Lemma 10.103.11. Suppose $R$ is a Noetherian local ring. Let $M$ be a Cohen-Macaulay module over $R$. For any prime $\mathfrak p \subset R$ the module $M_{\mathfrak p}$ is Cohen-Macaulay over $R_\mathfrak p$.

Proof. We may and do assume $\mathfrak p \not= \mathfrak m$ and $M$ not zero. Choose a maximal chain of primes $\mathfrak p = \mathfrak p_ c \subset \mathfrak p_{c - 1} \subset \ldots \subset \mathfrak p_1 \subset \mathfrak m$. If we prove the result for $M_{\mathfrak p_1}$ over $R_{\mathfrak p_1}$, then the lemma will follow by induction on $c$. Thus we may assume that there is no prime strictly between $\mathfrak p$ and $\mathfrak m$. Note that $\dim (\text{Supp}(M_\mathfrak p)) \leq \dim (\text{Supp}(M)) - 1$ because any chain of primes in the support of $M_\mathfrak p$ can be extended by one more prime (namely $\mathfrak m$) in the support of $M$. On the other hand, we have $\text{depth}(M_\mathfrak p) \geq \text{depth}(M) - \dim (R/\mathfrak p) = \text{depth}(M) - 1$ by Lemma 10.72.10 and our choice of $\mathfrak p$. Thus $\text{depth}(M_\mathfrak p) \geq \dim (\text{Supp}(M_\mathfrak p))$ as desired (the other inequality is Lemma 10.72.3). $\square$

Comment #6601 by WhatJiaranEatsTonight on

The last inequality should be $\text{depth}(M_{\mathfrak p})\geq \dim (\text{Supp}(M_{\mathfrak p}))$. The right bracket is omitted.

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