9. Tensor product and Hom

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1 9. Tensor prodct and Hom Starting from two R-modles we can define two other R-modles, namely M R N and Hom R (M, N), that are very mch related. The defining properties of these modles are simple, bt those same defining properties indce many, many different constrctions in the theory of R-modles. Understanding these constrctions gave rise to a whole new theory abot theories: the theory of categories and fnctors. In any theory with similar constrctions, we get similar conseqences. In this section we give a timid introdction. The defining property of the tensor prodct, gives rise to the idea of adjointness, which implies preservation of exactness of certain seqences. And indeed, everything becomes more and more abstract (giving a feeling of nease in the beginning), bt the proofs become more simple and easy to generalize to other theories (giving a good reason to persist). Getting sed to it is a bit like getting sed to the first isomorphism theorem in grop theory, and then in ring theory, and then in modle theory, and then in sheaf theory, and then in category theory... The first time was hard, the second time it was easier, by now it is (starting to become) trivial Defining property of tensor prodct. Let R be a ring and M, N, P three R-modles. An R-bilinear map β : M N P is a map which is bilinear in the two variables, i.e., β(r 1 m 1 + r 2 m 2, n) = r 1 β(m 1, n) + r 2 β(m 2, n) β(m, r 1 n 1 + r 2 n 2 ) = r 1 β(m, n 1 ) + r 2 β(m, n 2 ) for all r 1, r 2 R, m, m 1, m 2 M and n, n 1, n 2 N. We shall prove that all R-bilinear maps on M N are (niqely) prodced by one niversal R-blinear map. Theorem 9.1. Let R be a ring and M, N two R-modles. There is an R-modle, denoted M R N, and an R-bilinear map τ : M N M N with the following niversal property. For any R-modle P and any R-bilinear map β : M N P, there exists a niqe R-modle homomorphism f : M R N P sch that f τ = β. We shall give a constrction of the modle M R N (the tensor prodct of M and N over R) and the R-bilinear map τ in the proof of the theorem. The precise constrction is not so important, bt rather the fact that the pair exists and has the stated niversal property. The pair is niqe p to a niqe isomorphism in any case, as stated in the corollary. Corollary 9.1. Sppose the modle T and bilinear map σ : M N T also has the niversal property. Then there is a niqe R-modle isomorphism f : M R N T sch that f τ = σ. Proof. By the niversal property of the pair (M R N, τ) there is a niqe R-modle homomorphism f : M R N T sch that f τ = σ. By the niversal property of (T, σ) there is a niqe R-modle homomorphism g : T M R N sch that g σ = τ. So f g σ = f τ = σ. Bt also Id T σ = σ, so by the niqeness property f g = Id T. And g f τ = τ = 1 M N τ implies by the niqeness property, that g f = 1 M N. 33

2 34 We shall write m n := τ(m, n) M R N, called a pre tensor. Any element of M R N, sometimes called a tensor is a finite sm of pre tensors, which is what we prove next. Since τ is bilinear we have at least some relations among the pre tensors: (m 1 + m 2 ) n = m 1 n + m 2 n for all r R, m, m 1, m 2 M, n, n 1, n 2 N. m (n 1 + n 2 ) = m n 1 + m n 2 r(m n) = rm n = m rn Corollary 9.2. The R-modle M R N is generated by the image of τ, i.e., any element of M R N can be written in the form n i=1 m i n i. Proof. Write T for the R-sbmodle of M R N generated by the image of τ. So if we compose τ with the qotient map ν : M M/T we get the zero R-bilinear map. Bt since ν τ = 0 τ, the nicity property gives that ν = 0, i.e. M = T. So any tensor can be written as n n n r i (m i n i ) = (r i m i n i ) = (m i r i n i ). i=1 i=1 We now can at least write down a typical element of the tensor prodct, and maniplate somewhat the expressions sing bilinearity. Sppose now we want to define a homomorphism f : M R N P. At least we need to give the vale on the generators, i.e., what f(m n) P is. Bt there are many relations among the generators, so we have a problem of well-definedness. So sppose we have a formla depending on (m, n), say β(m, n), then the key in the definition of tensor prodcts is, that if the map (m, n) β(m, n) is R-bilinear then indeed the formla f(m n) = β(m, n) indces a well defined R- modle homomorphism f : M R N P. Example 9.1. We show that R R N N, for any R-modle N. The map R N N given by r, n rn is R-bilinear, so there is a niqe R-homomorphism f : R R N N sch that f(r n) = rn. On the other hand, define the map g : N R N by g(n) := 1 n, which is R-linear by the bilinear relations among the pre tensors. We check (g f)(r n) = g(rn) = 1 rn = r n, so g f is the identity on the generators, hence g f is the identity map. Likewise (f g)(n) = f(1 n) = 1 n = n, and so f g is the identity also. We conclde that f is an isomorphism. The two previos corollaries give a sggestion of the proof, which we will give now. Proof of Theorem 9.1. Let L be the free R-modle on the set M N, i.e., F is the collection of fnctions from M N to R that only finitely many times attains a non-zero vale L = {f : M N R; #{(m, n) M N; f(m, n) 0} < }. Write [m, n] for the fnction that takes vale 1 on (m, n) and vale 0 on any other cople (m, n ) M N. Than we can express f L as a finite sm f = (m,n) M N f(m, n) [m, n]. i=1

