Lemma 61.11.6. 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 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.
$\square$

## 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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #5952 by Owen on

Comment #6136 by Johan on