Isomorphic for free modules implies for projective

Question

I have if $F$ is a free $R$-module then $\mathrm{Hom}(L,M) \otimes F$ is isomorphic to $\mathrm{Hom}(L,M \otimes F)$. Then how can I conclude that $\mathrm{Hom}(L,M) \otimes P$ is isomorphic to $\mathrm{Hom}(L,M \otimes P)$ where $P$ is finitely generated projective?

Answer 1

$\DeclareMathOperator{\Hom}{Hom}$ If $P$ is a finitely generated projective $R$-module, then it is also finitely presented: Take a surjection $R^n\twoheadrightarrow P$ with kernel $Q$. Then $R^n = Q\oplus P$ as $P$ is projective. As the projection $R^n\rightarrow Q$ is surjective, $Q$ is also finitely generated. Thus, we obtain an exact sequence $R^m\rightarrow R^n\rightarrow P \rightarrow 0$.
We always have a natural morphism $\Hom(L,M)\otimes N\rightarrow \Hom(L,M\otimes N)$ for all $R$-modules $L, M, N$. The above exact sequence gives by functoriality a commutative diagram $$ \begin{array}{c} \Hom(L,M)\otimes R^m & \longrightarrow & \Hom(L,M)\otimes R^n & \longrightarrow & \Hom(L,M)\otimes P & \longrightarrow & 0\\ \downarrow & & \downarrow & & \downarrow & & \\ \Hom(L,M\otimes R^m) & \longrightarrow & \Hom(L,M\otimes R^n) & \longrightarrow & \Hom(L,M\otimes P) & \longrightarrow & 0 \end{array} $$ Since the left two vertical arrows are isomorphisms, the five Lemma (or direct computation) tells us that also the right vertical arrow is an isomorphism.

(Note: The morphism $\Hom(L,M\otimes R^n)\rightarrow \Hom(L,M\otimes P)$ in the second row is surjective because $R^n\rightarrow P$ is a split surjection.)