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)

Printable View

- Sep 28th 2009, 03:21 AMrobeulerTorsion free and divisible Z-modules
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) - Sep 28th 2009, 04:45 AMNonCommAlg
- Sep 28th 2009, 07:22 AMrobeuler
- Sep 28th 2009, 02:04 PMaliceinwonderland
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. - Sep 28th 2009, 07:45 PMNonCommAlg
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. (Wondering) - Sep 29th 2009, 03:02 AMrobeuler
Thank you for the tips. I think I might be able to prove it using the theorem that

A Z-module is injective if and only if it is a divisible group. I am working with some diagrams but I'm not sure. - Sep 30th 2009, 02:26 AMrobeuler
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.