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. $\square$

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