The Stacks project

Proof. Proof of (1). By Lemma 59.92.5 we have

By proper base change (in the form of Lemma 59.91.12) this is equal to the object

of $D(Y_{\acute{e}tale})$. Taking $R\Gamma (Y, -)$ on this object reproduces the left hand side of the equality in (1) by the Leray spectral sequence for $\text{pr}_2$. Thus we conclude by Lemma 59.92.4.

Proof of (2). This is exactly the same as the proof of (1) except that we use Lemmas 59.96.6, 59.92.3, and 59.96.5 as well as $\text{cd}(Y) < \infty $ by Lemma 59.96.2. $\square$

Proof. (If $\mathcal{F}$ is torsion, then the vanishing follows immediately from Lemma 59.95.7.) By Lemma 59.74.5 we can write $\mathcal{F}$ as a filtered colimit of constructible sheaves $\mathcal{F}_ i$ of $\mathbf{Z}$-modules whose supports $Z_ i \subset X$ are finite sets of closed points. By Proposition 59.46.4 such a sheaf is of the form $(Z_ i \to X)_*\mathcal{G}_ i$ where $\mathcal{G}_ i$ is a sheaf on $Z_ i$. As $K$ is separably closed, the scheme $Z_ i$ is a finite disjoint union of spectra of separably closed fields. Recall that $H^ q(Z_ i, \mathcal{G}_ i) = H^ q(X, \mathcal{F}_ i)$ by the Leray spectral sequence for $Z_ i \to X$ and vanising of higher direct images for this morphism (Proposition 59.55.2). By Lemmas 59.59.1 and 59.59.2 we see that $H^ q(Z_ i, \mathcal{G}_ i)$ is zero for $q > 0$ and that $H^0(Z_ i, \mathcal{G}_ i)$ generates $\mathcal{G}_ i$. We conclude the vanishing of $H^ q(X, \mathcal{F}_ i)$ for $q > 0$ and that $\mathcal{F}_ i$ is generated by global sections. By Theorem 59.51.3 we see that $H^ q(X, \mathcal{F}) = 0$ for $q > 0$. The proof is now done because a filtered colimit of globally generated sheaves of abelian groups is globally generated (details omitted). $\square$

Proof. The implication from left to right is trivial. Thus we need to prove the reverse implication.

Assume $Q$ is bounded below; this cases suffices for almost all applications. If $Q$ is not zero, then we can look at the smallest $i$ such that the cohomology sheaf $H^ i(Q)$ is nonzero. By Lemma 59.97.3 we have $H^ i(X, Q) = H^0(X, H^ i(Q)) \not= 0$ and we conclude.

General case. Let $\mathcal{B} \subset \mathop{\mathrm{Ob}}\nolimits (X_{\acute{e}tale})$ be the quasi-compact objects. By Lemma 59.97.3 the assumptions of Cohomology on Sites, Lemma 21.23.11 are satisfied. We conclude that $H^ q(U, Q) = H^0(U, H^ q(Q))$ for all $U \in \mathcal{B}$. In particular, this holds for $U = X$. Thus the conclusion by Lemma 59.97.3 as $Q$ is zero in $D(X_{\acute{e}tale})$ if and only if $H^ q(Q)$ is zero for all $q$. $\square$

Proof. Write $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}[n]$ where the colimit is over the multiplicative system of integers invertible in $K$. Since cohomology commutes with filtered colimits in our situation (for a precise reference see Lemma 59.86.3), it suffices to prove the lemma for $\mathcal{F}[n]$. Thus we may assume $\mathcal{F}$ is a sheaf of $\mathbf{Z}/n\mathbf{Z}$-modules for some $n$ invertible in $K$ (we will use this at the very end of the proof). In the proof we use the short hand $X \times _ K Y$ for the fibre product over $\mathop{\mathrm{Spec}}(K)$. We will prove the lemma by induction on $\dim (X) + \dim (Y)$. The lemma is trivial if $\dim (X) \leq 0$, since in this case $U$ is an open and closed subscheme of $X$. Choose a point $z \in X \times _ K Y$. We will show the stalk at $\overline{z}$ is an isomorphism.

