Lemma 60.11.6. Let $A \to B$ be a finite and finitely presented ring map. If $A$ is w-contractible, so is $B$.

Proof. We will use the criterion of Lemma 60.11.2. Set $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$. As $Y \to X$ is a finite morphism, we see that the set of closed points $Y_0$ of $Y$ is the inverse image of the set of closed points $X_0$ of $X$. Moreover, every point of $Y$ specializes to a unique point of $Y_0$ as (a) this is true for $X$ and (b) the map $X \to Y$ is separated. For every $y \in Y_0$ with image $x \in X_0$ we see that $\mathcal{O}_{Y, y}$ is strictly henselian by Algebra, Lemma 10.153.4 applied to $\mathcal{O}_{X, x} \to B \otimes _ A \mathcal{O}_{X, x}$. It remains to show that $Y_0$ is extremally disconnected. To do this we look at $X_0 \times _ X Y \to X_0$ where $X_0 \subset X$ is the reduced induced scheme structure. Note that the underlying topological space of $X_0 \times _ X Y$ agrees with $Y_0$. Now the desired result follows from Lemma 60.11.5. $\square$

Comment #5952 by Owen on

'Moreover, every point of $Y$ specializes to a unique point of $Y_0$ as (a) this is true for $X$ and (b) the map $X\to Y$ is separated.' It should be $Y\to X$, not $X\to Y$, but I still don't follow. It's possible for a specialization in $X$ to lift to more than one specialization along a separated map $Y\ra X$.

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