INTRO TO TENSOR PRODUCTS MATH 250B


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1 INTRO TO TENSOR PRODUCTS MATH 250B ADAM TOPAZ 1. Definition of the Tensor Product Throughout this note, A will denote a commutative ring. Let M, N be two Amodules. For a third Amodule Z, consider the set of Abilinear maps M N Z. Bilin(M N, Z) Proposition 1.1. The functor Z Bilin(M N, Z) is representable by an object which we denote by M A N. This object M A N is unique upto a unique isomorphism, and it satisfies the following universal property. First, there is a representing bilinear map φ : M N M A N. And for all modules Z, the following holds. For any Abilinear map f : M N Z, there exists a unique morphism M A N Z such that f is the composition M N φ M A N Z. Proof. The uniqueness statements all follow from general nonsense on representable functors. Thus, it suffices to determine the object M A N and the morphism φ. This is where the usual construction of the tensor product comes in: M A N = R(M N) relations where the relations are generated by (1) a(m, n) (am, n) (m, an). (2) (m + m, n) (m, n) + (m, n). (3) (m, n + n ) (m, n) + (m, n ). The map φ is given by (m, n) (m, n) =: m n M A N. Here are a few important properties of the tensor product. Proposition 1.2. The following hold: (1) is commutative and associated (upto canonical isomorphisms). (2) is distributive (with respect to arbitrary direct sums). (3) N is a rightexact functor. (4) One has Hom(M N, Z) = Hom(M, Hom(N, Z)) canonically, i.e. N is leftadjoint to Hom(N, ). Proof. All properties can be deduced from the construction of the tensor product. But a better proof uses the fact that the tensor product represents Bilinear maps. 1
2 Example 1.3. Let N be a submodule of N. Then one has a canonical isomorphism (N/N ) M = N M/ im(n M N M). In particular, if a is an ideal of A, then one has M/a M = (A/a) M We can formulate an analogous definition for nmultilinear maps, from M 1 M n to Z, but this will just yield the nfold tensor product M 1 A A M n which represents the set of nmultilinear maps to Z. This nfold tensor product makes sense since is associative. The details are left as an exercise. 2. Base Change Recall that an Aalgebra is a commutative ring B endowed with a morphism A B. We can consider B as an Amodule with respect to the natural action of a A on b B via the structure map and multiplication in B. For an Amodule M, the base change of M to B is the Bmodule B A M where the action of b B on (b m) is defined as b (b m) = (bb ) m. On the other hand, if N is a Bmodule, then we can consider N as an Amodule via the morphism A B. Proposition 2.1. The following hold: (1) One has Hom B (B A M, N) = Hom A (M, N). (2) One has N B (B A M) = N A M. Suppose that B and C are two Aalgebras. Then the tensor product B A C can be given the structure of an Aalgebra as follows: (1) The product in B C is given by (b c) (b c ) = (bb ) (cc ). (2) The map A B A C is given by a a 1 = 1 a. Note that one also has canonical morphisms of Aalgebras B B C and C B C given by b b 1 and c 1 c. Proposition 2.2. The tensor product of algebras B A C, endowed with the morphisms B B C and C B C, is the fibered coproduct in the category of Aalgebras. I.e. if E is an Aalgebra which fits into the commutative diagram: then the diagram can be completed by a unique morphism.... Example 2.3. Let B be an Aalgebra. Then one has B A A[t 1,..., t n ] = B[t 1,..., t n ] as Balgebras. This isn t hard to see from the construction of the tensor product. But let s try to give a different proof here. Namely, we know that A[t 1,..., t n ] represents the functor which sends an Aalgebra C to C n. But then by the properties of base change, B A A[t 1,..., t n ] sends a Balgebra C to C n by the proposition above. The claim follows. 2
