The Stacks project

Proof. We will use the criterion of Lemma 61.11.2. Set $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$ and denote $f : Y \to X$ the induced morphism. As $f : Y \to X$ 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 $X$. Let $y \in Y$ with image $x \in X$. Then $x$ specializes to a unique closed point $x_0 \in X$. Say $f^{-1}(\{ x_0\} ) = \{ y_1, \ldots , y_ n\} $ with $y_ i$ closed in $Y$. Since $R = \mathcal{O}_{X, x_0}$ is strictly henselian and since $f$ is finite, we see that $Y \times _{f, X} \mathop{\mathrm{Spec}}(R)$ is equal to $\coprod _{i = 1, \ldots , n} \mathop{\mathrm{Spec}}(R_ i)$ where each $R_ i$ is a local ring finite over $R$ whose maximal ideal corresponds to $y_ i$, see Algebra, Lemma 10.153.3 part (10). Then $y$ is a point of exactly one of these $\mathop{\mathrm{Spec}}(R_ i)$ and we see that $y$ specializes to exactly one of the $y_ i$. In other words, every point of $Y$ specializes to a unique point of $Y_0$. Thus $Y$ is w-local. 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 61.11.5. $\square$