Lemma 10.102.11. 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.**
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 $\mathfrak m$.

If $M_\mathfrak p = 0$, then the lemma holds. Assume $M_\mathfrak p \not= 0$. We have $\dim (\text{Supp}(M_\mathfrak p)) \leq \dim (\text{Supp}(M)) - 1$ as a chain of primes in the support of $M_\mathfrak p$ is a chain a primes in the support of $M$ not including $\mathfrak m$. Thus it suffices to show that the depth of $M_\mathfrak p$ is at least the depth of $M$ minus $1$. We will prove by induction on the depth of $M$ that there exists an $M$-regular sequence $f_1, \ldots , f_{\text{depth}(M) - 1}$ in $\mathfrak p$. This will prove the lemma since localization at $\mathfrak p$ is exact. Since $\text{depth}(M) = \dim ((\text{Supp}(M)) \geq \dim (\text{Supp}(M_\mathfrak p)) + 1 \geq 1$ we see that the base case happens when the depth of $M$ is $1$ and this case is trivial. Assume the depth of $M$ is at least $2$.

Let $I \subset R$ be the annihilator of $M$ such that $\mathop{\mathrm{Spec}}(R/I) = V(I) = \text{Supp}(M)$ (Lemma 10.39.5). By Lemmas 10.102.6 and 10.102.9 every maximal chain of primes in $V(I)$ has length $\geq 2$. Hence none of the minimal primes of $V(I)$ are equal to $\mathfrak p$. Thus we can use Lemma 10.14.2 to find a $f_1 \in \mathfrak p$ which is not contained in any of the minimal primes of $V(I)$. Then $f_1$ is a nonzerodivisor on $M$ and $M/f_1M$ has depth exactly one less by Lemma 10.102.3. By induction we can extend to an $M$-regular sequence $f_1, \ldots , f_ r \in \mathfrak p$ with $r = \text{depth}(M) - 1$ as desired. $\square$

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