Lemma 60.11.5. 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 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 58.71.3. Because $X$ is extremally disconnected, every constructible locally closed subset is open and closed, 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

$\xymatrix{ \coprod _{i = 1, \ldots , r} X' \ar[r] \ar[d] & Y \ar[d] \\ X' \ar[r] & X }$

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 60.5.6), 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. $\square$

Comment #5950 by Owen on

The reference to tag 0978 suggests that $\operatorname{Spec} A$ is presumed profinite; perhaps that hypothesis should be added to the statement of the lemma? (Definition 08YI of extremally disconnected space doesn't presume compact Hausdorff and I believe there are extremally disconnected spectral spaces that aren't profinite.)

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