For isolated singularity algebra, is every maximal Cohen-Macaulay module locally projective?

Question

Let $R$ be a Cohen-Macaulay noetherian local ring. Let $\Lambda$ be a noetherian $R$-algebra which is maximal Cohen-Macaulay as an $R$-module, where for every nonmaximal prime $\mathfrak{p}$, $\Lambda_{\mathfrak{p}}$ has finite global dimension. How can I prove that every $\Lambda$-module which is maximal Cohen-macaulay as an $R$-module, is locally projective on the punctured spectrum of $R$?

Answer 1

The answer is no. Because locally, you can have a module over $\Lambda$ which is mCM over $R$ but not free. There are examples of singular $R$ with a module $M$ such that $\Lambda= End_R(M)$ has finite global dimension. But $M$ itself is a $\Lambda$-module and if it is free, then $R$ is Morita equivalent to $\Lambda$, contradicting the fact that $R$ is not regular. For example, $R=k[x,y,z]/(xy-z^2)$ and $M = R\oplus (x,z)$.

To construct a concrete counter example, take $R= \mathbb C[[x,y,z,t]]/(xy-z^2)$ and $M = R\oplus (x,z)$ and $\Lambda = End_R(M)$. $M$ is not free at the point $(x,y,z)$. On the other hand, $\Lambda$ has finite global dimension on the punctured spectrum. (the singular locus is $V(x,y,z)$ so we only need to check at $(x,y,z)$.