Notation

In this paper, we use the following notation. Let $K$ be a nonarchimedean local field. Let ${\mathcal{O}}_{K}$ denote the ring of integers of $K$ , and let $k$ denote the residue field of $K$ . Let $p$ be the characteristic of $k$ . We fix a uniformizer $\unicode[STIX]{x1D71B}$ of $K$ . Let $q=|k|$ . We fix an algebraic closure $K^{\text{ac}}$ of $K$ . For any finite extension $F$ of $K$ in $K^{\text{ac}}$ , let $G_{F}$ denote the absolute Galois group of $F$ , let $W_{F}$ denote the Weil group of $F$ , and let $I_{F}$ denote the inertia subgroup of $W_{F}$ . The completion of $K^{\text{ac}}$ is denoted by $\mathbf{C}$ . Let ${\mathcal{O}}_{\mathbf{C}}$ be the ring of integers of $\mathbf{C}$ , and let $k^{\text{ac}}$ be the residue field of $\mathbf{C}$ . For an element $a\in {\mathcal{O}}_{\mathbf{C}}$ , we write $\bar{a}$ for the image of $a$ by the reduction map ${\mathcal{O}}_{\mathbf{C}}\rightarrow k^{\text{ac}}$ . Let $v(\cdot )$ denote the valuation of $\mathbf{C}$ such that $v(\unicode[STIX]{x1D71B})=1$ . Let $K^{\text{ur}}$ denote the maximal unramified extension of $K$ in $K^{\text{ac}}$ . The completion of $K^{\text{ur}}$ is denoted by $\widehat{K}^{\text{ur}}$ . For $a,b\in \mathbf{C}$ and a rational number $\unicode[STIX]{x1D6FC}\in \mathbb{Q}_{{\geqslant}0}$ , we write $a\equiv b\;(\text{mod}~\unicode[STIX]{x1D6FC})$ if we have $v(a-b)\geqslant \unicode[STIX]{x1D6FC}$ , and $a\equiv b\;(\text{mod}~\unicode[STIX]{x1D6FC}+)$ if we have $v(a-b)>\unicode[STIX]{x1D6FC}$ . For a curve $X$ over $k^{\text{ac}}$ , we denote by $X^{c}$ the smooth compactification of the normalization of $X$ , and the genus of $X$ means the genus of $X^{\text{c}}$ . For an affinoid $\mathbf{X}$ , we write $\overline{\mathbf{X}}$ for its reduction. The category of sets is denoted by $\mathbf{Set}$ . For a representation $\unicode[STIX]{x1D70F}$ of a group, the dual representation of $\unicode[STIX]{x1D70F}$ is denoted by $\unicode[STIX]{x1D70F}^{\ast }$ . We take rational powers of $\unicode[STIX]{x1D71B}$ compatibly as needed.

1.1 The universal deformation

Let $\unicode[STIX]{x1D6F4}$ denote a formal ${\mathcal{O}}_{K}$ -module of dimension $1$ and height $2$ over $k^{\text{ac}}$ , which is unique up to isomorphism. Let $n$ be a natural number. We define $K_{1}(\mathfrak{p}^{n})$ as in the introduction. In the following, we define the connected Lubin–Tate curve $\mathbf{X}_{1}(\mathfrak{p}^{n})$ with level $K_{1}(\mathfrak{p}^{n})$ .

Let ${\mathcal{C}}$ be the category of Noetherian complete local ${\mathcal{O}}_{\widehat{K}^{\text{ur}}}$ -algebras with residue field $k^{\text{ac}}$ . For $A\in {\mathcal{C}}$ , a formal ${\mathcal{O}}_{K}$ -module ${\mathcal{F}}=\text{Spf}\,A[[X]]$ over $A$ and an $A$ -valued point $P$ of ${\mathcal{F}}$ , the corresponding element of the maximal ideal of $A$ is denoted by $x(P)$ . We consider the functor

$$\begin{eqnarray}{\mathcal{A}}_{1}(\mathfrak{p}^{n}):{\mathcal{C}}\rightarrow \mathbf{Set};\qquad A\mapsto [({\mathcal{F}},\unicode[STIX]{x1D704},P)],\end{eqnarray}$$

where ${\mathcal{F}}$ is a formal ${\mathcal{O}}_{K}$ -module over $A$ with an isomorphism $\unicode[STIX]{x1D704}:\unicode[STIX]{x1D6F4}\simeq {\mathcal{F}}\otimes _{A}k^{\text{ac}}$ and $P$ is a $\unicode[STIX]{x1D71B}^{n}$ -torsion point of ${\mathcal{F}}$ such that

$$\begin{eqnarray}\mathop{\prod }_{a\in {\mathcal{O}}_{K}/\unicode[STIX]{x1D71B}^{n}{\mathcal{O}}_{K}}(X-x([a]_{{\mathcal{F}}}(P)))\biggm\vert[\unicode[STIX]{x1D71B}^{n}]_{{\mathcal{F}}}(X)\end{eqnarray}$$

in $A[[X]]$ . This functor is represented by a regular local ring ${\mathcal{R}}_{1}(\mathfrak{p}^{n})$ by [Reference Drinfel’dDr, Section 4.B) Lemma]. We write $\mathfrak{X}_{1}(\mathfrak{p}^{n})$ for $\text{Spf}\,{\mathcal{R}}_{1}(\mathfrak{p}^{n})$ . Its generic fiber is denoted by $\mathbf{X}_{1}(\mathfrak{p}^{n})$ , which we call the connected Lubin–Tate curve with level $K_{1}(\mathfrak{p}^{n})$ . The space $\mathbf{X}_{1}(\mathfrak{p}^{n})$ is a rigid analytic curve over $\widehat{K}^{\text{ur}}$ . We can define the Lubin–Tate curve $\text{LT}_{1}(\mathfrak{p}^{n})$ with level $n$ by changing ${\mathcal{C}}$ to be the category of ${\mathcal{O}}_{\widehat{K}^{\text{ur}}}$ -algebras where $\unicode[STIX]{x1D71B}$ is nilpotent, and $\unicode[STIX]{x1D704}$ to be a quasi-isogeny $\unicode[STIX]{x1D6F4}\otimes _{k^{\text{ac}}}A/\unicode[STIX]{x1D71B}A\rightarrow {\mathcal{F}}\otimes _{A}A/\unicode[STIX]{x1D71B}A$ . We consider $\text{LT}_{1}(\mathfrak{p}^{n})$ as a rigid analytic curve over $\widehat{K}^{\text{ur}}$ .

The ring ${\mathcal{R}}_{1}(1)$ is isomorphic to the ring of formal power series ${\mathcal{O}}_{\widehat{K}^{\text{ur}}}[[u]]$ . We simply write ${\mathcal{B}}(1)$ for $\text{Spf}\,{\mathcal{O}}_{\widehat{K}^{\text{ur}}}[[u]]$ . Let $B(1)$ denote an open unit ball such that $B(1)(\mathbf{C})=\{u\in \mathbf{C}\mid v(u)>0\}$ . The generic fiber of ${\mathcal{B}}(1)$ is equal to $B(1)$ . Then, the space $\mathbf{X}_{1}(1)$ is identified with $B(1)$ . Let ${\mathcal{F}}^{\text{univ}}$ denote the universal formal ${\mathcal{O}}_{K}$ -module over $\mathfrak{X}_{1}(1)$ .

