Proof. It suffices to assume that $\mathbb {G}_{m,K}^k\subset T_K$ and to deduce that $\mathbb {G}_{m,S}^k\subset T$. Let $\overline {\eta }$ be a geometric point over the generic point $\operatorname {Spec} K\overset {\eta }{\rightarrow } S$. We have $\mathscr {M}(T)=\operatorname {Hom}_{\overline {\eta }\text {-gr.}}(T_{\overline {\eta }}, \mathbb {G}_{m,\overline {\eta }})= \mathscr {M}(T_K)$. Note that $\mathbb {G}_{m,K}^k$ corresponds to a quotient lattice $\Lambda$ of $\mathscr {M}(T_K)$ such that $\Lambda$ is of rank $k$ with trivial $\pi _1^{\mathrm {\acute {e}t}}(K)$-action. On the other hand, by [Sta18, 0BQI], the natural map $\pi _1^{\mathrm {\acute {e}t}}(K)\twoheadrightarrow \pi _1^{\mathrm {\acute {e}t}}(S)$ is surjective. Therefore, $\mathscr {M}(T)$ has a quotient lattice that has rank $k$ with trivial $\pi _1^{\mathrm {\acute {e}t}}(S)$-action. This implies that $\mathbb {G}_{m,S}^k\subset T$.

Proof. Since every connected component of $S$ is open, we may assume that $S$ is connected. Then, $M$ is fpqc locally of the form $\mathscr {D}(H)$ for a finite-type abelian group $H$ (determined by $M$). For $P:= \underline {\operatorname {Isom}}_{S\text {-gr.}}(M,\mathscr {D}_S(H))$, our goal is to find a finite étale cover $S^{\prime }\rightarrow S$ such that $P(S^{\prime })\neq \emptyset$. By [SGA2, X, 5.8, 5.10 (i)], $P$ is representable by a clopen subscheme of $\underline {\operatorname {Hom}}_{S\text {-gr.}}(M,\mathscr {D}_S(H))$ and there is an étale surjective morphism $\widetilde {S}\rightarrow S$ such that $P_{\widetilde {S}}$ is a disjoint union of copies of $\widetilde {S}$. In particular, $P$ is $S$-étale. By [Reference Grothendieck and DieudonnéGD67, 18.8.15, 18.10.7], $\widetilde {S}$ is locally integral and geometrically unibranch. We prove the following.

Proof of the claim Let $\eta \in S$ be the generic point and let $\xi _i$ be the generic point of $P_i$. By [Reference Grothendieck and DieudonnéGD65, 2.3.4], the $S$-flatness of $P$ implies that every $\xi _i$ lies over $\eta$. Therefore, $(P_i)_{\eta }$ is the closure of $\xi _i$ in $P_{\eta }$. The quasi-finiteness of $P\rightarrow S$ implies that $P_{\eta }$ is discrete, so we have $(P_i)_{\eta }=\{\xi _i\}$. On the other hand, since $S$ is integral and geometrically unibranch, by [Reference Grothendieck and DieudonnéGD67, 18.10.7], all $P_i$ are geometrically unibranch, and

\[ \textstyle P=\bigsqcup_{\xi_i\in P_{\eta}}P_i. \]

Therefore, every $P_i$ is clopen in $P$. Since it suffices to show that each $(P_i)_{\widetilde {S}}$ is $\widetilde {S}$-finite, note that every connected component of $\widetilde {S}$ is open, we may assume that $\widetilde {S}$ is connected so that $P_{\widetilde {S}}\cong \bigsqcup _{\Psi }{\widetilde {S}}$ for a set $\Psi$. Each $P_i\subset P$ satisfies that $(P_i)_{\widetilde {S}}\cong \bigsqcup _{\Phi _i}\widetilde {S}$ for a subset $\Phi _i\subset \Psi$. As $(P_i)_{\eta }=\{\xi _i\}$ is a single point, this forces that $\Phi _i$ is finite. Consequently, the base change $(P_i)_{\widetilde {S}}$ is finite over $\widetilde {S}$, so $P_i$ is $S$-finite.

As $S$ is connected and all $P_i\rightarrow S$ are finite étale, take $S^{\prime }:= P_i$, whose image is $S$. The canonical embedding $S^{\prime }\hookrightarrow P$ then induces a section of $P_{S^{\prime }}\rightarrow S^{\prime }$, so we get $M_{S^{\prime }}\simeq \mathscr {D}_{S^{\prime }}(H)$, as desired.

Proof. Let $\Gamma := \operatorname {Aut}_X(Y)$ be the Galois group of $Y/X$, then $Y$ is a $\underline {\Gamma }_X$-torsor on $X$, and $Y^{\prime }$ is a $\underline {\Gamma }_{X^{\prime }}$-torsor on $X^{\prime }$. In particular, $\Gamma$ acts transitively on each $X^{\prime }$-fiber of $Y^{\prime }$, hence induces isomorphisms among connected components of $Y^{\prime }$. We choose a geometric point $\eta ^{\prime }\rightarrow Y^{\prime }$, and denote its composites as $\eta \rightarrow Y$, $\xi ^{\prime }\rightarrow X^{\prime }$, and $\xi \rightarrow X$, respectively. Recall [Sta18, 0BND] that the fiber functors $F_{\xi }\colon \text{F}{\mathrm {\acute {E}t}}_{X}\overset {\sim }{\longrightarrow } {\mathrm {Finite}\text {-}\pi _1^{\mathrm {\acute {e}t}}(X,\xi )\text {-sets}}$ and $F_{\xi ^{\prime }}\colon \mathrm {F\acute Et}_{X^{\prime }}\overset {\sim }{\longrightarrow } {\mathrm {Finite}\text {-}\pi _1^{\mathrm {\acute {e}t}}(X^{\prime },\xi ^{\prime })\text {-sets}}$ are equivalences of categories. In addition, $f$ induces a continuous homomorphism $f_{\ast }\colon \pi _1^{\mathrm {\acute {e}t}}(X^{\prime },\xi ^{\prime })\rightarrow \pi _1^{\mathrm {\acute {e}t}}(X,\xi )$ of profinite groups, fitting into the following commutative diagram.

Thus, we have $F_{\xi ^{\prime }}(Y^{\prime })=f_{\ast }F_{\xi }(Y)=F_{\xi }(Y)=\Gamma$ set-theoretically and the short exact sequence

\[ 1\rightarrow \pi_1^{\mathrm{\acute{e}t}}(Y,\eta)\rightarrow \pi_1^{\mathrm{\acute{e}t}}(X,\xi)\rightarrow \Gamma\cong\operatorname{Aut}_{\Gamma\text{-set}}(F_{\xi}(Y)) \rightarrow 1. \]

By the commutative diagram above, the $\pi _1^{\mathrm {\acute {e}t}}(X^{\prime },\xi ^{\prime })$-action on $F_{\xi ^{\prime }}(Y^{\prime })$ is equal to the $\pi _1^{\mathrm {\acute {e}t}}(X^{\prime },\xi ^{\prime })$-action on $F_{\xi }(Y)$ via the composite $\pi _1^{\mathrm {\acute {e}t}}(X^{\prime },\xi ^{\prime })\overset {f_{\ast }}{\rightarrow }\pi _1^{\mathrm {\acute {e}t}}(X,\xi )\twoheadrightarrow \Gamma$, whose image is denoted by $\Gamma ^{\prime }\subset \Gamma$. The surjection $\pi _1^{\mathrm {\acute {e}t}}(X^{\prime },\xi ^{\prime })\twoheadrightarrow \Gamma ^{\prime }$ gives rise to the $\pi _1^{\mathrm {\acute {e}t}}(X^{\prime },\xi ^{\prime })$-set structure on $F_{\xi ^{\prime }}(Y^{\prime })$. Precisely, the $\pi _1^{\mathrm {\acute {e}t}}(X^{\prime },\xi ^{\prime })$-action on $F_{\xi ^{\prime }}(Y^{\prime })$ is just the restriction $\Gamma ^{\prime }\times \Gamma \rightarrow \Gamma$ of $\Gamma \times \Gamma \rightarrow \Gamma$, leading to the coset decomposition for $\Gamma ^{\prime }\subset \Gamma$

\[ \textstyle \Gamma=\bigsqcup_{\gamma\in \Gamma^{\prime}\backslash \Gamma}(\Gamma^{\prime}\cdot \gamma) \]

so that all left $\Gamma ^{\prime }$-actions on $\Gamma ^{\prime }\cdot \gamma$ are simply transitive and all $\Gamma ^{\prime }\cdot \gamma$ have the same $\Gamma ^{\prime }$-set structure. Hence, the equivalence $F_{\xi ^{\prime }}\colon \mathrm {F\acute Et}_{X^{\prime }}\overset {\sim }{\longrightarrow } {\mathrm {Finite}\text {-}\pi _1^{\mathrm {\acute {e}t}}(X^{\prime },\xi ^{\prime })\text {-sets}}$ (combined with [Sta18, 03SF]) implies that $(\Gamma ^{\prime }\cdot \gamma )_{\gamma \in \Gamma ^{\prime }\backslash \Gamma }$ correspond to Galois covers $(Y^{\prime }_{\gamma })_{\gamma \in \Gamma ^{\prime }\backslash \Gamma }$ of $X^{\prime }$ that are isomorphic to each other. Further, the finite $\pi _1^{\mathrm {\acute {e}t}}(X^{\prime },\xi ^{\prime })$-set $F_{\xi ^{\prime }}(Y^{\prime })$ corresponds to $Y^{\prime }$, which decomposes into connected components

\[ \textstyle Y^{\prime}=\bigsqcup_{\gamma\in \Gamma^{\prime}\backslash \Gamma}Y^{\prime}_{\gamma}, \]

where $Y^{\prime }_{\gamma }$ are Galois covers of $X^{\prime }$ with Galois group $\Gamma ^{\prime }$. If $\eta ^{\prime }\rightarrow Y^{\prime }$ factors through $Y^{\prime }_{\gamma _0}$, then

\[ 1\rightarrow \pi_1^{\mathrm{\acute{e}t}}(Y_{{\gamma}_0}^{\prime}, \eta^{\prime})\rightarrow \pi_1^{\mathrm{\acute{e}t}}(X^{\prime},\xi^{\prime})\rightarrow \Gamma^{\prime}=\mathrm{Gal}(Y_{\gamma_0}^{\prime}/X^{\prime})\rightarrow 1 \]

is a short exact sequence. Since the torus $T$ induces a representation $\rho _T\colon \pi _1^{\mathrm {\acute {e}t}}(X,\xi )\rightarrow \operatorname {GL}(\textbf {Z}^n)$ with the image $\Gamma$, where $\textbf {Z}^n\simeq \operatorname {Hom}_{\xi \text {-gr.}}(T_{\xi },\mathbb {G}_m)$, its base change $T_{X^{\prime }}$ induces a representation $f_{\ast }\circ \rho _T\colon \pi _1^{\mathrm {\acute {e}t}}(X^{\prime },\xi ^{\prime })\rightarrow \operatorname {GL}(\textbf {Z}^n)$. By construction of $\Gamma ^{\prime }$, we have $\Gamma ^{\prime }=\mathrm {Im}(f_{\ast }\circ \rho _T)$. Thus, the desired minimality of $Y_{\gamma _0}^{\prime }$ amounts to the equality $\Gamma ^{\prime }=\pi _1^{\mathrm {\acute {e}t}}(X^{\prime }, \xi ^{\prime })/\pi _1^{\mathrm {\acute {e}t}}(Y_{\gamma _0}^{\prime },\eta ^{\prime })$, which follows from the last displayed short exact sequence.

Proof. Let $X=\operatorname {Spec} V$ and $Z=\operatorname {Spec}(V/\mathfrak {m}_V)$. Let $n\geq 1$ be the rank of $V$, then $X\backslash Z$ is the spectrum of a valuation ring of rank $n-1$. By excision [Reference MilneMil80, III, 1.28], we may replace $X$ by its Henselization $X^{\mathrm {h}}$. For a variable $X$-étale scheme $X^{\prime }$ with preimage $Z^{\prime }:= X^{\prime }\times _{X} Z$, let $\mathcal {H}^q_{Z}(-,F)$ be the étale sheafification of the presheaf $X^{\prime }\mapsto H^q_{Z^{\prime }}(X^{\prime }, F)$. By the local-to-global $E_2$ spectral sequence

\begin{align*} \text{$H^p_{\mathrm{\acute{e}t}}(X,\mathcal{H}^q_Z(X,F)) \Rightarrow H^{p+q}_Z(X,F)$$\quad$ {SGA4$_{\rm II}$}, V, 6.4]} \end{align*}

to show that $H^2_{Z}(X,F)=0$, it suffices to obtain the vanishings

\[ H^0_{\mathrm{\acute{e}t}}(X,\mathcal{H}^2_{Z}(X, F))=H^1_{\mathrm{\acute{e}t}}(X,\mathcal{H}^1_Z(X, F))=H^2_{\mathrm{\acute{e}t}}(X, \mathcal{H}^0_Z(X, F))=0. \]

Subsequently, in the following two paragraphs, we calculate $\mathcal {H}^q_Z(X,F)$ for $0\leq q\leq 2$.

Let $\overline {x}\rightarrow X$ be a geometric point. If $\overline {x}$ factors through $X\backslash Z$, then $\mathcal {H}^q_Z(X,F)_{\overline {x}}=0$. Now, we take $\overline {x}$ as a fixed geometric point over $\mathfrak {m}_V$, so $\mathcal {H}^q_Z(X,F)_{\overline {x}}=H^q_{\overline {\mathfrak {m}}_V}(V^{\mathrm {sh}},F)$, where $V^{\mathrm {sh}}$ is the strict Henselization of $V$ with the maximal ideal $\overline {\mathfrak {m}}_V$. The local map $V\rightarrow V^{\mathrm {sh}}$ of local rings is faithfully flat [Sta18, 07QM] and preserves value groups [Sta18, 0ASK]. Therefore, for the prime $\mathfrak {p}\subset V$ of height $n-1$, there is a unique prime ideal $\mathfrak {P}\subset V^{\mathrm {sh}}$ lying over $\mathfrak {p}$ (that is, $\mathfrak {p} V^{\mathrm {sh}}=\mathfrak {P}$). By [SGA4_II, V, 6.5], we have the exact sequence

(2.6.1)\begin{equation} \cdots\rightarrow H^i_{\mathrm{\acute{e}t}}(V^{\mathrm{sh}}, F) \rightarrow H^i_{\mathrm{\acute{e}t}}((V^{\mathrm{sh}})_{\mathfrak{P}},F)\rightarrow H^{i+1}_{\overline{\mathfrak{m}}_V}(V^{\mathrm{sh}},F)\rightarrow H^{i+1}_{\mathrm{\acute{e}t}}(V^{\mathrm{sh}},F)\rightarrow \cdots. \end{equation}

First, we compute $H^{q}_{\overline {\mathfrak {m}}_V}(V^{\mathrm {sh}},F)$ when $q=0$ and $2$. The injectivity of $H^0_{\mathrm {\acute {e}t}}(V^{\mathrm {sh}},F)\hookrightarrow H^0_{\mathrm {\acute {e}t}}((V^{\mathrm {sh}})_{\mathfrak {P}},F)$ and the vanishings of $H^1_{\mathrm {\acute {e}t}}((V^{\mathrm {sh}})_{\mathfrak {P}},F)$ and $H^i_{\mathrm {\acute {e}t}}(V^{\mathrm {sh}},F)$ for $i=1,2$ (see [Sta18, 03QO]) imply the following:

(2.6.2)\begin{equation} H^0_{\overline{\mathfrak{m}}_V}(V^{\mathrm{sh}},F)=H^2_{\overline{\mathfrak{m}}_V}(V^{\mathrm{sh}},F)=0. \end{equation}

This (2.6.2) leads to $\mathcal {H}^0_Z(X,F)=\mathcal {H}^2_Z(X,F)=0$, so we get $H^0_{\mathrm {\acute {e}t}}(X,\mathcal {H}^2_Z(X,F))=H^2_{\mathrm {\acute {e}t}}(X,\mathcal {H}^0_Z (X,F))=0$.

Next, we calculate $H^{1}_{\overline {\mathfrak {m}}_V}(V^{\mathrm {sh}},F)$. From (2.6.1) we obtain the following short exact sequence:

\[ 0\rightarrow H^0_{\mathrm{\acute{e}t}}(V^{\mathrm{sh}},F)\rightarrow H^0_{\mathrm{\acute{e}t}}((V^{\mathrm{sh}})_{\mathfrak{P}}, F)\rightarrow H^1_{\overline{\mathfrak{m}}_V}(V^{\mathrm{sh}}, F)\rightarrow H^1_{\mathrm{\acute{e}t}}(V^{\mathrm{sh}}, F)=0. \]

For the Cartier dual $\mathscr {D}_X(F)$ of $F$, let $\Lambda := \mathscr {D}_X(F)(V^{\mathrm {sh}})$ and $\Lambda ^{\vee }:= \operatorname {Hom}_{\textbf {Z}}(\Lambda,\textbf {Z})$. By Cartier duality,

\[ H^0_{\mathrm{\acute{e}t}}(V^{\mathrm{sh}},F)\cong F\left(V^{\mathrm{sh}}\right)\cong \mathscr{H}\! om_{\text{$V$-gr.}}(\mathscr{D}_X(F),\mathbb{G}_m)(V^{\mathrm{sh}})=\operatorname{Hom}_{\textbf{Z}}(\Lambda, ({V^{\mathrm{sh}}})^{\times})\cong\Lambda^{\vee}\otimes_{\textbf{Z}}({V^{\mathrm{sh}}})^{\times}, \]

and similarly,

\[ \text{$H^0_{\mathrm{\acute{e}t}}((V^{\mathrm{sh}})_{\mathfrak{P}},F)\cong \Lambda^{\vee}\otimes_{\textbf{Z}}(V^{\mathrm{sh}})_{\mathfrak{P}}^{\times}$.} \]

The value group $\Gamma _{V^{\mathrm {sh}}/\mathfrak {P}}$ of $V^{\mathrm {sh}}/\mathfrak {P}$, by Proposition A.2 (v), is isomorphic to $(V^{\mathrm {sh}})_{\mathfrak {P}}^{\times }/({V^{\mathrm {sh}}})^{\times }$. Therefore,

\[ H^1_{\overline{\mathfrak{m}}_V}(V^{\mathrm{sh}},F)=(\Lambda^{\vee}\otimes_{\textbf{Z}}(V^{\mathrm{sh}})_{\mathfrak{P}}^{\times})/(\Lambda^{\vee}\otimes_{\textbf{Z}}(V^{\mathrm{sh}})^{\times})\cong \Lambda^{\vee}\otimes_{\textbf{Z}}\Gamma_{V^{\mathrm{sh}}/\mathfrak{P}}. \]

Since $X$ is Henselian local and $\mathcal {H}^1_Z(X,F)$ is an abelian sheaf on $X$, by [SGA4_II, VIII, 8.6], we have

