Lemma 33.44.2. Let $k'/k$ be an extension of fields. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. Let $\mathcal{E}$ be a locally free $\mathcal{O}_ X$-module of constant rank $n$. Then the degree of $\mathcal{E}/X/k$ is equal to the degree of $\mathcal{E}_{k'}/X_{k'}/k'$.

**Proof.**
More precisely, set $X_{k'} = X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(k')$. Let $\mathcal{E}_{k'} = p^*\mathcal{E}$ where $p : X_{k'} \to X$ is the projection. By Cohomology of Schemes, Lemma 30.5.2 we have $H^ i(X_{k'}, \mathcal{E}_{k'}) = H^ i(X, \mathcal{E}) \otimes _ k k'$ and $H^ i(X_{k'}, \mathcal{O}_{X_{k'}}) = H^ i(X, \mathcal{O}_ X) \otimes _ k k'$. Hence we see that the Euler characteristics are unchanged, hence the degree is unchanged.
$\square$

## 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Your email address will not be published.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.

## Comments (0)

There are also: