A Primer on Homological Algebra

Transcription

1 A Primer on Homological Algebra Henry Y Chan July 12, Modules For people who have taken the algebra sequence, you can pretty much skip the first section Before telling you what a module is, you probably should know what a ring is Definition 11 A ring is a set R with two operations + and and two identities and 1 1 (R, +, ) is an abelian group 2 (Associativity) (x y) z = x (y z), for all x, y, z R 3 (Multiplicative Identity) x 1 = 1 x = x, for all x R 4 (Left Distributivity) x (y + z) = x y + x z, for all x, y, z R 5 (Right Distributivity) (x + y) z = x z + y z, for all x, y, z R A ring is commutative if is commutative Note that multiplicative inverses do not have to exist! Example 12 1 Z, Q, R, C with the standard addition, the standard multiplication,, and 1 2 Z/nZ with addition and multiplication modulo n,, and 1 3 R [x], the set of all polynomials with coefficients in R, where R is a ring, with the standard polynomial addition and multiplication 4 M n n, the set of all n-by-n matrices, with matrix addition and multiplication, n, and I n For convenience, from now on we only consider commutative rings Definition 13 Assume (R, + R, R, R, 1 R ) is a commutative ring A R-module is an abelian group (M, + M, M ) with an operation : R M M 1

2 1 r (m + M n) = (r m) + M (r n), for all r R, m, n M 2 (r + R s) m = (r m) + M (s m), for all r, s R, m M 3 (r R s) m = r (s m), for all r, s R, m M 4 1 R m = m Remark 14 When R is a field, an R-module is exactly a R-vector space Exercise 15 Show that every abelian group can be regarded as a Z-module Exercise 16 Define what a homomorphism between two rings means homomorphism between two R-modules means Define what a Definition 17 Let M, N be two R-modules, and ϕ : M N be a homomorphism The kernel of ϕ, denoted ker ϕ, is defined as ker ϕ = ϕ 1 ({ N }) Exercise 18 Show that ϕ is injective if and only if ker ϕ = { M } Remark 19 From simplicity, we use to denote the trivial subgroup of every group, ie, the subgroup containing only the identity element Definition 11 We say a homomorphism ϕ : M N is trivial if it maps everything to N 2 Exact Sequences From now on R will be a commutative ring Definition 21 An exact sequence of R-modules consists of a sequence of R-modules {M i } and homomorphisms {ϕ i } looking like ϕ 3 M 2 ϕ 2 M 1 ϕ 1 M ϕ M1 ϕ 1 M2 ϕ 2 Example 22 1 ker ϕ i = im ϕ i 1, i ϕ M 1 ϕ 1 Clearly both ϕ and ϕ 1 are trivial maps, so ker ϕ 1 = M 1, im ϕ = Because the sequence is exact, M 1 must equal to 2

3 2 We know that ϕ M 1 ϕ 1 M2 ϕ 2 im ϕ 1 = ker ϕ 2 = M 2 ker ϕ 1 = im ϕ =, 3 so ϕ 1 is both surjective and injective Hence it is an isomorphism, ie, M 1 = M2 ϕ M 1 ϕ 1 M2 ϕ 2 M3 ϕ 3 This is called a short exact sequence Similar to 2, we know that ϕ 2 is surjective and ϕ 1 is injective In fact, short exact sequences contains more information, but we need the following theorem first Definition 23 Let N M be a submodule The quotient of M by N, denoted M/N, is defined as M/, where is the equivalence relation defined as m n if m n N (Check that it is indeed an equivalence relation) Theorem 24 (First Isomorphism Theorem) If ϕ : M N is a homomorphism, then im ϕ = M/ ker ϕ Corollary 25 If is an exact sequence, then ϕ M 1 ϕ 1 M2 ϕ 2 M3 ϕ 3 M 3 = M2 /M 1 (Technically we should write M 3 = M2 /ϕ 1 (M 1 ), but ϕ 1 (M 1 ) = M 1 since ϕ 1 is injective) Remark 26 You might think that M 2 = M1 M 3 However, this is not necessarily true Check that the following is an exact sequence, but Z and Z Z/2Z are clearly not isomorphic n 2n Z Z Z/2Z m m + 2Z However, if R is a field, then M 2 = M1 M 3 is always true 3

