Finitely generated projective modules are isomorphic to their double dual.

Let $P$ be a finitely generated projective module. Prove that $P\cong \operatorname{Hom}_{R}(\operatorname{Hom}_{R}(P,R),R)$.

Solution 1:

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.

Solution 2:

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.