Lemma 61.11.5. Let $A \to B$ be a quasi-finite and finitely presented ring map. If the residue fields of $A$ are separably algebraically closed and $\mathop{\mathrm{Spec}}(A)$ is Hausdorff and extremally disconnected, then $\mathop{\mathrm{Spec}}(B)$ is extremally disconnected.
Proof. Set $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$. Choose a finite partition $X = \coprod X_ i$ and $X'_ i \to X_ i$ as in Étale Cohomology, Lemma 59.72.3. The map of topological spaces $\coprod X_ i \to X$ (where the source is the disjoint union in the category of topological spaces) has a section by Topology, Proposition 5.26.6. Hence we see that $X$ is topologically the disjoint union of the strata $X_ i$. Thus we may replace $X$ by the $X_ i$ and assume there exists a surjective finite locally free morphism $X' \to X$ such that $(X' \times _ X Y)_{red}$ is isomorphic to a finite disjoint union of copies of $X'_{red}$. Picture
The assumption on the residue fields of $A$ implies that this diagram is a fibre product diagram on underlying sets of points (details omitted). Since $X$ is extremally disconnected and $X'$ is Hausdorff (Lemma 61.5.6), the continuous map $X' \to X$ has a continuous section $\sigma $. Then $\coprod _{i = 1, \ldots , r} \sigma (X) \to Y$ is a bijective continuous map. By Topology, Lemma 5.17.8 we see that it is a homeomorphism and the proof is done. $\square$
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Your email address will not be published.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 (2)
Comment #5950 by Owen on
Comment #6135 by Johan on