Lemma 54.78.7. 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 $H^ q_{\acute{e}tale}(Z, \underline{M}) = 0$ for $q \geq 1$.

**Proof.**
We have seen that the result is true for $H^1$ in Lemma 54.78.6. We will prove the result for $q \geq 2$ by induction on $q$. Let $\xi \in H^ q_{\acute{e}tale}(Z, \underline{M})$.

Let $X = \mathop{\mathrm{Spec}}(R)$. Let $I \subset R$ be the set of elements $f \in R$ sch that $\xi |_{Z \cap D(f)} = 0$. All local rings of $Z$ are strictly henselian by Lemma 54.78.3 and Algebra, Lemma 10.150.16. Hence étale cohomology on $Z$ or open subschemes of $Z$ is equal to Zariski cohomology, see Lemma 54.78.1. In particular $\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 now assume that $f, g \in I$ and we show that $f + g \in I$. Note that

By Mayer-Vietoris (Cohomology, Lemma 20.9.2 which applies as étale cohomology on open subschemes of $Z$ equals Zariski cohomology) we have an exact sequence

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

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## Comments (2)

Comment #4216 by Fabrice Orgogozo on

Comment #4217 by Fabrice Orgogozo on