Let $(R,\mathfrak m)$ be a local Gorenstein domain of dimension $2$. Let $M$ be a finitely generated $1$-dimensional module with projective dimension $1$. Then by Auslander-Buchsbaum formula,
$\mathrm{depth}(M)=2-\mathrm{pd}(M)=1=\dim M$. Hence, $M$ is a Cohen-Macaulay module.
My question is:
Must it be true that $M_{\mathfrak p}$ is not free for some prime ideal $\mathfrak p\ne \mathfrak m$ ?
If $\mathfrak p\in \mathrm{Ass}(M)$, then $\dim(R/\mathfrak p)=\mathrm{depth}(M)$, so $\dim(R/\mathfrak p)=1$. It follows that $\mathrm{ht}(\mathfrak p)=1$. From Auslander-Buchsbaum we get $\mathrm{pd}_{R_{\mathfrak p}}(M_{\mathfrak p})+\mathrm{depth}(M_{\mathfrak p})=1$. Since $\mathfrak p\in \mathrm{Ass}(M)$ we have that $\mathrm{depth}(M_{\mathfrak p})=0$, and thus $\mathrm{pd}_{R_{\mathfrak p}}(M_{\mathfrak p})=1$, so $M_{\mathfrak p}$ is not free.
