to isomorphism. If there is 0 0 another choice of injective coresolutions Q∗ then Q∗ ' Q∗ which implies that F (Q∗) ' 0 F (Q∗) which implies that i ∼ i 0 H (F (Q∗)) = H (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) 28 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA Thus Theorem 9.2. The zero-th derived functor R0F is canonically isomorphic to F . In particular, 0 ∼ ExtR(A, X) = HomR(A, X) 1 Example 9.3. Compute ExtZ(Z3, Z2). Take an injective coresolution of Z2: j 0 → Z2 → Q/2Z −→ Q/Z → 0 This gives a left exact sequence

j] 0 → Hom(Z3, Z2) → Hom(Z3, Q/2Z) −→ Hom(Z3, Q/Z) 1 Then, by definition, ExtZ(Z3, Z2) is the cokernel of j] (which is a monomorphism since Hom(Z3, Z2) = 0). 1 Claim: This map is an isomorphism and, therefore, Ext (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 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, both of there groups have 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 is that they fit into a long exact sequence where some maps are induced from the given maps α, β in the short exact sequence and every third map is a mysterious morphism called the “connecting map”. The main technical detail is to construct this map. There are several ways to do this. We can “chase the diagram” (the traditional method) or we can “rotate” the exact sequence (a conceptual method which is more difficult but makes some sense) Theorem 9.4. Given any short exact sequence β 0 → A −→α B −→ C → 0 there is a sequence of morphisms (the connecting morphisms) n n+1 δn : R F (C) → R F (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 exact rows: α β 0 / A / B / C / 0

f g h α0 β0 0 / A0 / B0 / C0 / 0 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 29 we get a commuting square:

δn RnF (C) / Rn+1F (A)

h∗ f∗

δn RnF (C0) / Rn+1F (A0) 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 and RnF (K) ∼= Rn+1F (A) for all n ≥ 1.

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

j j0 j1 j2 0 → A −→ Q0 −→ Q1 −→ Q2 −→ Q3 → · · · ∼ ∼ ∼ Since coker j = im j0 = ker j1 = K, we can break this up into two exact sequences: j 0 → A −→ Q0 → K → 0

j1 j2 0 → K → Q1 −→ Q2 −→ Q3 → · · · The second exact sequence shows that the injective coresolution of K is the same as that 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:

(j1)∗ 0 → F (K) → F (Q1) −−→ 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 ﬁrst cohomology of the cochain 1 complex F (Q∗) which is equal to R F (A). So, we get the exact sequence π 1 F (Q0) → F (K) −→ R F (A) → 0

We already know that the kernel of F (Q0) → F (K) is F (A) so this proves the lemma. 30 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA The construction of the delta operator proceeds as follows. Start with any short exact sequence 0 → A → B → C → 0. Then choose an injective coresolution of A:

j j0 j1 j2 0 → A −→ Q0 −→ Q1 −→ Q2 −→ 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 commuting diagram:

α β (9.1) 0 / A / B / C / 0

idA f g j p 0 / A / Q0 / K / 0 n n The map g : C → K induces a map g∗ : R F (C) → R F (K) and the connecting morphism δn for n ≥ 1 can be deﬁned to be the composition: n g∗ n ∼ n+1 δn : R F (C) −→ R F (K) = R (A). When n = 0, the map on the right is an epimorphism, not an isomorphism: g∗ π 1 δ0 : F (C) −→ F (K) −→ R (A).

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

α β 0 / A / B / C / 0 h 0 f−f 0 g−g0 j { p 0 / A / Q0 / K / 0 h ◦ β = f − f 0 ⇒ p ◦ h = g − g0 α β Rn−1F (C) / RnF (A) / RnF (B) / RnF (C) / 0

