The Stacks project

Lemma 53.3.4. Let $k$ be a field. Let $X$ be a nonsingular proper curve over $k$. Let $(\mathcal{L}, V)$ be a $\mathfrak g^1_ d$ on $X$. Then the morphism $\varphi : X \to \mathbf{P}^1_ k$ of Lemma 53.3.2 has degree $\leq d$.

Proof. By Lemma 53.3.2 we see that $\mathcal{L}' = \varphi ^*\mathcal{O}_{\mathbf{P}^1_ k}(1)$ has a nonzero map $\mathcal{L}' \to \mathcal{L}$. Hence by Varieties, Lemma 33.43.12 we see that $\deg (\mathcal{L}') \leq d$. On the other hand, we have

\[ \deg (\mathcal{L}') = \deg (X/\mathbf{P}^1_ k) \deg (\mathcal{O}_{\mathbf{P}^1_ k}(1)) \]

by Varieties, Lemma 33.43.11. This finishes the proof as the degree of $\mathcal{O}_{\mathbf{P}^1_ k}(1)$ is $1$. $\square$


Comments (1)

Comment #5100 by Tongmu He on

It seems that the morphism in 53.3.4 might be constant. Since might not be algebraically closed, even if the global sections of are linearly independent over , they might not be algebraically independent over . A possible consequence is that the morphism induced by these two sections might factor through a -rational point of where is a nontrivial field extension.


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Your email address will not be published.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