A crash course in homological algebra

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1 Chapter 2 A crash course in homoloical alebra By the 194 s techniques from alebraic topoloy bean to be applied to pure alebra, ivin rise to a new subject. To bein with, recall that a cateory C consists of a set or class 1 of objects (e.. sets, roups, topoloical spaces) and morphisms (e.. functions, homomorphisms, continuous maps) between pairs of objects Hom C (A, B). We require an identity 1 A 2 Hom(A, A) for each object A, and associative composition law. In this section, we will focus on one particular example. Let R be an associative but possibly noncommutative rin with 1, and let Mod R be the cateory of left R-modules and homomorphisms. We write Hom R (, ) for the morphisms. It is worth notin that Mod Z is the cateory of abelian roups. These cateories have the followin features 1. Hom R (, ) is an abelian roup, and composition is distributive. 2. There is a zero object such that Hom R (,M)=Hom R (M,) =. 3. Every pair objects A, B has direct sum A B characterized by certain universal properties. 4. Morphisms have kernels and imaes, characterized by the appropriate universal properties. We will encounter other cateories satisfyin these conditions later on. Such cateories are called abelian. We have been a bit vaue about the precise axioms; see Weibel s Homoloical Alebra for this. 1 Iwillmostlyinoresettheoreticissues,butyoushouldatleastbeawarethatformin the set of all sets or roups or spaces leads immediately to paradoxes. The way around this in Gödel-Bernays set theory is to allow two types of constructions, sets and classes. Sets are classes, but there are classes which too bi to be sets... 5

2 O > 2..7 Diaram Chasin Asequence A f! B! C is called exact if ker =imf. A useful skill in this business is to be able to prove thins by diaram chasin. Exercise 1. Given a commutative diaram with exact rows / A / B / C / f / A / B / C / Show that is an isomorphism if f and h are isomorphisms. Theorem 2..8 (Snake Lemma). If / A / B / C / f / A / B / C is a commutative diaram with exact rows, then there is an exact sequence! ker f! ker! ker coker f! coker! coker h The only part of the sequence which is not obvious is the so called connectin This will be explained in class Hom Functors A (covariant) functor F from one cateory to another is a function takin objects to objects and morphisms to morphism such that if f : A! B then F (f) : F (A)! F (B), F (1 A )=1 F (A) and F (f ) =F (f) F (). A contravariant functor reverses direction in the sense that F (f) :F (B)! F (A), F (1 A )=1 F (A) and F (f ) =F () F (f). Here are two basic examples: If M 2 Mod R,then F ( )=Hom R (M, ) is a covariant functor from Mod R to Mod Z. A f / B M F (f)=f When R commutative, F ( ) is naturally an R-module, but not otherwise. Similarly, Hom R (,M) is a contravariant functor from Mod R to Mod Z (or Mod R when R is commutative). h h 6

3 Lemma Suppose that is exact. Then! A! B! C! (a) (b)! Hom(M,A)! Hom(M,B)! Hom(M,C)! Hom(C, M)! Hom(B,M)! Hom(A, M) are both exact. The proof is straihtforward and will be omitted. Exercise 2. Prove the lemma. Exercise 3. Prove that and! Hom(M,A)! Hom(M,B)! Hom(M,C)!! Hom(C, M)! Hom(B,M)! Hom(A, M)! are exact when the sequence! A! B p! C! is split exact. This means that there exists a map s : C! B, called a splittin, such that p s =1 C. A (contravariant) functor is called exact if it preserves exact sequences. The lemma says that the Hom functors have the weaker property left exactness. They are not exact, in eneral: Example Let R = Z M = Z/2. NotethatHom(M,Z) = and Hom(M,M) =Z/2. SoHom(M, ) applied to yields the sequence The last map is certainly not onto.! Z 2! Z! Z/2!!!! Z/2 Exercise 4. Find an example for which Hom(,M) isn t exact. Lemma If M is a free module, then Hom(M, ) is exact. 7

