The Stacks project

Lemma 37.36.4. Let $X \to S$ be a smooth morphism of schemes. Let $x \in X$ with image $s \in S$. Then

  1. The number of geometric branches of $X$ at $x$ is equal to the number of geometric branches of $S$ at $s$.

  2. If $\kappa (x)/\kappa (s)$ is a purely inseparable1 extension of fields, then number of branches of $X$ at $x$ is equal to the number of branches of $S$ at $s$.

Proof. Follows immediately from More on Algebra, Lemma 15.106.8 and the definitions. $\square$

[1] In fact, it would suffice if $\kappa (x)$ is geometrically irreducible over $\kappa (s)$. If we ever need this we will add a detailed proof.

Comments (0)


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Your email address will not be published.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 0DQ2. Beware of the difference between the letter 'O' and the digit '0'.