4 Theorem 27 (Splitting Lemma) For the exact sequence, the following statements are equivalent M 1 ϕ M2 ψ M3 1 (Left Split) There exists a homomorphism f : M 2 M 1 f ϕ = id M1 2 (Right Split) There exists a homomorphism g : M 3 M 2 ψ g = id M3 3 M 2 = M1 M 3 We say the exact sequence splits if the above conditions hold The following is by far the most important theorem regarding exact sequences! Theorem 28 (The Five Lemma) Given two exact sequences A B C D E, A B C D E and five homomorphisms α, β, γ, δ, ɛ the following diagram commutes (That means all possible compositions of homomorphisms from X to Y must be the same) A B C D E Then α β A B C D E 1 If β and δ are surjective, and ɛ is injective, then γ is surjective 2 If β and δ are injective, and α is surjective, then γ is injective In particular, if α, β, δ, ɛ are all isomorphisms, then γ is an isomorphism γ δ ɛ 3 Chain Complex and Homology Definition 31 A chain complex C consists of a sequence of R-modules {C i } and boundary homomorphisms {d i } looking like d 3 C 2 d 2 C 1 d 1 C d C1 d 1 C2 d 2 d i d i 1 =, i The condition is called the boundary condition 4

5 Exercise 32 Show that the boundary condition is equivalent to im d i 1 ker d i, i Hence every exact sequence is a chain complex extremely uninteresting chain complexes) (We will see that exact sequences are Definition 33 Given a chain complex C The homology of this chain complex, denoted H (C ) is defined as H i (C ) = ker d i / im d i 1 Exercise 34 Show that the homology of an exact sequence is all zero Moreover, show that if a chain complex has zero homology, then it is an exact sequence Definition 35 Given a homomorphism ϕ : M N, the cockerel of ϕ is defined as coker ϕ = N/ im ϕ Exercise 36 Show that coker ϕ = if and only if ϕ is surjective Exercise 37 Show that every homomorphism ϕ : M N induces the following exact sequence ker ϕ i M ϕ q N coker ϕ Where i : ker ϕ M and q : N coker ϕ are inclusion map and quotient map, respectively Theorem 38 (The Snake Lemma) Given two exact sequences A B C, A B C and homomorphisms α, β, γ the following diagram commutes A f B g C α A f B g C β γ Then there is an exact sequence ker α f ker β g ker γ δ coker α f coker β g coker γ 5

6 , where δ is called the connecting homomorphism It is called Snake Lemma because the induced exact sequence zig zags through the original diagram ker α ker β ker γ A B C A B C coker α coker β coker γ Definition 39 We say that A B C is a short exact sequence of chain complexes if there exist homomorphisms f i : A i B i and g i : B i C i the following diagram commutes and every row is an exact sequence f 2 A 2 g 2 B 2 f 1 A 1 g 1 B 1 f A g B C 2 C 1 C 6

7 Corollary 31 A short exact sequence of chain complexes A B C induces a long exact sequence in their homologies δ 3 H 2 (A ) f 2 H2 (B ) g 2 H2 (C ) δ 2 H 1 (A ) f 1 H1 (B ) g 1 H1 (C ) δ 1 H (A ) f H (B ) g H (C ) δ 4 Cochain Complex and Cohomology Cochain complexes and cohomologies are nothing special but chain complexes and homologies with arrows reversed Definition 41 A cochain complex C consists of a sequence of R-modules {C i } and coboundary homomorphisms {d i } looking like d 3 C 2 d 2 C 1 d 1 C d C 1 d 1 C 2 d 2 d i d i+1 =, i The condition is called the coboundary condition Definition 42 Given a cochain complex C The cohomology of this cochain complex, denoted H (C ) is defined as H i (C ) = ker d i / im d i+1 7