The Stacks project

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. $\square$