$\text{Hom}_k(P,P)$ is isomophic to $P\otimes_k \text{Hom}_k(P,k)$

Question

Let $P$ be a finitely generated projective right module over $k$. Then, $$\text{Hom}_k(P,P)\cong P \otimes_k \text{Hom}_k(P,k).$$

I was able to show the congruence assuming $P$ is a vector space over $k$.

The map was $f \colon P\otimes_k \text{Hom}_k(P,k) \to \text{Hom}_k(P,P)$ such that $$f(p,g)(p_0)=p.g(p_0)$$ for $p,p_0\in P$ and $g\in \text{Hom}_k(P,k)$. Then using the vector space dimension argument would give us the isomorphism.

But I don't have much idea as how to prove it for projective modules.

Answer 1

I will assume that $k$ is just a commutative ring with unity. Also, I will write $P^\vee$ for $\text{Hom}_k(P,k)$, and for every $k$-module $M$ denote by $\varphi_M$ to the map $M \otimes_k P^\vee \to \text{Hom}_k(P,M)$ such that $m \otimes g \mapsto g(\_)m$ (so, $\varphi_P$ is the map that you describes).

Note that if $f \colon M_1 \to M_2$ is any $k$-module homomorphism, we have a commutative diagram $$ \require{AMScd} \begin{CD} M_1 \otimes_k P^\vee @>{f \otimes 1}>> M_2 \otimes_k P^\vee \\ @VV{\varphi_{M_1}}V @VV{\varphi_{M_2}}V \\ \text{Hom}_k(P,M_1) @>>{f_*}> \text{Hom}_k(P,M_2) \end{CD} $$ in which $f_*$ is pre-composition with $f$.

Now, if $P$ is projective and finitely generated, $P$ is a direct summand of a free and finitely generated $k$-module $F$. Therefore, we have maps $i \colon P \to F$ and $\pi \colon F \to P$ such that $\pi \circ i = 1$.

Thus, from the diagram $$ \require{AMScd} \begin{CD} P \otimes_k P^\vee @>{i \otimes 1}>> F \otimes_k P^\vee @>{\pi \otimes 1}>> P \otimes_k P^\vee \\ @VV{\varphi_P}V @VV{\varphi_F}V @VV{\varphi_P}V \\ \text{Hom}_k(P,P) @>>{i_*}> \text{Hom}_k(P,F) @>>{\pi_*}> \text{Hom}_k(P,P) \end{CD} $$ we observe the following:

1. Since $(\pi \otimes 1) \circ (i \otimes 1) = 1$ and $\pi_* \circ i_* = 1$, the two horizontal arrows of the first (resp. second) square are injective (resp. surjective).
2. If $\varphi_F$ is injective, then so is $\varphi_F \circ (i \otimes 1) = i_* \circ \varphi_P$, which implies that $\varphi_P$ is injective too.
3. If $\varphi_F$ is surjective, then so is $\pi_* \circ \varphi_F = \varphi_P \circ (\pi \otimes 1)$, which implies that $\varphi_P$ is surjective too.

Hence, if we prove that $\varphi_F$ is an isomorphism, we’re done. Indeed, since $F = k^n$ for some $n \in \Bbb N$, consider the projections $p_1,\dots,p_n \colon F \to k$, and note that the diagram $$ \require{AMScd} \begin{CD} F \otimes_k P^\vee @>{\varphi_F}>> \text{Hom}_k(P,F) \\ @V{(a_1,\dots,a_n) \otimes g \,\mapsto\, (a_1g,\dots,a_ng)\ }VV @VV{f \,\mapsto\, (p_1 \circ f,\dots,p_n \circ f)}V \\ (P^\vee)^n @= (P^\vee)^n \end{CD} $$ commutes. Because the vertical arrows are isomorphisms, then so is $\varphi_F$.