Countable submodule of free module

Question

Is it true that if $F_0$ is a countable submodule of free $\mathbb{Z}$-module $F$ then $F/F_0$ doesn't contain any uncountable injective submodule ?

Answer 1

Yes, this is true. Let $B$ be a basis for $F$; then there is a countable subset $B_0\subseteq B$ such that $F_0$ is contained in the subgroup $G$ generated by $B_0$. Let $H$ be the subgroup generated by $B\setminus B_0$; we then have $F=G\oplus H$ and $F/F_0=G/F_0\oplus H$. If $I\subseteq F/F_0$ is injective, then its projection onto $H$ must be trivial, as any homomorphism from an injective $\mathbb{Z}$-module to a free $\mathbb{Z}$-module is trivial (since injective modules are divisible, and a free module contains no nonzero divisible elements). Thus in fact $I\subseteq G/F_0$, and so $I$ is countable since $G$ is countable.