TITLE: Is $\text{Hom}(\prod_p \Bbb Z/p\Bbb Z, \Bbb Q) = 0$ possible without choice?
QUESTION [15 upvotes]: 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?

REPLY [5 votes]: I'm upgrading my comment to an answer, since it offers a similar, albeit different, argument than OohAah's answer.

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.