The Stacks project

Lemma 37.53.3. Let $f : X \to S$ be a morphism of schemes. Let $s \in S$. Then $X_ s$ is geometrically connected, if and only if for every étale neighbourhood $(U, u) \to (S, s)$ the base change $X_ U \to U$ has connected fibre $X_ u$.

Proof. If $X_ s$ is geometrically connected, then any base change of it is connected. On the other hand, suppose that $X_ s$ is not geometrically connected. Then by Varieties, Lemma 33.7.11 we see that $X_ s \times _{\mathop{\mathrm{Spec}}(\kappa (s))} \mathop{\mathrm{Spec}}(k)$ is disconnected for some finite separable field extension $k/\kappa (s)$. By Lemma 37.35.2 there exists an affine étale neighbourhood $(U, u) \to (S, s)$ such that $\kappa (u)/\kappa (s)$ is identified with $k/\kappa (s)$. In this case $X_ u$ is disconnected. $\square$

Comments (2)

Comment #7357 by Yijin Wang on

Typo in lemma 37.52.3: In the forth line 'we see that X_s×Spec(κ(s) Spec(k) is disconnected ' should be 'we see that X_s×Spec(κ(s)) Spec(k) is disconnected '

There are also:

  • 3 comment(s) on Section 37.53: Stein factorization

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Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03GZ. Beware of the difference between the letter 'O' and the digit '0'.