Let $A$ be an abelian group, and consider the abelian group $\text{Hom}(A,\mathbb{Z})$ of homomorphisms from $A$ to $\mathbb{Z}$. What can be said about this group?
Since $\mathbb{Z}$ is torsion-free, so is $\text{Hom}(A,\mathbb{Z})$.
If $A$ is finitely generated, then so is $\text{Hom}(A,\mathbb{Z})$. It follows that for finitely generated $A$, $\text{Hom}(A,\mathbb{Z})$ is free of finite rank, i.e. isomorphic to $\mathbb{Z}^n$ for some non-negative integer $n$.
When $A = \bigoplus_{n=1}^{\infty} \mathbb{Z}$, then $\text{Hom}(A,\mathbb{Z}) \cong \prod_{n=1}^{\infty} \mathbb{Z}$, which is not a free group (as proved by Baer).
When $A = \prod_{n=1}^{\infty} \mathbb{Z}$, then $\text{Hom}(A,\mathbb{Z}) \cong \bigoplus_{n=1}^{\infty} \mathbb{Z}$ (proved by Specker).
Since $\mathbb{Z}$ is reduced (contains no divisible elements aside from $0$), so is $\text{Hom}(A,\mathbb{Z})$. Slightly more is true: every non-zero element of $\text{Hom}(A,\mathbb{Z})$ is a multiple of an element that is only divisible by $\pm 1$.
Is there anything more that can be said about $\text{Hom}(A,\mathbb{Z})$? In particular, is $\text{Hom}(A,\mathbb{Z})$ always a product of free abelian groups? If not, what kind of isomorphism types of abelian groups arise as $\text{Hom}(A,\mathbb{Z})$?
Best Answer
Abelian groups of the form $\operatorname{Hom}(A,\mathbb{Z})$ are called "dual groups" and have been studied quite a lot. In particular, the book
Eklof, Paul C.; Mekler, Alan H., Almost free modules. Set-theoretic methods., North-Holland Mathematical Library 65. Amsterdam: North-Holland (ISBN 0-444-50492-3/hbk). xxi, 597 p. (2002). ZBL1054.20037.
contains quite a lot about them, including a counterexample for the question.
The "Reid class" is the smallest class of nonzero abelian groups that contains $\mathbb{Z}$ and is closed under direct sums and direct products. So in particular, any direct product of free abelian groups is in the Reid class.
Let $C(\mathbb{Q},\mathbb{Z})$ be the group of continuous functions (not group homomorphisms) $\mathbb{Q}\to\mathbb{Z}$, and let $C(\mathbb{Q},\mathbb{Z})^\ast$ be its dual.
2.6D in Eklof and Mekler's book states that $C(\mathbb{Q},\mathbb{Z})^\ast$ is not in the Reid class (and so is not a direct product of free abelian groups), which answers the question asked here.
But not being in the Reid class is a much stronger condition than not being a direct product of free abelian groups, so there may well be much simpler counterexamples for the original question. For example, @Bart Michel's suggestion in comments of an infinite direct product of infinite direct sums of copies of $\mathbb{Z}$ seems likely (to me) to be an example.