In this subsection, we choose a parametrization of $\mathfrak{X}_{1}(1)\simeq {\mathcal{B}}(1)$ such that the universal formal ${\mathcal{O}}_{K}$ -module has a simple form. Let ${\mathcal{F}}$ be a formal ${\mathcal{O}}_{K}$ -module of dimension $1$ over a flat ${\mathcal{O}}_{K}$ -algebra $R$ . For a nontrivial invariant differential $\unicode[STIX]{x1D714}$ on ${\mathcal{F}}$ , a logarithm of ${\mathcal{F}}$ means a unique isomorphism $F:{\mathcal{F}}\stackrel{{\sim}}{\rightarrow }\mathbb{G}_{a}$ over $R\otimes K$ with $dF=\unicode[STIX]{x1D714}$ (cf. [Reference Gross and HopkinsGH, 3]). In the following, we always take an invariant differential $\unicode[STIX]{x1D714}$ on ${\mathcal{F}}$ so that a logarithm $F$ has the following form:

$$\begin{eqnarray}F(X)=X+\mathop{\sum }_{i\geqslant 1}f_{i}X^{q^{i}}\quad \text{with}~f_{i}\in R\otimes K.\end{eqnarray}$$

Let $F(X)=\sum _{i\geqslant 0}f_{i}X^{q^{i}}\in K[[u,X]]$ be the universal logarithm over ${\mathcal{O}}_{K}[[u]]$ . By [Reference Gross and HopkinsGH, (5.5), (12.3), Proposition 12.10], the coefficients $\{f_{i}\}_{i\geqslant 0}$ satisfy $f_{0}=1$ and $\unicode[STIX]{x1D71B}f_{i}=\sum _{0\leqslant j\leqslant i-1}f_{j}v_{i-j}^{q^{j}}$ for $i\geqslant 1$ , where $v_{1}=u$ , $v_{2}=1$ and $v_{i}=0$ for $i\geqslant 3$ . Hence, we have the following:

(1.1) $$\begin{eqnarray}\begin{array}{@{}c@{}}f_{0}=1,\qquad f_{1}={\displaystyle \frac{u}{\unicode[STIX]{x1D71B}}},\qquad f_{2}={\displaystyle \frac{1}{\unicode[STIX]{x1D71B}}}\biggl(1+{\displaystyle \frac{u^{q+1}}{\unicode[STIX]{x1D71B}}}\biggr),\\ f_{3}={\displaystyle \frac{1}{\unicode[STIX]{x1D71B}^{2}}}\biggl(u+u^{q^{2}}+{\displaystyle \frac{u^{q^{2}+q+1}}{\unicode[STIX]{x1D71B}}}\biggr),\qquad \cdots \,.\end{array}\end{eqnarray}$$

By [Reference Gross and HopkinsGH, Proposition 5.7] or [Reference HazewinkelHa, 21.5], if we set

(1.2) $$\begin{eqnarray}{\mathcal{F}}^{\text{univ}}(X,Y)=F^{-1}(F(X)+F(Y)),\quad [a]_{{\mathcal{F}}^{\text{univ}}}(X)=F^{-1}(aF(X))\end{eqnarray}$$

for $a\in {\mathcal{O}}_{K}$ , it is known that these power series have coefficients in ${\mathcal{O}}_{K}[[u]]$ and define the universal formal ${\mathcal{O}}_{K}$ -module ${\mathcal{F}}^{\text{univ}}$ over ${\mathcal{O}}_{\widehat{K}^{\text{ur}}}[[u]]$ of dimension $1$ and height $2$ with logarithm $F(X)$ . We have the following approximation formula for $[\unicode[STIX]{x1D71B}]_{\text{u}}(X)$ .

Lemma 1.1. We have the following congruence:

$$\begin{eqnarray}\displaystyle [\unicode[STIX]{x1D71B}]_{{\mathcal{F}}^{\text{univ}}}(X) & \equiv & \displaystyle \unicode[STIX]{x1D71B}X+uX^{q}+X^{q^{2}}-{\displaystyle \frac{u}{\unicode[STIX]{x1D71B}}}\{(uX^{q}+X^{q^{2}})^{q}-u^{q}X^{q^{2}}-X^{q^{3}}\}\nonumber\\ \displaystyle & & \displaystyle \quad \hspace{0.2em}\bmod \hspace{0.2em}(\unicode[STIX]{x1D71B}^{2}X^{q},u\unicode[STIX]{x1D71B}X^{q},\unicode[STIX]{x1D71B}X^{q^{2}},X^{q^{3}+1}).\nonumber\end{eqnarray}$$

In the following, ${\mathcal{F}}^{\text{univ}}$ means the universal formal ${\mathcal{O}}_{K}$ -module with the identification $\mathfrak{X}_{1}(1)\simeq {\mathcal{B}}(1)$ given by (1.2), and we simply write $[a]_{\text{u}}$ for $[a]_{{\mathcal{F}}^{\text{univ}}}$ . The reduction of (1.2) gives a simple model of $\unicode[STIX]{x1D6F4}$ such that

(1.3) $$\begin{eqnarray}X+_{\unicode[STIX]{x1D6F4}}Y=X+Y,\qquad [\unicode[STIX]{x1D701}]_{\unicode[STIX]{x1D6F4}}(X)=\bar{\unicode[STIX]{x1D701}}X\quad \text{for }\unicode[STIX]{x1D701}\in \unicode[STIX]{x1D707}_{q-1}({\mathcal{O}}_{K}),\quad [\unicode[STIX]{x1D71B}]_{\unicode[STIX]{x1D6F4}}(X)=X^{q^{2}}\!.\end{eqnarray}$$

We put

$$\begin{eqnarray}\mathfrak{A}_{n}={\mathcal{O}}_{\widehat{K}^{\text{ur}}}[[u,X_{n}]]/([\unicode[STIX]{x1D71B}^{n}]_{\text{u}}(X_{n})/[\unicode[STIX]{x1D71B}^{n-1}]_{\text{u}}(X_{n})).\end{eqnarray}$$

Then, there is a natural identification

(1.4) $$\begin{eqnarray}\mathfrak{X}_{1}(\mathfrak{p}^{n})\simeq \text{Spf}\,\mathfrak{A}_{n}\end{eqnarray}$$

that is compatible with the identification $\mathfrak{X}_{1}(1)\simeq {\mathcal{B}}(1)$ . The Lubin–Tate curve $\mathbf{X}_{1}(\mathfrak{p}^{n})$ is identified with the generic fiber of the right-hand side of (1.4). We set $X_{i}=[\unicode[STIX]{x1D71B}^{n-i}]_{\text{u}}(X_{n})$ for $1\leqslant i\leqslant n-1$ . We write $\mathfrak{X}(1)$ for $\mathfrak{X}_{1}(1)$ .

1.2 Action of a division algebra on $\mathfrak{X}_{1}(\mathfrak{p}^{n})$

Let $D$ be the central division algebra over $K$ of invariant $1/2$ . We write ${\mathcal{O}}_{D}$ for the ring of integers of $D$ . In this subsection, we recall the left action of ${\mathcal{O}}_{D}^{\times }$ on the space $\mathfrak{X}_{1}(\mathfrak{p}^{n})$ .

Let $K_{2}$ be the unramified quadratic extension of $K$ . Let $k_{2}$ be the residue field of $K_{2}$ , and let $\unicode[STIX]{x1D70E}\in \text{Gal}(K_{2}/K)$ be the nontrivial element. The ring ${\mathcal{O}}_{D}$ has the following description: ${\mathcal{O}}_{D}={\mathcal{O}}_{K_{2}}\oplus \unicode[STIX]{x1D711}{\mathcal{O}}_{K_{2}}$ , with $\unicode[STIX]{x1D711}^{2}=\unicode[STIX]{x1D71B}$ and $a\unicode[STIX]{x1D711}=\unicode[STIX]{x1D711}a^{\unicode[STIX]{x1D70E}}$ for $a\in {\mathcal{O}}_{K_{2}}$ . We define an action of ${\mathcal{O}}_{D}$ on $\unicode[STIX]{x1D6F4}$ by $\unicode[STIX]{x1D701}(X)=\bar{\unicode[STIX]{x1D701}}X$ for $\unicode[STIX]{x1D701}\in \unicode[STIX]{x1D707}_{q^{2}-1}({\mathcal{O}}_{K_{2}})$ and $\unicode[STIX]{x1D711}(X)=X^{q}$ . Then, this gives an isomorphism ${\mathcal{O}}_{D}\simeq \text{End}(\unicode[STIX]{x1D6F4})$ by [Reference Gross and HopkinsGH, Proposition 13.10].

