The Stacks project

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

\[ H^1_{\acute{e}tale}(D(f + g), \mathcal{F}) \longrightarrow H^1_{\acute{e}tale}(D(f(f + g)), \mathcal{F}) \oplus H^1_{\acute{e}tale}(D(g(f + g)), \mathcal{F}) \]

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

\[ D(f + g) = D(f(f + g)) \cup D(g(f + g)) \]

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

\[ \xymatrix{ H^{i - 1}_{\acute{e}tale}(D(fg(f + g)), \mathcal{F}) \ar[d] \\ H^ i_{\acute{e}tale}(D(f + g), \mathcal{F}) \ar[d] \\ H^ i_{\acute{e}tale}(D(f(f + g)), \mathcal{F}) \oplus H^ i_{\acute{e}tale}(D(g(f + g)), \mathcal{F}) } \]

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


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0GY0. Beware of the difference between the letter 'O' and the digit '0'.