Let $A$ be an abelian group that is countable. Is it true that the injective hull $E(A)$ of $A$ (as $\mathbb{Z}$-module) is also countable?
For example, $E(\mathbb{Z})=\mathbb{Q}$ and for every prime $p$, $E(\mathbb{Z}/p^n)=\mathbb{Z}(p^{\infty})$ is the Prüfer $p$-group. Therefore, it is true when $A$ is finitely generated.
Another instance $\mathbb{Q}/\mathbb{Z}$ is countable, injective $\mathbb{Z}$-module, but not finitely generated.
These examples motivate my question.
Yes. It is enough to embed a countable abelian group $A$ into a countable divisible (i.e., injective) abelian group $D$, because such a $D$ will contain an isomorphic copy of the injective hull of $A$.
Since $A$ is countable, there is a surjection $\phi: \bigoplus_{n \in \mathbb{N}} \mathbb{Z} \to A$, with kernel $K$ say. Take the pushout of this map with the obvious injection $i: \bigoplus_{n \in \mathbb{N}} \mathbb{Z} \to \bigoplus_{n \in \mathbb{N}} \mathbb{Q}$. The pushout of an injection along any map $\phi$ is again an injection, so we get an injection $j: A \to D$. We claim $D$ is countable and divisible. Indeed, $D$ is a quotient of $\bigoplus_{n \in \mathbb{N}} \mathbb{Q}$ (because the pushout of a surjection along any map $i$ is again a surjection), and the quotient of a divisible group is again divisible (and of course a quotient of a countable group is countable).