Lemma 10.122.5. Let $R \to S$ be a finite type ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. Let $f \in R$, $f \not\in \mathfrak p$ and $g \in S$, $g \not\in \mathfrak q$. Then $R \to S$ is quasi-finite at $\mathfrak q$ if and only if $R_ f \to S_{fg}$ is quasi-finite at $\mathfrak qS_{fg}$.
Proof. The fibre of $\mathop{\mathrm{Spec}}(S_{fg}) \to \mathop{\mathrm{Spec}}(R_ f)$ is homeomorphic to an open subset of the fibre of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$. Hence the lemma follows from part (1) of the equivalent conditions of Lemma 10.122.2. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)