Suppose that $z \mapsto x \in X$ and assume $\text{trdeg}_ K(\kappa (x)) > 0$. Set $X' = \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^{sh})$ and denote $U' \subset X'$ the inverse image of $U$. Consider the base change

of our diagram by $X' \to X$. Observe that $X' \to X$ is a filtered colimit of étale morphisms. By smooth base change in the form of Lemma 59.89.3 the pullback of $q^{-1}Rj_*\mathcal{F} \to Rh_*p^{-1}\mathcal{F}$ to $X'$ to $Y \times _ K X'$ is the map $(q')^{-1}Rj'_*\mathcal{F}' \to Rj'_*(p')^{-1}\mathcal{F}'$ where $\mathcal{F}'$ is the pullback of $\mathcal{F}$ to $U'$. (In this step it would suffice to use étale base change which is an essentially trivial result.) So it suffices to show that $(q')^{-1}Rj'_*\mathcal{F}' \to Rj'_*(p')^{-1}\mathcal{F}'$ is an isomorphism in order to prove that our original map is an isomorphism on stalks at $\overline{z}$. By Lemma 59.95.5 there is a separably closed field $L/K$ such that $X' = \mathop{\mathrm{lim}}\nolimits X_ i$ with $X_ i$ affine of finite type over $L$ and $\dim (X_ i) < \dim (X)$. For $i$ large enough there exists an open $U_ i \subset X_ i$ restricting to $U'$ in $X'$. We may apply the induction hypothesis to the diagram

over the field $L$ and the pullback of $\mathcal{F}$ to these diagrams. By Lemma 59.86.3 we conclude that the map $(q')^{-1}Rj'_*\mathcal{F}' \to Rj'_*(p')^{-1}\mathcal{F}$ is an isomorphism.

Suppose that $z \mapsto y \in Y$ and assume $\text{trdeg}_ K(\kappa (y)) > 0$. Let $Y' = \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^{sh})$. By Lemma 59.95.5 there is a separably closed field $L/K$ such that $Y' = \mathop{\mathrm{lim}}\nolimits Y_ i$ with $Y_ i$ affine of finite type over $L$ and $\dim (Y_ i) < \dim (Y)$. In particular $Y'$ is a scheme over $L$. Denote with a subscript $L$ the base change from schemes over $K$ to schemes over $L$. Consider the commutative diagrams

and observe the top and bottom rows are the same on the left and the right. By smooth base change we see that $f^{-1}Rh_*p^{-1}\mathcal{F} = Rh'_*(f')^{-1}p^{-1}\mathcal{F}$ (similarly to the previous paragraph). By smooth base change for $\mathop{\mathrm{Spec}}(L) \to \mathop{\mathrm{Spec}}(K)$ (Lemma 59.90.1) we see that $Rj_{L, *}\mathcal{F}_ L$ is the pullback of $Rj_*\mathcal{F}$ to $X_ L$. Combining these two observations, we conclude that it suffices to prove the base change map for the upper square in the diagram on the right is an isomorphism in order to prove that our original map is an isomorphism on stalks at $\overline{z}$¹. Then using that $Y' = \mathop{\mathrm{lim}}\nolimits Y_ i$ and argueing exactly as in the previous paragraph we see that the induction hypothesis forces our map over $Y' \times _ K X$ to be an isomorphism.

Thus any counter example with $\dim (X) + \dim (Y)$ minimal would only have nonisomorphisms $q^{-1}Rj_*\mathcal{F} \to Rh_*p^{-1}\mathcal{F}$ on stalks at closed points of $X \times _ K Y$ (because a point $z$ of $X \times _ K Y$ is a closed point if and only if both the image of $z$ in $X$ and in $Y$ are closed). Since it is enough to prove the isomorphism locally, we may assume $X$ and $Y$ are affine. However, then we can choose an open dense immersion $Y \to Y'$ with $Y'$ projective. (Choose a closed immersion $Y \to \mathbf{A}^ n_ K$ and let $Y'$ be the scheme theoretic closure of $Y$ in $\mathbf{P}^ n_ K$.) Then $\dim (Y') = \dim (Y)$ and hence we get a “minimal” counter example with $Y$ projective over $K$. In the next paragraph we show that this can't happen.

Consider a diagram as in the statement of the lemma such that $q^{-1}Rj_*\mathcal{F} \to Rh_*p^{-1}\mathcal{F}$ is an isomorphism at all non-closed points of $X \times _ K Y$ and such that $Y$ is projective. The restriction of the map to $(X \times _ K Y)_{K^{sep}}$ is the corresponding map for the diagram of the lemma base changed to $K^{sep}$. Thus we may and do assume $K$ is separably algebraically closed. Choose a distinguished triangle

in $D((X \times _ K Y)_{\acute{e}tale})$. Since $Q$ is supported in closed points we see that it suffices to prove $H^ i(X \times _ K Y, Q) = 0$ for all $i$, see Lemma 59.97.4. Thus it suffices to prove that $q^{-1}Rj_*\mathcal{F} \to Rh_*p^{-1}\mathcal{F}$ induces an isomorphism on cohomology. Recall that $\mathcal{F}$ is annihilated by $n$ invertible in $K$. By the Künneth formula of Lemma 59.97.2 we have

