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

$R \to S'$ is integral,

$R \to S$ is finite type,

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

$g$ 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.
$\square$

## Comments (2)

Comment #2871 by Ko Aoki on

Comment #2933 by Johan on