Dual of a finitely generated projective module is again finitely generated and projective.

Question

Let $M$ be a module over the commutative ring $k$. Put $M^* = \operatorname{Hom}_k(M,k)$.

If $M$ is finitely generated and projective, is the same true for $M^*$?

I tried using a finite dual base to construct a finite dual base for $M^*$ but it does not work out. How can I show this?

Answer 1

Suppose $M \oplus N \cong k^n$, then $M^* \oplus N^* \cong \mathrm{Hom}(M \oplus N,k) \cong \mathrm{Hom}(k^n,k)=k^n$. Thus $M^*$ is projective. As we have a surjection $k^n \to M^*$, $M^*$ is finitely generated.