If A is a torsion free Z module (abelian group) and D is a divisble Z module, then Hom(A,D) is divisible.
I'm also wondering if it is necessary to assume that A is torsion free. (I think so but I can't think of a counter example)
Below is my attempt and I'll assume A is a finitely generated torsion free Z module.
Lemma 1. Let A, B, and be modules over . Then there are isomorphisms of abelian groups:
Proof. (Hungerford, "Algebra" p201).
Lemma 2. If M is a module over , then there is an isomorphism of -modules
Proof. (Hungerford, "Algebra" p202).
Lemma 3. Any product of (even infinitely many) injective modules is injective (link).
Lemma 4. An abelian group (Z-module) is injective if and only if it is divisible.
By the structure theorem for finitely generated modules over PID, a finitely generated torsion free -module A can be denoted as:
, where n is a rank of A.
Thus, by lemma 1 and 2, , where , , and n is a rank of A. Thus, Hom(A,D) is divisible by lemma 3 and 4.
There are certain issues if A is an infinite product of torsion-free Z modules. Then, A is not isomorphic to a direct sum of Z's, so lemma 1 cannot be applied. This issue is discussed in (Hungerford, p201) but there is no explanation about it in the book. I don't have an answer for this issue yet.
if is finitely generated, the problem can be solved much easier because we know that any finitely generated torsion free module is free. now let be a basis for and suppose
that and are given. then for every there exists such that because is divisible. define by
note that is well-defined because is a basis for it's clear that and therefore is divisible. Q.E.D.
for the general case, i.e. if is not finitely generated, i'm not convinced that the claim remains true, although i have no counter-example yet.
I think I have it!
Prove a lemma that a Z module is injective if and only if it is a divisible group.
Then look at the sequence
0 -> A -> A
Where the map from A to A is multiplication by n. It is injective because A is torsion free.
Lift the middle A to D by a homomorphism f and see that there exists a map from the right A to D, call it g such that g(nx)=f(x). Since we are working with Z modules we can pull the n out. And since our choice of f was arbitrary and f is divisible we see that Hom(A,D) is divisible.