# Localization of Cohen-Macaulay module of finite projective dimension at non-maximal prime ideal

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 $$\mathfrak p\ne \mathfrak m$$ ?

• "Gorenstein domain" can be replaced by "Cohen-Macaulay ring". Commented Feb 22, 2022 at 3:17

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.