The Stacks project

Proof. Let $X_{\overline{y}}^{sh}$ denote the fiber product $X \times _ Y \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, \overline{y}}^{sh})$. By Theorem 59.53.1 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 59.32.5 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 59.55.1 implies that $(R^ qf_*\mathcal{F})_{\overline{y}} = 0$ for $q > 0$ which implies (2) by Theorem 59.29.10. Part (1) follows from the corresponding statement of Lemma 59.55.1. $\square$