Lemma 10.29.2. Let $\varphi : R \to S$ be a ring map. The induced continuous map $f : \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is quasi-compact. For any constructible set $E \subset \mathop{\mathrm{Spec}}(R)$ the inverse image $f^{-1}(E)$ is constructible in $\mathop{\mathrm{Spec}}(S)$.
Proof. We first show that the inverse image of any quasi-compact open $U \subset \mathop{\mathrm{Spec}}(R)$ is quasi-compact. By Lemma 10.29.1 we may write $U$ as a finite open of standard opens. Thus by Lemma 10.17.4 we see that $f^{-1}(U)$ is a finite union of standard opens. Hence $f^{-1}(U)$ is quasi-compact by Lemma 10.29.1 again. The second assertion now follows from Topology, Lemma 5.15.3. $\square$
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Your email address will not be published.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)
There are also: