Lemma 10.122.11. 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 $R$. By Lemma 10.122.4 the assumption means that $S \otimes _ R \kappa (\mathfrak p) = S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$ is a finite $\kappa (\mathfrak p)$-algebra. Thus we may find monic polynomials $P_ i \in R_{\mathfrak p}[X]$ such that $P_ i(x_ i)$ maps to zero in $S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$. Since $\mathfrak p$ is a minimal prime, $\mathfrak pR_{\mathfrak p}$ is a locally nilpotent ideal, see Lemma 10.25.1. Hence $\mathfrak pS_{\mathfrak p}$ is a locally nilpotent ideal, see Lemma 10.32.3. Thus 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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