Lemma 10.121.10. Let $R \to S$ be a ring map of finite type. Let $\mathfrak p \subset R$ be a minimal prime. Assume that there are at most finitely many primes of $S$ lying over $\mathfrak p$. Then there exists a $g \in R$, $g \not\in \mathfrak p$ such that the ring map $R_ g \to S_ g$ is finite.

**Proof.**
Let $x_1, \ldots , x_ n$ be generators of $S$ over $R$. Since $\mathfrak p$ is a minimal prime we have that $\mathfrak pR_{\mathfrak p}$ is a locally nilpotent ideal, see Lemma 10.24.1. Hence $\mathfrak pS_{\mathfrak p}$ is a locally nilpotent ideal, see Lemma 10.31.3. By assumption the finite type $\kappa (\mathfrak p)$-algebra $S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$ has finitely many primes. Hence (for example by Lemmas 10.60.3 and 10.114.4) $\kappa (\mathfrak p) \to S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$ is a finite ring map. Thus we may find monic polynomials $P_ i \in R_{\mathfrak p}[X]$ such that $P_ i(x_ i)$ maps to zero in $S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$. By what we said above there exist $e_ i \geq 1$ such that $P(x_ i)^{e_ i} = 0$ in $S_{\mathfrak p}$. Let $g_1 \in R$, $g_1 \not\in \mathfrak p$ be an element such that $P_ i$ has coefficients in $R[1/g_1]$ for all $i$. Next, let $g_2 \in R$, $g_2 \not\in \mathfrak p$ be an element such that $P(x_ i)^{e_ i} = 0$ in $S_{g_1g_2}$. Setting $g = g_1g_2$ we win.
$\square$

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