and

This finishes the proof. $\square$

Proof. The question is local on $X$, hence we may assume $X$ is affine. By Limits, Lemma 32.7.2 we can write $X = \mathop{\mathrm{lim}}\nolimits X_ i$ as a cofiltered limit with affine transition morphisms of schemes $X_ i$ of finite type over $K$. Denote $X'_ i = X_ i \times _{\mathop{\mathrm{Spec}}(K)} S'$ and $Y_ i = X'_ i \times _{S'} T$. By Lemma 59.86.3 it suffices to prove the statement for the squares with corners $X_ i, Y_ i, S_ i, T_ i$. Thus we may assume $X$ is of finite type over $K$. Similarly, we may write $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}[n]$ where the colimit is over the multiplicative system of integers invertible in $K$. The same lemma used above reduces us to the case where $\mathcal{F}$ is a sheaf of $\mathbf{Z}/n\mathbf{Z}$-modules for some $n$ invertible in $K$.

We may replace $K$ by its algebraic closure $\overline{K}$. Namely, formation of direct image commutes with base change to $\overline{K}$ according to Lemma 59.90.1 (works for both $g$ and $h$). And it suffices to prove the agreement after restriction to $X'_{\overline{K}}$. Next, we may replace $X$ by its reduction as we have the topological invariance of étale cohomology, see Proposition 59.45.4. After this replacement the morphism $X \to \mathop{\mathrm{Spec}}(K)$ is flat, finite presentation, with geometrically reduced fibres and the same is true for any base change, in particular for $X' \to S'$. Hence $(f')^{-1}g_*\mathcal{F} \to Rh_*e^{-1}\mathcal{F}$ is an isomorphism by Lemma 59.87.2.

At this point we may apply Lemma 59.90.3 to see that it suffices to prove: given a commutative diagram

with both squares cartesian, where $S'$ is affine, integral, and normal with algebraically closed function field $K$, then $R^ qh_*(\mathbf{Z}/d\mathbf{Z})$ is zero for $q > 0$ and $d | n$. Observe that this vanishing is equivalent to the statement that

is an isomorphism, because the left hand side is zero for example by Lemma 59.80.5.

Write $S' = \mathop{\mathrm{Spec}}(B)$ so that $L$ is the fraction field of $B$. Write $B = \bigcup _{i \in I} B_ i$ as the union of its finite type $K$-subalgebras $B_ i$. Let $J$ be the set of pairs $(i, g)$ where $i \in I$ and $g \in B_ i$ nonzero with ordering $(i', g') \geq (i, g)$ if and only if $i' \geq i$ and $g$ maps to an invertible element of $(B_{i'})_{g'}$. Then $L = \mathop{\mathrm{colim}}\nolimits _{(i, g) \in J} (B_ i)_ g$. For $j = (i, g) \in J$ set $S_ j = \mathop{\mathrm{Spec}}(B_ i)$ and $U_ j = \mathop{\mathrm{Spec}}((B_ i)_ g)$. Then

Thus we may apply Lemma 59.86.3 to see that it suffices to prove base change holds in the diagrams on the right which is what we proved in Lemma 59.97.5. $\square$

Proof. This lemma is a generalization of Lemma 59.97.6 to objects of the derived category; the assertion of our lemma is true because in Lemma 59.97.6 the scheme $X$ over $K$ is arbitrary. We strongly urge the reader to skip the laborious proof (alternative: read only the last paragraph).

We may represent $E$ by a bounded below K-flat complex $\mathcal{E}^\bullet $ consisting of flat $\mathbf{Z}/n\mathbf{Z}$-modules. See Cohomology on Sites, Lemma 21.46.4. Choose an integer $b$ such that $H^ i(F) = 0$ for $i < b$. Choose a large integer $N$ and consider the short exact sequence

of stupid truncations. This produces a distinguished triangle $E'' \to E \to E' \to E''[1]$ in $D(X_{\acute{e}tale}, \mathbf{Z}/n\mathbf{Z})$. For fixed $F$ both sides of the arrow in the statement of the lemma are exact functors in $E$. Observe that

are sitting in degrees $\geq N + b$. Hence, if we can prove the lemma for the object $E'$, then we see that the lemma holds in degrees $\leq N + b$ and we will conclude. Some details omitted. Thus we may assume $E$ is represented by a bounded complex of flat $\mathbf{Z}/n\mathbf{Z}$-modules. Doing another argument of the same nature, we may assume $E$ is given by a single flat $\mathbf{Z}/n\mathbf{Z}$-module $\mathcal{E}$.