3 35 We define K L to be the R-sbmodle generated by the elements [m 1 + m 2, n] [m 1, n] [m 2, n]; [m, n 1 + n 2 ] [m, n 1 ] [m, n 2 ]; r [m, n] [rm, n]; r [m, n] [m, rn], for all possible m, m 1, m 2 M, n, n 1, n 2 N and r R. We define M R N := L/K and τ : M N L/K by τ(m, n) = [m, n] + K =: m n. The form of the generators of K force the map τ to be R-bilinear. Now let Q : M N P be any R-bilinear map. We first can extend to the niqe R-linear map F : L P defined by F (f) = (m,n) M N f(m, n)β(m, n), in particlar [m, n] β(m, n). For example [m 1 + m 2, n] [m 1, n] [m 2, n] β(m 1 + m 2, n) β(m 1, n) β(m 2, n) = 0, and more generally any generator of K is mapped to zero. So there is a niqe R-modle homomorphism f : L/K P sch that F = f ν K, where ν K : L L/K is the qotient map. So f(m n) = β(m, n) indeed, and f is niqe since f is totally determined by the vale of the generators Tensor prodct as fnctor. We shall say that F is a fnctor from R-modles to R-modles, if for any modle M we are given a modle F (M) and for any homomorphism f : M we are given a homomorphism F (f) : F ( ) F (M) and where we reqire that F (g f) = F (g) F (f) for g : M. Taking tensor prodct with a fixed modle N is sch a fnctor, according to the next reslt. Corollary 9.3. Let f : M be an R-modle homomorphism, and N an R-modle. There is a niqe R-modle homomorphism, noted f 1 : R N M R N, sch that f(m n) = f(m ) n. If g : M is a second homomorphism, then (g 1) (f 1) = (g f) 1. Proof. The map β : N M R N : m, n f(m ) n is bilinear, so by the defining property of the tensor prodct there is a niqe R-homomorpism sch that m n f(m ) n. Hence f 1 exists. The homomorphism (g 1) (f 1) and (g f) 1 both map the generators m n to g(f(m)) n, so they are eqal. Fix a modle N then we define F (M) = M R N and F (f) = f 1, where f : M. Then F is a fnctor. Let D be a mltiplicatively closed sbset of R, then localization (M D 1 M and f D 1 f) is also a fnctor. This fnctor transforms exact seqences to exact seqences ( localization is an exact fnctor ). This last property is no longer tre for R N. We will prove a little bit later (right after Proposition 9.1) that the fnctor R N is right exact, bt in general does not preserve injections. Theorem 9.2. Let v 0

4 36 be exact and N an R-modle. Then is also exact. R N 1 R N v 1 R N 0 Example 9.2. Let s give an example where injectivity is not preserved. Let f : Z Z be mltiplication by 2, i.e., f(n) = 2n. Take for N the Z-modle Z/2Z. Then Z Z Z/2Z f 1 Z Z Z/2Z maps n m to 2n m = n 2m = n 2m = n 0 = n 00 = 0n 0 = 0 0 = 0. So f 1 is the zero-map, which is non-injective, since Z Z Z/2Z Z/2Z The Hom R (, ) fnctor. There are other important fnctors. We recall that for two R-modles N and M the collection of R-modle homomorphisms from N to M, denoted by Hom R (N, M), is an R-modle with external mltiplication (rf)(n) := r f(n) = f(r n). If f : M and φ Hom R (N, M) then f φ Hom R (N, ), or we get a map Hom R (N, ) f Hom R (N, M) which is an R-modle homomorphism. If g : M is another homomorphism then ((g f) ) = (g ) (f ). In particlar, for fixed N we get another fnctor G(M) := Hom R (N, M) and G(f) = (f ). Which is left exact, bt we will prove more in the next two theorems. Similarly if f : N N and g : N N are R-modle homomorphisms, we get homomorphism Hom R (N, M) g Hom R (N, M) f Hom R (N, M) and ( (g f)) = ( f) ( g). Theorem 9.3. The seqence of R-modles v 0 is exact if and only if for any R-modle N the seqence is exact. 0 Hom R (, N) v Hom R (M, N) Hom R (, N) Proof. (i) We first prove in one direction. Sppose v 0 is exact. And let N be any R-modle. In particlar Im() Ker(v), i.e., v = 0. So for every φ Hom R (, N) it holds that i.e., Im( v) Ker( ). [( ) ( v)](φ) = φ v = 0,

