## 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.

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.

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

$1 = a^ n + a_1 a^{n - 1}b + a_2 a^{n - 2}b^2 + \ldots + a_ n b^ n$

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

$B = A[y]/(y^ n + a_1 y^{n - 1} + a_2 y^{n - 2} + \ldots + a_ n)$

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

$0 = y^ n + a_1 y^{n - 1} + a_2 y^{n - 2} + \ldots + a_ n = b^{-n}(a^ n + a_1 a^{n - 1}b + a_2 a^{n - 2}b^2 + \ldots + a_ nb^ n) = b^{-n}$

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

$A = \mathbf{Z}[\zeta ] = \mathbf{Z}[x]/(x^{\ell - 1} + x^{\ell - 2} + \ldots + 1)$

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

$\pi = \zeta - 1$

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

$G = \mathop{\mathrm{Spec}}(A[s, \frac{1}{\pi s + 1}])$

with group law given by the comultiplication

$\mu : A[s, \frac{1}{\pi s + 1}] \longrightarrow A[s, \frac{1}{\pi s + 1}] \otimes _ A A[s, \frac{1}{\pi s + 1}],\quad s \longmapsto \pi s \otimes s + s \otimes 1 + 1 \otimes s$

With this choice we have

$\mu (\pi s + 1) = (\pi s + 1) \otimes (\pi s + 1)$

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

$H = \mathop{\mathrm{Spec}}(A[t, \frac{1}{\pi ^\ell t + 1}])$

with group law given by the comultiplication

$\mu : A[t, \frac{1}{\pi ^\ell t + 1}] \longrightarrow A[t, \frac{1}{\pi ^\ell t + 1}] \otimes _ A A[t, \frac{1}{\pi ^\ell t + 1}],\quad t \longmapsto \pi ^\ell t \otimes t + t \otimes 1 + 1 \otimes t$

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

$\Phi (s) = \frac{(\pi s + 1)^\ell - 1}{\pi ^\ell }$

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

$A[t, \frac{1}{\pi ^\ell t + 1}] \longrightarrow A[s, \frac{1}{\pi s + 1}],\quad t \longmapsto \Phi (s)$

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

$f : G \to H$

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

$A[s, \frac{1}{\pi s + 1}] = \left(A[t, \frac{1}{\pi ^\ell t + 1}]\right)[s]/(\Phi (s) - t)$

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

$\sigma _ i : s = \frac{\zeta ^ i - 1}{\pi } = \frac{\zeta ^ i - 1}{\zeta - 1} = \zeta ^{i - 1} + \zeta ^{i - 2} + \ldots + 1,\quad i = 0, 1, \ldots , \ell - 1$

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

$\underline{\mathbf{Z}/\ell \mathbf{Z}}_ A \longrightarrow \mathop{\mathrm{Ker}}(f)$

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

59.81.2.1
$$\label{etale-cohomology-equation-ses} 0 \to \underline{\mathbf{Z}/\ell \mathbf{Z}}_ A \to G \to H \to 0$$

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. $1 + \pi s$ is a unit,

2. $s \equiv 0 \bmod xR$, and

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

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

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

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$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

$U \cap V = A \amalg B$

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] Because $1 + \pi x$ is congruent to $1$ modulo $\pi$, congruent to $1$ modulo $x$, and congruent to $1 + \pi = \zeta$ modulo $x - 1$ and because we have $(\pi ) \cap (x) \cap (x - 1) = (\pi x (x - 1))$ in $A[x]$.
[2] Modulo calculation errors we have $\tau = \sum i \tau _ i$.

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