Lemma 10.123.13. Let R \to S be a finite type ring map. The set of points \mathfrak q of \mathop{\mathrm{Spec}}(S) at which S/R is quasi-finite is open in \mathop{\mathrm{Spec}}(S).
Proof. Let \mathfrak q \subset S be a point at which the ring map is quasi-finite. By Theorem 10.123.12 there exists an integral ring extension R \to S', S' \subset S and an element g \in S', g\not\in \mathfrak q such that S'_ g \cong S_ g. Since S and hence S_ g are of finite type over R we may find finitely many elements y_1, \ldots , y_ N of S' such that S''_ g \cong S_ g where S'' \subset S' is the sub R-algebra generated by g, y_1, \ldots , y_ N. Since S'' is finite over R (see Lemma 10.36.4) we see that S'' is quasi-finite over R (see Lemma 10.122.4). It is easy to see that this implies that S''_ g is quasi-finite over R, for example because the property of being quasi-finite at a prime depends only on the local ring at the prime. Thus we see that S_ g is quasi-finite over R. By the same token this implies that R \to S is quasi-finite at every prime of S which lies in D(g). \square
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Comment #3242 by Dario Weißmann on
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