The Stacks project

Proof. Let $\epsilon : S_{\acute{e}tale}\to S_{Zar}$ be the morphism of sites given by the inclusion functor. The Zariski sheaf $R^ p\epsilon _*\mathcal{F}$ is the sheaf associated to the presheaf $U \mapsto H^ p_{\acute{e}tale}(U, \mathcal{F})$. Thus the stalk at $x \in X$ is $\mathop{\mathrm{colim}}\nolimits H^ p_{\acute{e}tale}(U, \mathcal{F}) = H^ p_{\acute{e}tale}(\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}), \mathcal{G}_ x)$ where $\mathcal{G}_ x$ denotes the pullback of $\mathcal{F}$ to $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$, see Lemma 59.51.5. Thus the higher direct images of $R^ p\epsilon _*\mathcal{F}$ are zero by Lemma 59.55.1 and we conclude by the Leray spectral sequence. $\square$

Proof. By Lemma 59.80.1 we see that étale cohomology of $\mathcal{F}$ agrees with Zariski cohomology on any open of $\mathop{\mathrm{Spec}}(R)$. We will prove by induction on $i$ the statement: for $h \in R$ we have $H^ q_{\acute{e}tale}(D(h), \mathcal{F}) = 0$ for $1 \leq q \leq i$. The base case $i = 0$ is trivial. Assume $i \geq 1$.

Let $\xi \in H^ q_{\acute{e}tale}(D(h), \mathcal{F})$ for some $1 \leq q \leq i$ and $h \in R$. If $q < i$ then we are done by induction, so we assume $q = i$. After replacing $R$ by $R_ h$ we may assume $\xi \in H^ i_{\acute{e}tale}(\mathop{\mathrm{Spec}}(R), \mathcal{F})$; some details omitted. Let $I \subset R$ be the set of elements $f \in R$ such that $\xi |_{D(f)} = 0$. Since $\xi $ is Zariski locally trivial, it follows that for every prime $\mathfrak p$ of $R$ there exists an $f \in I$ with $f \not\in \mathfrak p$. Thus if we can show that $I$ is an ideal, then $1 \in I$ and we're done. It is clear that $f \in I$, $r \in R$ implies $rf \in I$. Thus we assume that $f, g \in I$ and we show that $f + g \in I$. If $q = i = 1$, then this is exactly the assumption of the lemma! Whence the result for $i = 1$. For $q = i > 1$, note that

By Mayer-Vietoris (Cohomology, Lemma 20.8.2 which applies as étale cohomology on open subschemes of $\mathop{\mathrm{Spec}}(R)$ equals Zariski cohomology) we have an exact sequence

and the result follows as the first group is zero by induction. $\square$

Proof. Condition (1) implies that the underlying topological space of $S$ is profinite, see Algebra, Lemma 10.26.5. Thus the higher cohomology groups of an abelian sheaf on the topological space $S$ (i.e., Zariski cohomology) is trivial, see Cohomology, Lemma 20.22.3. The local rings are strictly henselian by Algebra, Lemma 10.153.10. Thus étale cohomology of $S$ is computed by Zariski cohomology by Lemma 59.80.1 and the proof is done. $\square$

Proof. Let $R$ be a normal domain whose fraction field is separably algebraically closed. Let $R \to A$ be an étale ring map. Then $A \otimes _ R K$ is as a $K$-algebra a finite product $\prod _{i = 1, \ldots , n} K$ of copies of $K$. Let $e_ i$, $i = 1, \ldots , n$ be the corresponding idempotents of $A \otimes _ R K$. Since $A$ is normal (Algebra, Lemma 10.163.9) the idempotents $e_ i$ are in $A$ (Algebra, Lemma 10.37.12). Hence $A = \prod Ae_ i$ and we may assume $A \otimes _ R K = K$. Since $A \subset A \otimes _ R K = K$ (by flatness of $R \to A$ and since $R \subset K$) we conclude that $A$ is a domain. By the same argument we conclude that $A \otimes _ R A \subset (A \otimes _ R A) \otimes _ R K = K$. It follows that the map $A \otimes _ R A \to A$ is injective as well as surjective. Thus $R \to A$ defines an open immersion by Morphisms, Lemma 29.10.2 and Étale Morphisms, Theorem 41.14.1.