4 Proof. Let M = L S R,whereS miht be infinite. Given f : B! C surjective, We have Hom(M,B) / Hom(M,C) = Q Q S B f / Q = S C The horizontal map on the bottom is clearly surjective. Given a module M, let R (M) = M m2m R This is a very bi free module which maps onto M by sendin the 1 in the mth copy of R to m. Let K(M) bethekernel.wehaveacanonical exact sequence Ext Inductively, define! K(M)! R (M)! M! (2.1) Ext 1 R(M,N) = coker[hom(r (M),N)! Hom(K(M),N)] Ext i+1 R (M,N) =Exti R(K(M),N) This is not the way these roups are usually defined, but we will et to that later. These are clearly covariant functors in the second variable. Theorem Ext i ( a short exact sequence we have an infinte lon exact sequence,n) is a covariant functor in the first variable. Given! A! B! C!! Hom(C, N)! Hom(B,N)! Hom(A, N)! Ext 1 (C, N)! Ext 1 (B,N)!... Proof. We prove the second part about the exact sequence. The first part is 8

5 ( 6 similar. One constructs a commutative diaram with exact rows and columns / K(A) / K(B) / K(C) / / R (A) / R (B) / R (C) / / A / B / C / Note that the middle column is split exact. Hom the top two rows into N to et / Hom(R (C),N) / Hom(R (B),N) / Hom(R (A),N) / / Hom(K(C),N) / Hom(K(B),N) / Hom(K(A),N) We use the split exactness to see that the top row is exact. Now the snake lemma ives the first 6 terms of the exact sequence. Applyin this to yields! K(A)! K(B)! K(C)!...Hom(K(C),N)! Ext 2 (A, N)... We need to show that this map factors throuh Ext 1 (C, N). To see this, use the fact that z below is zero because b is surjective. Hom(R (B),N) b / Hom(R (C),N) / Hom(K(C),N) / Ext 1 (C, N) z Ext 2 (A, N) It follows that the oriinal 6 term sequence can be continued to a 9 term sequence. This can be continued indefinitely. 9

6 Lemma If M is free then Ext 1 (M,N) =for any N. Proof. If M is free, then we can choose a basis m i. Let s : M! R (M) be the homomorphism which sends m i to 1 in the m i th copy of R. This ives a splittin of (2.1). It follows that is surjective. Therefore Ext 1 vanishes. Hom(R (M),N)! Hom(K(M),N) Lemma If is exact, then! K! F! M! Ext 1 (M,N) = coker[hom(f, N)! Hom(K, N)] Proof. This follows from the previous lemma and theorem This is useful for doin computations. Exercise 5. Usin the sequence compute Ext 1 Z (Z/n, Z/m).! Z n! Z! Z/n! Ext via free resolutions We can now compare our definition with the more conventional one. resolution of M is a possibly infinite exact sequence A free...f F! M! (2.2) where the F i are all free. Exactness implies 2 =. So by functoriality, if we Hom this into another module N, we still et a complex Hom(F,N)! Hom(F 1,N)!... Theorem Ext i (M,N) = H i (Hom(F,N)) Proof. We break (2.2) into short exact sequences! K! F! M!! K 1! F 1! K! etc. Usin the left exactness of Hom, we can see that im[hom(f,n)! Hom(F 1,N)] = im[hom(f,n)! Hom(K,N)] ker[hom(f 1,N)! Hom(F 2,N)] = ker Hom[(F 1,N)! Hom(K 1,N)] = Hom(K,N) 1

7 / The theorem for i = 1 follows from this and lemma We have a free resolution...f 2! F 1! K! Denote the tail by F 1. The previous case implies that Ext 2 (M,N) = Ext 1 (K,N) = H 1 (Hom(F 1,N)) = H 2 (Hom(F,N)) This implies the theorem for i =2etc. Note that in the usual approach, it is not a priori clear that H i (Hom(F,N)) is well defined. This would require proof. Here is an example which shows the utility of this description. Exercise 6. Let R = k[x, y] and M = R/(x, y) = k. Construct a free resolution of M (which is a special case of the Koszul complex)! 1 y A x! R 2 x y! R! M! where the maps are iven by the indicated matrices. Usin this, we can see that 8 >< k 2 if i =1 Ext i (M,M) = k if i =2 >: if i>2 This description can be also be used to clarify meanin of connectin maps. We merely outline the idea. Suppose that we have an exact sequence! A! B! C! Then we can construct free resolutions fittin into a commutative diaram...f 1 f F / A v...g 1 v / G / B We build a new complex called the mappin cone of v with maps C(v) =...G 2 F 1!G 1 F!G (a, b) =((a) ± v(b),f(b)) We can map C (v) =G to C by composin G! B and B! C. It can be shown that this ives a free resolution of C. Note that by construction, we have a map C i (v)!f i 1 which induces a map on cohomoloy H i (Hom(C (v),n))! H i+1 (Hom(F,N)) which is precisely the connectin homomorphism. 11