16 MATH 131B: ALGEBRA II PART A: 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 ob- ject of an additive with kernels and enough projectives) then a projective of M is defined toA be a long 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 ,wegetalongexactsequenceofabeliangroups:

(P, Pn+1) (P, Pn) (P, Pn 1) (P, P0) (P, M) 0 ···A !A !A !···!A !A ! where we use the abbreviation (X, Y ) = Hom (X, Y ). One application of this longA exact sequence isA the “uniqueness” of projective resolu- tions: Given any other projective resolution P 0 P 0 M,thereisasequence ··· ! 1 ! 0 ! of f : P 0 P making the following diagram commute: i i ! i d d 20 10 "0 / P 0 / P 0 / P 0 / M ··· 2 1 0 f2 f1 f0 =

✏ d2 ✏ d1 ✏ " ✏ / P / P / P / M ··· 2 1 0 The f : P 0 P exists since " : P M is an . The morphism 0 0 ! 0 0 ! f d0 : P 0 P goes to the kernel of " : P M.SinceP maps epimorphically onto 0 1 1 ! 0 0 ! 1 ker ",wegetaliftingf : P 0 P ,andsoon.(f d0 (P 0,P )mapsto0 (P 0,M) 1 1 ! 1 0 1 2A 1 0 2A 1 and therefore lifts to (P10,P1).) The (right) projectiveA dimension of M is the smallest integer n 0sothatthere is a projective resolution of the form ✏ 0 Pn Pn 1 P0 M 0 ! ! !···! ! ! We write pd(M)=n.Ifthereisnofiniteprojectiveresolutionthenthepd(M)= . 1 If pd(M) = 0 we have, in a general additive category, an epimorphism: " : P0 ⇣ M with kernel 0. This does not necessarily imply that " is an isomorphism. For example, in (Z)wesawthatmultiplicationby2isanepimorphismZ Z with kernel zero. We needP the following additional assumption on : ! A ( ) Every epimorphism is the cokernel of its kernel. ⇤ The (right) global dimension of the R written gl dim(R)isthemaximum 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 1sinceeverysubmod- ule of a 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 Q Q ! ! 0 ! 1 !··· 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:Weshowedthat every module M embeds in an Q0 and there is a canonical choice of Q0 called the injective envelope of M.ThenexttermQ1 is the injective envelope of (or any injective module which contains) the cokernel of M, Q0 and so on. The injective dimension id(M)isthesmallestinteger!n so that there is an injective coresolution of the form 0 M Q Q Q 0 ! ! 0 ! 1 !···! n ! We will see later that the maximum injective dimension is equal to the maximum pro- jective 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 Ialreadymentionedseveraltimes.Weskippedthisinclass. Theorem 7.3. Suppose that R is a PID and E is a free R-module. Then every sub- module 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,letE be the free submodule✓ of E ✓ J generated by the subset J of I.LetFJ = F EJ .ThenF = FI . Step 1: Define a poset: \ P = (J, w):J I and F is free with basis w { ✓ J } The partial ordering is (J, w) (J 0,w0)ifJ 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 F = (x, 0, x) is freely generated by w = e e .So,(J, e e ) P . J { } 1 3 1 3 2 Step 2: Show that each tower (J↵,w↵) in P has an upper bound. There is an obvious upper bound{ given in} the usual way by (J, w)=( J , w ) [↵ ↵ [↵ ↵ 18 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA This clearly has the property that (J ,w ) (J, w)forall↵.Weneedtoverifythat ↵ ↵  (J, w)isanelementoftheposetP .Certainly,J = J↵ is a subset of I.So,itremains to check that [

(1) w is linearly independent over R,i.e.,ifafinitesum wiri =0thenallri =0. (2) w spans FJ ,i.e.,w FJ and every element of FJ is an R-linear combination of elements of w. ⇢ P (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 w :Ifx , ,x w then each ↵ 1 ··· n 2 xi is contained in some w↵i .Let↵ be the largest ↵i.Thenw↵ contains all the xi.Since w↵ is a basis for FJ↵ ,theseelementsarelinearlyindependent.