h 0 ∗ 0 0 (f−f )∗ g∗−g∗=0 ∼ w n−1 = if n≥2 n j n p n R F (K) / / R F (A) / R F (Q0) = 0 / R F (K) / 0 0 0 In the case n = 0, g∗ − g∗ = F (g − g ) still factors through F (Q0), which is not zero 1 in general. But this is OK since F (Q0) goes to the kernel of π : F (K) → R F (A). So, 0 0 π∗(g∗ − g∗) = 0 making δ0 = π∗ ◦ g∗ = π∗ ◦ g∗. MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 31 9.3. connecting morphism: standard method. We will go over the standard method of 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 “rotating” the short exact sequence. To make this comparison between diﬀerent δ∗’s we need the formula for δ∗ in each case. Recall the rotation construction: Given a short exact sequence β 0 → A −→α B −→ C → 0 we take an injective coresolution of A:

j j0 j1 j2 0 → A −→ Q0 −→ Q1 −→ Q2 −→· · ·

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

α β (9.2) A / B / / C j0

g ' 0 idA f Q0 0 j0 j p g ' 0 A / Q0 / / K 0 Q1 j 0 ' g + Q1 1 j1 ' Q2 The morphism g induces the connecting morphism:

i i R F (gi) i ∼ i+1 δi : R FC −−−−→ R FK = R RA 0 where gi : Qi → 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.

f∗ g∗ Lemma 9.7 (Long exact sequence). Let A∗ −→ B∗ −→ C∗ be a short exact sequence of chain complexes or cochain complexes (or complexes indexed by all integers). I.e., for each n, An → Bn → Cn is a short exact sequence. Then we get a long exact sequence in homology:

Hn(f∗) Hn(g∗) δ∗ · · · → Hn(A∗) −−−−→ Hn(B∗) −−−−→ Hn(C∗) −→ 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 to z˜ ∈ Bn and letting x ∈ An−1 be the unique element which maps to dn(˜z) ∈ Bn−1:

fn gn 0 / An / ∃z˜ ∈ Bn / z ∈ Cn / 0

dn dn dn

fn−1 gn−1 0 / x ∈ An−1 / dn(˜z) ∈ Bn−1 / 0 ∈ Cn−1 / 0 32 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA

You need to verify that δ∗ is well-deﬁned and that, together with the induced maps Hn(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 of them are exact, so is the third.

Lemma 9.9 (Horseshoe Lemma). Given injective coresolutions A → Q∗ of A and C → 0 Q∗ of C, there is an injective coresolution of B of the following form: 0 0 0 B −→ Q0 ⊕ Q0 −→ Q1 ⊕ Q1 −→ Q2 ⊕ Q2 −→· · · and the following commuting diagram with exact rows and columns:

α β 0 / A / B / C / 0

 f  j   j0 1 j0 ◦ β   0 [0,1] 0 0 0 / Q0 / Q0 ⊕ Q0 / Q0 / 0

  j0 h0 0 j0    0  j0 1 0 j0   0 [0,1] 0 0 0 / Q1 / Q1 ⊕ Q1 / Q1 / 0

  j1 h1 0 j1    0  j1 1 0 j1   0 [0,1] 0 0 0 / Q2 / Q2 ⊕ Q2 / Q2 / 0

i+1 Where f : B → Q0 and hi = (−1) gi 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, 0 j0 h0 f 0 j0 ◦ f + h0 ◦ j ◦ β 0 0 = = 0 0 0 j0 j ◦ β 0 j0 ◦ j ◦ β 0 Looking at diagram (9.2) we see that j0 ◦ f = g0 ◦ j ◦ β. So, h0 = −g0 makes this work. 0 0 (Also j0 ◦ j = 0 is given.) And 0 j1 h1 j0 h0 0 0 j1 ◦ j0 j1 ◦ h0 + h1 ◦ j0 0 0 = = 0 0 0 j1 0 j0 0 0 0 j1 ◦ j0 0 By diagram (9.2) we see that j1 ◦g0 = g1 ◦j0. So, h0 = −g0, h1 = g1 makes this work. MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 33

Proof of Theorem 9.4. When we apply the functor F to the three coresolutions Q∗ → 0 0 Q∗ ⊕ Q∗ → Q∗, using Lemma 9.7 we get the long exact sequence: 0 → FA → FB → FC −→δ0 R1FA → R1FB → R1FC −→δ1 R2FA → · · · i+1 By the formula hi = (−1) gi, the connecting morphism δi is the same, up to sign, as the one given by rotating the exact sequence. 9.4. Left derived functors. There are two cases when we use projective 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 = − ⊗R B.