5 Sppose ψ Ker( ), i.e., ψ : M N sch that the composition ψ : N is the zero map. Or Im Ker ψ and we get a factorization of ψ. With ν : M M/ Im the qotient map, there is a homomorphism f : M/ Im N sch that ψ = f ν. By assmption, we have Im = Ker v and v is srjective, so there is an isomorphism h : M/ Im sch that v = h ν. Then ψ = f ν = (f h 1 ) (h ν) = (f h 1 ) v and ψ Im( v). We conclde that Im( v) Ker( ). Note that we needed srjectivity of v to prove this inclsion. It remains to prove the injectivity of ( v). Let φ Ker( v). This means that φ : N sch that the composition φ v : M N is the zero map. Let m be any element. By the srjectivity of v there exists an m M sch that v(m) = m. We get that φ(m ) = (φ v)(m) = 0, or that φ is the zero map. We showed that ( v) is injective. (ii) Now sppose for any R-modle N the seqence 0 Hom R (, N) v Hom R (M, N) Hom R (, N) is exact. We want to show that the seqence v 0 is exact. We start proving that v necessarily srjective. Consider the qotient map ν : / Im v. The composition ν v : M / Im v is the zero map. When we take N = / Im v we get an exact seqence 0 Hom R (, / Im v) v Hom R (M, / Im v) and we reinterprete the statement above as saying that ν Ker( v). We conclde from exactness that ν = 0. This means that for any m we have ν(m ) = 0, i.e., m Im v, or that v is srjective. To show that Im Ker v or that v = 0 consider the identity map Id : and take the exact seqence corresponding to the choice N = M. Then for Id Hom R (, ) we get from the exactness that [( ) ( v)](id) = 0 Hom R (, M) or v = Id v = 0, indeed. Finally, we mst show that Ker v Im. This time we consider the qotient map ν : M M/ Im having composition ν = 0 or ν Ker( ) = Im( v. So there is a φ : M/ Im sch that ν = φ v. So ν(ker v) = φ(v(ker v)) = 0 or Ker v Ker ν = Im. Theorem 9.4. The seqence of R-modles 0 N N is exact if and only if for any R-modle N the seqence is exact. v N 0 Hom R (M, N ) Hom R (M, N) v Hom R (M, N) 37 Proof. Exercise.

6 Adjoint fnctors. For a fixed modle N the fnctors R N and Hom R (N, ) are in some sense dal, or rather adjoint, to each other. Let F and G be two fnctors from R-modles to R-modles. We shall say that (F, G) is an adjoint cople if for any pair of modles (M, P ) we are given an isomorphism α M,P Hom R (F (M), P ) Hom R (M, G(P )) that is natral in the following technical sense. For any f : M and φ Hom R (F (M), P ) we have α M,P (φ) f = α,p (φ F (f)) and for any g : P P and ψ Hom R (F (M), P ) we have α M,P (g ψ) = G(g) α M,P (ψ). The following reslt says that ( R N, Hom R (N, )) is indeed an adjoint cople. This follows almost directly from the defining property of a tensor prodct in terms of bilinear maps. Lemma 9.1. Let R be a ring and M, N, P three R-modles. (i) The map α M,N,P : Hom R (M R N, P ) Hom R (M, Hom R (N, P )) defined by α M,N,P (φ)(m)(n) = φ(m n) is an isomorphism. The inverse is given by β M,N,P : ψ (m n ψ(m)(n)). (ii) Let f : M and φ Hom R (M R N, P ), then α M,N,P (φ) f = α,n,p (φ (f 1)) (iii) Let g : P P and ψ Hom R (M R N, P ), then α M,N,P (g ψ) = (g ) α M,N,P (ψ). Proof. (i) We compose the two maps and obtain identities: (β α)(φ) = β(α(φ)) = (m n α(φ)(m)(n)) = (m n φ(m n)) = φ. (α β)(ψ) = α(β(ψ)) = (m (n β(ψ)(m n))) = (m (n ψ(m)(n))) = ψ. (ii) Let f : M and φ Hom R (M R N, P ). Then α M,N,P (φ) f and α,n,p (φ (f 1)) both are homomorphism from to Hom R (N, P ). Let m and n N. Then (α M,N,P (φ) f)(m )(n) = (α M,N,P (φ))(f(m ))(n) = φ(f(m ) n), and α,n,p (φ (f 1))(m )(n) = (φ (f 1))(m n) = φ(f(m ) n). We get eqality. (iii) Exercise.

