The Stacks project

Lemma 29.57.3. If $f$ is a finite locally free morphism of degree $d$, then $d$ bounds the degree of the fibres of $f$.

Proof. This is true because any base change of $f$ is finite locally free of degree $d$ (Lemma 29.48.4) and hence the fibres of $f$ all have degree $d$. $\square$

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Your email address will not be published.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 0CC2. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 29.57.3 as pdf