12 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA. 4. Tensor product

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1 12 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA Here s an outlne of what I dd: (1) categorcal defnton (2) constructon (3) lst of basc propertes (4) dstrbutve property (5) rght exactness (6) localzaton s flat (7) extenson of scalars (8) applcatons 4. Tensor product 4.1. defnton. Frst I gave the categorcal defnton and then I gave an explct constructon unversal condton. Tensor product s usually defned by the followng unversal condton. Defnton 4.1. If E, F are two modules over a commutatve rng R, ther tensor product E F s defned to be the R-module havng the followng unversal property. Frst, there exsts an R-blnear mappng f : E F E F. Second, ths mappng s unversal n the sense that, for any other R- module M and blnear mappng g : E F M, there exsts a unque R-module homomorphsm h : E F M makng the followng dagram commute. E F g f E F!h M As wth all unversal condtons, ths defnton only gves the unqueness of E F up to somorphsm. For the exstence we need a constructon constructon of E F. The mappng f : E F E F s not onto. However, the mage must generate E F otherwse we get a contradcton. The elements n the mage of f are denoted f(x, y) = x y. Defnton 4.2. The tensor product E F s defned to be the R module whch s generated by the symbols x y for all x E, y F modulo the followng condtons

2 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA 13 (1) x s R-blnear. I.e. (a) x ry = r(x y) for all r R (b) x (y + z) = (x y) + (x z) (2) y s R-blnear. I.e., (a) rx y = r(x y) for all r R (b) (x + y) z = (x z) + (y z) I ponted out that these condtons requre R to be commutatve snce rs(x y) = r(sx y) = sx ry = s(x ry) = sr(x y). Proposton 4.3. E F as gven n the second defnton satsfes the unversal condton of the frst defntons and therefore, the tensor product exsts and s unque up to somorphsm. Proof. I sad n class that ths s obvous. If there s a blnear mappng g : E F M, the nduced mappng h : E F M must take the generators x y to g(x, y). Otherwse the dagram wll not commute. Therefore, h s gven on the generators and s thus unque. The only thng we need s to show that h s a homomorphsm. But ths s equvalent to showng that the elements of the form rx y r(x y) and elements correspondng to the other three condtons n the second defnton go to zero n M. But ths element goes to g(rx, y) rg(x, y) = 0 snce g s R-blnear and smlarly for the other three elements. So, h s an R-module homomorphsm and we are done functoral propertes of tensor product. The frst propertes I mentoned were the categorcal propertes whch follow drectly from the defnton. Proposton 4.4. For a fxed R-module M, tensor product wth M s a functor M : R-Mod R-Mod. What ths means s that, gven an homomorphsm f : A B there s an R-module homomorphsm whch satsfes two condtons: (1) 1 d A = d M A 1 f : M A M B

3 14 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA (2) 1 fg = (1 f)(1 g). The defnton s (1 f)(x y) = x f(y). Ths gves a homomorphsm snce the mappng M A M B gven by (x, y) x f(y) s blnear and therefore nduces the desred mappng 1 f. More generally, gven two homomorphsms f : M N, g : A B we get a homomorphsm by the formula f g : M A N B (f g)(x y) = f(x) g(y) exact functors and flat modules. Flat modules are those for whch the functor M s exact. An exact functor s one that takes short exact sequences to short exact sequences. So, frst I explaned the defntons. Defnton 4.5. An exact sequence s a sequence of modules and homomorphsms so that the mage of each map s equal to the kernel of the next map. A short exact sequence s an exact sequence of the followng form: 0 A α B β C 0. In other words, α : A B s a monomorphsm, β : B C s an epmorphsm and m α = ker β or: C = B/αA. Sometmes short exact sequences are wrtten: A B C. Defnton 4.6. A functor F : R-Mod R-Mod s called exact f t takes short exact sequences to short exact sequences. Thus the short exact sequence above should gve the short exact sequence 0 F A F α F B F β F C 0. Defnton 4.7. An R-module M s called flat f M s an exact functor. I.e., 0 M A 1 α M B 1 β M C 0 s exact for all short exact sequences A B C. One of the man results (whch we wll see s actually trval) s that S 1 R s flat for any multplcatve set S. I.e., localzaton s exact.