Let $d=d_{1}+\unicode[STIX]{x1D711}d_{2}\in {\mathcal{O}}_{D}^{\times }$ , where $d_{1}\in {\mathcal{O}}_{K_{2}}^{\times }$ and $d_{2}\in {\mathcal{O}}_{K_{2}}$ . By the definition of the action of ${\mathcal{O}}_{D}$ on $\unicode[STIX]{x1D6F4}$ , we have

(1.5) $$\begin{eqnarray}d(X)\equiv \bar{d}_{1}X+(\bar{d}_{2}X)^{q}\quad \hspace{0.2em}\bmod \hspace{0.2em}(X^{q^{2}}).\end{eqnarray}$$

We take a lifting $\tilde{d}(X)\in {\mathcal{O}}_{K_{2}}[[X]]$ of $d(X)\in k_{2}[[X]]$ . Let ${\mathcal{F}}_{\tilde{d}}$ be the formal ${\mathcal{O}}_{K}$ -module defined by

$$\begin{eqnarray}{\mathcal{F}}_{\tilde{d}}(X,Y)=\tilde{d}({\mathcal{F}}^{\text{univ}}(\tilde{d}^{-1}(X),\tilde{d}^{-1}(Y))),\qquad [a]_{{\mathcal{F}}_{\tilde{d}}}(X)=\tilde{d}([a]_{\text{u}}(\tilde{d}^{-1}(X)))\end{eqnarray}$$

for $a\in {\mathcal{O}}_{K}$ . Then, we have an isomorphism

$$\begin{eqnarray}\tilde{d}:{\mathcal{F}}^{\text{univ}}\stackrel{{\sim}}{\longrightarrow }{\mathcal{F}}_{\tilde{d}};\qquad (u,X)\mapsto (u,\tilde{d}(X)).\end{eqnarray}$$

By [Reference Gross and HopkinsGH, Proposition 14.7], the formal ${\mathcal{O}}_{K}$ -module ${\mathcal{F}}_{\tilde{d}}$ with

$$\begin{eqnarray}\unicode[STIX]{x1D6F4}\stackrel{d^{-1}}{\longrightarrow }\unicode[STIX]{x1D6F4}\stackrel{\unicode[STIX]{x1D704}}{\longrightarrow }{\mathcal{F}}^{\text{univ}}\otimes k^{\text{ac}}\xrightarrow[]{\{}{\mathcal{F}}_{\tilde{d}}\otimes k^{\text{ac}}\end{eqnarray}$$

gives an isomorphism

(1.6) $$\begin{eqnarray}d:\mathfrak{X}(1)\rightarrow \mathfrak{X}(1),\end{eqnarray}$$

which is independent of the choice of a lifting $\tilde{d}$ , such that there is the unique isomorphism

$$\begin{eqnarray}j:d^{\ast }{\mathcal{F}}^{\text{univ}}\stackrel{{\sim}}{\longrightarrow }{\mathcal{F}}_{\tilde{d}};\qquad (u,X)\mapsto (u,j(X))\end{eqnarray}$$

satisfying $j(X)\equiv X\hspace{0.2em}\bmod \hspace{0.2em}(\unicode[STIX]{x1D71B},u)$ , where $d^{\ast }{\mathcal{F}}^{\text{univ}}$ denotes the pullback of ${\mathcal{F}}^{\text{univ}}$ over $\mathfrak{X}(1)$ by the map (1.6). Hence, we have

(1.7) $$\begin{eqnarray}[\unicode[STIX]{x1D71B}]_{d^{\ast }{\mathcal{F}}^{\text{univ}}}(j^{-1}(X))=j^{-1}([\unicode[STIX]{x1D71B}]_{{\mathcal{F}}_{\tilde{d}}}(X)).\end{eqnarray}$$

On the other hand, we have the following isomorphism:

$$\begin{eqnarray}d^{\ast }{\mathcal{F}}^{\text{univ}}\stackrel{{\sim}}{\rightarrow }{\mathcal{F}}^{\text{univ}};\qquad (u,X^{\prime })\mapsto (d(u),X^{\prime }).\end{eqnarray}$$

Furthermore, we consider the following isomorphism under the identification (1.4):

(1.8) $$\begin{eqnarray}\unicode[STIX]{x1D713}_{d}:\mathfrak{X}_{1}(\mathfrak{p}^{n})\longrightarrow \mathfrak{X}_{1}(\mathfrak{p}^{n});\qquad (u,X_{n})\mapsto (d(u),j^{-1}(\tilde{d}(X_{n}))),\end{eqnarray}$$

which depends only on $d$ as in [Reference Gross and HopkinsGH, Proposition 14.7]. We put

$$\begin{eqnarray}d^{\ast }(X)=j^{-1}(\tilde{d}(X)).\end{eqnarray}$$

We define a left action of $d$ on $\mathfrak{X}_{1}(\mathfrak{p}^{n})$ by

$$\begin{eqnarray}[({\mathcal{F}},\unicode[STIX]{x1D704},P)]\mapsto [({\mathcal{F}},\unicode[STIX]{x1D704}\circ d^{-1},P)].\end{eqnarray}$$

Then, this action coincides with $\unicode[STIX]{x1D713}_{d}$ by the definition.

By (1.5), we have

(1.9) $$\begin{eqnarray}\tilde{d}^{-1}(X)=d_{1}^{-1}X-d_{1}^{-(q+1)}d_{2}^{q}X^{q}\quad \hspace{0.2em}\bmod \hspace{0.2em}(\unicode[STIX]{x1D71B},X^{q^{2}})\end{eqnarray}$$

in ${\mathcal{O}}_{K_{2}}[[X]]$ . We use the following lemma later to compute the ${\mathcal{O}}_{D}^{\times }$ -action on the stable reduction of $\mathbf{X}_{1}(\mathfrak{p}^{3})$ .

Lemma 1.2. We assume $v(u)=1/(2q)$ . Let $d=d_{1}+\unicode[STIX]{x1D711}d_{2}\in {\mathcal{O}}_{D}^{\times }$ . We set $u^{\prime }=d(u)$ . We change variables as $u=\unicode[STIX]{x1D71B}^{1/(2q)}\tilde{u}$ and $u^{\prime }=\unicode[STIX]{x1D71B}^{1/(2q)}\tilde{u} ^{\prime }$ . Then, we have the following:

(1.10) $$\begin{eqnarray}\displaystyle u^{\prime } & \equiv & \displaystyle d_{1}^{-(q-1)}u(1+d_{1}^{-q}d_{2}u)\quad \hspace{0.2em}\bmod \hspace{0.2em}(\unicode[STIX]{x1D71B},u^{3}),\end{eqnarray}$$

(1.11) $$\begin{eqnarray}\displaystyle j^{-1}(X) & \equiv & \displaystyle X+d_{1}^{-q}d_{2}uX\quad \hspace{0.2em}\bmod \hspace{0.2em}(\unicode[STIX]{x1D71B},u^{2}X,uX^{2}).\end{eqnarray}$$

Proof. We set $d^{-1}=d_{1}^{\prime }+\unicode[STIX]{x1D711}d_{2}^{\prime }$ . Then,

$$\begin{eqnarray}d_{1}^{\prime }\equiv d_{1}^{-1},\qquad d_{2}^{\prime }\equiv -d_{1}^{-(q+1)}d_{2}\quad \left(\text{mod}~1\right).\end{eqnarray}$$

First, we prove (1.10). If $v(u)=1/(2q)$ , the function $w(u)$ in [Reference Gross and HopkinsGH, (25.11)] is well approximated by a function $\unicode[STIX]{x1D71B}u(\unicode[STIX]{x1D71B}+u^{q+1})^{-1}$ . By [Reference Gross and HopkinsGH, (25.13)], we have

$$\begin{eqnarray}\displaystyle {\displaystyle \frac{\unicode[STIX]{x1D71B}u^{\prime }}{\unicode[STIX]{x1D71B}+u^{\prime q+1}}} & \equiv & \displaystyle {\displaystyle \frac{d_{1}^{\prime q}\unicode[STIX]{x1D71B}u(\unicode[STIX]{x1D71B}+u^{q+1})^{-1}+\unicode[STIX]{x1D71B}d_{2}^{\prime q}}{d_{2}^{\prime }\unicode[STIX]{x1D71B}u(\unicode[STIX]{x1D71B}+u^{q+1})^{-1}+d_{1}^{\prime }}}\nonumber\\ \displaystyle & \equiv & \displaystyle {\displaystyle \frac{\unicode[STIX]{x1D71B}u(d_{1}-d_{2}^{q}u^{q})}{d_{1}^{q}(\unicode[STIX]{x1D71B}+u^{q+1})-d_{2}\unicode[STIX]{x1D71B}u}}\quad \left(\text{mod}~1+\right).\nonumber\end{eqnarray}$$

Hence, we acquire the following by $u=\unicode[STIX]{x1D71B}^{1/(2q)}\tilde{u}$ and $u^{\prime }=\unicode[STIX]{x1D71B}^{1/(2q)}\tilde{u} ^{\prime }$ :

(1.12) $$\begin{eqnarray}\displaystyle {\displaystyle \frac{\tilde{u} ^{\prime }}{\tilde{u} ^{\prime q+1}+\unicode[STIX]{x1D71B}^{(q-1)/(2q)}}}\equiv {\displaystyle \frac{\tilde{u} (d_{1}-\unicode[STIX]{x1D71B}^{1/2}d_{2}^{q}\tilde{u} ^{q})}{d_{1}^{q}\tilde{u} ^{q+1}+\unicode[STIX]{x1D71B}^{(q-1)/(2q)}d_{1}^{q}-\unicode[STIX]{x1D71B}^{1/2}d_{2}\tilde{u} }}\quad \left(\text{mod}~{\displaystyle \frac{1}{2}}+\right). & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

