It is known that the existence of nontrivial injective abelian groups is independent of choice in ZF (or, rather, ZFA). In particular, $\mathbb{Q}$ is not provably injective, much less $\mathbb{R}$, so let's work in ZF and suppose injectivity of $\mathbb{Q}$. It is also known, though (and this may just reflect my ignorance of homological algebra) seems to be more folkloric, that (at least with choice) the injective abelian groups are precisely the direct sums of $\mathbb{Q}$'s and Prüfer $p$-groups. In particular, as every vector space has a basis (with choice!) and $\mathbb{R}$ is always a vector space over $\mathbb{Q}$, $\mathbb{R}$ is injective with choice. My question then is "how much choice" is needed - say, can we get injectivity as long as $\mathbb{Q}$ is injective? or do we need more machinery?