4 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA lst of propertes. I explaned that the exactness of localzaton was one of the key deas. However, the explanaton requred an understandng of the basc propertes of tensor product. So, I went back to the begnnng wth ths lst. (0) (unty) R M = M. (1) (commutatve) M N = N M (2) (dstrbutve) N M = (N M ) (3) (assocatve) (A B) C = A (B C) (4) (rght exactness) M s rght exact,.e., a short exact sequence A B C gves an exact sequence M A 1 α M B 1 β M C 0 (5) (localzaton s exact) I.e., we get an exact sequence: 0 S 1 A S 1 B S 1 C 0. (6) (extenson of scalars) Gven a rng homomorphsm R S, every R-module M gves an S module S R M Grothedeck rng. I dd not prove propertes (1) and (3). I sad they were obvous. However, I put the frst three condtons nto a conceptual framework by pontng out that these are the axoms of a rng. The only thng that we don t have s an addtve nverse. The algebrac constructon s as follows. Frst, you take the set of all somorphsm classes of fnntely generated R-modules [M]. Ths set has addton and multplcaton gven by [M] + [N] = [M N] [M][N] = [M N] Addton and multplcaton are assocatve and commutatve and have unts: [0] s the addtve unt and [R] s the multplcatve unt. It just doesn t have addtve nverses. So, Grothendeck sad to just put n formal nverses: [M] [N] whch are defned lke fractons: [M] [N] = [A] [B] f there exsts another module C so that M B C = N A C. Ths gves a rng whose name s G(R). The notaton K 0 (R) s for the rng of formal dfferences of f.g. projectve R-modules.

5 16 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA R M = M. After usng ths formula many tmes n the lecture, I decded I should prove t. I put the proof at the begnnng n the notes where t belongs. Theorem 4.8. R M = M for any R-module M. Proof. Snce the mappng R M M gven by (r, x) rx s blnear t nduces a mappng µ : R M M so that µ(r x) = rx. The nverse mappng φ : M R M s gven by φ(x) = 1 x. We carefully checked that these are nverse to each other: φµ(r x) = φ(rx) = 1 rx = r(1 x) = r x µφ(x) = µ(1 x) = 1x = x. So, these maps are both somorphsms of R-modules dstrbutve property. I gave a category theory proof of the dstrbutvty of tensor product over drect sum. Frst I ponted out that the followng formal characterzaton of drect sum. Lemma 4.9. M s the drect sum of modules M 1,, M n f and only f there are ncluson maps s : M M and projecton maps p : M M so that (1) p j s = δ j,.e., equal to the dentty mappng on M f = j and equal to 0 f j. (2) s p = d M. I drew the followng dagrams to llustrate these equatons. δ j M M j s p j M p j s = δ j n M M =1 s p = d M p s M Ths lemma was proved n any preaddtve category n Part B, Theorem 7.4.

6 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA 17 Theorem If M = n =1 M then n N M = N M = =1 n (N M ). =1 Proof. Consder the homomorphsms: N M 1 s 1 p j N M N M j a) (1 p j )(1 s ) = 1 p j s = 1 δ j = δ j (1 1). b) (1 s )(1 p ) = 1 s p = 1 1 = d N M. These condtons mply that N M = N M by the above lemma. Remark Ths proof works n any preaddtve category to show that any lnear functor dstrbutes over drect sum rght exactness of tensor product. I ddn t prove the rght exactness of tensor product ths frst tme because the elementary proof s messy and not very nstructve. I just explaned that ths s a specal case of a much more general prncple that: All lnear left adjont functors are rght exact. I wll explan ths later. The statement of the theorem s the followng. Theorem Tensor product wth any R-module M sends any exact sequence of R-modules of the form: A α B β C 0 to another exact sequence of the same form: M A M B M C 0. Ths statement appears stronger than the orgnal statement snce the hypothess s weaker. But I explaned that the frst statement mples ths second verson. Suppose that we know that M sends short exact sequences to rght exact sequences as above. Then how can we conclude that t sends the more general rght exact sequences A B C 0 to rght exact sequences? The frst statement mples that M takes epmorphsms to epmorphsms. (In fact ths s obvous snce the generators x y M C come from generators x ỹ M B.) Therefore M A maps onto M α(a). If we assume the weaker condton that the functor M

