Finitely generated projective modules are isomorphic to their double dual. Let $P$ be a finitely generated projective module. Prove that $P\cong \operatorname{Hom}_{R}(\operatorname{Hom}_{R}(P,R),R)$.
 A: Assume $P_R$ is a finitely generated projective right $R$-module; then you have $P\oplus Q=R^n$ for some right $R$-module $Q$ and $n \in \mathbb{N}$. Now
$$
\def\Hom{\operatorname{Hom}}
(R^n)^*=\Hom_R(R^n,R)\cong\Hom_R(P,R)\oplus\Hom_R(Q,R)=P^*\oplus Q^*
$$
as left $R$-modules (yes, this can be done for arbitrary rings). Apply the dual again:
$$
(R^n)^{**}\cong\Hom_R((R^n)^*,R)\cong\Hom_R(P^*\oplus Q^*,R)\cong P^{**}\oplus Q^{**}
$$
It is clear that the canonical morphism $R\to R^{**}$ is an isomorphism, and this transfers easily to $R^n\to (R^n)^{**}$. Now just check that
$$\require{AMScd}
\begin{CD}
R^n @= P\oplus Q\\
@VVV @V(f_P,f_Q)VV \\
(R^n)^{**} @>>> P^{**}\oplus Q^{**}
\end{CD}
$$
is commutative, where the map $(f_P,f_Q)$ is obviously defined. This map is an isomorphism, so both components are.
A: We have a morphism $\phi : P\to P^{**}$, it's enough to show that it is an isomorphism every time you localize at a prime $q$ of $R$. Projectivity is stable under localization (it's easy using the universal property), so you may suppose that $R$ is local. But if $R$ is local and $P$ is finitely generated, $P$ is free, and in this case the proposition is trivial.
