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
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