(2.6.3)\begin{equation} H^1_{\mathrm{\acute{e}t}}(X,\mathcal{H}^1_Z(X,F)) \cong H^1(\pi_1^{\mathrm{\acute{e}t}}(V),H^1_{\overline{\mathfrak{m}}_V}(V^{\mathrm{sh}},F)) \cong H^1(\pi_1^{\mathrm{\acute{e}t}}(V), \mathrm{Hom}_{\textbf{Z}}(\Lambda,\Gamma_{V^{\mathrm{sh}}/\mathfrak{P}})). \end{equation}

To see the action of $\pi _1^{\mathrm {\acute {e}t}}(V)$ on $\mathrm {Hom}_{\textbf {Z}}(\Lambda, \Gamma _{V^{\mathrm {sh}}/\mathfrak {P}})$, by Lemma 2.3, we first note that the $\pi _1^{\mathrm {\acute {e}t}}(V)$-action on $\Lambda$ factors through its quotient $\operatorname {Gal}(Y/X)$, where $Y$ is the minimal Galois cover of $X$ splitting $F$. In addition,

so $\pi _1^{\mathrm {\acute {e}t}}(V)$ acts trivially on $\Gamma _{V^{\mathrm {sh}}/\mathfrak {P}}\cong \operatorname {Frac}(V/\mathfrak {p})^{\times }/(V/\mathfrak {p})^{\times }$. Thus, the $\pi _1^{\mathrm {\acute {e}t}}(V)$-action on $\mathrm {Hom}_{\textbf {Z}}(\Lambda, \Gamma _{V/\mathfrak {p}})$ factors through $\operatorname {Gal}(Y/X)$. Since $\pi _1^{\mathrm {\acute {e}t}}(V)$ is a projective limit of finite groups $\operatorname {Gal}(X_{\alpha }/X)$, where $X_{\alpha }$ ranges over Galois covers of $X$, a limit argument [Reference SerreSer02, I, § 2.2, Corollary 1] reduces (2.6.3) to

(2.6.4)\begin{equation} \textstyle H^1_{\mathrm{\acute{e}t}}(X,\mathcal{H}^1_Z(X,F))\simeq \varinjlim_{\alpha} H^1\big(\operatorname{Gal}(X_{\alpha}/X), \mathrm{Hom}_{\textbf{Z}}(\Lambda,\Gamma_{V/\mathfrak{p}})^{\pi_1^{\mathrm{\acute{e}t}}(X_{\alpha})}\big). \end{equation}

We express $\Gamma _{V/\mathfrak {p}}$ as a direct limit of finite type $\textbf {Z}$-submodules $(\Gamma _i)_{i\in I}$. Since $\Lambda$ is $\textbf {Z}$-finitely presented,

(2.6.5)\begin{equation} \textstyle \varinjlim_{i\in I}\operatorname{Hom}_{\textbf{Z}}(\Lambda, \Gamma_{i})\overset{\sim}{\longrightarrow} \operatorname{Hom}_{\textbf{Z}}(\Lambda,\Gamma_{V/\mathfrak{p}}). \end{equation}

Combining the isomorphism (2.6.5) with a limit argument [Reference SerreSer02, I, § 2.2, Proposition 8], we reduce (2.6.4) to

\begin{align*} &\textstyle \varinjlim_{\alpha}H^1\big(\operatorname{Gal}(X_{\alpha}/X), \varinjlim_{i\in I}\operatorname{Hom}_{\textbf{Z}}(\Lambda,\Gamma_i)^{\pi_1^{\mathrm{\acute{e}t}}(X_{\alpha})}\big)\\ &\quad = \varinjlim_{\alpha}\varinjlim_{i\in I}H^1\big(\operatorname{Gal}(X_{\alpha}/X), \operatorname{Hom}_{\textbf{Z}}(\Lambda,\Gamma_i)^{\pi_1^{\mathrm{\acute{e}t}}(X_{\alpha})}\big). \end{align*}

It suffices to calculate for a large $\alpha _{0}$ such that $X_{\alpha _0}$ splits $F$. In this situation, $\pi _1^{\mathrm {\acute {e}t}}(X_{\alpha _0})$ acts trivially on $\mathrm {Hom}_{\textbf {Z}}(\Lambda,\Gamma _{i})$. Since $F$ is a flasque torus, its character group $\Lambda$ is a flasque $\operatorname {Gal}(X_{\alpha _0}/X)$-module. As aforementioned, $\operatorname {Gal}(X_{\alpha _0}/X)$ acts trivially on $\Gamma _{V/\mathfrak {p}}$, so the $\Gamma _{i}$ are finite-type $\textbf {Z}$-lattices with trivial $\operatorname {Gal}(X_{\alpha _0}/X)$-action. The example in § 2.5 implies $H^1(\operatorname {Gal}(X_{\alpha _0}/X), \operatorname {Hom}_{\textbf {Z}}(\Lambda,\Gamma _i)) =0$, which verifies that

\[ H^1_{\mathrm{\acute{e}t}}(X, \mathcal{H}^1_Z(X,F))=0. \]

Proof. As $V$ is a filtered direct union of valuation subrings of finite rank [Reference Bhatt and MathewBM21, 2.22], a limit argument [SGA4_II, VII, 5.7] reduces us to the case when $V$ has finite rank $n$. Note that for a quasitrivial $V$-torus $P$, we have $P\simeq \prod _{S^{\prime }_i}\mathrm {Res}_{S^{\prime }_i/\operatorname {Spec} V}\mathbb {G}_{m}$ for finite étale connected $V$-schemes $S_{i}^{\prime }$, so [Reference Gille and PoloGP, XIX, 8.4] gives an isomorphism $H^{1}_{\mathrm {\acute {e}t}}(V,P)\cong \prod _{S_{i}^{\prime }}H^{1}_{\mathrm {\acute {e}t}}(S_{i}^{\prime }, \mathbb {G}_m)$. The Grothendieck–Hilbert 90 [SGA2, VIII, 4.5] identifies $H^1_{\mathrm {\acute {e}t}}(S_i^{\prime },\mathbb {G}_{m})\simeq H^1_{\mathrm {Zar}}(S_i^{\prime }, \mathbb {G}_m)$, which are trivial by [Reference BourbakiBou98, II, § 5, no. 3, Proposition 5]. Thus, we have

\[ \text{$H^1_{\mathrm{\acute{e}t}}(V,P)=\{\ast\}$$\quad$ for every quasitrivial $V$-torus $P$.} \]

(i) First, we reduce to the case for flasque tori. By the short exact sequence [Reference Colliot-Thélène and SansucC-TS87, 1.3.2]

\[ \text{$1\rightarrow M\rightarrow F\rightarrow P\rightarrow 1$,} \]

where $F$ is flasque and $P$ is quasitrivial, we obtain the commutative diagram with exact rows

where $H^{1}_{\mathrm {fpqc}}(V,P)=H^{1}_{\mathrm {\acute {e}t}}(V,P)=\{\ast \}$. Hence, it suffices to prove the assertion for the flasque $F$.

