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I'm able to prove it for finitely generated modules, by appealing to the characterization of a projective module as a summand of a free module, and the fact that finite-rank free modules are isomorphic to their duals.

Is it true for all modules? I have seen seemingly conflicting evidence both ways (mostly against, by observations like the dual of the direct sum of countably many copies of $\mathbb{Z}$ is not free (but could it still be projective?).)

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    $\begingroup$ No, it can't be projective because for abelian groups, projective is equivalent to free (since a subgroup of a free abelian group is always free, hence a direct summand of a free abelian group is itself free abelian). $\endgroup$ – Arturo Magidin Feb 22 '12 at 1:31
  • $\begingroup$ Okay, so then let's move away from $\mathbb{Z}$-modules to arbitrary $\mathbb{R}$-modules. $\endgroup$ – JeremyKun Feb 22 '12 at 2:07
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    $\begingroup$ Oh, right, I guess that's a counterexample then. $\endgroup$ – JeremyKun Feb 22 '12 at 2:08
  • $\begingroup$ This is rather curious. You get several Google hits for the phrase "dual of a projective module is projective" but I guess they all work under additional assumptions...? $\endgroup$ – Qiaochu Yuan Feb 22 '12 at 2:53
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    $\begingroup$ @QiaochuYuan: It's true for finitely generated (as noted by Bean); that would account for lots of Google hits in and of itself. $\endgroup$ – Arturo Magidin Feb 22 '12 at 3:14
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Let $P = \bigoplus_{\mathbb{N}}\mathbb{Z}$. Then the dual $\text{Hom}(\bigoplus_{\mathbb{N}}\mathbb{Z},\mathbb{Z})$ is not free.

Assume it is projective,and hence there is some $B$ such that $\text{Hom}(\bigoplus_{\mathbb{N}}\mathbb{Z},\mathbb{Z}) \oplus B$ is free. As Arturo points out subgroups of free Abelian groups are free and so $\text{Hom}(\bigoplus_{\mathbb{N}}\mathbb{Z},\mathbb{Z})$ must be free - which is a contradiction.

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