# Is $\text{Hom}(A,\mathbb{Z})$ a product of free abelian groups for all abelian groups $A$?

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 $$\text{Hom}(A,\mathbb{Z})$$?

• $\text{Hom}(\mathbb{Q},\mathbb{Z}) = 0$ since a homomorphism sends divisible elements to divisible elements, all of $\mathbb{Q}$ is divisible, and only $0$ is divisible in $\mathbb{Z}$. More generally, $\text{Hom}(A,\mathbb{Z})$ is isomorphic to $\text{Hom}(A/(T(A)+D(A)),\mathbb{Z})$, where $T$ and $D$ denote the torsion subgroup and the subgroup of divisible elements. So for the purpose of the question, it is enough to consider groups $A$ that are torsion-free and reduced. Commented Apr 27 at 6:51
• I think a counter example is an infinite direct product of infinite direct sums of $Z$, because its dual is in infinite direct sum of infinite direct products and this is probably not a product of free groups. Commented Apr 28 at 14:51
• @BartMichels I think a countable direct product of countable direct sums is isomorphic to a countable direct sum of countable direct products (write the entries in a square and take the "transpose" map) -- maybe you can avoid this by varying the cardinalities Commented Apr 29 at 12:04
• @hunter For the countable direct product of countable direct sums, each row of your square has finitely many nonzero entries (but the number of nonzero entries may not be bounded), but for the countable direct sum of countable direct products, there are only finitely many nonzero rows. So they are not transpose situations. That doesn't prove that they're not isomorphic, but in fact they're not: see the answer to this MSE question. Commented Apr 30 at 7:15
• @BartMichels It seems that the dual of $\prod_{n=1}^{\infty}\bigoplus_{n=1}^{\infty} \mathbb{Z}$ is indeed isomorphic to $\bigoplus_{n=1}^{\infty}\prod_{n-1}^{\infty} \mathbb{Z}$. This follows from Corollary 2.10 in Fuchs' Abelian groups. Commented Apr 30 at 9:58

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.