Lemma 59.80.1. Let $S$ be a scheme all of whose local rings are strictly henselian. Then for any abelian sheaf $\mathcal{F}$ on $S_{\acute{e}tale}$ we have $H^ i(S_{\acute{e}tale}, \mathcal{F}) = H^ i(S_{Zar}, \mathcal{F})$.

## 59.80 Schemes with strictly henselian local rings

In this section we collect some results about the étale cohomology of schemes whose local rings are strictly henselian. For example, here is a fun generalization of Lemma 59.55.1.

**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.
$\square$

Lemma 59.80.2. Let $R$ be a ring all of whose local rings are strictly henselian. Let $\mathcal{F}$ be a sheaf on $\mathop{\mathrm{Spec}}(R)_{\acute{e}tale}$. Assume that for all $f, g \in R$ the kernel of

is zero. Then $H^ q_{\acute{e}tale}(\mathop{\mathrm{Spec}}(R), \mathcal{F}) = 0$ for $q > 0$.

**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. $\square$

Lemma 59.80.3. Let $S$ be an affine scheme such that (1) all points are closed, and (2) all residue fields are separably algebraically closed. Then for any abelian sheaf $\mathcal{F}$ on $S_{\acute{e}tale}$ we have $H^ i(S_{\acute{e}tale}, \mathcal{F}) = 0$ for $i > 0$.

**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.
$\square$

The spectrum of an absolutely integrally closed ring is an example of a scheme all of whose local rings are strictly henselian, see More on Algebra, Lemma 15.14.7. It turns out that normal domains with separably closed fraction fields have an even stronger property as explained in the following lemma.

Lemma 59.80.4. Let $X$ be an integral normal scheme with separably closed function field.

A separated étale morphism $U \to X$ is a disjoint union of open immersions.

All local rings of $X$ are strictly henselian.

**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. $\square$

Lemma 59.80.5. Let $f : X \to Y$ be a morphism of schemes where $X$ is an integral normal scheme with separably closed function field. Then $R^ qf_*\underline{M} = 0$ for $q > 0$ and any abelian group $M$.

**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.
$\square$

Lemma 59.80.6. Let $X$ be an affine integral normal scheme with separably closed function field. Let $Z \subset X$ be a closed subscheme. Let $V \to Z$ be an étale morphism with $V$ affine. Then $V$ is a finite disjoint union of open subschemes of $Z$. If $V \to Z$ is surjective and finite étale, then $V \to Z$ has a section.

**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. $\square$

Lemma 59.80.7. Let $X$ be a normal integral affine scheme with separably closed function field. Let $Z \subset X$ be a closed subscheme. For any finite abelian group $M$ we have $H^1_{\acute{e}tale}(Z, \underline{M}) = 0$.

**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.
$\square$

Lemma 59.80.8. Let $X$ be a normal integral affine scheme with separably closed function field. Let $Z \subset X$ be a closed subscheme. For any finite abelian group $M$ we have $H^ q_{\acute{e}tale}(Z, \underline{M}) = 0$ for $q \geq 1$.

**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.
$\square$

Lemma 59.80.9. Let $X$ be an affine scheme.

There exists an integral surjective morphism $X' \to X$ such that for every closed subscheme $Z' \subset X'$, every finite abelian group $M$, and every $q \geq 1$ we have $H^ q_{\acute{e}tale}(Z', \underline{M}) = 0$.

For any closed subscheme $Z \subset X$, finite abelian group $M$, $q \geq 1$, and $\xi \in H^ q_{\acute{e}tale}(Z, \underline{M})$ there exists a finite surjective morphism $X' \to X$ of finite presentation such that $\xi $ pulls back to zero in $H^ q_{\acute{e}tale}(X' \times _ X Z, \underline{M})$.

**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. $\square$

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