Lemma 59.81.1. Let $A$ be a ring. Let $a, b \in A$ such that $aA + bA = A$ and $a \bmod bA$ is a root of unity. Then there exists a monogenic extension $A \subset B$ and an element $y \in B$ such that $u = a - by$ is a unit.

## 59.81 Absolutely integrally closed vanishing

Recall that we say a ring $R$ is absolutely integrally closed if every monic polynomial over $R$ has a root in $R$ (More on Algebra, Definition 15.14.1). In this section we prove that the étale cohomology of $\mathop{\mathrm{Spec}}(R)$ with coefficients in a finite torsion group vanishes in positive degrees (Proposition 59.81.5) thereby slightly improving the earlier Lemma 59.80.8. We suggest the reader skip this section.

**Proof.**
Say $a^ n \equiv 1 \bmod bA$. In particular $a^ i$ is a unit modulo $b^ mA$ for all $i, m \geq 1$. We claim there exist $a_1, \ldots , a_ n \in A$ such that

Namely, since $1 - a^ n \in bA$ we can find an element $a_1 \in A$ such that $1 - a^ n - a_1 a^{n - 1} b \in b^2A$ using the unit property of $a^{n - 1}$ modulo $bA$. Next, we can find an element $a_2 \in A$ such that $1 - a^ n - a_1 a^{n - 1} b - a_2 a^{n - 2} b^2 \in b^3A$. And so on. Eventually we find $a_1, \ldots , a_{n - 1} \in A$ such that $1 - (a^ n + a_1 a^{n - 1}b + a_2 a^{n - 2}b^2 + \ldots + a_{n - 1} ab^{n - 1}) \in b^ nA$. This allows us to find $a_ n \in A$ such that the displayed equality holds.

With $a_1, \ldots , a_ n$ as above we claim that setting

works. Namely, suppose that $\mathfrak q \subset B$ is a prime ideal lying over $\mathfrak p \subset A$. To get a contradiction assume $u = a - by$ is in $\mathfrak q$. If $b \in \mathfrak p$ then $a \not\in \mathfrak p$ as $aA + bA = A$ and hence $u$ is not in $\mathfrak q$. Thus we may assume $b \not\in \mathfrak p$, i.e., $b \not\in \mathfrak q$. This implies that $y \bmod \mathfrak q$ is equal to $a/b \bmod \mathfrak q$. However, then we obtain

a contradiction. This finishes the proof. $\square$

In order to explain the proof we need to introduce some group schemes. Fix a prime number $\ell $. Let

In other words $A$ is the monogenic extension of $\mathbf{Z}$ generated by a primitive $\ell $th root of unity $\zeta $. We set

A calculation (omitted) shows that $\ell $ is divisible by $\pi ^{\ell - 1}$ in $A$. Our first group scheme over $A$ is

with group law given by the comultiplication

With this choice we have

and hence we indeed have an $A$-algebra map as indicated. We omit the verification that this indeed defines a group law. Our second group scheme over $A$ is

with group law given by the comultiplication

The same verification as before shows that this defines a group law. Next, we observe that the polynomial

is in $A[s]$ and of degree $\ell $ and monic in $s$. Namely, the coefficicient of $s^ i$ for $0 < i < \ell $ is equal to ${\ell \choose i}\pi ^{i - \ell }$ and since $\pi ^{\ell - 1}$ divides $\ell $ in $A$ this is an element of $A$. We obtain a ring map

which the reader easily verifies is compatible with the comultiplications. Thus we get a morphism of group schemes

The following lemma in particular shows that this morphism is faithfully flat (in fact we will see that it is finite étale surjective).

Lemma 59.81.2. We have

In particular, the Hopf algebra of $G$ is a monogenic extension of the Hopf algebra of $H$.

**Proof.**
Follows from the discussion above and the shape of $\Phi (s)$. In particular, note that using $\Phi (s) = t$ the element $\frac{1}{\pi ^\ell t + 1}$ becomes the element $\frac{1}{(\pi s + 1)^\ell }$.
$\square$

Next, let us compute the kernel of $f$. Since the origin of $H$ is given by $t = 0$ in $H$ we see that the kernel of $f$ is given by $\Phi (s) = 0$. Now observe that the $A$-valued points $\sigma _0, \ldots , \sigma _{\ell - 1}$ of $G$ given by

are certainly contained in $\mathop{\mathrm{Ker}}(f)$. Moreover, these are all pairwise distinct in **all** fibres of $G \to \mathop{\mathrm{Spec}}(A)$. Also, the reader computes that $\sigma _ i +_ G \sigma _ j = \sigma _{i + j \bmod \ell }$. Hence we find a closed immersion of group schemes

sending $i$ to $\sigma _ i$. However, by construction $\mathop{\mathrm{Ker}}(f)$ is finite flat over $\mathop{\mathrm{Spec}}(A)$ of degree $\ell $. Hence we conclude that this map is an isomorphism. All in all we conclude that we have a short exact sequence

of group schemes over $A$.

Lemma 59.81.3. Let $R$ be an $A$-algebra which is absolutely integrally closed. Then $G(R) \to H(R)$ is surjective.

**Proof.**
Let $h \in H(R)$ correspond to the $A$-algebra map $A[t, \frac{1}{\pi ^\ell t + 1}] \to R$ sending $t$ to $a \in A$. Since $\Phi (s)$ is monic we can find $b \in A$ with $\Phi (b) = a$. By Lemma 59.81.2 sending $s$ to $b$ we obtain a unique $A$-algebra map $A[s, \frac{1}{\pi s + 1}] \to R$ compatible with the map $A[t, \frac{1}{\pi ^\ell t + 1}] \to R$ above. This in turn corresponds to an element $g \in G(R)$ mapping to $h \in H(R)$.
$\square$

Lemma 59.81.4. Let $R$ be an $A$-algebra which is absolutely integrally closed. Let $I, J \subset R$ be ideals with $I + J = R$. There exists a $g \in G(R)$ such that $g \bmod I = \sigma _0$ and $g \bmod J = \sigma _1$.

**Proof.**
Choose $x \in I$ such that $x \equiv 1 \bmod J$. We may and do replace $I$ by $xR$ and $J$ by $(x - 1)R$. Then we are looking for an $s \in R$ such that

$1 + \pi s$ is a unit,

$s \equiv 0 \bmod xR$, and

$s \equiv 1 \bmod (x - 1)R$.

The last two conditions say that $s = x + x(x - 1)y$ for some $y \in R$. The first condition says that $1 + \pi s = 1 + \pi x + \pi x (x - 1) y$ needs to be a unit of $R$. However, note that $1 + \pi x$ and $\pi x (x - 1)$ generate the unit ideal of $R$ and that $1 + \pi x$ is an $\ell $th root of $1$ modulo $\pi x (x - 1)$^{1}. Thus we win by Lemma 59.81.1 and the fact that $R$ is absolutely integrally closed.
$\square$

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

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

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

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 $\tau _ i$^{2}. 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

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

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

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

(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

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

$g_2 \in G(V)$ 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

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

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