59.55 Vanishing of finite higher direct images

The next goal is to prove that the higher direct images of a finite morphism of schemes vanish.

Lemma 59.55.1. Let $R$ be a strictly henselian local ring. Set $S = \mathop{\mathrm{Spec}}(R)$ and let $\overline{s}$ be its closed point. Then the global sections functor $\Gamma (S, -) : \textit{Ab}(S_{\acute{e}tale}) \to \textit{Ab}$ is exact. In fact we have $\Gamma (S, \mathcal{F}) = \mathcal{F}_{\overline{s}}$ for any sheaf of sets $\mathcal{F}$. In particular

$\forall p\geq 1, \quad H_{\acute{e}tale}^ p(S, \mathcal{F})=0$

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

Proof. If we show that $\Gamma (S, \mathcal{F}) = \mathcal{F}_{\overline{s}}$ then $\Gamma (S, -)$ is exact as the stalk functor is exact. Let $(U, \overline{u})$ be an étale neighbourhood of $\overline{s}$. Pick an affine open neighborhood $\mathop{\mathrm{Spec}}(A)$ of $\overline{u}$ in $U$. Then $R \to A$ is étale and $\kappa (\overline{s}) = \kappa (\overline{u})$. By Theorem 59.32.4 we see that $A \cong R \times A'$ as an $R$-algebra compatible with maps to $\kappa (\overline{s}) = \kappa (\overline{u})$. Hence we get a section

$\xymatrix{ \mathop{\mathrm{Spec}}(A) \ar[r] & U \ar[d]\\ & S \ar[ul] }$

It follows that in the system of étale neighbourhoods of $\overline{s}$ the identity map $(S, \overline{s}) \to (S, \overline{s})$ is cofinal. Hence $\Gamma (S, \mathcal{F}) = \mathcal{F}_{\overline{s}}$. The final statement of the lemma follows as the higher derived functors of an exact functor are zero, see Derived Categories, Lemma 13.16.9. $\square$

Proposition 59.55.2. 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 $R^ q f_*\mathcal{F} = 0$ for $\mathcal{F}$ in $\textit{Ab}(X_{\acute{e}tale})$.

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. $\square$

Lemma 59.55.3. Consider a cartesian square

$\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y }$

of schemes with $f$ a finite morphism. For any sheaf of sets $\mathcal{F}$ on $X_{\acute{e}tale}$ we have $f'_*(g')^{-1}\mathcal{F} = g^{-1}f_*\mathcal{F}$.

Proof. In great generality there is a pullback map $g^{-1}f_*\mathcal{F} \to f'_*(g')^{-1}\mathcal{F}$, see Sites, Section 7.45. It suffices to check on stalks (Theorem 59.29.10). Let $\overline{y}' : \mathop{\mathrm{Spec}}(k) \to Y'$ be a geometric point. We have

\begin{align*} (f'_*(g')^{-1}\mathcal{F})_{\overline{y}'} & = \prod \nolimits _{\overline{x}' : \mathop{\mathrm{Spec}}(k) \to X',\ f' \circ \overline{x}' = \overline{y}'} ((g')^{-1}\mathcal{F})_{\overline{x}'} \\ & = \prod \nolimits _{\overline{x}' : \mathop{\mathrm{Spec}}(k) \to X',\ f' \circ \overline{x}' = \overline{y}'} \mathcal{F}_{g' \circ \overline{x}'} \\ & = \prod \nolimits _{\overline{x} : \mathop{\mathrm{Spec}}(k) \to X,\ f \circ \overline{x} = g \circ \overline{y}'} \mathcal{F}_{\overline{x}} \\ & = (f_*\mathcal{F})_{g \circ \overline{y}'} \\ & = (g^{-1}f_*\mathcal{F})_{\overline{y}'} \end{align*}

The first equality by Proposition 59.55.2. The second equality by Lemma 59.36.2. The third equality holds because the diagram is a cartesian square and hence the map

