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Let $M$ be a projective $\mathbb{Z}$-module. Must $M$ be free?

It is easy to see that the answer is yes if $M$ is finitely generated, but I do not know about the general case.

If the answer is "yes" (which would surprise me), is the same true for Dedekind domains?

user26857
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Jana
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    Any subgroup of a free abelian group is free. – egreg Nov 02 '14 at 20:33
  • @egreg: Is that true for non-finitely-generated subgroups? One has to be careful here; for instance, direct limits of free abelian groups need not be free. – Jana Nov 02 '14 at 20:38
  • That's well known: http://en.wikipedia.org/wiki/Free_abelian_group#Subgroups The result requires the axiom of choice, of course. – egreg Nov 02 '14 at 20:40
  • @egreg: OK, thanks. The portion of my question about Dedekind domains remains open. – Jana Nov 02 '14 at 20:57
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    See http://math.stackexchange.com/a/671635/62967 Infinintely generated projective modules over a noetherian domain are free. This is generally false for finitely generated projective modules over Dedekind domains, because any ideal is projective as a module. – egreg Nov 02 '14 at 21:13
  • @egreg: You're right, in my haste I had forgotten to assume that said Dedekind domain has trivial ideal class group. Pete Clark's answer to that question finishes off my question. – Jana Nov 02 '14 at 21:17

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