By taking an inverse of the congruence (1.12), we obtain

(1.13) $$\begin{eqnarray}\displaystyle (\tilde{u} ^{\prime }-d_{1}^{-(q-1)}\tilde{u} )^{q} & \equiv & \displaystyle \unicode[STIX]{x1D71B}^{(q-1)/(2q)}\biggl({\displaystyle \frac{\tilde{u} ^{\prime }-d_{1}^{-(q-1)}\tilde{u} }{d_{1}^{-(q-1)}\tilde{u} \tilde{u} ^{\prime }}}\biggr)\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D71B}^{1/2}(d_{1}^{q-2}d_{2}^{q}\tilde{u} ^{2q}-d_{1}^{-1}d_{2})\quad \left(\text{mod}~{\displaystyle \frac{1}{2}}+\right).\end{eqnarray}$$

Now, we set $\tilde{u} ^{\prime }-d_{1}^{-(q-1)}\tilde{u} =\unicode[STIX]{x1D71B}^{1/(2q)}x$ . By substituting this into (1.13) and dividing it by $\unicode[STIX]{x1D71B}^{1/2}$ , we obtain

$$\begin{eqnarray}(x-d_{1}^{1-2q}d_{2}\tilde{u} ^{2})^{q}\equiv d_{1}^{2q-2}\tilde{u} ^{-2}(x-d_{1}^{1-2q}d_{2}\tilde{u} ^{2})\quad \left(\text{mod}~0+\right).\end{eqnarray}$$

Since $x$ is an analytic function of $\tilde{u}$ , a congruence $x\equiv d_{1}^{1-2q}d_{2}\tilde{u} ^{2}\;(\text{mod}~0+)$ must hold. Hence, we have

$$\begin{eqnarray}\tilde{u} ^{\prime }\equiv d_{1}^{-(q-1)}\tilde{u} (1+\unicode[STIX]{x1D71B}^{1/(2q)}d_{1}^{-q}d_{2}\tilde{u} )\quad \left(\text{mod}~{\displaystyle \frac{1}{2q}}+\right)\end{eqnarray}$$

using $\tilde{u} ^{\prime }-d_{1}^{q-1}\tilde{u} =\unicode[STIX]{x1D71B}^{1/(2q)}x$ . This implies (1.10), because $u^{\prime }$ is an analytic function of $u$ .

By Lemma 1.1, (1.7) and (1.9), we have

$$\begin{eqnarray}u^{\prime }j^{-1}(X)^{q}\equiv j^{-1}(ud_{1}^{-(q-1)}X^{q})\quad \hspace{0.2em}\bmod \hspace{0.2em}(\unicode[STIX]{x1D71B},X^{q^{2}}).\end{eqnarray}$$

Hence, the assertion (1.11) follows from (1.10) and $j^{-1}(X)\equiv X\hspace{0.2em}\bmod \hspace{0.2em}(\unicode[STIX]{x1D71B},u)$ .◻

4.1 Definitions of several subspaces in $\mathbf{X}_{1}(\mathfrak{p}^{3})$

In this subsection, we define several subspaces of $\mathbf{X}_{1}(\mathfrak{p}^{3})$ . Let $(u,X_{3})$ be the parameter of $\mathbf{X}_{1}(\mathfrak{p}^{3})$ given by the identification (1.4).

Let $\mathbf{Y}_{1,2}$ , $\mathbf{Y}_{2,1}$ and $\mathbf{Z}_{1,1}^{0}$ be the subspaces of $\mathbf{X}_{1}(\mathfrak{p}^{3})$ defined by the following conditions.

$$\begin{eqnarray}\displaystyle & \displaystyle \mathbf{Y}_{1,2}:v(u)={\displaystyle \frac{1}{q+1}},\quad v(X_{1})={\displaystyle \frac{q}{q^{2}-1}},\qquad v(X_{2})={\displaystyle \frac{1}{q(q^{2}-1)}}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \qquad v(X_{3})={\displaystyle \frac{1}{q^{3}(q^{2}-1)}}. & \displaystyle \nonumber\\ \displaystyle & \displaystyle \mathbf{Y}_{2,1}:v(u)={\displaystyle \frac{1}{q(q+1)}},\quad v(X_{1})={\displaystyle \frac{q^{2}+q-1}{q(q^{2}-1)}},\qquad v(X_{2})={\displaystyle \frac{1}{q^{2}-1}}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle v(X_{3})={\displaystyle \frac{1}{q^{2}(q^{2}-1)}}. & \displaystyle \nonumber\\ \displaystyle & \displaystyle \mathbf{Z}_{1,1}^{0}:v(u)={\displaystyle \frac{1}{2q}},\quad v(X_{1})={\displaystyle \frac{2q-1}{2q(q-1)}},\qquad v(X_{2})={\displaystyle \frac{1}{2q(q-1)}}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \qquad v(X_{3})={\displaystyle \frac{1}{2q^{3}(q-1)}}. & \displaystyle \nonumber\end{eqnarray}$$

We write down the following possible cases for $(u,X_{1},X_{2})$ :

(4.1) $$\begin{eqnarray}\displaystyle \begin{array}{@{}l@{}}(1)\quad 0<v(u)<{\displaystyle \frac{1}{q+1}},\quad v(X_{1})={\displaystyle \frac{1-v(u)}{q-1}},\qquad v(X_{2})={\displaystyle \frac{1-qv(u)}{q(q-1)}};\\ (2)\quad 0<v(u)<{\displaystyle \frac{1}{q+1}},\quad v(X_{1})={\displaystyle \frac{1-v(u)}{q-1}},\qquad v(X_{2})={\displaystyle \frac{v(u)}{q(q-1)}};\\ (3)\quad v(u)={\displaystyle \frac{1}{q+1}},\quad v(X_{1})={\displaystyle \frac{q}{q^{2}-1}},\qquad v(X_{2})={\displaystyle \frac{1}{q(q^{2}-1)}};\\ (4)\quad 0<v(u)<{\displaystyle \frac{q}{q+1}},\quad v(X_{1})={\displaystyle \frac{v(u)}{q(q-1)}},\qquad v(X_{2})={\displaystyle \frac{v(u)}{q^{3}(q-1)}};\\ (5)\quad v(u)\geqslant {\displaystyle \frac{q}{q+1}},\quad v(X_{1})={\displaystyle \frac{1}{q^{2}-1}},\qquad v(X_{2})={\displaystyle \frac{1}{q^{2}(q^{2}-1)}};\\ (6)\quad {\displaystyle \frac{1}{q+1}}<v(u)<{\displaystyle \frac{q}{q+1}},\quad v(X_{1})={\displaystyle \frac{1-v(u)}{q-1}},\qquad v(X_{2})={\displaystyle \frac{1-v(u)}{q^{2}(q-1)}}.\end{array} & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Next, we consider the following possible cases for $(X_{2},X_{3})$ :

(4.2) $$\begin{eqnarray}\displaystyle \begin{array}{@{}c@{}}(1^{\prime })\quad v(X_{3}^{q^{2}})=v(X_{2})<v(uX_{3}^{q}),\qquad (2^{\prime })\quad v(uX_{3}^{q})=v(X_{2})<v(X_{3}^{q^{2}}),\\ (3^{\prime })\quad v(X_{2})>v(X_{3}^{q^{2}})=v(uX_{3}^{q}),\qquad (4^{\prime })\quad v(X_{2})=v(X_{3}^{q^{2}})=v(uX_{3}^{q}).\end{array} & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Let $\mathbf{W}_{i,j^{\prime }}$ be the subspace of $\mathbf{X}_{1}(\mathfrak{p}^{3})$ defined by the conditions $1\leqslant i\leqslant 6$ in (4.1) and $1^{\prime }\leqslant j^{\prime }\leqslant 4^{\prime }$ in (4.2). We note that $\mathbf{W}_{3,1^{\prime }}=\mathbf{Y}_{1,2}$ and $\mathbf{W}_{1,4^{\prime }}=\mathbf{Y}_{2,1}$ . Let $\mathbf{W}_{1,1^{\prime }}^{+}$ and $\mathbf{W}_{1,1^{\prime }}^{-}$ be the subspaces of $\mathbf{W}_{1,1^{\prime }}$ defined by $1/(2q)<v(u)<1/(q+1)$ and $1/(q(q+1))<v(u)<1/(2q)$ respectively.

