Tag 03QP
Chapter 53: Étale Cohomology > Section 53.54: Vanishing of finite higher direct images
Proposition 53.54.2. Let $f : X \to Y$ be a finite morphism of schemes.
- 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 $\textit{Ab}(X_{\acute{e}tale})$.
- 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 53.52.1 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 53.32.5 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 53.54.1 implies that $(R^qf_*\mathcal{F})_{\overline{y}} = 0$ for $q > 0$ which implies (2) by Theorem 53.29.10. Part (1) follows from the corresponding statement of Lemma 53.54.1. $\square$
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\begin{proposition}
\label{proposition-finite-higher-direct-image-zero}
Let $f : X \to Y$ be a finite morphism of schemes.
\begin{enumerate}
\item For any geometric point $\overline{y} : \Spec(k) \to Y$ we have
$$
(f_*\mathcal{F})_{\overline{y}} =
\prod\nolimits_{\overline{x} : \Spec(k) \to X,\ f(\overline{x}) =
\overline{y}} \mathcal{F}_{\overline{x}}.
$$
for $\mathcal{F}$ in $\Sh(X_\etale)$ and
$$
(f_*\mathcal{F})_{\overline{y}} =
\bigoplus\nolimits_{\overline{x} : \Spec(k) \to X,\ f(\overline{x}) =
\overline{y}} \mathcal{F}_{\overline{x}}.
$$
for $\mathcal{F}$ in $\textit{Ab}(X_\etale)$.
\item For any $q \geq 1$ we have $R^q f_*\mathcal{F} = 0$.
\end{enumerate}
\end{proposition}
\begin{proof}
Let $X_{\overline{y}}^{sh}$ denote the fiber product
$X \times_Y \Spec(\mathcal{O}_{Y, \overline{y}}^{sh})$.
By Theorem \ref{theorem-higher-direct-images}
the stalk of $R^qf_*\mathcal{F}$ at $\overline{y}$ is computed by
$H_\etale^q(X_{\overline{y}}^{sh}, \mathcal{F})$.
Since $f$ is finite, $X_{\bar y}^{sh}$ is finite over
$\Spec(\mathcal{O}_{Y, \overline{y}}^{sh})$, thus
$X_{\bar y}^{sh} = \Spec(A)$ for some ring $A$
finite over $\mathcal{O}_{Y, \bar y}^{sh}$.
Since the latter is strictly henselian,
Lemma \ref{lemma-finite-over-henselian}
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 $X_{\overline{y}}^{sh} = \coprod_{i = 1}^r \Spec(A_i)$.
The vanishing of
Lemma \ref{lemma-vanishing-etale-cohomology-strictly-henselian}
implies that $(R^qf_*\mathcal{F})_{\overline{y}} = 0$ for $q > 0$
which implies (2) by Theorem \ref{theorem-exactness-stalks}.
Part (1) follows from the corresponding statement of
Lemma \ref{lemma-vanishing-etale-cohomology-strictly-henselian}.
\end{proof}Comments (0)
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