Lemma 10.129.2. Suppose that $R \to S$ is a ring map which is finite type, flat. Let $d$ be an integer such that all fibres $S \otimes _ R \kappa (\mathfrak p)$ are Cohen-Macaulay and equidimensional of dimension $d$. Let $f_1, \ldots , f_ i$ be elements of $S$. The set

\[ \{ \mathfrak q \in V(f_1, \ldots , f_ i) \mid f_1, \ldots , f_ i \text{ are a regular sequence in } S_{\mathfrak q}/\mathfrak p S_{\mathfrak q} \text{ where }\mathfrak p = R \cap \mathfrak q \} \]

is open in $V(f_1, \ldots , f_ i)$.

**Proof.**
Write $\overline{S} = S/(f_1, \ldots , f_ i)$. Suppose $\mathfrak q$ is an element of the set defined in the lemma, and $\mathfrak p$ is the corresponding prime of $R$. We will use relative dimension as defined in Definition 10.125.1. First, note that $d = \dim _{\mathfrak q}(S/R) = \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) + \text{trdeg}_{\kappa (\mathfrak p)}\ \kappa (\mathfrak q)$ by Lemma 10.116.3. Since $f_1, \ldots , f_ i$ form a regular sequence in the Noetherian local ring $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ general dimension theory tells us that $\dim (\overline{S}_{\mathfrak q}/\mathfrak p\overline{S}_{\mathfrak q}) = \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) - i$. By the same Lemma 10.116.3 we then conclude that $\dim _{\mathfrak q}(\overline{S}/R) = \dim (\overline{S}_{\mathfrak q}/\mathfrak p\overline{S}_{\mathfrak q}) + \text{trdeg}_{\kappa (\mathfrak p)}\ \kappa (\mathfrak q) = d - i$. By Lemma 10.125.6 we have $\dim _{\mathfrak q'}(\overline{S}/R) \leq d - i$ for all $\mathfrak q' \in V(f_1, \ldots , f_ i) = \mathop{\mathrm{Spec}}(\overline{S})$ in a neighbourhood of $\mathfrak q$. Thus after replacing $S$ by $S_ g$ for some $g \in S$, $g \not\in \mathfrak q$ we may assume that the inequality holds for all $\mathfrak q'$. The result follows from Lemma 10.129.1.
$\square$

## Comments (1)

Comment #6277 by nkym on

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