4.2 Reductions of the affinoid spaces $\mathbf{Y}_{1,2}$ and $\mathbf{Y}_{2,1}$

In this subsection, we compute the reductions of the affinoid spaces $\mathbf{Y}_{1,2}$ and $\mathbf{Y}_{2,1}$ . The reductions of $\mathbf{Y}_{2,1}$ and $\mathbf{Y}_{1,2}$ are isomorphic to the affine curve defined by $x^{q}y-xy^{q}=1$ . These curves have genus $q(q-1)/2$ .

Proof. We change variables as $u=\unicode[STIX]{x1D71B}^{1/(q+1)}\tilde{u}$ , $X_{1}=\unicode[STIX]{x1D71B}^{q/(q^{2}-1)}x_{1}$ , $X_{2}=\unicode[STIX]{x1D71B}^{1/(q(q^{2}-1))}x_{2}$ and $X_{3}=\unicode[STIX]{x1D71B}^{1/(q^{3}(q^{2}-1))}x_{3}$ . By Lemma 1.1, we have

(4.3) $$\begin{eqnarray}\tilde{u} \equiv -x_{1}^{-(q-1)},\qquad x_{1}\equiv \tilde{u} x_{2}^{q}+x_{2}^{q^{2}},\qquad x_{2}\equiv x_{3}^{q^{2}}\quad \left(\text{mod}~0+\right).\end{eqnarray}$$

Then, we have $\tilde{u} =-x_{1}^{-(q-1)}+F_{0}(\tilde{u} ,x_{1})$ for some function $F_{0}(\tilde{u} ,x_{1})$ satisfying $v(F_{0}(\tilde{u} ,x_{1}))>v(\tilde{u} )$ . Substituting $\tilde{u} =-x_{1}^{-(q-1)}+F_{0}(\tilde{u} ,x_{1})$ into $F_{0}(\tilde{u} ,x_{1})$ and repeating it, we see that $\tilde{u}$ is written as a function of $x_{1}$ . Similarly, by $x_{2}\equiv x_{3}^{q^{2}}\;(\text{mod}~0+)$ , we can see that $x_{2}$ is written as a function of $x_{1}$ and $x_{3}$ . By (4.3), we acquire

(4.4) $$\begin{eqnarray}1\equiv {\displaystyle \frac{x_{3}^{q^{4}}}{x_{1}}}-{\displaystyle \frac{x_{3}^{q^{3}}}{x_{1}^{q}}}\quad \left(\text{mod}~0+\right).\end{eqnarray}$$

By setting $1+x_{1}^{-1}x_{3}^{q^{2}}=x_{3}^{q^{3}}t_{1}^{-1}$ and substituting this into (4.4), we obtain $t_{1}^{q}\equiv x_{1}\;(\text{mod}~0+)$ and hence $(1+x_{3}^{q}t_{1}^{-1})^{q}\equiv x_{3}^{q^{3}}t_{1}^{-1}\;(\text{mod}~0+)$ . By setting $1+x_{3}^{q}t_{1}^{-1}=x_{3}^{q^{2}}t_{2}^{-1}$ , we obtain $t_{2}^{q}\equiv t_{1}\;(\text{mod}~0+)$ . Hence,

$$\begin{eqnarray}(1+x_{3}t_{2}^{-1})^{q}\equiv x_{3}^{q^{2}}t_{2}^{-1}\quad \left(\text{mod}~0+\right).\end{eqnarray}$$

Finally, by setting $x=x_{3}$ and $1+x_{3}t_{2}^{-1}=x_{3}^{q}y$ , we acquire $y^{q}\equiv t_{2}^{-1}\;(\text{mod}~0+)$ . Hence, we have $x^{q}y-xy^{q}\equiv 1\;(\text{mod}~0+)$ . Note that

(4.5) $$\begin{eqnarray}x=x_{3},\quad y={\displaystyle \frac{x_{1}(1+x_{3}^{q(q^{2}-1)}+x_{3}^{(q+1)(q^{2}-1)})+x_{3}^{q^{2}}}{x_{1}x_{3}^{q^{3}+q^{2}-1}}},\end{eqnarray}$$

which we will use later. ◻

We put

$$\begin{eqnarray}\unicode[STIX]{x1D6FE}_{i}=\unicode[STIX]{x1D71B}^{(q-1)/(2q^{i})}\end{eqnarray}$$

for $1\leqslant i\leqslant 4$ . We choose an element $c_{0}$ such that $c_{0}^{q}-\unicode[STIX]{x1D6FE}_{1}^{2}c_{0}+1=0$ . Note that we have $c_{0}\equiv -1\;(\text{mod}~0+)$ . Further, we choose a $q$ th root $c_{0}^{1/q}$ of $c_{0}$ .

Proof. We change variables as

$$\begin{eqnarray}\displaystyle & \displaystyle u=\unicode[STIX]{x1D71B}^{1/(q(q+1))}\tilde{u} ,\qquad X_{1}=\unicode[STIX]{x1D71B}^{(q^{2}+q-1)/(q(q^{2}-1))}x_{1},\qquad X_{2}=\unicode[STIX]{x1D71B}^{1/(q^{2}-1)}x_{2} & \displaystyle \nonumber\\ \displaystyle & \displaystyle \qquad \text{and}\qquad X_{3}=\unicode[STIX]{x1D71B}^{1/(q^{2}(q^{2}-1))}x_{3}. & \displaystyle \nonumber\end{eqnarray}$$

By Lemma 1.1, we have

(4.6) $$\begin{eqnarray}\displaystyle \tilde{u} & \equiv & \displaystyle -x_{1}^{-(q-1)}\quad \left(\text{mod}~{\displaystyle \frac{q^{2}-1}{q^{2}}}+\right),\end{eqnarray}$$

(4.7) $$\begin{eqnarray}\displaystyle x_{1} & \equiv & \displaystyle \tilde{u} x_{2}^{q}+\unicode[STIX]{x1D6FE}_{1}^{2}(x_{2}^{q^{2}}+x_{2})\quad \left(\text{mod}~{\displaystyle \frac{q^{2}-1}{q^{2}}}+\right),\end{eqnarray}$$

(4.8) $$\begin{eqnarray}\displaystyle x_{2} & \equiv & \displaystyle x_{3}^{q^{2}}+\tilde{u} x_{3}^{q}\quad \left(\text{mod}~{\displaystyle \frac{q-1}{q^{2}}}+\right).\end{eqnarray}$$

By (4.6) and (4.8), we can see that $\tilde{u}$ is written as a function of $x_{1}$ , and that $x_{2}$ is written as a function of $x_{1}$ and $x_{3}$ . We define a parameter $t$ by

(4.9) $$\begin{eqnarray}{\displaystyle \frac{x_{2}}{x_{1}}}=c_{0}+\unicode[STIX]{x1D6FE}_{2}^{2}{\displaystyle \frac{x_{2}^{q}}{t}}.\end{eqnarray}$$

We note that $v(t)=0$ . By considering $x_{1}^{-1}\times$ (4.7), we have

(4.10) $$\begin{eqnarray}\biggl({\displaystyle \frac{x_{2}}{x_{1}}}\biggr)^{q}+1-\unicode[STIX]{x1D6FE}_{1}^{2}{\displaystyle \frac{x_{2}}{x_{1}}}\equiv \unicode[STIX]{x1D6FE}_{1}^{2}{\displaystyle \frac{x_{2}^{q^{2}}}{x_{1}}}\quad \left(\text{mod}~{\displaystyle \frac{q^{2}-1}{q^{2}}}+\right).\end{eqnarray}$$

By substituting (4.9) into the left-hand side of the congruence (4.10), and dividing it by $\unicode[STIX]{x1D6FE}_{1}^{2}x_{2}^{q^{2}}$ , we acquire

(4.11) $$\begin{eqnarray}x_{1}\equiv t^{q}\biggl(1-\unicode[STIX]{x1D6FE}_{2}^{2}{\displaystyle \frac{t^{q-1}}{x_{2}^{q(q-1)}}}\biggr)^{-1}\quad \left(\text{mod}~{\displaystyle \frac{q-1}{q^{2}}}+\right).\end{eqnarray}$$

By this congruence, we can see that $x_{1}$ is written as a function of $t$ and $x_{3}$ . By considering $x_{1}^{-1}\times$ (4.8), we acquire

