The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03QP. Beware of the difference between the letter 'O' and the digit '0'.



View Proposition 54.54.2 as pdf