$\{ \overline{x}' : \mathop{\mathrm{Spec}}(k) \to X',\ f' \circ \overline{x}' = \overline{y}'\} \longrightarrow \{ \overline{x} : \mathop{\mathrm{Spec}}(k) \to X,\ f \circ \overline{x} = g \circ \overline{y}'\}$

sending $\overline{x}'$ to $g' \circ \overline{x}'$ is a bijection. The fourth equality by Proposition 59.55.2. The fifth equality by Lemma 59.36.2. $\square$

Lemma 59.55.4. Consider a cartesian square

$\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y }$

of schemes with $f$ an integral morphism. For any sheaf of sets $\mathcal{F}$ on $X_{\acute{e}tale}$ we have $f'_*(g')^{-1}\mathcal{F} = g^{-1}f_*\mathcal{F}$.

Proof. The question is local on $Y$ and hence we may assume $Y$ is affine. Then we can write $X = \mathop{\mathrm{lim}}\nolimits X_ i$ with $f_ i : X_ i \to Y$ finite (this is easy in the affine case, but see Limits, Lemma 32.7.3 for a reference). Denote $p_{i'i} : X_{i'} \to X_ i$ the transition morphisms and $p_ i : X \to X_ i$ the projections. Setting $\mathcal{F}_ i = p_{i, *}\mathcal{F}$ we obtain from Lemma 59.51.9 a system $(\mathcal{F}_ i, \varphi _{i'i})$ with $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits p_ i^{-1}\mathcal{F}_ i$. We get $f_*\mathcal{F} = \mathop{\mathrm{colim}}\nolimits f_{i, *}\mathcal{F}_ i$ from Lemma 59.51.7. Set $X'_ i = Y' \times _ Y X_ i$ with projections $f'_ i$ and $g'_ i$. Then $X' = \mathop{\mathrm{lim}}\nolimits X'_ i$ as limits commute with limits. Denote $p'_ i : X' \to X'_ i$ the projections. We have

\begin{align*} g^{-1}f_*\mathcal{F} & = g^{-1} \mathop{\mathrm{colim}}\nolimits f_{i, *}\mathcal{F}_ i \\ & = \mathop{\mathrm{colim}}\nolimits g^{-1}f_{i, *}\mathcal{F}_ i \\ & = \mathop{\mathrm{colim}}\nolimits f'_{i, *}(g'_ i)^{-1}\mathcal{F}_ i \\ & = f'_*(\mathop{\mathrm{colim}}\nolimits (p'_ i)^{-1}(g'_ i)^{-1}\mathcal{F}_ i) \\ & = f'_*(\mathop{\mathrm{colim}}\nolimits (g')^{-1}p_ i^{-1}\mathcal{F}_ i) \\ & = f'_*(g')^{-1} \mathop{\mathrm{colim}}\nolimits p_ i^{-1}\mathcal{F}_ i \\ & = f'_*(g')^{-1}\mathcal{F} \end{align*}

as desired. For the first equality see above. For the second use that pullback commutes with colimits. For the third use the finite case, see Lemma 59.55.3. For the fourth use Lemma 59.51.7. For the fifth use that $g'_ i \circ p'_ i = p_ i \circ g'$. For the sixth use that pullback commutes with colimits. For the seventh use $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits p_ i^{-1}\mathcal{F}_ i$. $\square$

The following lemma is a case of cohomological descent dealing with étale sheaves and finite surjective morphisms. We will significantly generalize this result once we prove the proper base change theorem.

Lemma 59.55.5. Let $f : X \to Y$ be a surjective finite morphism of schemes. Set $f_ n : X_ n \to Y$ equal to the $(n + 1)$-fold fibre product of $X$ over $Y$. For $\mathcal{F} \in \textit{Ab}(Y_{\acute{e}tale})$ set $\mathcal{F}_ n = f_{n, *}f_ n^{-1}\mathcal{F}$. There is an exact sequence

$0 \to \mathcal{F} \to \mathcal{F}_0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \ldots$

on $X_{\acute{e}tale}$. Moreover, there is a spectral sequence

$E_1^{p, q} = H^ q_{\acute{e}tale}(X_ p, f_ p^{-1}\mathcal{F})$

converging to $H^{p + q}(Y_{\acute{e}tale}, \mathcal{F})$. This spectral sequence is functorial in $\mathcal{F}$.