Next, we induct on the rank $n$ of $V$. The case of $V=\operatorname {Frac} V$ is trivial, so when $n\geq 1$, for the prime $\mathfrak {p}$ of $V$ of height $n-1$, we assume that the assertion holds for $V_{\mathfrak {p}}$ (which has rank $n-1$). Let $X=\operatorname {Spec} V$ and $Z=\operatorname {Spec}(V/\mathfrak {m}_V)$. By [SGA4_II, V, 6.5], we have the long exact sequence:

(2.7.1)\begin{equation} \text{$\cdots \rightarrow H^2_{Z}(X, F)\rightarrow H^2_{\mathrm{fpqc}}(X,F)\rightarrow H^2_{\mathrm{fpqc}}(X-Z,F)\rightarrow H^3_{Z}(X,F)\rightarrow \cdots$.} \end{equation}

We conclude by the induction hypothesis and $H^2_{Z}(X,F)=0$ proved in Lemma 2.6.

(ii) We first reduce to the case when $M$ is a torus. The isotriviality of $M$ yields a short exact sequence

\[ 1\rightarrow T\rightarrow M\rightarrow \mu\rightarrow 1, \]

where $T$ is a $V$-torus and $\mu$ is a finite multiplicative type $V$-group. For the commutative diagram

with exact rows, the valuative criterion for properness of $\mu$ leads to $\mu (V)=\mu (\operatorname {Frac} V)$ and the injectivity of $H^1_{\mathrm {fpqc}}(V,\mu )\hookrightarrow H^1_{\mathrm {fpqc}}(\operatorname {Frac} V,\mu )$. Thus, a diagram chase reduces us to showing that

\[ \text{$H^1_{\mathrm{\acute{e}t}}(V,T)\rightarrow H^1_{\mathrm{\acute{e}t}}(\operatorname{Frac} V,T)$$\quad$ is injective.} \]

A flasque resolution of $T$ as (2.5.1) leads to the following commutative diagram with exact rows

where $H^{1}_{\mathrm {\acute {e}t}}(V,P)=\{\ast \}$. Since the map $H^2_{\mathrm {\acute {e}t}}(V,F)\hookrightarrow H^2_{\mathrm {\acute {e}t}}(\operatorname {Frac} V,F)$ is injective by part (i), the map

\[ \text{$H^1_{\mathrm{\acute{e}t}}(V,T)\hookrightarrow H^1_{\mathrm{\acute{e}t} }(\operatorname{Frac} V,T)$$\quad$ is injective.}\]

Proof. The injectivity follows from Proposition 2.7 (ii). A limit argument reduces us to the case when $V$ has finite rank, then we iteratively use Lemma 2.6 with the exact sequence (cf. 2.7.1)

\[ H^1_{\mathrm{\acute{e}t}}(V,F)\rightarrow H^1_{\mathrm{\acute{e}t}}(\operatorname{Spec} V\backslash \{\mathfrak{m}_V\},F)\rightarrow H^2_{\mathfrak{m}_V}(V,F)=0, \]

to reduce the rank of valuation rings by removing closed points, so the surjectivity follows.

Proof. For parts (i) and (ii), note that by Proposition A.4, $K$ is Hausdorff so $K^{\times }\subset K$ is open. It is clear that the inversion on $K^{\times }$ is continuous for the subspace topology. It suffices to use Lemma 3.2 to topologize $X(K)$; moreover, if $X$ is separated over $K$, then $X(K)$ is Hausdorff for the valuation topology. Assertion (iii) follows from Lemma 3.3 and Proposition A.4 that the ball $V\subset K$ is closed and open.

For assertion (iv), we recall (Appendix A.11) that the topology on $K$ is indeed $a$-adic for an $a\in V$ such that $\sqrt {(a)}$ is of height one. Thus $\widehat {K}$ is the $a$-adic completion ${\widehat{K^a}}$. We then apply [Reference Bouthier and ČesnavičiusBC22, 2.2.10 (iii)] and check the following conditions.

• – Let the topological ring $B$ be $K$ with $a$-adic topology. Then $\widehat {B}={\widehat{K^a}}$ and $({\widehat{K^a}} )^{\times }\subset {\widehat{K^a}}$ is an open subring with continuous inversion.

• – Let the nonunital open subring $B^{\prime }$ be the ideal $(a)$ of the valuation ring $V$. The induced topology on $(a)$ has an open neighborhood base of zero consisting of ideals $(a^n)_{n\geq 1}\subset (a)$ (Proposition A.10(i)).

• – The nonunital ring $(a)$ is Henselian in the sense of Gabber [Reference Bouthier and ČesnavičiusBC22, 2.2.1], that is, every polynomial $f(T)=T^N(T-1)+a_NT^N+\cdots +a_1T+a_0$ where $a_i\in (a)$ and $N\geq 1$ has a (unique) root in $1+(a)$. Because $V$ is Henselian, by [Sta18, 0DYD], the pair $(V,(a))$ is also Henselian. Hence, Gabber's criterion shows that $(a)$ is Henselian, so the conditions in [Reference Bouthier and ČesnavičiusBC22, 2.2.10 (iii)] are satisfied.

Proof. For part (i), see [Reference Gabber, Gille and Moret-BaillyGGM-B14, 3.1.4] and note that the ‘topological Henselianity’ there yields the desired openness by [Reference Gabber, Gille and Moret-BaillyGGM-B14, 3.1.2]. For part (ii), see [Reference ČesnavičiusCes15, 4.3 (a) and 2.8 (2)], where $R$ is our $F$.

Proof. Since $\overline {S}\cdot H\subset \overline {S\cdot H}$, it suffices to see that $S\cdot H=\overline {S\cdot H}$. The subset $G\backslash (S\cdot H)$ is a union of $g_iH$ for some $g_i\in G$, hence is open. In particular, $S\cdot H$ is closed, so the assertion follows.

