Lemma 10.145.1. Let $R \to S' \to S$ be ring maps. Let $\mathfrak p \subset R$ be a prime. Let $g \in S'$ be an element. Assume

1. $R \to S'$ is integral,

2. $R \to S$ is finite type,

3. $S'_ g \cong S_ g$, and

4. $g$ invertible in $S' \otimes _ R \kappa (\mathfrak p)$.

Then there exists a $f \in R$, $f \not\in \mathfrak p$ such that $R_ f \to S_ f$ is finite.

Proof. By assumption the image $T$ of $V(g) \subset \mathop{\mathrm{Spec}}(S')$ under the morphism $\mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(R)$ does not contain $\mathfrak p$. By Section 10.41 especially, Lemma 10.41.6 we see $T$ is closed. Pick $f \in R$, $f \not\in \mathfrak p$ such that $T \cap D(f) = \emptyset$. Then we see that $g$ becomes invertible in $S'_ f$. Hence $S'_ f \cong S_ f$. Thus $S_ f$ is both of finite type and integral over $R_ f$, hence finite. $\square$

Comment #2871 by Ko Aoki on

Typo in the proof: "$T \cap V(f) = \emptyset$" should be replaced by "$T \cap D(f) = \emptyset$".

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).