7 39 Proposition 9.1. Let (F, G) be an adjoint cople of fnctors from R-modles to R-modles. (i) Sppose v 0 is an exact seqence. Then the seqence is also exact. (ii) Sppose is an exact seqence. Then the seqence is also exact. Proof. (i) Since the seqence of R-modles F ( ) F () F (M) F (v) F ( ) 0 0 P P v P 0 G(P ) G() G(P ) G(v) G(P ) v 0 is exact, by Theorem 9.3 for all R-modle P the seqence 0 Hom R (, G(P )) v Hom R (M, G(P )) Hom R (, G(P )) is exact. Now applying the isomorphism α we get for all P an exact seqence 0 Hom R (F ( ), P ) F (v) Hom R (F (M), P ) F () Hom R (F ( ), P ) The natrality conditions ensre that the maps are indeed F (v) and F (). We conclde by Theorem 9.3 again that F ( ) F () F (M) F (v) F ( ) 0 is also exact. (ii) Exercise. Proof of Theorem 9.2. The theorem follows since it is the left part of an adjoint cople Tor and Ext. Althogh R N is not qite exact, we still have a handle. In homological algebra it is then shown that there are other fnctors L i F (the left-derived fnctors for all non-negative integers i 0 sch that L 0 F = F with the property that if 0 v 0 is a short exact seqence, then there is a long exact seqence (with natral δ s)... L 3 F ( ) δ 2 L 2 F ( ) L2 F () L 2 F (M) L2 F (v) L 2 F ( ) δ 1 δ 1 L 1 F ( ) L1 F () L 1 F (M) L1 F (v) L 1 F ( ) δ 0 F ( ) F () F (M) F (v) F ( ) 0 And for all non-negative integers i 0 we get right-derived fnctors R i F sch that R 0 F = F with the property that if 0 v 0

8 40 is a short exact seqence, then there is a long exact seqence (with natral δ s) 0 G( ) G() G(M) G(v) G( ) δ 0 R 1 G( ) R1 G() R 1 G(M) R1 G(v) R 1 G( ) δ 1 δ 1 R 2 G( ) R2 G() R 2 G(M) R2 G(v) R 2 G( ) δ 2 R 3 G( ) For or cople special names were given. For F = R N we have L i F (M) = Tor R i (M, N) and for G = Hom R (N, ) we have R i G(P ) = Ext i R (N, P ). For special modles N the fnctor R N can still be exact. In that case N is called flat. If Hom R (N, ) is still exact, N is called projective. If Hom R (, N) preserves exactness, then N is called injective. As yo can gess, there is a whole theory of projective, injective and flat modles. Example 9.3. If D is a mltiplicatively closed sbset then D 1 R is a flat R-modle. This follows from the exactness of localization and the following easy lemma. Lemma 9.2. Let D R be a mltiplicatively closed sbset and M an R-modle. Then α M : D 1 R R M D 1 M : r d m rm d is an isomorphism of D 1 R-modles. If f : M is a homomorphism, then D 1 f α = α M (1 f). So localization fnctor is p to an isomorphism the same thing as R D 1 R, and so D 1 R is a flat R-modle. There are many tensor prodct identities that hold and shold be verified, like M N N N of (N M) P N (M P ), bt we will stop here, and skip to a different topic Coming back to Hilbert s theorems from the 1890 s. For R a polynomial ring over over a field with n variables. Hilbert s deep syzygy theorem is a forernner of homological algebra, implying among others that we have vanishing theorems Tor R >n(m, N) = 0 = Ext >n R (M, N). This sonds very technical, bt there are many nexpected conseqences and motivations to look for similar things in other mathematical theories. Hilbert s theorem was a very deep first exploration in homological algebra.

9 41 Département de mathématiqes et de statistiqe, Université de Montréal, C.P. 6128, sccrsale Centre-ville, Montréal (Qébec), Canada H3C 3J7 address: