The Stacks project

Lemma 33.44.11. Let $k$ be a field. Let $f : X \to Y$ be a nonconstant morphism of proper curves over $k$. Let $\mathcal{E}$ be a locally free $\mathcal{O}_ Y$-module. Then

\[ \deg (f^*\mathcal{E}) = \deg (X/Y) \deg (\mathcal{E}) \]

Proof. The degree of $X$ over $Y$ is defined in Morphisms, Definition 29.51.8. Thus $f_*\mathcal{O}_ X$ is a coherent $\mathcal{O}_ Y$-module of rank $\deg (X/Y)$, i.e., $\deg (X/Y) = \dim _{\kappa (\xi )} (f_*\mathcal{O}_ X)_\xi $ where $\xi $ is the generic point of $Y$. Thus we obtain

\begin{align*} \chi (X, f^*\mathcal{E}) & = \chi (Y, f_*f^*\mathcal{E}) \\ & = \chi (Y, \mathcal{E} \otimes f_*\mathcal{O}_ X) \\ & = \deg (X/Y) \deg (\mathcal{E}) + n \chi (Y, f_*\mathcal{O}_ X) \\ & = \deg (X/Y) \deg (\mathcal{E}) + n \chi (X, \mathcal{O}_ X) \end{align*}

as desired. The first equality as $f$ is finite, see Cohomology of Schemes, Lemma 30.2.4. The second equality by projection formula, see Cohomology, Lemma 20.52.2. The third equality by Lemma 33.44.5. $\square$


Comments (0)

There are also:

  • 4 comment(s) on Section 33.44: Degrees on curves

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Your email address will not be published.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