# A submodule of a free module over a PID

$R$ is a principal ideal domain and $F$ is a free module over $R$ of infinite rank with basis $\{e_1,...,e_n,...\}$.

Is it true that the $R$-submodule of $F$ spanned by $\{e_1,...,e_n\}$ has a complement in $F$?

Note that the $R$-submodule of $F$ spanned by $\{e_1,...,e_n\}$ is free, because $\{e_1,...,e_n\}$ is a basis for it (however, a submodule of a free module over a PID is always free).

But a free module is always projective; we call $M$ the module spanned by $\{e_1,...,e_n\}$. Then we have an exact sequence $$0 \rightarrow K \rightarrow F \rightarrow M \rightarrow 0$$ where the maps are the obvious ones, i.e. the projection $\phi : F \to M$ over the first $n$ coordinates, and $K$ is the kernel of $\phi$ with the injection $i : K \to F$.

But $M$ is projective, so the sequence splits, and so $$F \cong K \oplus M$$ and so $M$ has a complement.