If $ \operatorname{Ass}(M)= \operatorname{Assh}(M)$, then $M$ is a Cohen-Macaulay $R$-module?

Question

Let $R$ be a local commutative Noetherian ring and $M$ a finitely generated $R$-module. We denote by $ \operatorname{Assh}(M)=\{ \mathfrak{p}\in \operatorname{Ass}(M) \mid \dim R/\mathfrak{p}=\dim M\}.$ Assume that $\operatorname{Ass}(M)=\operatorname{Assh}(M).$ Can we conclude that $M$ is a Cohen-Macaulay $R$-module?

Thank you very much.

Answer 1

No, this is not true. When $M=R/I$ for some ideal $I$, the condition $\operatorname{Assh}(R/I)=\operatorname{Ass}(R/I)$ means that the ideal $I$ is unmixed. This is a weaker condition than the Cohen-Macaulay property, for instance $k[x,y,z,t]/(x,y) \cap (z,t)$ satisfies your property, but it is not Cohen-Macaulay.