Is $\text{Hom}(\prod_p \Bbb Z/p\Bbb Z, \Bbb Q) = 0$ possible without choice?

Question

That divisible abelian groups are precisely the injective groups is equivalent to choice; indeed, there are some models of ZF with no injective groups at all. Now, given that $\Bbb Q$ is injective, one immediately has that $\text{Hom}(\prod_p \Bbb Z/p\Bbb Z, \Bbb Q)$ is nontrivial (pick an element of infinite order in the former group; then the definition of injective gives a nonzero map to $\Bbb Q$ factoring the inclusion $\Bbb Z \hookrightarrow \Bbb Q$.)

Now, assuming that $\Bbb Q$ is not injective does not obviously prove that $\text{Hom}(\prod_p \Bbb Z/p\Bbb Z, \Bbb Q)$ is trivial, whence: is there a model of ZF in which this is true?

Answer 1

Work in ZF + DC + "every set of reals has the Baire property". Put $G = \prod_p \mathbb{Z}_p$. Suppose $\phi:\prod_p \mathbb{Z}_p \to \mathbb{Q}$ is a non constant additive function. Let $H = ker(\phi)$ (so $H$ has Baire property). Since all subgroups of $\mathbb{Q}$ are infinite, $H$ must have infinitely many cosets in $G$. By Pettis' theorem, $H$ is clopen. Since $G$ is compact, $H$ has only finitely many cosets: A contradiction.

Addendum: This is also true for any compact Polish group by the same argument.

Answer 2

A related question is the following: does there exist a model of ZF where choice fails but the hom-set is nontrivial? The following is an answer by Andreas Blass.

The answer to that is yes. The reason is that the failure of choice can be such as to involve only very large sets, much larger than the groups in the question. That is, one can start with a model of ZFC, where the Hom-set in question is nonzero, and then one can force over this model to add new subsets of some cardinal kappa that is much bigger than the cardinal of the continuum, and then one can pass to a symmetric inner model to falsify the axiom of choice; all of this can be done in such a way as not to adjoin any new elements in the product of Z_p's, so that product is the same in the new model as in the original, and all its non-zero homomorphisms to Q in the original ZFC model remain perfectly good nonzero homomorphisms in the new model that violates choice.

Answer 3

I'm upgrading my comment to an answer, since it offers a similar, albeit different, argument than OohAah's answer.

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Consider a model of $\sf ZF+DC+BP$ (where $\sf BP$ is the statement that every set of reals have the Baire property), this theory was shown consistent by Solovay and Shelah (the former did it from an inaccessible cardinal, and the latter removed that requirement).

In such model every homomorphism from a Polish group to $\Bbb Q$ is continuous. Therefore a homomorphism from a compact Polish group to $\Bbb Q$ must have a compact image, but the only compact subgroup of $\Bbb Q$ is $\{0\}$, since compact subsets of $\Bbb Q$ are bounded and nontrivial subgroups are not.