Proving Module Homomorphisms: A x B to M & M to A x B

[JdotAckdot]

If you can help, that would be great.

Let R be a commutative ring, and A,B,M be R-modules. Prove:

a) HomR(A x B, M) is isomorphic to HomR(A, M) x HomR(B, M)
b) HomR(M, A x B) is isomorphic to HomR(M, A) x HomR(M, B)

 

[mathwonk]

step one: try to write down a map from one to the other.

e.g. given a pair of homomorphisms f:M-->A and g:M-->B. how would you construct, in the simplest most natural way, a homomorphism M-->AxB?

conversely, given a homomorphism M-->AxB, how would you construct homomorphisms M-->A and M-->B?

 

[M98Ranger]

a) To prove that HomR(A x B, M) is isomorphic to HomR(A, M) x HomR(B, M), we need to show that there exists a bijective homomorphism between the two sets. Let f: HomR(A x B, M) -> HomR(A, M) x HomR(B, M) be defined as f(phi) = (phi1, phi2) where phi1(a) = phi(a, 0) and phi2(b) = phi(0, b) for all a in A and b in B. It is clear that f is a homomorphism since for any (a,b) in A x B, we have f(phi)(a,b) = (phi1(a), phi2(b)) = (phi(a,0), phi(0,b)) = phi(a,b). Thus, f is a homomorphism.

To show that f is bijective, we will show that f is injective and surjective. To prove injectivity, let phi1, phi2 be two elements in HomR(A, M) and HomR(B, M) respectively such that f(phi1) = f(phi2). This means that phi1(a) = phi2(a) and phi1(b) = phi2(b) for all a in A and b in B. Then, for any (a,b) in A x B, we have f(phi1)(a,b) = (phi1(a), phi1(b)) = (phi2(a), phi2(b)) = f(phi2)(a,b). Thus, phi1 = phi2 and f is injective.

To prove surjectivity, let (phi1, phi2) be an element in HomR(A, M) x HomR(B, M). Then, for any (a,b) in A x B, we have f(phi1, phi2)(a,b) = (phi1(a), phi2(b)) = (phi(a,0), phi(0,b)) = phi(a,b). Thus, f is surjective and hence bijective.

Therefore, HomR(A x B, M) is isomorphic to HomR(A, M) x HomR(B, M).

b) To prove that HomR(M, A x B) is isomorphic to HomR(M, A) x HomR(M, B), we will use a similar