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)
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.
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.