In both cases we take a projective resolution P∗ → A → 0 and define the left derived n n n functors to be L F (A) = H (F (P∗)) in the first case and L F (A) = Hn(F (P∗)) in the second case. i (Whether the contravariant functor ExtR(−,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, or as a right-exact functor Mod-R → (Mod-Z)op as in [Lang].) n Definition 9.10. The left derived functors of F (A) = A ⊗R B are called L F (A) = R Torn (A, B) 9.4.1. review of tensor product. Following Lang, I will take tensor products only over commutative rings. The advantage is that A ⊗R B will be an R-module. The tensor product is defined by a universal condition. Definition 9.11. Suppose that A, B are modules over a commutative ring R. Then a map g : A × B → C from the Cartesian product A × B to a third R-module C is called R-bilinear if it is an R-homomorphism in each variable. I.e., for each a ∈ A, the mapping b 7→ g(a, b) is a homomorphism B → C and similarly g(−, b) ∈ HomR(A, C) for all a ∈ A. A ⊗R B is defined 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 → C there 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 existence we need to construct it. But this is easy. You just take A ⊗R B to be the free R-module generated 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 + a0) ⊗ b = a ⊗ b + a0 ⊗ b (3) a ⊗ rb = r(a ⊗ b) (4) a ⊗ (b + b0) = a ⊗ b + a ⊗ b0 34 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 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 is called the left adjoint and G is called the right adjoint and F,G are called adjoint functors. Lemma 9.12. Any left adjoint functor is right exact. In particular, tensor product is right exact. Also, any right adjoint functor is left exact. Proof. In the first case, suppose that F is a left adjoint functor and β (9.3) 0 → A −→α A0 −→ A00 → 0 is a short exact sequence. Then for any C, the left exactness of HomR(−,G(C)) gives an exact sequence 00 0 0 → HomR(A ,G(C)) → HomR(A ,G(C)) → HomR(A, G(C)) By adjunction, this is equivalent to an exact sequence 00 0 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 following sequence by definition of coker F (α): F (α) F (A) −−→ F (A0) → F (A00) → 0 The left exactness of G is analogous. (Also, the proof uses the left exactness of Hom so the second case is dumb.) MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 35 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) and ∼ A ⊗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 = idA , P i 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 is commutative. 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 g is 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. P Conversely, take any element ri ⊗ bi ∈ R ⊗ B. Then X X X X 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 = Z and A = Z/n. Then a projective resolution of A is given by ·n 0 → Z −→ Z → Z/n → 0 Since this sequence is exact it gives the following right exact sequence for any abelian group B: ·n Z ⊗ B −→ Z ⊗ B → Z/nZ ⊗ B → 0 ∼ Using the lemma that R ⊗R B = B this becomes: ·n B −→ 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 ∼ r A = Z ⊕ Z/t1 ⊕ Z/t2 ⊕ · · · ⊕ Z/tn 36 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA and, since tensor product distributes over direct sum we get: r A ⊗Z B = B ⊕ B/t1B ⊕ B/t2B ⊕ · · · ⊕ B/jnB Z The derived functor Tor1 (Z/n, B) is by definition the kernel of the map ·n Z ⊗ B −→ Z ⊗ B Since Z ⊗ B = B this is just the map B → B given by multiplication by n. So, Z Tor1 (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 homomorphism of free R-modules f Rn −→ Rm n is given by a matrix M(f) = (aij) as follows. If the basis elements of R are ej and the m basis elements of R are ei then m X f(ej) = eiaij i=1 for some aij ∈ R. These numbers determine f since, for an arbitrary element x = P n ejxj ∈ R we have ! X X f(x) = f ejxj = eiaijxj j i,j This can be written in matrix form:   P    x1 a1jxj x1 P x2   a2jxj  x2  f   =   = (aij)   ···  ···  ··· P xn anjxj xn n n n When you tensor with S you get R ⊗R S = (R ⊗R S) = S

