$Hom_R(P,P)$ is projective if $P$ is projective.

Question

Let $R$ be a commutative ring so that $Hom_R(P, P)$ has a left $R$-module structure. Then show $Hom_R(P,P)$ is projective if $P$ is finitely generated and projective.
The module structure defined on $Hom_R(P,P)$ is $(r \cdot f)(x) = f(r\cdot x) $
I don't see any relation between $Hom_R(P, P)$ and $P$ when $P$ is projective. I showed that $P$ is projective iff $\oplus P_i$ is projective, and I am thinking of applying this result somehow in $Hom_R(P,P)$. Maybe use another result that $Hom_R(P^n,P)\cong \bigoplus_i^n Hom_R(P,P)$?

Answer 1

For the moment I only know a proof for $P$ being finitely generated and projective. Im still posting this, since this answer might help anyways. You can use the result that $\text{Hom}_R(\bigoplus_{i \in I} M_i,N) \cong \prod_{i \in I}\text{Hom}_R(M_i,N).$ Since $P$ is projective there exists a set $I$ and a $R$-Module $N$ such that $$P \oplus N \cong \bigoplus_{i \in I}R.$$

This gives you $$\text{Hom}_R(P,P) \oplus \text{Hom}_R(N,P) \cong \text{Hom}_R(P \oplus N,P) \cong \text{Hom}_R(R^{(I)},P) \cong \prod_{i \in I} \text{Hom}_R(R,P) \cong \prod_{i \in I} P. $$ Now, as far as I know, direct products of projective modules are not necessarily projective, however in the case where $P$ is finitely generated, we would have a finite direct sum, e.g. $$\text{Hom}_R(P,P) \oplus \text{Hom}_R(N,P) \cong \bigoplus_{i=1}^n P,$$ thus $\text{Hom}_R(P,P)$ is projective too. I hope that this might help, despite it only proving a special case.

Edit: The direct sum is finite in this case, since there exists a $n \in \mathbb{N}$ and a surjective $R$-linear map, $\varphi$, such that the sequence $$ 0 \xrightarrow{} \text{ker}(\varphi) \xrightarrow{} R^n \xrightarrow{} P \xrightarrow{} 0$$ is exact. Since $P$ is projective this sequence splits and we obtain $P \oplus \text{ker}(\varphi) \cong R^n$. If we now set $N=\text{ker}(\varphi)$ we obtain a finite direct sum.

Answer 2

Hint: if $M$ is a direct summand of $N$, then for any $A$, $\hom(M,A)$ is a direct summand of $\hom(N,A)$.