Lemma 10.130.4. Let $R$ be a ring. Let $R \to S$ be of finite presentation and flat. For any $d \geq 0$ the set
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
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
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. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)