Lemma 10.130.5. Let $R$ be a ring. Let $R \to S$ be of finite presentation and flat. For any $d \geq 0$ the set

$\left\{ \begin{matrix} \mathfrak q \in \mathop{\mathrm{Spec}}(S) \text{ such that setting }\mathfrak p = R \cap \mathfrak q \text{ the fibre ring} \\ S_{\mathfrak q}/\mathfrak pS_{\mathfrak q} \text{ is Cohen-Macaulay} \text{ and } \dim _{\mathfrak q}(S/R) = d \end{matrix} \right\}$

is open in $\mathop{\mathrm{Spec}}(S)$.

Proof. Let $\mathfrak q$ be an element of the set indicated, with $\mathfrak p$ the corresponding prime of $R$. We have to find a $g \in S$, $g \not\in \mathfrak q$ such that all fibre rings of $R \to S_ g$ are Cohen-Macaulay. During the course of the proof we may (finitely many times) replace $S$ by $S_ g$ for a $g \in S$, $g \not\in \mathfrak q$. Thus by Lemma 10.125.2 we may assume there is a quasi-finite ring map $R[t_1, \ldots , t_ d] \to S$ with $d = \dim _{\mathfrak q}(S/R)$. Let $\mathfrak q' = R[t_1, \ldots , t_ d] \cap \mathfrak q$. By Lemma 10.130.1 we see that the ring map

$R[t_1, \ldots , t_ d]_{\mathfrak q'} / \mathfrak p R[t_1, \ldots , t_ d]_{\mathfrak q'} \longrightarrow S_{\mathfrak q}/\mathfrak p S_{\mathfrak q}$

is flat. Hence by the critère de platitude par fibres Lemma 10.128.8 we see that $R[t_1, \ldots , t_ d]_{\mathfrak q'} \to S_{\mathfrak q}$ is flat. Hence by Theorem 10.129.4 we see that for some $g \in S$, $g \not\in \mathfrak q$ the ring map $R[t_1, \ldots , t_ d] \to S_ g$ is flat. Replacing $S$ by $S_ g$ we see that for every prime $\mathfrak r \subset S$, setting $\mathfrak r' = R[t_1, \ldots , t_ d] \cap \mathfrak r$ and $\mathfrak p' = R \cap \mathfrak r$ the local ring map $R[t_1, \ldots , t_ d]_{\mathfrak r'} \to S_{\mathfrak r}$ is flat. Hence also the base change

$R[t_1, \ldots , t_ d]_{\mathfrak r'} / \mathfrak p' R[t_1, \ldots , t_ d]_{\mathfrak r'} \longrightarrow S_{\mathfrak r}/\mathfrak p' S_{\mathfrak r}$

is flat. Hence by Lemma 10.130.1 applied with $k = \kappa (\mathfrak p')$ we see $\mathfrak r$ is in the set of the lemma as desired. $\square$

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