Let $f : U \to X$ be a separated étale morphism. Let $\eta \in X$ be the generic point and let $f^{-1}(\{ \eta \} ) = \{ \xi _ i\} _{i \in I}$. The result of the previous paragraph shows the following: For any affine open $U' \subset U$ whose image in $X$ is contained in an affine we have $U' = \coprod _{i \in I} U'_ i$ where $U'_ i$ is the set of point of $U'$ which are specializations of $\xi _ i$. Moreover, the morphism $U'_ i \to X$ is an open immersion. It follows that $U_ i = \overline{\{ \xi _ i\} }$ is an open and closed subscheme of $U$ and that $U_ i \to X$ is locally on the source an isomorphism. By Morphisms, Lemma 29.49.7 the fact that $U_ i \to X$ is separated, implies that $U_ i \to X$ is injective and we conclude that $U_ i \to X$ is an open immersion, i.e., (1) holds.

Part (2) follows from part (1) and the description of the strict henselization of $\mathcal{O}_{X, x}$ as the local ring at $\overline{x}$ on the étale site of $X$ (Lemma 59.33.1). It can also be proved directly, see Fundamental Groups, Lemma 58.12.2. $\square$

Proof. Recall that $R^ qf_*\underline{M}$ is the sheaf associated to the presheaf $V \mapsto H^ q_{\acute{e}tale}(V \times _ Y X, M)$ on $Y_{\acute{e}tale}$, see Lemma 59.51.6. If $V$ is affine, then $V \times _ Y X \to X$ is separated and étale. Hence $V \times _ Y X = \coprod U_ i$ is a disjoint union of open subschemes $U_ i$ of $X$, see Lemma 59.80.4. By Lemma 59.80.1 we see that $H^ q_{\acute{e}tale}(U_ i, M)$ is equal to $H^ q_{Zar}(U_ i, M)$. This vanishes by Cohomology, Lemma 20.20.2. $\square$

Proof. By Algebra, Lemma 10.143.10 we can lift $V$ to an affine scheme $U$ étale over $X$. Apply Lemma 59.80.4 to $U \to X$ to get the first statement.

The final statement is a consequence of the first. Let $V = \coprod _{i = 1, \ldots , n} V_ i$ be a finite decomposition into open and closed subschemes with $V_ i \to Z$ an open immersion. As $V \to Z$ is finite we see that $V_ i \to Z$ is also closed. Let $U_ i \subset Z$ be the image. Then we have a decomposition into open and closed subschemes

where the disjoint union is over $\{ 1, \ldots , n\} = A \amalg B$ where $A$ has at least one element. Each of the strata is contained in a single $U_ i$ and we find our section. $\square$

Proof. By Cohomology on Sites, Lemma 21.4.3 an element of $H^1_{\acute{e}tale}(Z, \underline{M})$ corresponds to a $\underline{M}$-torsor $\mathcal{F}$ on $Z_{\acute{e}tale}$. Such a torsor is clearly a finite locally constant sheaf. Hence $\mathcal{F}$ is representable by a scheme $V$ finite étale over $Z$, Lemma 59.64.4. Of course $V \to Z$ is surjective as a torsor is locally trivial. Since $V \to Z$ has a section by Lemma 59.80.6 we are done. $\square$

Proof. Write $X = \mathop{\mathrm{Spec}}(R)$ and $Z = \mathop{\mathrm{Spec}}(R')$ so that we have a surjection of rings $R \to R'$. All local rings of $R'$ are strictly henselian by Lemma 59.80.4 and Algebra, Lemma 10.156.4. Furthermore, we see that for any $f' \in R'$ there is a surjection $R_ f \to R'_{f'}$ where $f \in R$ is a lift of $f'$. Since $R_ f$ is a normal domain with separably closed fraction field we see that $H^1_{\acute{e}tale}(D(f'), \underline{M}) = 0$ by Lemma 59.80.7. Thus we may apply Lemma 59.80.2 to $Z = \mathop{\mathrm{Spec}}(R')$ to conclude. $\square$

Proof. Write $X = \mathop{\mathrm{Spec}}(A)$. Write $A = \mathbf{Z}[x_ i]/J$ for some ideal $J$. Let $R$ be the integral closure of $\mathbf{Z}[x_ i]$ in an algebraic closure of the fraction field of $\mathbf{Z}[x_ i]$. Let $A' = R/JR$ and set $X' = \mathop{\mathrm{Spec}}(A')$. This gives an example as in (1) by Lemma 59.80.8.

Proof of (2). Let $X' \to X$ be the integral surjective morphism we found above. Certainly, $\xi $ maps to zero in $H^ q_{\acute{e}tale}(X' \times _ X Z, \underline{M})$. We may write $X'$ as a limit $X' = \mathop{\mathrm{lim}}\nolimits X'_ i$ of schemes finite and of finite presentation over $X$; this is easy to do in our current affine case, but it is a special case of the more general Limits, Lemma 32.7.3. By Lemma 59.51.5 we see that $\xi $ maps to zero in $H^ q_{\acute{e}tale}(X'_ i \times _ X Z, \underline{M})$ for some $i$ large enough. $\square$