Proof. If we prove the first statement of the lemma, then we obtain a spectral sequence with $E_1^{p, q} = H^ q_{\acute{e}tale}(Y, \mathcal{F})$ converging to $H^{p + q}(Y_{\acute{e}tale}, \mathcal{F})$, see Derived Categories, Lemma 13.21.3. On the other hand, since $R^ if_{p, *}f_ p^{-1}\mathcal{F} = 0$ for $i > 0$ (Proposition 59.55.2) we get

$H^ q_{\acute{e}tale}(X_ p, f_ p^{-1}\mathcal{F}) = H^ q_{\acute{e}tale}(Y, f_{p, *}f_ p^{-1} \mathcal{F}) = H^ q_{\acute{e}tale}(Y, \mathcal{F}_ p)$

by Proposition 59.54.2 and we get the spectral sequence of the lemma.

To prove the first statement of the lemma, observe that $X_ n$ forms a simplicial scheme over $Y$, see Simplicial, Example 14.3.5. Observe moreover, that for each of the projections $d_ j : X_{n + 1} \to X_ n$ there is a map $d_ j^{-1} f_ n^{-1}\mathcal{F} \to f_{n + 1}^{-1}\mathcal{F}$. These maps induce maps

$\delta _ j : \mathcal{F}_ n \to \mathcal{F}_{n + 1}$

for $j = 0, \ldots , n + 1$. We use the alternating sum of these maps to define the differentials $\mathcal{F}_ n \to \mathcal{F}_{n + 1}$. Similarly, there is a canonical augmentation $\mathcal{F} \to \mathcal{F}_0$, namely this is just the canonical map $\mathcal{F} \to f_*f^{-1}\mathcal{F}$. To check that this sequence of sheaves is an exact complex it suffices to check on stalks at geometric points (Theorem 59.29.10). Thus we let $\overline{y} : \mathop{\mathrm{Spec}}(k) \to Y$ be a geometric point. Let $E = \{ \overline{x} : \mathop{\mathrm{Spec}}(k) \to X \mid f(\overline{x}) = \overline{y}\}$. Then $E$ is a finite nonempty set and we see that

$(\mathcal{F}_ n)_{\overline{y}} = \bigoplus \nolimits _{e \in E^{n + 1}} \mathcal{F}_{\overline{y}}$

by Proposition 59.55.2 and Lemma 59.36.2. Thus we have to see that given an abelian group $M$ the sequence

$0 \to M \to \bigoplus \nolimits _{e \in E} M \to \bigoplus \nolimits _{e \in E^2} M \to \ldots$

is exact. Here the first map is the diagonal map and the map $\bigoplus _{e \in E^{n + 1}} M \to \bigoplus _{e \in E^{n + 2}} M$ is the alternating sum of the maps induced by the $(n + 2)$ projections $E^{n + 2} \to E^{n + 1}$. This can be shown directly or deduced by applying Simplicial, Lemma 14.26.9 to the map $E \to \{ *\}$. $\square$

Remark 59.55.6. In the situation of Lemma 59.55.5 if $\mathcal{G}$ is a sheaf of sets on $Y_{\acute{e}tale}$, then we have

$\Gamma (Y, \mathcal{G}) = \text{Equalizer}( \xymatrix{ \Gamma (X_0, f_0^{-1}\mathcal{G}) \ar@<1ex>[r] \ar@<-1ex>[r] & \Gamma (X_1, f_1^{-1}\mathcal{G}) } )$

This is proved in exactly the same way, by showing that the sheaf $\mathcal{G}$ is the equalizer of the two maps $f_{0, *}f_0^{-1}\mathcal{G} \to f_{1, *}f_1^{-1}\mathcal{G}$.

Comment #4090 by Tongmu on

Typos: in the statement of lemma 59.55.5, in the beginning of the fourth line, it should be "on $Y_{\acute{e}tale}$"; in the proof of lemma 59.55.5, in the beginning of the second line, it should be "$E_1^{p, q} = H^ q_{\acute{e}tale}(Y, \mathcal{F}^p)$".

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