# Submodule of free module over a p.i.d. 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?

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Rotman has it in Advanced Modern Algebra, page 651! –  Dedalus Jun 25 '12 at 18:29
(The result still holds) –  Dedalus Jun 25 '12 at 18:29
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 $R$.