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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.

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    $\begingroup$ Since a free $M$ has all the properties, I presume you want $M$ to be non-free. $\endgroup$
    – Mohan
    Commented Apr 3, 2023 at 5:24
  • $\begingroup$ @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 $1$) $\endgroup$
    – Alex
    Commented Apr 3, 2023 at 8:01
  • $\begingroup$ 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 $m = 3$. $\endgroup$
    – math54321
    Commented Apr 4, 2023 at 5:39

1 Answer 1

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

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