Proposition 59.81.5. Let $R$ be an absolutely integrally closed ring. Let $M$ be a finite abelian group. Then $H^ i_{\acute{e}tale}(\mathop{\mathrm{Spec}}(R), \underline{M}) = 0$ for $i > 0$.

Proof. Since any finite abelian group has a finite filtration whose subquotients are cyclic of prime order, we may assume $M = \mathbf{Z}/\ell \mathbf{Z}$ where $\ell$ is a prime number.

Observe that all local rings of $R$ are strictly henselian, see More on Algebra, Lemma 15.14.7. Furthermore, any localization of $R$ is also absolutely integrally closed by More on Algebra, Lemma 15.14.3. Thus Lemma 59.80.2 tells us it suffices to show that the kernel of

$H^1_{\acute{e}tale}(D(f + g), \mathbf{Z}/\ell \mathbf{Z}) \longrightarrow H^1_{\acute{e}tale}(D(f(f + g)), \mathbf{Z}/\ell \mathbf{Z}) \oplus H^1_{\acute{e}tale}(D(g(f + g)), \mathbf{Z}/\ell \mathbf{Z})$

is zero for any $f, g \in R$. After replacing $R$ by $R_{f + g}$ we reduce to the following claim: given $\xi \in H^1_{\acute{e}tale}(\mathop{\mathrm{Spec}}(R), \mathbf{Z}/\ell \mathbf{Z})$ and an affine open covering $\mathop{\mathrm{Spec}}(R) = U \cup V$ such that $\xi |_ U$ and $\xi |_ V$ are trivial, then $\xi = 0$.

Let $A = \mathbf{Z}[\zeta ]$ as above. Since $\mathbf{Z} \subset A$ is monogenic, we can find a ring map $A \to R$. From now on we think of $R$ as an $A$-algebra and we think of $\mathop{\mathrm{Spec}}(R)$ as a scheme over $\mathop{\mathrm{Spec}}(A)$. If we base change the short exact sequence (59.81.2.1) to $\mathop{\mathrm{Spec}}(R)$ and take étale cohomology we obtain

$G(R) \to H(R) \to H^1_{\acute{e}tale}(\mathop{\mathrm{Spec}}(R), \mathbf{Z}/\ell \mathbf{Z}) \to H^1_{\acute{e}tale}(\mathop{\mathrm{Spec}}(R), G)$

Please keep this in mind during the rest of the proof.

Let $\tau \in \Gamma (U \cap V, \mathbf{Z}/\ell \mathbf{Z})$ be a section whose boundary in the Mayer-Vietoris sequence (Lemma 59.50.1) gives $\xi$. For $i = 0, 1, \ldots , \ell - 1$ let $A_ i \subset U \cap V$ be the open and closed subset where $\tau$ has the value $i \bmod \ell$. Thus we have a finite disjoint union decomposition

$U \cap V = A_0 \amalg \ldots \amalg A_{\ell - 1}$

such that $\tau$ is constant on each $A_ i$. For $i = 0, 1, \ldots , \ell - 1$ denote $\tau _ i \in H^0(U \cap V, \mathbf{Z}/\ell \mathbf{Z})$ the element which is equal to $1$ on $A_ i$ and equal to $0$ on $A_ j$ for $j \not= i$. Then $\tau$ is a sum of multiples of the $\tau _ i$1. Hence it suffices to show that the cohomology class corresponding to $\tau _ i$ is trivial. This reduces us to the case where $\tau$ takes only two distinct values, namely $1$ and $0$.

Assume $\tau$ takes only the values $1$ and $0$. Write

$U \cap V = A \amalg B$

where $A$ is the locus where $\tau = 0$ and $B$ is the locus where $\tau = 1$. Then $A$ and $B$ are disjoint closed subsets. Denote $\overline{A}$ and $\overline{B}$ the closures of $A$ and $B$ in $\mathop{\mathrm{Spec}}(R)$. Then we have a “banana”: namely we have

$\overline{A} \cap \overline{B} = Z_1 \amalg Z_2$

with $Z_1 \subset U$ and $Z_2 \subset V$ disjoint closed subsets. Set $T_1 = \mathop{\mathrm{Spec}}(R) \setminus V$ and $T_2 = \mathop{\mathrm{Spec}}(R) \setminus U$. Observe that $Z_1 \subset T_1 \subset U$, $Z_2 \subset T_2 \subset V$, and $T_1 \cap T_2 = \emptyset$. Topologically we can write

$\mathop{\mathrm{Spec}}(R) = \overline{A} \cup \overline{B} \cup T_1 \cup T_2$

We suggest drawing a picture to visualize this. In order to prove that $\xi$ is zero, we may and do replace $R$ by its reduction (Proposition 59.45.4). Below, we think of $A$, $\overline{A}$, $B$, $\overline{B}$, $T_1$, $T_2$ as reduced closed subschemes of $\mathop{\mathrm{Spec}}(R)$. Next, as scheme structures on $Z_1$ and $Z_2$ we use

$Z_1 = \overline{A} \cap (\overline{B} \cup T_1) \quad \text{and}\quad Z_2 = \overline{A} \cap (\overline{B} \cup T_2)$

(scheme theoretic unions and intersections as in Morphisms, Definition 29.4.4).

Denote $X$ the $G$-torsor over $\mathop{\mathrm{Spec}}(R)$ corresponding to the image of $\xi$ in $H^1(\mathop{\mathrm{Spec}}(R), G)$. If $X$ is trivial, then $\xi$ comes from an element $h \in H(R)$ (see exact sequence of cohomology above). However, then by Lemma 59.81.3 the element $h$ lifts to an element of $G(R)$ and we conclude $\xi = 0$ as desired. Thus our goal is to prove that $X$ is trivial.

Recall that the embedding $\mathbf{Z}/\ell \mathbf{Z} \to G(R)$ sends $i \bmod \ell$ to $\sigma _ i \in G(R)$. Observe that $\overline{A}$ is the spectrum of an absolutely integrally closed ring (namely a qotient of $R$). By Lemma 59.81.4 we can find $g \in G(\overline{A})$ with $g|_{\overline{A} \cap Z_1} = \sigma _0$ and $g|_{\overline{A} \cap Z_2} = \sigma _1$ (scheme theoretically). Then we can define

1. $g_1 \in G(U)$ which is $g$ on $\overline{A} \cap U$, which is $\sigma _0$ on $\overline{B} \cap U$, and $\sigma _0$ on $T_1$, and

2. $g_2 \in G(V)$ which is $g$ on $\overline{A} \cap V$, which is $\sigma _1$ on $\overline{B} \cap V$, and $\sigma _1$ on $T_2$.

Namely, to find $g_1$ as in (1) we glue the section $\sigma _0$ on $\Omega = (\overline{B} \cup T_1) \cap U$ to the restriction of the section $g$ on $\Omega ' = \overline{A} \cap U$. Note that $U = \Omega \cup \Omega '$ (scheme theoretically) because $U$ is reduced and $\Omega \cap \Omega ' = Z_1$ (scheme theoretically) by our choice of $Z_1$. Hence by Morphisms, Lemma 29.4.6 we have that $U$ is the pushout of $\Omega$ and $\Omega '$ along $Z_1$. Thus we can find $g_1$. Similarly for the existence of $g_2$ in (2). Then we have

$\tau = g_2|_{A \cup B} - g_1|_{A \cup B} \quad (\text{addition in group law})$

and we see that $X$ is trivial thereby finishing the proof. $\square$

[1] Modulo calculation errors we have $\tau = \sum i \tau _ i$.

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