Lemma 69.22.9. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\overline{y}$ be a geometric point of $Y$. Assume
$Y$ locally Noetherian,
$f$ is proper, and
$\dim (X_{\overline{y}}) = d$.
For proper maps, stalks of higher direct images are trivial in degrees larger than the dimension of the fibre.
Then for any coherent sheaf $\mathcal{F}$ on $X$ we have $(R^ pf_*\mathcal{F})_{\overline{y}} = 0$ for all $p > d$.
Proof. Let $\kappa (\overline{y})$ be the residue field of the local ring of $\mathcal{O}_{Y, \overline{y}}$. As in Lemma 69.22.7 we set $X_{\overline{y}} = X_1 = \mathop{\mathrm{Spec}}(\kappa (\overline{y})) \times _ Y X$. Moreover, the underlying topological space of each infinitesimal neighbourhood $X_ n$ is the same as that of $X_{\overline{y}}$. Hence $H^ p(X_ n, \mathcal{F}_ n) = 0$ for all $p > d$ by Lemma 69.10.1. Hence we see that $(R^ pf_*\mathcal{F})_{\overline{y}}^\wedge = 0$ by Lemma 69.22.7 for $p > d$. Note that $R^ pf_*\mathcal{F}$ is coherent by Lemma 69.20.2 and hence $R^ pf_*\mathcal{F}_{\overline{y}}$ is a finite $\mathcal{O}_{Y, \overline{y}}$-module. By Algebra, Lemma 10.97.1 this implies that $(R^ pf_*\mathcal{F})_{\overline{y}} = 0$. $\square$
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Comment #1280 by Johan Commelin on