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

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.

• What's $\text{Ass}(M)$ ? I'm at work so I can't Google it.
– Joe
Feb 19, 2019 at 15:18
• The set $Ass(M)$ is the set of associated primes of $M.$ Feb 19, 2019 at 15:27

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.
• Thanks. If, in addition, we assume that $R$ is a Cohen-Macaulay ring, then is the above statement true? Feb 21, 2019 at 2:52
• No, in the example above $R$ is a polynomial ring Feb 21, 2019 at 13:15