
Proposition 54.54.2. Let $f : X \to Y$ be a finite morphism of schemes.

1. For any geometric point $\overline{y} : \mathop{\mathrm{Spec}}(k) \to Y$ we have

$(f_*\mathcal{F})_{\overline{y}} = \prod \nolimits _{\overline{x} : \mathop{\mathrm{Spec}}(k) \to X,\ f(\overline{x}) = \overline{y}} \mathcal{F}_{\overline{x}}.$

for $\mathcal{F}$ in $\mathop{\mathit{Sh}}\nolimits (X_{\acute{e}tale})$ and

$(f_*\mathcal{F})_{\overline{y}} = \bigoplus \nolimits _{\overline{x} : \mathop{\mathrm{Spec}}(k) \to X,\ f(\overline{x}) = \overline{y}} \mathcal{F}_{\overline{x}}.$

for $\mathcal{F}$ in $\textit{Ab}(X_{\acute{e}tale})$.

2. For any $q \geq 1$ we have $R^ q f_*\mathcal{F} = 0$.

Proof. Let $X_{\overline{y}}^{sh}$ denote the fiber product $X \times _ Y \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, \overline{y}}^{sh})$. By Theorem 54.52.1 the stalk of $R^ qf_*\mathcal{F}$ at $\overline{y}$ is computed by $H_{\acute{e}tale}^ q(X_{\overline{y}}^{sh}, \mathcal{F})$. Since $f$ is finite, $X_{\bar y}^{sh}$ is finite over $\mathop{\mathrm{Spec}}(\mathcal{O}_{Y, \overline{y}}^{sh})$, thus $X_{\bar y}^{sh} = \mathop{\mathrm{Spec}}(A)$ for some ring $A$ finite over $\mathcal{O}_{Y, \bar y}^{sh}$. Since the latter is strictly henselian, Lemma 54.32.5 implies that $A$ is a finite product of henselian local rings $A = A_1 \times \ldots \times A_ r$. Since the residue field of $\mathcal{O}_{Y, \overline{y}}^{sh}$ is separably closed the same is true for each $A_ i$. Hence $A_ i$ is strictly henselian. This implies that $X_{\overline{y}}^{sh} = \coprod _{i = 1}^ r \mathop{\mathrm{Spec}}(A_ i)$. The vanishing of Lemma 54.54.1 implies that $(R^ qf_*\mathcal{F})_{\overline{y}} = 0$ for $q > 0$ which implies (2) by Theorem 54.29.10. Part (1) follows from the corresponding statement of Lemma 54.54.1. $\square$

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