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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.

Martin Sleziak
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Hana Bailey
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    Projective modules are direct summand of free modules. In particular they are submodules of a free module and hence they are free because the ring is a PID (this last theorem is usually proved in the finitely generated case but is true in general). – nowhere dense Oct 14 '17 at 12:31

2 Answers2

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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.

Martin Sleziak
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Pete L. Clark
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  • Pete, do you know of any characterization of rings over which all projectives are free? Clearly this class must be huge since it contains all local rings. Do you have any algebraic intuition as to why "submodule of a free module is free" is so strong (PID) but "direct summand of a free module is free" is so much weaker? – Alex Youcis Feb 11 '14 at 06:34
  • @Alex: There is almost certainly no simple characterization of this class: the celebrated Serre's Conjecture (aka Quillen-Suslin Theorem) shows that polynomial rings in finitely many variables over a field satisfy this property: that was a very deep result. Note also that even the difference between stably free modules and free modules is the birth of algebraic K-theory: this gives the group $K_0(R)$. These are deep waters... – Pete L. Clark Feb 11 '14 at 07:10
  • As to your second question: every finitely generated torsionfree module over a domain is a submodule of a free module ($\S$ 21.3 of my notes), so there is all the difference in the world between a submodule and a direct summand. Whether that counts as intuition I'm not sure: it seems rather like a fact of life. – Pete L. Clark Feb 11 '14 at 07:11
  • Ah, of course, the fact that Quillen-Suslin would be encompassed by any such characterization is something that slipped my mind. I didn't know that second fact, it's an interesting one! Thanks! – Alex Youcis Feb 11 '14 at 08:02
  • Oops: in my above comment I should have said "the difference between projective modules and stably free modules", since stably free modules are precisely those projective modules which represent $0$ in $K_0(R)$. The difference between stably free modules and free modules is harder than alegebraic K-theory.... – Pete L. Clark Feb 11 '14 at 19:01
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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.

Alex Shpilkin
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Qiaochu Yuan
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