3 HEre is an arithmetic example: Example 2.4. Let L K be a finite Galois extension of fields with Galois group G. Then one has L K L = σ G L where G acts on the lefthand side by acting on the first component of L K L and on the right hand side by acting as permutation of the components in the direct product. This equality follows from the primitive element theorem for Galois extensions. In fact, this fact can be used to construct the Galois closure of a finite separable extension L K discussed in class. 3. Flat Modules Since M is rightexact, it makes sense to study the extent to which is fails to be exact. We say that M is a flat module provided that A M is an exact functor. Proposition 3.1. The following hold: (1) Free modules are flat. (2) The tensor product of flat modules is flat. (3) If B is an Aalgebra and M is a flat Amodule then the base change B A M is flat over B. (4) If B is a flat Aalgebra, then every flat Bmodule is also flat over A. Example 3.2. Let M = Z/n as a Zmodule n 2. Then M is not flat over Z. Namely, tensoring the injection Z n Z with M yields a noninjective map. Theorem 3.3. Let A be an aribtary ring and let M be an Amodule. Then the following are equivalent: (1) M is flat. (2) For all ideals I of A, the canonical map I A M IM is an isomorphism. Proof. (1) implies (2) is trivial. Let us assume (2). Let N N be an injective morphism of Amodules. We must show that M preserves injectivity. First assume that N is free of finite rank, and we proceed by induiction on the rank. The case rank = 1 is the assumption. For higher rank, one has N = N 1 N 2 with both N i N. Let N 1 = N 1 N and N 2 the image of N in N 2 = N/N 1. We have a diagram: 0 N 1 N N 2 0 with exact rows and injective vertical maps. Now tensor this with M to get 0 N 1 N N 2 0 N 1 M N M N 2 M 0 0 N 1 M N M N 2 M 0 3
4 by assumption the left and right vertical morphisms are injective and the bottom row is exact since N 1 is a direct factor of N. Now the snake lemma implies that the middle morphism is injective as well. Now suppose that N is an aribtrary free module, and let N 0 be a direct factor of N of finite rank. By the above, the map N N 0 M N 0 M is injective. But every x N M comes from N N 0 M for some N 0 as above. Thus, the map N M N M is injective. Now let N be arbitrary, and take a free presentation F N of N with kernel K. Let F be the preimage of N in F. Then one has 0 K F N 0 with exact rows and injective vertical maps. Tensoring with M we get 0 K F N 0 0 K M F M N M 0 0 K M F M N M 0 whose middle map is injective. by the snake lemma, the right map is also injective. Corollary 3.4. Let A be a PID. Then an Amodule M is flat if and only if M is torsionfree. Proof. exercise. 4. Tor Functors We define Tor i A(M, ) as L i (M A ) the leftderived functors of M A. Recall that to compute this, we take a projective resolution P of M, and define Here are a few immediate properties: Tor i A(M, N) = H i (P A N). Proposition 4.1. (1) A module M is flat if and only if Tor i (M, N) = 0 for all i 1 and all N. (2) One has canonical (functorial) isomorphisms Tor (M, N) = Tor (N, M). (3) Tor is additive (in both arguments). (4) A short exact sequence of modules yields a long exact sequence of Tor s. Since flat modules are acyclic with respect to Tor, we can calculate Tor s using flat resolutions. Namely, if F is a flat resolution of M, then one has Tor (M, N) = H i (F N). A particularly important case of this is using free resolustions which are, of course, flat. So in practice, one always uses a free resolution of M in order to compute Tor (M, ). Let s look at some examples: (1) Suppose that f is not a zerodivisori in A, then Tor 1 (A/(f), M) = f M the ftorsion in M. I.e. f M = {m M : fm = 0}. (2) In particular, one has Tor 1 (Z/n, M) is the ntorsion of M. 4
5 (3) One has Tor 1 (Q/Z, M) is the torsion subgroup of M. (4) One has Tor q (Q p, Z p, M) is the ppwoer torsion group of M. (5) One has Tor (A, M) = 0 for all M and all 1 since A is free! (6) Using the above calculations, it is easy to calculate Tor (T, M) for any finitely generated abelian group T, using the fundamental theorem of finitely generated abelian groups. 5
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