The Stacks project

Lemma 10.122.9. Let $A \to B$ and $B \to C$ be ring homomorphisms such that $A \to C$ is of finite type. Let $\mathfrak r$ be a prime of $C$ lying over $\mathfrak q \subset B$ and $\mathfrak p \subset A$. If $A \to C$ is quasi-finite at $\mathfrak r$, then $B \to C$ is quasi-finite at $\mathfrak r$.

Proof. Observe that $B \to C$ is of finite type (Lemma 10.6.2) so that the statement makes sense. Let us use characterization (3) of Lemma 10.122.2. If $A \to C$ is quasi-finite at $\mathfrak r$, then there exists some $c \in C$ such that

\[ \{ \mathfrak r' \subset C \text{ lying over }\mathfrak p\} \cap D(c) = \{ \mathfrak {r}\} . \]

Since the primes $\mathfrak r' \subset C$ lying over $\mathfrak q$ form a subset of the primes $\mathfrak r' \subset C$ lying over $\mathfrak p$ we conclude $B \to C$ is quasi-finite at $\mathfrak r$. $\square$

Comments (2)

Comment #3952 by Manuel Hoff on

I think, that the Lemma only needs the assumption that is of finite type (and this form of the Lemma is needed in 00Q9).

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Your email address will not be published.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 0C6H. Beware of the difference between the letter 'O' and the digit '0'.