A question over Cohen–Macaulay local rings

Question

Let $R$ be a Cohen–Macaulay local ring and $M$ be a finitely generated $R$-module. If ${\rm Hom}_R(M , R)=R$ then can we conclude that $M=R$ ?

Answer 1

Take $R=k[[x,y]]$ and $M$ be the maximal ideal.

Answer 2

No, this isn't true. And it doesn’t really have anything to do with being Cohen-Macaulay or local.

Take $R = \mathbb{Z}_{(2)}$ and $M = R \oplus \mathbb{F}_2$. We have $\mbox{Hom}_R(M,R) = R$ since $R$ has no zerodivisors, and for the same reason, $R \not\cong M$.

It may be easier to think about the non-local version of this: $R = \mathbb{Z}$ and $M = \mathbb{Z} \oplus \mathbb{F}_2$.