Lemma 10.122.10. 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 10.25.1. Hence \mathfrak pS_{\mathfrak p} is a locally nilpotent ideal, see Lemma 10.32.3. 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 10.61.3 and 10.115.4) \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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