Proof. An isomorphism [SGA2, XII, 4.7 c] of schemes $\underline {\mathrm {Tor}}(G)\simeq \underline {\mathrm {Tor}}(G/Z)$ reduces us to the semisimple adjoint case, where the maximal tori of $G$ are exactly the subgroups of type (C) [SGA2, XIV, 3.18]. These subgroups are bijectively assigned by $D\mapsto \mathrm {Lie}(D)$ to the Cartan subalgebras of $\mathfrak {g}:= \mathrm {Lie}(G)$, see [SGA2, XIV, 3.9]. It suffices to lift a Cartan subalgebra $\mathfrak {c}_{\kappa }\subset \mathfrak {g}_{\kappa }$ to that of $\mathfrak {g}$. Since $\sharp \kappa \geq \dim (G/Z)=\dim (G)$, by [Reference BarnesBar67, Theorem 1], $\mathfrak {c}_{\kappa }$ is of the form $\mathrm {Nil}(a_{\kappa }):= \bigcup _{n}\ker (\mathrm {ad}(a_{\kappa }^n))$ for some $a_{\kappa }\in \mathfrak {c}_{\kappa }$. Hence, [SGA2, XIII, 5.7] implies that each $a_{\kappa }\in \mathfrak {c}_{\kappa }$ is a regular element of $\mathfrak {g}_{\kappa }$. We take a section $a$ of $\mathfrak {g}$ passing through $a_{\kappa }$ and claim that $\mathcal {V}:= \{s\in \operatorname {Spec} R \,|\, a_{s}\in \mathfrak {g}_{s}\ \text {is regular}\}$ is an open subset of $\operatorname {Spec} R$. We may assume that $R$ is reduced. Since the nilpotent rank of $\mathfrak {g}$ is locally constant, the Killing polynomial of $\mathfrak {g}$ at every $s\in \operatorname {Spec} R$ is uniformly of the form $P_{\mathfrak {g}_s}(t)=t^r(t^{n-r}+(c_1)_st^{n-r-1}+\cdots +(c_{n-r})_{s})$ such that $(c_{n-r})_s$ is nonzero. Thus, the regular locus in $\mathfrak {g}$ is the principle open subset $\{c_{n-r}\neq 0\}\subset \textbf {W}(\mathfrak {g})$. The morphism $\textbf {W}(\mathfrak {g})\rightarrow \operatorname {Spec} R$ is flat, so $\mathcal {V}\neq \emptyset$ is open, forcing that $\mathcal {V}=\operatorname {Spec} R$. In particular, the regular elements $a_{\kappa }\in \mathfrak {c}_{\kappa }$ lifts to a quasi-regular section $a\in \mathfrak {g}$, which by [Reference Gille and PoloGP, XIV, 3.7], is regular. By definition of regular sections, there is a Cartan subalgebra of $\mathfrak {g}$ containing $a$ and is the desired lifting of $\mathfrak {c}_{\kappa }$.

Proof. Taking $a$-adic completion of $V{[\frac {1}{a}]}$ yields the surjection map

\[ \textstyle \mathcal{A}:= \varinjlim_{m\geq 0}\mathrm{Cauchy}^{\geq m}(V{[\frac{1}{a}]})\twoheadrightarrow {\widehat{K^a}}, \]

whose kernel is denoted by $I$. For any sequence $(b_N)_N\in I$, its tail lies in $\mathrm {Im}(a^mV\rightarrow V{[\frac {1}{a}]})$ for all $m>0$, so the tail of $(1+b_N)_N$ consists of units in $V$ that lie in $\mathrm {Im}((1+a^mV)\rightarrow V{[\frac {1}{a}]})$. Since $V{[\frac {1}{a}]}$ is local, the tail of $(1+b_N)_N$ is termwise invertible in $V{[\frac {1}{a}]}$ and the inverses form a Cauchy sequence. Since $I\subset \mathcal {A}$ is an ideal such that $\mathcal {A}/I$ is a field and $1+I$ is invertible, $\mathcal {A}$ is a local ring with residue field ${\widehat{K^a}}$.

Proof. As shown in Lemma 3.13, the ring $\varinjlim _{m\geq 0}\mathrm {Cauchy}^{\geq m}(V{[\frac {1}{a}]})$ is local with residue field ${\widehat{K^a}}$. Since $\underline {\mathrm {Tor}}(G)$ is finitely presented and affine over $V{[\frac {1}{a}]}$, the lifting Lemma 3.10 leads to a surjection as follows:

\[ \textstyle\varinjlim_{m\geq 0}\big(\underline{\mathrm{Tor}}(G)(\mathrm{Cauchy}^{\geq m}(V{[\frac{1}{a}]}))\big)\simeq \underline{\mathrm{Tor}}(G)\big(\varinjlim_{m\geq 0}(\mathrm{Cauchy}^{\geq m}(V{[\frac{1}{a}]}))\big)\twoheadrightarrow \underline{\mathrm{Tor}}(G)({\widehat{K^a}}). \]

Due to this surjection, all elements in $\underline {\mathrm {Tor}}(G)({\widehat{K^a}} )$ are limits of Cauchy sequences in $\underline {\mathrm {Tor}}(G)(V[\frac {1}{a}])$, hence the image of the map $\underline {\mathrm {Tor}}(G)(V{[\frac {1}{a}]})\rightarrow \underline {\mathrm {Tor}}(G)({\widehat{K^a}} )$ is $a$-adically dense in $\underline {\mathrm {Tor}}(G)({{\widehat{K^a}} })$.

Proof. (i) The existence of a minimal Galois cover $R/V[\frac {1}{a}]$ splitting $T$ follows from Lemma 2.3. Since $R$ is a finite flat $V{[\frac {1}{a}]}$-module, it is free and we have ${\widehat{R^a}} \simeq R\otimes _{V{[\frac {1}{a}]}}{\widehat{K^a}} \simeq \prod _{i=1}^rL_i$, where $L_i$ are $a$-adically complete fields. By Proposition 2.4 and § 2.1 we conclude.

