Lemma 37.52.6. Let $\varphi : A \to B$ be a local ring map of local rings. Let $V \subset \mathop{\mathrm{Spec}}(B)$ be an open subscheme which contains at least one prime not lying over $\mathfrak m_ A$. Assume $A$ is Noetherian, $\varphi$ essentially of finite type, and $A/\mathfrak m_ A \subset B/\mathfrak m_ B$ is finite. Then there exists a $\mathfrak q \in V$, $\mathfrak m_ A \not= \mathfrak q \cap A$ such that $A \to B/\mathfrak q$ is the localization of a quasi-finite ring map.

Proof. Since $A$ is Noetherian and $A \to B$ is essentially of finite type, we know that $B$ is Noetherian too. By Properties, Lemma 28.6.4 the topological space $\mathop{\mathrm{Spec}}(B) \setminus \{ \mathfrak m_ B\}$ is Jacobson. Hence we can choose a closed point $\mathfrak q$ which is contained in the nonempty open

$V \setminus \{ \mathfrak q \subset B \mid \mathfrak m_ A = \mathfrak q \cap A\} .$

(Nonempty by assumption, open because $\{ \mathfrak m_ A\}$ is a closed subset of $\mathop{\mathrm{Spec}}(A)$.) Then $\mathop{\mathrm{Spec}}(B/\mathfrak q)$ has two points, namely $\mathfrak m_ B$ and $\mathfrak q$ and $\mathfrak q$ does not lie over $\mathfrak m_ A$. Write $B/\mathfrak q = C_{\mathfrak m}$ for some finite type $A$-algebra $C$ and prime ideal $\mathfrak m$. Then $A \to C$ is quasi-finite at $\mathfrak m$ by Algebra, Lemma 10.122.2 (2). Hence by Algebra, Lemma 10.123.13 we see that after replacing $C$ by a principal localization the ring map $A \to C$ is quasi-finite. $\square$

## 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Your email address will not be published.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 05GT. Beware of the difference between the letter 'O' and the digit '0'.