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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.

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

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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.

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

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