Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?).)

share|cite|improve this question
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). – Arturo Magidin Feb 22 '12 at 1:31
Okay, so then let's move away from $\mathbb{Z}$-modules to arbitrary $\mathbb{R}$-modules. – JeremyKun Feb 22 '12 at 2:07
Oh, right, I guess that's a counterexample then. – JeremyKun Feb 22 '12 at 2:08
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...? – Qiaochu Yuan Feb 22 '12 at 2:53
@QiaochuYuan: It's true for finitely generated (as noted by Bean); that would account for lots of Google hits in and of itself. – Arturo Magidin Feb 22 '12 at 3:14

1 Answer 1

up vote 1 down vote accepted

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.

share|cite|improve this answer
I think the hardest part would be to show that $\prod_{\mathbb{N}} \mathbb{Z}$ not free! – Juan S Feb 22 '12 at 2:21
Looks like there's a proof here… – JeremyKun Feb 22 '12 at 3:58
You can also take a look at this answer. – Pierre-Yves Gaillard Feb 22 '12 at 4:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.