# non-free module of finite projective dimension and locally free on punctured spectrum on a local non-Cohen-macaulay ring

Consider the local ring $$R=\mathbb C [[x,y,z]]/(xy, yz)$$. My question is: Does there exist a non-free finitely generated $$R$$-module $$M$$ of finite projective dimension such that $$M_P$$ is $$R_P$$-free for every non-maximal prime ideal $$P$$ of $$R$$?

Thoughts: I think $$R$$ has dimension $$2$$ and depth $$1$$. Moreover, I believe $$R_P$$ is regular local every non-maximal prime ideal $$P$$ of $$R$$. I still have no idea how to construct a module as in the question.

• Since a free $M$ has all the properties, I presume you want $M$ to be non-free. Commented Apr 3, 2023 at 5:24
• @Mohan: Yes yes, I actually want $M$ to be non-free not just non-zero... essentially I kind of want a non-trivial vector bundle on punctured spectrum (but I guess Horrocks' kind of equivalence between vector bundles and modules might not actually work here as the ring has depth $1$)
– Alex
Commented Apr 3, 2023 at 8:01
• By Auslander-Buchsbaum any such $M$ has projective dimension $1$, i.e. is the cokernel of an injection of free modules $\varphi : R^n \hookrightarrow R^m$. If $n = 1$ and the entries of $\varphi$ generate the maximal ideal, then locally at any non-maximal prime $P$ one of the entries of $\varphi$ will be a unit, so $\varphi$ splits locally and $M_P$ will be free. The answer below is the simplest example of this with the least possible $m = 3$. Commented Apr 4, 2023 at 5:39

Take $$M$$ to be the quotient, $$0\to R\stackrel{(x,y,z)}{\to} R^3\to M\to 0$$.