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

Shaun
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Lukas Lewark
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  • I don't have a clear answer, but here's some food for thought. First, $\hom(-, \mathbb{Z})$ sends colimits to limits (https://ncatlab.org/nlab/show/hom-functor+preserves+limits), and that might help. Second, have you thought about what happens with $\mathbb{Q}$? – Antonio Lorenzin Apr 26 '24 at 12:08
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    $\text{Hom}(\mathbb{Q},\mathbb{Z}) = 0$ 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. – Lukas Lewark Apr 27 '24 at 06:51
  • I haven't got a clue on your question but wanted to thank you for it, I've unlearned a lot of wrong things I "knew" about free abelian groups in the last hour! Two examples I've had fun thinking about are $\prod_p \mathbb{F}p / \oplus_p \mathbb{F}_p$ and $\prod\mathbb{N} \mathbb{Z} / \oplus_\mathbb{N} \mathbb{Z}$. – hunter Apr 28 '24 at 14:21
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    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. – Bart Michels Apr 28 '24 at 14:51
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    @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 – hunter Apr 29 '24 at 12:04
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    @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. – Jeremy Rickard Apr 30 '24 at 07:15
  • @BartMichels Is it clear that the dual of an infinite direct product of infinite direct sums of copies of $\mathbb{Z}$ is an infinite direct sum of infinite direct products of copies of $\mathbb{Z}$? The other way round is clear: that the dual of an infinite direct sum of infinite direct products of copies of $\mathbb{Z}$ is an infinite direct product of infinite direct sums of copies of $\mathbb{Z}$, at least for countable sums and products. – Jeremy Rickard Apr 30 '24 at 07:19
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    @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. – Jeremy Rickard Apr 30 '24 at 09:58

1 Answers1

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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.

Jeremy Rickard
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