n n f⊗idS m m R ⊗R S = S −−−→ R ⊗R S = S

The claim is that M(f ⊗idS) = M(f). So, f∗ = f ⊗idS is obtained from f by “extending scalars” to S. If you have an integer matrix, you just take the same matrix and consider it as a real matrix. MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 37 9.4.5. two definitions of Ext. The last topic in Homological Algebra is the proof that n the two definitions of ExtR(A, B) that we now had are equivalent. Similarly, the two R definitions of Torn (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 injective resolution of B then n ∼ n H (Hom(P∗,B)) = H (Hom(A, Q∗)) n So, either formula gives ExtR(A, B). Proof. The theorem is true in the case when n = 0 because both sides are isomorphic to HomR(A, B). So, suppose n ≥ 1. I will give the proof in the case n = 2. n The idea is to give a description of ExtR(A, B) which uses both P∗ and Q∗. Take P∗, Q∗ ﬁxed and consider all possibilities for the following commuting diagram.

d3 d2 d1 d0 (9.4) P3 / P2 / P1 / P0 / A / 0

f3 f2 f1 f0 f 0 / B / Q0 / Q1 / Q2 / Q3 j0 j1 j2

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 both n n H (Hom(P∗,B)) and H (Hom(A, Q∗)).

Proof: The ﬁrst step is to show that f determines the fi, i ≥ 1 uniquely up to homotopy. This follows from Theorem 8.4 since P∗ is a projective resolution of A and 0 → B → Q0 → Q1 → · · · → Qn is exact. But the choice of f up to homotopy is:

ker((jn)] : HomR(A, Qn) → HomR(A, Qn+1)) n [f] ∈ = H (Hom(A, Q∗)) im((jn−1)] : HomR(A, Qn−1) → HomR(A, Qn))

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

Similarly, the morphism fn : Pn → B determines the maps fn−1 : Pn−1 → Q0, etc down to f : A → Qn uniquely up to homotopy since the Qi are injective. And the homotopy class of fn is given by ] ker(dn+1 : HomR(Pn,B) → HomR(Pn+1,B)) n [fn] ∈ ] = H (Hom(P∗,B)). im(dn : HomR(Pn−1,B) → HomR(Pn,B))

So, the group of homotopy classes of chain maps f∗ :(P∗ → A) → (B → Q∗) shifted n in degree by n is isomorphic to H (Hom(P∗,B)). Therefore, n ∼ n H (Hom(A, Q∗)) = H (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 38 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA Call these n-fold extension of B by A. Deﬁne two such sequences to be equivalent if there is a chain map between them which is the identity on both A and B:

0 / B / E1 / E2 / ··· / En / A / 0

= = 0 0 0 0 / B / E1 / E2 / ··· / En / A / 0 Theorem 9.17. The set of equivalence classes of n-fold extension of B by A is in n bijection with the set ExtR(A, B). I won’t go through the details. Just to say that, given any n-fold extension as above, n the corresponding homotopy class [g] ∈ H (Hom(P∗,B)) is given as follows. Since P∗ is a projective resolution of A, there is, by Theorem 8.4, a unique homotopy class of chain maps from P∗ to B → E1 → · · · → En covering the identity on A. On level n this gives the map g : Pn → B. Similarly, since Q∗ is an injective coresolution of B, there is a unique homotopy class of cochain complexes: E1 → · · · → En → A → 0 to Q∗ under the identity on B. At n level n this gives a map f : A → Qn representing [f] ∈ H (Hom(A, Q∗)). The diﬃculty with this construction is that one cannot easily see how to add two n-fold extensions. However, there is one advantage. There is a clear notion of iteration of extension given a map: n m n+m ExtR(A, B) ⊗ ExtR (B,C) → ExtR (A, C) And this notion is clearly associative and makes L Extn(A, A) into an associative ring! 0 Exercise 9.18. Show that the identity in this ring is idA ∈ HomR(A, A) = Ext (A, A).