Lemma 10.36.14. Let R \to S be a ring map. Let f_1, \ldots , f_ n \in R generate the unit ideal.
If each R_{f_ i} \to S_{f_ i} is integral, so is R \to S.
If each R_{f_ i} \to S_{f_ i} is finite, so is R \to S.
Lemma 10.36.14. Let R \to S be a ring map. Let f_1, \ldots , f_ n \in R generate the unit ideal.
If each R_{f_ i} \to S_{f_ i} is integral, so is R \to S.
If each R_{f_ i} \to S_{f_ i} is finite, so is R \to S.
Proof. Proof of (1). Let s \in S. Consider the ideal I \subset R[x] of polynomials P such that P(s) = 0. Let J \subset R denote the ideal (!) of leading coefficients of elements of I. By assumption and clearing denominators we see that f_ i^{n_ i} \in J for all i and certain n_ i \geq 0. Hence J contains 1 and we see s is integral over R. Proof of (2) omitted. \square
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