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 $D(g)$. $\square$

Comment #3242 by Dario Weißmann on

Typo in the proof: $S''_g \cong S$ should be $S_g$

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