(4.12) $$\begin{eqnarray}c_{0}+\unicode[STIX]{x1D6FE}_{2}^{2}{\displaystyle \frac{x_{2}^{q}}{t}}\equiv {\displaystyle \frac{x_{3}^{q^{2}}}{x_{1}}}-\biggl({\displaystyle \frac{x_{3}}{x_{1}}}\biggr)^{q}\quad \left(\text{mod}~{\displaystyle \frac{q-1}{q^{2}}}+\right)\end{eqnarray}$$

by (4.9). Substituting (4.11) into (4.12), we have

(4.13) $$\begin{eqnarray}\biggl(c_{0}^{1/q}-{\displaystyle \frac{x_{3}^{q}}{t}}+{\displaystyle \frac{x_{3}}{x_{1}}}\biggr)^{q}\equiv -\unicode[STIX]{x1D6FE}_{2}^{2}{\displaystyle \frac{(x_{2}+x_{3})^{q^{2}}}{tx_{2}^{q(q-1)}}}\quad \left(\text{mod}~{\displaystyle \frac{q-1}{q^{2}}}+\right).\end{eqnarray}$$

By (4.9) and $c_{0}\equiv -1\;(\text{mod}~0+)$ , we have $x_{2}\equiv -x_{1}\;(\text{mod}~0+)$ . Therefore, we acquire

$$\begin{eqnarray}(x_{2}+x_{3})^{q}\equiv x_{2}^{q-1}x_{3}^{q^{2}}\quad \left(\text{mod}~0+\right)\end{eqnarray}$$

by (4.6) and (4.8). In particular, we obtain $v(x_{2}+x_{3})=0$ . We introduce a new parameter $t_{1}$ as

(4.14) $$\begin{eqnarray}c_{0}^{1/q}-{\displaystyle \frac{x_{3}^{q}}{t}}+{\displaystyle \frac{x_{3}}{x_{1}}}=-\unicode[STIX]{x1D6FE}_{3}^{q}{\displaystyle \frac{(x_{2}+x_{3})^{q}}{t_{1}x_{2}^{q-1}}}.\end{eqnarray}$$

Substituting this into the left-hand side of the congruence (4.13), and dividing it by $-\unicode[STIX]{x1D6FE}_{2}^{2}x_{2}^{-q(q-1)}(x_{2}+x_{3})^{q^{2}}$ , we acquire $t\equiv t_{1}^{q}\;(\text{mod}~0+)$ . By this congruence, we can see that $t$ is written as a function of $t_{1}$ and $x_{3}$ . By (4.14), we obtain

$$\begin{eqnarray}x_{3}\equiv t_{1}^{q^{2}}(1+x_{3}t_{1}^{-1})^{q}\quad \left(\text{mod}~0+\right)\end{eqnarray}$$

using $t\equiv t_{1}^{q}\;(\text{mod}~0+)$ and $x_{1}\equiv t^{q}\;(\text{mod}~0+)$ . Hence, by setting $x=t_{1}^{-1}$ and $~y=t_{1}^{q}(1+x_{3}t_{1}^{-1})$ , we acquire $x^{q}y-yx^{q}\equiv 1\;(\text{mod}~0+)$ .◻

4.3 Reduction of the affinoid space $\mathbf{Z}_{1,1}^{0}$

In this subsection, we calculate the reduction of the affinoid space $\mathbf{Z}_{1,1}^{0}$ . We define ${\mathcal{S}}_{1}$ as in the introduction. The reduction $\overline{\mathbf{Z}}_{1,1}^{0}$ is isomorphic to the affine curve defined by $Z^{q}+x_{3}^{q^{2}-1}+x_{3}^{-(q^{2}-1)}=0$ . This affine curve has genus $0$ and singularities at $x_{3}\in {\mathcal{S}}_{1}$ .

We put

$$\begin{eqnarray}\unicode[STIX]{x1D714}_{i}=\unicode[STIX]{x1D71B}^{1/(2q^{i}(q-1))},\qquad \unicode[STIX]{x1D716}_{i}={\displaystyle \frac{1}{2q^{i}}}\end{eqnarray}$$

for $1\leqslant i\leqslant 4$ . We change variables as $u=\unicode[STIX]{x1D714}_{1}^{q-1}\tilde{u}$ , $X_{1}=\unicode[STIX]{x1D714}_{1}^{2q-1}x_{1}$ , $X_{2}=\unicode[STIX]{x1D714}_{1}x_{2}$ and $X_{3}=\unicode[STIX]{x1D714}_{3}x_{3}$ . By Lemma 1.1, we have

(4.15) $$\begin{eqnarray}\displaystyle \tilde{u} & \equiv & \displaystyle -x_{1}^{-(q-1)}\quad \left(\text{mod}~{\textstyle \frac{1}{2}}+\right),\end{eqnarray}$$

(4.16) $$\begin{eqnarray}\displaystyle x_{1} & \equiv & \displaystyle \tilde{u} x_{2}^{q}+\unicode[STIX]{x1D6FE}_{1}x_{2}^{q^{2}}+\unicode[STIX]{x1D6FE}_{1}^{2}x_{2}\quad \left(\text{mod}~{\textstyle \frac{1}{2}}+\right),\end{eqnarray}$$

(4.17) $$\begin{eqnarray}\displaystyle x_{2} & \equiv & \displaystyle x_{3}^{q^{2}}+\unicode[STIX]{x1D6FE}_{2}\tilde{u} x_{3}^{q}\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{1}+\right).\end{eqnarray}$$

Note that we have $v(\unicode[STIX]{x1D6FE}_{1}^{2})>1/2$ if $q\neq 2$ . By (4.15) and (4.17), we can see that $\tilde{u}$ is written as a function of $x_{1}$ , and that $x_{2}$ is written as a function of $x_{1}$ and $x_{3}$ . We define a parameter $t$ by

(4.18) $$\begin{eqnarray}{\displaystyle \frac{x_{2}}{x_{1}}}=-1+\unicode[STIX]{x1D6FE}_{2}{\displaystyle \frac{x_{2}^{q}}{t}}.\end{eqnarray}$$

By considering $x_{1}^{-1}\times$ (4.16), we acquire

(4.19) $$\begin{eqnarray}\biggl({\displaystyle \frac{x_{2}}{x_{1}}}+1\biggr)^{q}\equiv \unicode[STIX]{x1D6FE}_{1}{\displaystyle \frac{x_{2}^{q^{2}}}{x_{1}}}\biggl(1+{\displaystyle \frac{\unicode[STIX]{x1D6FE}_{1}}{x_{2}^{q^{2}-1}}}\biggr)\quad \left(\text{mod}~{\displaystyle \frac{1}{2}}+\right)\end{eqnarray}$$

by (4.15). Substituting (4.18) into (4.19), and dividing it by $\unicode[STIX]{x1D6FE}_{1}x_{2}^{q^{2}}$ , we obtain

(4.20) $$\begin{eqnarray}x_{1}\equiv t^{q}\biggl(1+{\displaystyle \frac{\unicode[STIX]{x1D6FE}_{1}}{x_{2}^{q^{2}-1}}}\biggr)\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{1}+\right).\end{eqnarray}$$

Therefore, we have $v(t)=0$ . By considering $x_{1}^{-1}\times$ (4.17), we acquire

(4.21) $$\begin{eqnarray}\biggl(1+{\displaystyle \frac{x_{3}^{q}}{t}}\biggr)^{q}-\unicode[STIX]{x1D6FE}_{1}{\displaystyle \frac{x_{3}^{q^{2}}}{t^{q}x_{2}^{q^{2}-1}}}\equiv \unicode[STIX]{x1D6FE}_{2}\biggl({\displaystyle \frac{x_{2}^{q}}{t}}+\biggl({\displaystyle \frac{x_{3}}{x_{1}}}\biggr)^{q}\biggr)\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{1}+\right)\end{eqnarray}$$

by (4.15), (4.18) and (4.20). We define a parameter $Z_{0}$ by

(4.22) $$\begin{eqnarray}1+{\displaystyle \frac{x_{3}^{q}}{t}}=\unicode[STIX]{x1D6FE}_{3}Z_{0}.\end{eqnarray}$$

We note that $v(Z_{0})\geqslant 0$ . Substituting this into (4.21), and dividing it by $\unicode[STIX]{x1D6FE}_{2}$ , we obtain

(4.23) $$\begin{eqnarray}Z_{0}^{q}\equiv {\displaystyle \frac{x_{2}^{q}}{t}}+\biggl({\displaystyle \frac{x_{3}}{x_{1}}}\biggr)^{q}+\unicode[STIX]{x1D6FE}_{2}^{q-1}{\displaystyle \frac{x_{3}^{q^{2}}}{t^{q}x_{2}^{q^{2}-1}}}\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{2}+\right).\end{eqnarray}$$

