Submodule of free module over a p.i.d. is free even when the module is not finitely generated?

I have heard that any submodule of a free module over a p.i.d. is free.

I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so it does not apply to modules that are not finitely generated.

Does the result still hold? What's the argument?

Solution 1:

Let $F$ be a free $R$-module, where $R$ is a PID, and $U$ be a submodule. Then $U$ is also free (and the rank is at most the rank of $F$). Here is a hint for the proof.

Let $\{e_i\}_{i \in I}$ be a basis of $F$. Choose a well-ordering $\leq$ on $I$ (this requires the Axiom of Choice). Let $p_i : F \to R$ be the projection on the $i$th coordinate. Let $F_i$ be the submodule of $F$ generated by the $e_j$ with $j \leq i$. Let $U_i = U \cap F_i$. Then $p_i(U_i)$ is a submodule of $R$, i.e. has the form $R a_i$. Choose some $u_i \in U_i$ with $p_i(u_i)=a_i$. If $a_i=0$, we may also choose $u_i=0$.

Now show that the $u_i \neq 0$ constitute a basis of $U$. Hint: Transfinite induction.

The same proof shows the more general result: If $R$ is a hereditary ring (every ideal of $R$ is projective over $R$), then any submodule of a free $R$-module is a direct sum of ideals of $R$.