The Stacks project

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 $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ Lemma 10.60.13 tells us that $\dim (\overline{S}_{\mathfrak q}/\mathfrak p\overline{S}_{\mathfrak q}) = \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) - i$. We 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 10.116.3. 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. $\square$