(ii) First, we prove that $U$ is $a$-adically open. For the norm map $\mathrm {Res}_{L_0/\widehat {K}^a}(T_{L_0})\rightarrow T_{\widehat {K}^a}$, its kernel $\mathcal {T}$ is a torus: after some base change, $T_{\widehat {K}^a}$ splits as $\mathbb {G}_m^k$, so the associated $\textbf {Z}$-module of the corresponding base change of $\mathcal {T}$ is the following $\textbf {Z}$-lattice with a trivial Galois action:

\[ \textstyle \text{$\mathrm{Coker}\big(\textbf{Z}^k\rightarrow \textbf{Z}[\operatorname{Gal}(L_0/{\widehat{K^a}})]^k, (n_i)\mapsto (n_i\cdot \operatorname{id})\big)\simeq \textbf{Z}[\operatorname{Gal}(L_0/{\widehat{K^a}})-\{\operatorname{id}\}]^k$}. \]

Thus, by [SGA2, IX, 2.1 e], as a torus, the kernel $\mathcal {T}$ is ${\widehat{K^a}}$-smooth. By Lemma 3.6(ii), the map

\[ \textstyle\text{$N_{L_0/\widehat{K}^a}\colon T(L_0)\rightarrow T({\widehat{K^a}})$,$\quad$ i.e. $\quad$ $(\mathrm{Res}_{L_0/\widehat{K}^a}T_{L_0})({\widehat{K^a}})\rightarrow \big((\mathrm{Res}_{L_0/\widehat{K}^a}T_{L_0})/\mathcal{T}\big)({\widehat{K^a}})$} \]

is $a$-adically open so the image $U= N_{L_0/\widehat {K}^a}(T(L_0))\subset T({\widehat{K^a}} )$ is $a$-adically open.

Next, we prove that $U\subset \overline {T(V{[\frac {1}{a}]})}$. The isomorphism ${\widehat{R^a}} \cong \prod _{i=1}^{r}L_i$ obtained in part (i) implies that the image of $R^{\times }\rightarrow \prod _{i=1}^{r}L_i^{\times }$ is $a$-adically dense. As $T_{R}$ is split, the image of the composite

\[ \textstyle T(R)\rightarrow \prod_{j=1}^{r}T(L_j)\overset{\mathrm{pr}_1}{\rightarrow} T(L_1)\cong T(L_0) \]

is $a$-adically dense. Composing this with $N_{L_0/\widehat {K}^a}$, we see that $T(R)$ has dense image in $U=N_{L_0/\widehat {K}^a}(T(L_0))$. The composite $T(R)\rightarrow T(L_0) \rightarrow T({\widehat{K^a}} )$ factors through the norm map $N_{R/V{[\frac {1}{a}]}}\colon T(R)\rightarrow T(V{[\frac {1}{a}]})$, so the image of $T(V{[\frac {1}{a}]})$ is dense in $U$, that is, $\textstyle U\subset \overline {T(V{[\frac {1}{a}]})}$.

Proof. The $a$-adically open aspect of the assertion follows from Lemma 3.16(ii) because the arguments there, by base change, apply to all ${\widehat{K^a}}$-tori as well. The proof for $U\subset \overline {G(V{[\frac {1}{a}]})}$ proceeds as follows.

(i) Since ${\widehat{K^a}}$ is Henselian, by a criterion for openness Lemma 3.6(ii), the following map from (3.9.1) is a-adically open:

\[ \textstyle \text{$\phi\colon G({\widehat{K^a}})\rightarrow \underline{\mathrm{Tor}}(G)({\widehat{K^a}}),$} $\quad$ {g\mapsto gTg^{-1}$.} \]

Consequently, $\phi$ sends every $a$-adically open neighborhood $W$ of $\operatorname {id}\in G({\widehat{K^a}} )$ to an $a$-adically open neighborhood of $T$. The density lemma (Lemma 3.15) of $\underline {\mathrm {Tor}}(G)(V{[\frac {1}{a}]})$ in $\underline {\mathrm {Tor}}(G)({\widehat{K^a}} )$ implies that

\[ \textstyle \phi(W)\cap \mathrm{Im}(\underline{\mathrm{Tor}}(G)(V{[\frac{1}{a}]})\rightarrow \underline{\mathrm{Tor}}(G)({\widehat{K^a}}))\neq \emptyset. \]

Hence, there are a torus $T^{\prime }\in \underline {\mathrm {Tor}}(G)(V{[\frac {1}{a}]})$ and a $g\in W$ such that $gTg^{-1}=T^{\prime }_{\widehat {K}^a}\in \phi (W)$.

(ii) For any $u\in U$, the map $\sigma _u\colon G({\widehat{K^a}} )\rightarrow G({\widehat{K^a}} )$ defined by $g\mapsto g^{-1}ug$ is continuous. Let $W:= \sigma _u^{-1}(U)$. By the construction in part (i), there are a $w\in W$ and a torus $T^{\prime }\in \underline {\mathrm {Tor}}(G)(V{[\frac {1}{a}]})$ such that $w T w^{-1}=T^{\prime }_{\widehat {K}^a}$. Note that $u\in w U w^{-1}=\gamma N_{L_0/\widehat {K}^a}(T(L_0))\gamma ^{-1}$, which by transport of structure, is equal to $N_{L_0/\widehat {K}^a}(T^{\prime }_{\widehat {K}^a}(L_0))$. By Lemma 3.16, the last term is contained in $\overline {\operatorname {Im}(T^{\prime }(V{[\frac {1}{a}]})\rightarrow T^{\prime }({\widehat{K^a}} ))}$, so is contained in $\overline {G(V{[\frac {1}{a}]})}$.

Proof. By the argument (i) for Lemma 3.17, $\phi (W)\cap \underline {\mathrm {Tor}}(G)({\widehat{V^a}} )$ is an $a$-adically open neighborhood of $T_{\widehat {K}^a}\in \underline {\mathrm {Tor}}(G)({\widehat{K^a}} )$. Since $V\simeq V{[\frac {1}{a}]}\times _{\widehat {K}^a}{\widehat{V^a}}$ (Proposition A.10(vii)) and $\underline {\mathrm {Tor}}(G)$ is affine, we get

\[ \textstyle \underline{\mathrm{Tor}}(G)(V)\overset{\sim}{\longrightarrow} \underline{\mathrm{Tor}}(G)(V{[\frac{1}{a}]})\times_{\underline{\mathrm{Tor}}(G)(\widehat{K}^a)}\underline{\mathrm{Tor}}(G)({\widehat{V^a}}). \]

By Lemma 3.15, the image of $\underline {\mathrm {Tor}}(G)(V{[\frac {1}{a}]})\rightarrow \underline {\mathrm {Tor}}(G)({\widehat{K^a}} )$ is $a$-adically dense, so we have

\[ \textstyle \phi(W)\cap \underline{\mathrm{Tor}}(G)({\widehat{V^a}})\cap \mathrm{Im}(\underline{\mathrm{Tor}}(G)(V{[\frac{1}{a}]}))\neq \emptyset, \]

giving a maximal torus $T_0\in \underline {\mathrm {Tor}}(G)(V)$ and $g\in W$ such that $(T_0)_{\widehat {K}^a}=gT_{\widehat {K}^a}g^{-1}\in \phi (W)$.

