Let $(R, \mathfrak m)$ be a local ring of dimension $d>0$. Let $M$ be a finitely generated Cohen-Macaulay $R$-module (i.e., $\operatorname{depth}M=\dim M$). Then each localisation of $M$ is also Cohen-Macaulay (https://stacks.math.columbia.edu/tag/0AAG).
My question is:
Let $M$ be a Cohen-Macaulay $R$-module of positive dimension. If $P$ is a minimal prime ideal of $M$, then is $P$ also a minimal prime of $R$? I.e. if $\dim M_P=0$, then is $\dim R_P=0$?
It is well known that $\dim R/\mathfrak p=\dim M$ for every $\mathfrak{p}\in\operatorname{Ass}M$.
If $\dim M=\dim R$, that is, $M$ is MCM, then $\dim R/\mathfrak p=\dim R$, and thus $\mathfrak p$ is minimal in $R$.
If $\dim M<\dim R$, then $\dim R/\mathfrak p<\dim R$, and $\mathfrak p$ can be non-minimal. A counterexample was provided in the comments, and can be generalized as follows: let $R$ be a regular local ring with $\dim R\ge2$, and $p\in R$ a prime element; then $M=R/(p)$ has $P=(p)$ as a minimal prime, and $P$ is non minimal in $R$.