Next, we use the same arguments for the variable $F$ to reduce to the case where $F$ is given by a single sheaf of $\mathbf{Z}/n\mathbf{Z}$-modules $\mathcal{F}$. Say $\mathcal{F}$ is annihilated by an integer $m | n$. If $\ell $ is a prime number dividing $m$ and $m > \ell $, then we can look at the short exact sequence $0 \to \mathcal{F}[\ell ] \to \mathcal{F} \to \mathcal{F}/\mathcal{F}[\ell ] \to 0$ and reduce to smaller $m$. This finally reduces us to the case where $\mathcal{F}$ is annihilated by a prime number $\ell $ dividing $n$. In this case observe that

by the flatness of $\mathcal{E}$. Similarly for the other term. This reduces us to the case where we are working with sheaves of $\mathbf{F}_\ell $-vector spaces which is discussed

Assume $\ell $ is a prime number invertible in $K$. Assume $\mathcal{E}$, $\mathcal{F}$ are sheaves of $\mathbf{F}_\ell $-vector spaces on $X_{\acute{e}tale}$ and $T_{\acute{e}tale}$. We want to show that

is an isomorphism for every $q \geq 0$. This question is local on $X$ hence we may assume $X$ is affine. We can write $\mathcal{E}$ as a filtered colimit of constructible sheaves of $\mathbf{F}_\ell $-vector spaces on $X_{\acute{e}tale}$, see Lemma 59.73.2. Since tensor products commute with filtered colimits and since higher direct images do too (Lemma 59.51.7) we may assume $\mathcal{E}$ is a constructible sheaf of $\mathbf{F}_\ell $-vector spaces on $X_{\acute{e}tale}$. Then we can choose an integer $m$ and finite and finitely presented morphisms $\pi _ i : X_ i \to X$, $i = 1, \ldots , m$ such that there is an injective map

See Lemma 59.74.4. Observe that the direct sum is a constructible sheaf as well (Lemma 59.73.9). Thus the cokernel is constructible too (Lemma 59.71.6). By dimension shifting, i.e., induction on $q$, on the category of constructible sheaves of $\mathbf{F}_\ell $-vector spaces on $X_{\acute{e}tale}$, it suffices to prove the result for the sheaves $\pi _{i, *}\mathbf{F}_\ell $ (details omitted; hint: start with proving injectivity for $q = 0$ for all constructible $\mathcal{E}$). To prove this case we extend the diagram of the lemma to

with all squares cartesian. In the equations below we are going to use that $R\pi _{i, *} = \pi _{i, *}$ and similarly for $\pi '_ i$, $\rho _ i$, we are going to use the Leray spectral sequence, we are going to use Lemma 59.55.3, and we are going to use Lemma 59.96.6 (although this lemma is almost trivial for finite morphisms) for $\pi _ i$, $\pi '_ i$, $\rho _ i$. Doing so we see that

Similarly, we have

Simce $R^ qh_{i, *} \rho _ i^{-1}e^{-1}\mathcal{F} = (\pi '_ i)^{-1} (f')^{-1}R^ qg_*\mathcal{F}$ by Lemma 59.97.6 we conclude. $\square$

Proof. We will reduce this to Lemma 59.97.7 using that our functors commute with direct sums. We suggest the reader skip the proof. Recall that derived tensor product commutes with direct sums. Recall that (derived) pullback commutes with direct sums. Recall that $Rh_*$ and $Rg_*$ commute with direct sums, see Lemmas 59.96.2 and 59.96.3 (this is where we use our schemes are of finite type over $K$).

To finish the proof we can argue as follows. First we write $E = \text{hocolim} \tau _{\leq N} E$. Since our functors commute with direct sums, they commute with homotopy colimits. Hence it suffices to prove the lemma for $E$ bounded above. Similarly for $F$ we may assume $F$ is bounded above. Then we can represent $E$ by a bounded above complex $\mathcal{E}^\bullet $ of sheaves of $\mathbf{Z}/n\mathbf{Z}$-modules. Then

(stupid truncations). Thus we may assume $\mathcal{E}^\bullet $ is a bounded complex of sheaves of $\mathbf{Z}/n\mathbf{Z}$-modules. For $F$ we choose a bounded above complex of flat(!) sheaves of $\mathbf{Z}/n\mathbf{Z}$-modules. Then we reduce to the case where $F$ is represented by a bounded complex of flat sheaves of $\mathbf{Z}/n\mathbf{Z}$-modules. At this point Lemma 59.97.7 kicks in and we conclude. $\square$

Proof. By Lemma 59.97.8 we have

We conclude by Lemma 59.96.5 which we may use because $\text{cd}(X) < \infty $ by Lemma 59.96.2. $\square$