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
R \to S is of finite type,
\dim _{\mathfrak q}(S/R) = n, and
\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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