Proof. In the proof, all open subsets without the word ‘Zariski’ refer to $a$-adically open subsets.

(i) Fix a maximal torus $T\subset G_{\widehat {K}^a}$. We denote by $\mathfrak {g}$ the Lie algebra of $G_{\widehat {K}^a}$ and by $\mathfrak {h}$ the Lie algebra of $T$. For each $g\in G_{\widehat {K}^a}$ and the subspace $\mathfrak {g}^{\mathrm {ad}(g)}\subset \mathfrak {g}$ fixed by $\mathrm {ad}(g)$, by [SGA2, XIII, 2.6 b], $\dim \mathfrak {g}^{\mathrm {ad}(g)}\geq \dim T$. Let regular locus $G^{\mathrm {reg}}\subset G_{\widehat {K}^a}$ be the subscheme of all $g\in G_{\widehat {K}^a}$ that satisfy $\dim (\mathfrak {g}^{\mathrm {ad}(g)})=\dim T$. By [SGA2, XIII, 2.7], $G^{\mathrm {reg}}$ is Zariski open. By the equation

\[ \dim(\mathfrak{g}^{\mathrm{ad}(g)})=\dim(\mathfrak{h}^{\mathrm{ad}(g)})+\dim((\mathfrak{g}/\mathfrak{h})^{\mathrm{ad}(g)}), \]

an element $t\in T$ is regular in $G_{\widehat {K}^a}$ (namely, $t\in T^{\mathrm {reg}}:= G^{\mathrm {reg}}\cap T$) if and only if $(\mathfrak {g}/\mathfrak {h})^{\mathrm {ad}(t)}=0$.

(ii) Recall $L_0$ and the open subgroup $U\subset T({\widehat{K^a}} )$ in Lemma 3.17, we claim that ${U\cap T^{\mathrm {reg}}({\widehat{K^a}} )\neq \emptyset}$. Consider the norm map $\mathrm {Nm}\colon \operatorname {Res}_{L_0/\widehat {K}^a}(T_{L_0})\rightarrow T$. Note that $T_{L_0}\simeq \mathbb {G}_{m,L_0}^k$ is isomorphic to a Zariski-dense open subset of $\mathbb {A}^k_{L_0}$, so $\operatorname {Res}_{L_0/\widehat {K}^a}(T_{L_0})$ is also a Zariski-dense open subset of $\mathbb {A}^{mk}_{\widehat {K}^a}$ for $m:= [L_0:{\widehat{K^a}} ]$. The field ${\widehat{K^a}}$ is infinite, so we have $({\operatorname {Res}_{L_0/\widehat {K}^a}(T_{L_0})})({\widehat{K^a}} )\cap \mathrm {Nm}^{-1}(T^{\mathrm {reg}})({\widehat{K^a}} )\neq \emptyset$. Applying $\mathrm {Nm}$ to this nonempty intersection, we proved our claim that $U\cap T^{\mathrm {reg}}({\widehat{K^a}} )\neq \emptyset$.

(iii) For a fixed $t_{0}\in U\cap T^{\mathrm {reg}}({\widehat{K^a}} )$, by (i), we have $(\mathfrak {g}/\mathfrak {h})^{\mathrm {ad}(t_0)}=0$. So [SGA2, XIII, 2.2] implies that

\[ f\colon G_{\widehat{K}^a}\times T\rightarrow G_{\widehat{K}^a}, \quad (g,t)\mapsto gtg^{-1} \]

is smooth at $(\operatorname {id}, t_0)$. Thus, there is a Zariski-open neighborhood $W$ of $(\operatorname {id},t_0)$ such that $f|_W\colon W\rightarrow G_{\widehat {K}^a}$ is smooth. By Lemma 3.6 (i), $W({\widehat{K^a}} )\rightarrow G({\widehat{K^a}} )$ is open. Thus, the open neighborhood $W^{\prime }:= W({\widehat{K^a}} )\cap (G({\widehat{K^a}} )\times U)$ of $(\operatorname {id}, t_0)$ has open image under $f_{\mathrm {top}}$. The $G_{\widehat {K}^a}$-translations $\tau _{h}\colon (g,t)\mapsto (hg,t)$ for $h\in G_{\widehat {K}^a}$ induce automorphisms of $G_{\widehat {K}^a}\times T$, so $f$ is also smooth at $(h,t_0)$. Similar to the above, all $G({\widehat{K^a}} )$-translations of $W^{\prime }$ have open images under $f_{\mathrm {top}}$. Thus, there is an open subset $U_0\subset U$ such that $E:= f(G({\widehat{K^a}} )\times U_0)$ is open. Let $N$ be the subgroup of $G({\widehat{K^a}} )$ generated by $E$. The openness of $E$ implies that $N$ is an open subgroup of $G({\widehat{K^a}} )$.

(iv) As $E$ is stable under $G({\widehat{K^a}} )$-conjugation, $N$ is normal in $G({\widehat{K^a}} )$. For each $g\in G({\widehat{K^a}} )$, we let $T^{g}:= gTg^{-1}$. Then $U^{g}:= N_{L_0/\widehat {K}^a}(T^g(L_0))$ satisfies $U^{g}=gUg^{-1}$. Lemma 3.17 applies to $T^{g}$ and gives $U^{g}\subset \overline {G(V{[\frac {1}{a}]})}$. Thus, $E\subset \bigcup _{g\in G(\widehat {K}^a)} U^g \subset \overline {G(V{[\frac {1}{a}]})}$. Since $E$ generates $N$, we obtain

\[ \textstyle N\subset \overline{G(V{[\frac{1}{a}]})}. \]

Proof. The image of $G(V{[\frac {1}{a}]})\rightarrow G({\widehat{K^a}} )$ is a subgroup of $G({\widehat{K^a}} )$, hence so is its closure $\overline {G(V{[\frac {1}{a}]})}$. Since $\overline {G(V{[\frac {1}{a}]})}$ contains the open subset $N$, it is an open subgroup of $G({\widehat{K^a}} )$. Recall Example 3.14 that the subgroup $G({\widehat{V^a}} )\subset G({\widehat{K^a}} )$ is open and closed. By Lemma 3.7, the desired equation follows.