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Math 131B: Algebra Ii Part A: Homological Algebra

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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 IIIx3: 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 homo- morphisms. E.g., mod-Z is the category of finitely generated abelian groups. (3) The category tor-Z of finite abelian groups. (4) The category P(Z) of finitely generated free abelian groups (groups isomorphic to Zn for some finite n). The first and third categories are \abelian" but the other two are not. I don't 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 9!~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 9!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 cat- egory. 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 ? 9g~ 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 2 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 2 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 2 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 2 M and the corresponding generator [x] 2 F . Using the universal property of free modules, the homomorphism p is defined by the equation p[x] = x.
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