By (4.22) and (4.23), we acquire

(4.24) $$\begin{eqnarray}\biggl(Z_{0}+{\displaystyle \frac{x_{2}}{x_{3}}}-{\displaystyle \frac{x_{3}}{x_{1}}}\biggr)^{q}\equiv \unicode[STIX]{x1D6FE}_{3}\biggl({\displaystyle \frac{x_{2}}{x_{3}}}\biggr)^{q}Z_{0}+\unicode[STIX]{x1D6FE}_{2}^{q-1}{\displaystyle \frac{x_{3}^{q^{2}}}{t^{q}x_{2}^{q^{2}-1}}}\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{2}+\right).\end{eqnarray}$$

We introduce a new parameter $Z$ as

(4.25) $$\begin{eqnarray}Z_{0}+{\displaystyle \frac{x_{2}}{x_{3}}}-{\displaystyle \frac{x_{3}}{x_{1}}}=\unicode[STIX]{x1D6FE}_{4}{\displaystyle \frac{x_{2}}{x_{3}}}Z.\end{eqnarray}$$

We note that $v(Z)\geqslant 0$ . Substituting this into the left-hand side of the congruence (4.24), and dividing it by $\unicode[STIX]{x1D6FE}_{3}(x_{2}/x_{3})^{q}$ , we acquire

(4.26) $$\begin{eqnarray}Z^{q}\equiv Z_{0}+\unicode[STIX]{x1D6FE}_{3}^{q^{2}-q-1}{\displaystyle \frac{x_{3}^{q(q+1)}}{t^{q}x_{2}^{q^{2}+q-1}}}\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{3}+\right).\end{eqnarray}$$

By substituting (4.25) into (4.26), we obtain

(4.27) $$\begin{eqnarray}Z^{q}+x_{3}^{q^{2}-1}(1-\unicode[STIX]{x1D6FE}_{4}Z)+x_{3}^{-(q^{2}-1)}\equiv -\unicode[STIX]{x1D6FE}_{3}^{q^{2}-q-1}x_{3}^{-q(q^{2}-1)(q+1)}\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{3}+\right)\end{eqnarray}$$

by (4.17), (4.20) and (4.22). Note that we have $v(\unicode[STIX]{x1D6FE}_{3}^{q^{2}-q-1})>\unicode[STIX]{x1D716}_{3}$ , if $q\neq 2$ .

4.4 Analysis of the singular residue classes of $\mathbf{Z}_{1,1}^{0}$

In this subsection, we analyze the singular residue classes $\{{\mathcal{D}}_{\unicode[STIX]{x1D701}}\}_{\unicode[STIX]{x1D701}\in {\mathcal{S}}_{1}}$ of $\mathbf{Z}_{1,1}^{0}$ . If $q$ is odd, the space ${\mathcal{D}}_{\unicode[STIX]{x1D701}}$ is a basic wide open space with an underlying affinoid $\mathbf{X}_{\unicode[STIX]{x1D701}}$ , whose reduction $\overline{\mathbf{X}}_{\unicode[STIX]{x1D701}}$ is isomorphic to the affine curve defined by $z^{q}-z=w^{2}$ . On the other hand, if $q$ is even, the situation is slightly complicated, because the space ${\mathcal{D}}_{\unicode[STIX]{x1D701}}$ is not basic wide open. Hence, we have to cover ${\mathcal{D}}_{\unicode[STIX]{x1D701}}$ by smaller basic wide open spaces. As a result, in ${\mathcal{D}}_{\unicode[STIX]{x1D701}}$ , we find an affinoid $\mathbf{P}_{\unicode[STIX]{x1D701}}^{0}$ , whose reduction is isomorphic to the affine curve defined by $z_{f+1}^{2}=w_{1}(w_{1}^{q-1}-1)^{2}$ . This affine curve has $q-1$ singular points at $w_{1}\in k^{\times }$ . Then, by analyzing the tubular neighborhoods of these singular points, we find an affinoid $\mathbf{X}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }}\subset \mathbf{P}_{\unicode[STIX]{x1D701}}^{0}$ for each $\unicode[STIX]{x1D701}^{\prime }\in k^{\times }$ , whose reduction is isomorphic to the affine curve defined by $z^{2}+z=w^{3}$ .

4.4.1 $q$ odd

We assume that $q$ is odd. For each $\unicode[STIX]{x1D701}\in \unicode[STIX]{x1D707}_{2(q^{2}-1)}(k^{\text{ac}})$ , we define an affinoid $\mathbf{X}_{\unicode[STIX]{x1D701}}\subset {\mathcal{D}}_{\unicode[STIX]{x1D701}}$ and compute its reduction $\overline{\mathbf{X}}_{\unicode[STIX]{x1D701}}$ .

For $\unicode[STIX]{x1D704}\in \unicode[STIX]{x1D707}_{2}(k^{\text{ac}})$ , we choose an element $c_{1,\unicode[STIX]{x1D704}}^{\prime }\in {\mathcal{O}}_{K^{\text{ac}}}^{\times }$ such that $\bar{c}_{1,\unicode[STIX]{x1D704}}^{\prime }=-2\unicode[STIX]{x1D704}$ and $c_{1,\unicode[STIX]{x1D704}}^{\prime 2q}=4(1-\unicode[STIX]{x1D6FE}_{4}c_{1,\unicode[STIX]{x1D704}}^{\prime })$ . We take $\unicode[STIX]{x1D701}\in \unicode[STIX]{x1D707}_{2(q^{2}-1)}(k^{\text{ac}})$ . We put $c_{1,\unicode[STIX]{x1D701}}=c_{1,\unicode[STIX]{x1D701}^{q^{2}-1}}^{\prime }$ , and define $c_{2,\unicode[STIX]{x1D701}}\in {\mathcal{O}}_{K^{\text{ac}}}^{\times }$ by $c_{2,\unicode[STIX]{x1D701}}^{q^{2}-1}=-2c_{1,\unicode[STIX]{x1D701}}^{-q}$ and $\bar{c}_{2,\unicode[STIX]{x1D701}}=\unicode[STIX]{x1D701}$ . We put

$$\begin{eqnarray}a_{\unicode[STIX]{x1D701}}=\unicode[STIX]{x1D714}_{4}^{q-1}c_{2,\unicode[STIX]{x1D701}}^{q+1},\qquad b_{\unicode[STIX]{x1D701}}=-2\unicode[STIX]{x1D701}^{q^{2}-1}\unicode[STIX]{x1D714}_{3}^{(q-1)/2}c_{1,\unicode[STIX]{x1D701}}^{-q}c_{2,\unicode[STIX]{x1D701}}^{(q+3)/2}.\end{eqnarray}$$

Note that we have $v(a_{\unicode[STIX]{x1D701}})=1/(2q^{4})$ and $v(b_{\unicode[STIX]{x1D701}})=1/(4q^{3})$ .

For an element $\unicode[STIX]{x1D701}\in \unicode[STIX]{x1D707}_{2(q^{2}-1)}(k^{\text{ac}})$ , we define an affinoid $\mathbf{X}_{\unicode[STIX]{x1D701}}$ by $v(x_{3}-c_{2,\unicode[STIX]{x1D701}})\geqslant 1/(4q^{3})$ . We change variables as

$$\begin{eqnarray}Z=a_{\unicode[STIX]{x1D701}}z+c_{1,\unicode[STIX]{x1D701}},\qquad x_{3}=b_{\unicode[STIX]{x1D701}}w+c_{2,\unicode[STIX]{x1D701}}.\end{eqnarray}$$

Then, we acquire

$$\begin{eqnarray}a_{\unicode[STIX]{x1D701}}^{q}(z^{q}-z-w^{2})\equiv 0\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{3}+\right)\end{eqnarray}$$

by (4.27). Dividing this by $a_{\unicode[STIX]{x1D701}}^{q}$ , we have $z^{q}-z=w^{2}\;(\text{mod}~0+)$ . Hence, the reduction of $\mathbf{X}_{\unicode[STIX]{x1D701}}$ is isomorphic to the affine curve defined by $z^{q}-z=w^{2}$ .

Proposition 4.7. For each $\unicode[STIX]{x1D701}\in \unicode[STIX]{x1D707}_{2(q^{2}-1)}(k^{\text{ac}})$ , the reduction $\overline{\mathbf{X}}_{\unicode[STIX]{x1D701}}$ is isomorphic to the affine curve defined by $z^{q}-z=w^{2}$ , and the complement ${\mathcal{D}}_{\unicode[STIX]{x1D701}}\setminus \mathbf{X}_{\unicode[STIX]{x1D701}}$ is an open annulus.

