# A question about finitely generated projective modules

Let $A$ be a commutative ring with unity and let $P$ be a finitely generated projective $A$-module. For $any$ $A$-module $M$, how does one show that $\operatorname{Hom}_A(P,A) \otimes_A M \simeq \operatorname{Hom}_A(P,M)$?

## 1 Answer

This is because a finitely generated projective module is finitely presented, so that for any prime ideal $\mathfrak p\in\operatorname{Spec} A$, $$\bigl(\operatorname{Hom}_A(P,A)\bigl)_{\mathfrak p}\simeq\operatorname{Hom}_{A_\mathfrak p}(P_{\mathfrak p},A_{\mathfrak p})$$ and also $\;(E\otimes_AF)_{\mathfrak p}\simeq E_{\mathfrak p}\otimes_{A_\mathfrak p}F_{\mathfrak p}$.

Now, if $A$ is a local ring, a finitely generated projective module is free, and for a finitely generated free $A$-module $L$, we have: $$\operatorname{Hom}_A(L,A)\otimes_A M\simeq \operatorname{Hom}_A(L,M)$$ (It is trivially true if $L=A$, and the functors $\operatorname{Hom}$ and $\otimes$ commute with direct sums).

Thus the canonical map: \begin{align*}\operatorname{Hom}_A(P,A)\otimes_A M &\to \operatorname{Hom}_A(P,M)\\u\otimes m&\mapsto(x\mapsto u(x) m)\end{align*} is locally an isomorphism for each $\mathfrak p\in\operatorname{Spec} A$. This implies it is globally an isomorphism.