Lemma 10.125.3. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p$ of $R$. Assume

1. $R \to S$ is of finite type,

2. $\dim _{\mathfrak q}(S/R) = n$, and

3. $\text{trdeg}_{\kappa (\mathfrak p)}\kappa (\mathfrak q) = r$.

Then there exist $f \in R$, $f \not\in \mathfrak p$, $g \in S$, $g \not\in \mathfrak q$ and a quasi-finite ring map

$\varphi : R_ f[x_1, \ldots , x_ n] \longrightarrow S_ g$

such that $\varphi ^{-1}(\mathfrak qS_ g) = (\mathfrak p, x_{r + 1}, \ldots , x_ n)R_ f[x_{r + 1}, \ldots , x_ n]$

Proof. After replacing $S$ by a principal localization we may assume there exists a quasi-finite ring map $\varphi : R[t_1, \ldots , t_ n] \to S$, see Lemma 10.125.2. Set $\mathfrak q' = \varphi ^{-1}(\mathfrak q)$. Let $\overline{\mathfrak q}' \subset \kappa (\mathfrak p)[t_1, \ldots , t_ n]$ be the prime corresponding to $\mathfrak q'$. By Lemma 10.115.6 there exists a finite ring map $\kappa (\mathfrak p)[x_1, \ldots , x_ n] \to \kappa (\mathfrak p)[t_1, \ldots , t_ n]$ such that the inverse image of $\overline{\mathfrak q}'$ is $(x_{r + 1}, \ldots , x_ n)$. Let $\overline{h}_ i \in \kappa (\mathfrak p)[t_1, \ldots , t_ n]$ be the image of $x_ i$. We can find an element $f \in R$, $f \not\in \mathfrak p$ and $h_ i \in R_ f[t_1, \ldots , t_ n]$ which map to $\overline{h}_ i$ in $\kappa (\mathfrak p)[t_1, \ldots , t_ n]$. Then the ring map

$R_ f[x_1, \ldots , x_ n] \longrightarrow R_ f[t_1, \ldots , t_ n]$

becomes finite after tensoring with $\kappa (\mathfrak p)$. In particular, $R_ f[t_1, \ldots , t_ n]$ is quasi-finite over $R_ f[x_1, \ldots , x_ n]$ at the prime $\mathfrak q'R_ f[t_1, \ldots , t_ n]$. Hence, by Lemma 10.123.13 there exists a $g \in R_ f[t_1, \ldots , t_ n]$, $g \not\in \mathfrak q'R_ f[t_1, \ldots , t_ n]$ such that $R_ f[x_1, \ldots , x_ n] \to R_ f[t_1, \ldots , t_ n, 1/g]$ is quasi-finite. Thus we see that the composition

$R_ f[x_1, \ldots , x_ n] \longrightarrow R_ f[t_1, \ldots , t_ n, 1/g] \longrightarrow S_{\varphi (g)}$

is quasi-finite and we win. $\square$

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