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 20.54.2. The third equality by Lemma 33.44.5. $\square$
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