Freeness of endomorphism module

Question

Let $A$ be a Noetherian local domain, $M$ a finite type $A$-module. If $\mathrm{Hom}_A(M,M)$ is a free $A$-module, what can we conclude about $M$? In particular, is $M$ torsion free?

Answer 1

$M$ is certainly torsion free. Let $T\subset M$ be the torsion submodule and let $N=M/T$. If $T\neq 0$, we can localize at an associated prime of $T$ and the hypothesis is still true and so we may assume that $k\subset T$ where $k$ is the residue field. Further, it is clear that $N\neq 0$.The freeness says that $\mathrm{Hom}(M,T)=0$, since this is torsion, contained in $\mathrm{Hom}(M,M)$. But $\mathrm{Hom}(N,T)\neq 0$, by Nakayama. Thus $\mathrm{Hom}(M,T)\neq 0$, a contradiction.