Projective module over a PID is free?

Question

A common result is that finitely generated modules over a PID $R$ are projective iff they are free.

Is the same true that an arbitrary projective module over a PID is free? I can't find this fact anywhere, so I suspect it is false, but I can't construct an example.

Does anyone have an example of a projective module over a PID which is not free? Thank you.

Answer 1

The truth is (to me) quite surprising: Kaplansky showed that an infinitely generated projective module over any Dedekind domain $D$ is free! (The corresponding statement for finitely generated projective modules is equivalent to $D$ having trivial class group.) This is referenced, for example, here.

Answer 2

A proof that every projective module over a PID is free occurs in $\S$ 3.9 of my commutative algebra notes.

As Qiaochu Yuan mentions, infinitely generated projective modules long to be free. A generalization of Kaplansky's result is a 1963 theorem of H. Bass: let $R$ be a connected (i.e., without nontrivial idempotents) Noetherian ring. Then every infinitely generated projective $R$-module is free. There is also a result of Gabel that every infinitely generated stably free module is free. Both statements appear in $\S$ 6.5.1 of my notes; Gabel's theorem is proved (following notes of Keith Conrad); Bass's Theorem is not.

[Connectedness is necessary to rule out cheap examples of nonfree projective modules like $\{0\} \times R_2$ over the ring $R_1 \times R_2$. It is analogous to the fact that every disconnected topological space admits cheap examples of nontrivial vector bundles.]

I suspect that one can deduce Kaplansky's theorem from some of the other results in my notes, especially the very striking theorem (of Kaplansky) that any projective module (over any commutative ring!) is a direct sum of countably generated submodules. This reduces one to the countable case. Then I would like to say that one can show that a countably generated projective module over a Dedekind domain is free by using the fact that every finitely generated submodule is of the form $R^n \oplus I$ for an ideal $I$ of $R$ and "pushing the ideal $I$ off to infinity". (I will try to take a look at this when I have the chance. If anyone wants to help me out by supplying details, please feel free!)

Finally, $\S$ 6.5.1 also contains an example of an infinitely generated projective module over a non-Noetherian connected ring which is not free. This example is due to...Kaplansky.