(2) is also easy: w↵ FJ↵ FJ .So,theunionw = w↵ is contained in FJ . Any element x F has only⇢ a finite✓ number of nonzero coordinates[ which all lie in some J . 2 J ↵ So x FJ↵ which is spanned by w↵. Therefore,2 by Zorn’s Lemma, P has a maximal element (J ,w ). It remains to: 1 1 Step 3: Show J = I and thus w is a free basis for F = FI . To prove that J 1 = I suppose J 1is strictly smaller. Then there exists an element 1 1 k of I which is not in J .LetJ 0 = J k . Then we want to find a basis w0 for FJ 1 1 [{ } 0 so that (J ,w ) < (J 0,w0), i.e., the basis w0 needs to contain w .Thiswouldgivea contradiction1 to1 the maximality of (J ,w ). There are two cases.1 1 1 Case 1. FJ0 = FJ .Inthatcasetakew0 = w and we are done. 1 1 Case 2. FJ0 = FJ .ThismeansthatthereisatleastoneelementofFJ0 whose k- coordinate is nonzero.6 1 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 map p F E k R. J0 ✓ J0 !

So, A is an ideal in R.SinceR is a PID, A = a0 = a0R.Letx FJ0 so that pk(x)=a0. Then I claim that h i 2 w0 = w x 1 [{ } 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. 1 (b) w0 spans FJ0 :Givenanyz FJ0 we must have pk(z) A = a0R.Sopk(z)=a0r and p (z xr)=0.Thisimpliesthat2 z xr F which is2 spanned by w.Soz is xr k 2 J plus an R-linear combination of elements of w .So,w0 spans FJ0 and we are done. 1 ⇤ Example 7.4. (1) Z is a PID. (2) K[t]isaPIDforanyfieldK. (3) K[s, t]isnotaPID. (4) Z[t]isnotaPID. Corollary 7.5. Every subgroup of a free 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.TheproofforanyPIDissimilar.) 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.WeviewT as an endomorphism ⇥ T : Kn Kn ! of Kn.ThismakesM = Kn into a K[t]modulebylettingt act by T (tv := Tv). Since K[t]isaPID,M is a direct sum of cyclic modules: M = A B . The first reduction is that any direct sum decomposition of theK···[t]moduleM gives a block decomposition of T .IfM = A B then 1 TA 0 STS = 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]/ f(t) . We can assume f(t)ismonic: h i n n 1 n 2 f(t)=t + b t + b t + + b 1 2 ··· n We took n =4forsimplicityofnotation.Then,afterchangingbasis,Iclaimedthatthe matrix of T is (rational canonical form): 000 b 4 1 100 b3 STS = 2010b 3 2 6001 b17 6 7 This is the matrix of T with respect to the4 following basis5 for M = K[t]/ f(t) : h 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: t4 = b b t b t2 t t3. 4 3 2 1 In case K = C, every polynomial is a product of linear polynomials: f(t)= (t a )mi i The first step is to isolate each factor (t Ya )m1 using the following. i Theorem 7.8 (Chinese remainder theorem). If f(t)=g(t)h(t) where g(t), h(t) are relatively prime then K[t]/ f(t) = K[t]/ g(t) K[t]/ h(t) h i ⇠ h i h i 20 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA

This reduces us to the case when M = C[t]/ (t a)m .Forexample,whenm =4, we get the following Jordan canonical form h i a 100 1 0 a 10 STS = 200a 13 6000a7 6 7 4 5 We did the case m =2.ThenM = C2 and (t a)2 =0onM.LetV =(t a)M. This is one dimensional. Take e V .Then(t a)e =0meansTe = ae.Take 1 2 1 1 1 v M,v / V .Then(t a)v = be1 for some b =0inC.Takee2 = v/b.Thenthe 2 2 6 matrix of T with respect to e1,e2 is

1 a 1 STS = 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 is an additive category. Then a chain complex over is an infinite sequence of objectsA and morphisms (called boundary maps): A

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 ( ) be the category of chain complexes and chain maps over an additive category C.⇤ ThenA ( ) is an additive category. If has kernels, resp. cokernels then so doesA ( ). C⇤ A A C⇤ A Proof. The direct sum: C D is the chain complex with objects Cn Dn and boundary maps ⇤ ⇤ C D C D d = d d : C D C D n n n n n ! n+1 n+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

dn0 :kerfn ker fn 1 ! C C D induced by the morphism dn : Cn Cn 1.(Sincefn 1 dn = dn fn =0onkerfn,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 ,achain 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.