The Stacks project

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. $\square$

Comments (2)

Comment #2871 by Ko Aoki on

Typo in the proof: "" should be replaced by "".

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00UI. Beware of the difference between the letter 'O' and the digit '0'.