Is the dual of a projective module always projective?

Question

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?).)

Answer 1

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.