7 18 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA takes short exact sequences to rght exact sequences, then t wll take the short exact sequence to an exact sequence 0 α(a) B β C 0 M α(a) M B 1 β M C 0 Ths says that M α(a) maps onto the kernel of 1 β. But M A maps onto M α(a). So, M A also maps onto ker(1 β). So, we get an exact sequence M A 1 α M B 1 β M C 0. Here s an example of how ths s used. Corollary Suppose that I R s an deal. Then R/I M = M/IM where IM s the submodule of M generated by all products of the form ax where a I and x M. In partcular, when I = (p) s prncpal, we have R/(p) M = M/pM where pm = {px x M}. Proof. Suppose that I s generated by elements a. Then we have an epmorphsm of R modules R I sendng (r ) R to r a I. Ths gves an exact sequence R α R R/I 0. Tensor wth M to gve R M α 1 R M R/I M 0. Usng the somorphsms µ : R M = M and φ : M = R M we get an exact sequence M µ(α 1)φ M R/I M 0 where µ(α 1)φ sends (x ) M to a x M. The mage s equal to IM by defnton. So, R/I M = M/IM as clamed.

8 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA 19 For fntely generated modules over a PID we can now compute the tensor product: ( M R n ) R/(p n ) = M n M/p n M localzaton s exact. Recall that a multplcatve set s a subset S R whch s closed under multplcaton, contans 1 and does not contan 0. The localzaton S 1 R was defned to be the rng of all fractons r/s where r R and s S modulo the equvalence relaton r s r s f there s an element t S so that rs t = r st. Ths rng s also an R-module snce we have an acton of R gven by r x s = rx s. Proposton For any R-module M let S 1 M be the set of equvalence classes of fractons x/s where x M, s S modulo the equvalence relaton x/s y/s f there s a t S so that ts x = tsy. Then S 1 M s an R-module wth acton of R gven by r(x/s) = rx/s and S 1 M = S 1 R M. Proof. There s an obvous map S 1 R M S 1 M sendng r/s x to rx/s. The nverse map sends x/s to 1/s x. To show that ths s well-defned, take an equvalent element tx/ts. Ths goes to 1 ts tx = t ( 1 ts x The rest of the proof s straghtforward. ) = t ts x = 1 s x. Theorem S 1 R s a flat R-module. Equvalently, every short exact sequence of R-modules A B C nduces an exact sequence 0 S 1 A S 1 B S 1 C 0. Proof. Snce tensor product s rght exact, t suffces to show that S 1 A S 1 B s a monomorphsm. Ths s easy. We can assume that A B and suppose that a A and s S so that the element a/s S 1 A goes to zero n S 1 B. Ths means a s 0 s

9 20 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA n S 1 B. By defnton ths s equvalent to sayng that there exsts t S so that tsa = 0. But ths same equaton mples that a/s = 0/1 n S 1 A. So, we are done. The rng S 1 R acts on the module S 1 M n the obvous way: r x s t = rx st. Ths makes S 1 M nto a module over S 1 R. Ths s an example of extenson of scalars extenson of scalars. We had a concept before called restrcton of scalars. That was when we had a subrng S of R or, more generally, a rng homomorphsm φ : S R and we got an nduced map φ : R- Mod S- Mod whch sent an R-module M to the same thng wth the acton of S gven by s x = φ(s)x. I.e., we restrcted the acton of the rng to S. Extenson of scalars goes the other way. Proposton Gven a rng homomorphsm φ : R S and an R-module M, S R M s an S-module wth acton of S gven by s(t x) = st x. The module s sometmes wrtten as S φ M because the R-module structure s gven by r(s x) = (φ(r)s) x = s rx. Ths s the R-module structure nduced from the S-module structure by restrcton of scalars. Proof. Multplcaton by elements of S gves an R-lnear map S S and therefore gves an R-lnear map S M S M by naturalty of tensor product. Ths gves a sequence of rng homomorphsms S End R (S) End R (S M) whch defnes the S-module structure on S M. One specal case of ths s when R s a doman and F = Q(R) s the feld of fractons. Defnton Suppose that M s a module over a doman R. Then the rank of M s defned to be the dmenson of Q(R) M as a vector space over the feld Q(R). r(m) = dm Q(R) Q(R) M.

10 MATH 101A: ALGEBRA I, PART C: MULTILINEAR ALGEBRA 21 Theorem For a f.g. module M over a PID R, f M = R r R/(p n ), the number r s equal to the rank of M and s therefore unquely determned. Proof. Ths s a calculaton usng the fact that snce aq(r) = Q(R) for a 0: R/(a) Q(R) = Q(R)/aQ(R) = 0 Q(R) M = Q(R) R r Q(R)/p n Q(R) = Q(R) r. are unquely deter- It stll remans to show that the numbers p n mned.