Proof. We have already proved the first assertion. We prove the second assertion. We change variables as

$$\begin{eqnarray}Z=z^{\prime }+c_{1,\unicode[STIX]{x1D701}},\qquad x_{3}=w^{\prime }+c_{2,\unicode[STIX]{x1D701}}\end{eqnarray}$$

with $0<v(w^{\prime })<1/(4q^{3})$ . Substituting them into (4.27), we obtain

$$\begin{eqnarray}z^{\prime q}\equiv w^{\prime 2}\quad \left(\text{mod}~2v(w^{\prime })+\right).\end{eqnarray}$$

Note that we have $0<v(z^{\prime })<1/(2q^{4})$ . By setting $w^{\prime }=z^{\prime \prime }z^{\prime (q-1)/2}$ , we acquire

$$\begin{eqnarray}z^{\prime \prime 2}\equiv z^{\prime }\quad \left(\text{mod}~v(z^{\prime })+\right).\end{eqnarray}$$

Hence, we can see that $z^{\prime }$ is written as a function of $z^{\prime \prime }$ . Then, $w^{\prime }$ is also written as a function of $z^{\prime \prime }$ . Therefore, $({\mathcal{D}}_{\unicode[STIX]{x1D701}}\setminus \mathbf{X}_{\unicode[STIX]{x1D701}})(\mathbf{C})$ is identified with $\{z^{\prime \prime }\in \mathbf{C}\mid 0<v(z^{\prime \prime })<1/(4q^{4})\}$ .◻

4.4.2 $q$ even

We assume that $q$ is even. We put

$$\begin{eqnarray}Z_{1}=x_{3}^{q^{2}-1}.\end{eqnarray}$$

Then, the congruence (4.27) has the following form:

(4.28) $$\begin{eqnarray}Z^{q}+Z_{1}(1-\unicode[STIX]{x1D6FE}_{4}Z)+Z_{1}^{-1}\equiv -\unicode[STIX]{x1D6FE}_{3}^{q^{2}-q-1}Z_{1}^{-q(q+1)}\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{3}+\right).\end{eqnarray}$$

1. Projective lines For each $\unicode[STIX]{x1D701}\in k_{2}^{\times }$ , we define a subaffinoid $\mathbf{P}_{\unicode[STIX]{x1D701}}^{0}\subset {\mathcal{D}}_{\unicode[STIX]{x1D701}}$ by $v(Z)\geqslant 1/(4q^{4})$ . We change variables as

$$\begin{eqnarray}Z=\unicode[STIX]{x1D71B}^{1/(4q^{4})}w_{1},\qquad Z_{1}=1+\unicode[STIX]{x1D71B}^{1/(8q^{3})}z_{1}.\end{eqnarray}$$

Substituting these into (4.28) and dividing it by $\unicode[STIX]{x1D71B}^{1/(4q^{3})}$ , we acquire

(4.29) $$\begin{eqnarray}\displaystyle & & \displaystyle \quad (z_{1}+w_{1}^{q/2})^{2}+\unicode[STIX]{x1D71B}^{1/(8q^{3})}z_{1}^{3}+\unicode[STIX]{x1D71B}^{1/(4q^{3})}z_{1}^{4}+\unicode[STIX]{x1D71B}^{(q-1)/(4q^{4})}w_{1}\nonumber\\ \displaystyle & & \displaystyle \quad \qquad +\,\unicode[STIX]{x1D71B}^{(3q-2)/(8q^{4})}z_{1}w_{1}\equiv \unicode[STIX]{x1D71B}^{(2q-3)/(4q^{3})}\quad \left(\text{mod}~{\displaystyle \frac{1}{4q^{3}}}+\right).\end{eqnarray}$$

We can check that $v(z_{1})\geqslant 0$ . We set $q=2^{f}$ and put

$$\begin{eqnarray}l_{i}={\displaystyle \frac{(2^{i}-1)q}{2^{i}}},\qquad m_{i}={\displaystyle \frac{1}{2^{i+2}q^{3}}}\end{eqnarray}$$

for $1\leqslant i\leqslant f+1$ . Furthermore, we define parameters $z_{i}$ for $2\leqslant i\leqslant f+1$ by

(4.30) $$\begin{eqnarray}z_{i}+w_{1}^{l_{i}}=\unicode[STIX]{x1D71B}^{m_{i+1}}z_{i+1}\quad \text{for }1\leqslant i\leqslant f.\end{eqnarray}$$

Lemma 4.8. We assume that $v(Z)\geqslant 1/(4q^{4})$ . Then, we have

(4.31) $$\begin{eqnarray}\displaystyle z_{f+1}^{2}+w_{1}^{2q-1}+w_{1}+\unicode[STIX]{x1D71B}^{1/(8q^{4})}z_{f+1}w_{1}^{q}\equiv (q/2)\unicode[STIX]{x1D71B}^{1/(4q^{4})}\quad \left(\text{mod}~{\displaystyle \frac{1}{4q^{4}}}+\right). & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Proof. If $q=2$ , we can check that

(4.32) $$\begin{eqnarray}\displaystyle z_{2}^{2}+w_{1}^{3}+w_{1}+\unicode[STIX]{x1D71B}^{1/128}z_{2}w_{1}^{2}\equiv \unicode[STIX]{x1D71B}^{1/64}(w_{1}z_{2}^{2}+z_{1}^{4}+w_{1}^{2}+1)\quad \left(\text{mod}~{\displaystyle \frac{1}{64}}+\right) & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

by

$$\begin{eqnarray}z_{1}=-w_{1}+\unicode[STIX]{x1D71B}^{1/128}z_{2}.\end{eqnarray}$$

We have $v(z_{2}^{2}+w_{1}^{3}+w_{1})>0$ . Therefore, we obtain

$$\begin{eqnarray}w_{1}z_{2}^{2}+z_{1}^{4}+w_{1}^{2}\equiv w_{1}(z_{2}^{2}+w_{1}^{3}+w_{1})\equiv 0\quad \left(\text{mod}~0+\right).\end{eqnarray}$$

Hence, the required assertion in this case follows from (4.32). Assume that $f\geqslant 2$ . For $1\leqslant i\leqslant f+1$ , we put

$$\begin{eqnarray}n_{i}={\displaystyle \frac{q-2^{i-1}}{2^{i+1}q^{4}}}.\end{eqnarray}$$

We prove

(4.33) $$\begin{eqnarray}(z_{i}+w_{1}^{l_{i}})^{2}+\unicode[STIX]{x1D71B}^{m_{i}}z_{i}w_{1}^{q}+\unicode[STIX]{x1D71B}^{n_{i}}w_{1}\equiv 0\quad \left(\text{mod}~{\displaystyle \frac{1}{2^{i+1}q^{3}}}+\right)\end{eqnarray}$$

for $2\leqslant i\leqslant f+1$ by induction on $i$ . Eliminating $z_{1}$ from (4.29) by (4.30) and dividing it by $\unicode[STIX]{x1D71B}^{1/(8q^{4})}$ , we obtain

$$\begin{eqnarray}\displaystyle & & \displaystyle (z_{2}+w_{1}^{3q/4})^{2}+\unicode[STIX]{x1D71B}^{1/(16q^{3})}z_{2}w_{1}^{q}+\unicode[STIX]{x1D71B}^{(q-2)/(8q^{4})}w_{1}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\unicode[STIX]{x1D71B}^{1/(8q^{3})}w_{1}^{q/2}(z_{2}+w_{1}^{3q/4})^{2}\equiv 0\quad \left(\text{mod}~{\displaystyle \frac{1}{8q^{3}}}+\right).\nonumber\end{eqnarray}$$

This shows

$$\begin{eqnarray}v\Bigl(z_{2}+w_{1}^{3q/4}\Bigr)\geqslant {\displaystyle \frac{1}{32q^{3}}}.\end{eqnarray}$$

Hence, we have (4.33) for $i=2$ . Assume (4.33) for $i$ . Eliminating $z_{i}$ from (4.33) by (4.30) and dividing it by $\unicode[STIX]{x1D71B}^{m_{i}}$ , we obtain (4.33) for $i+1$ . Hence, we have (4.33) for $f+1$ , which is equivalent to (4.31).◻

Proposition 4.9. For each $\unicode[STIX]{x1D701}\in k_{2}^{\times }$ , the reduction $\overline{\mathbf{P}}_{\unicode[STIX]{x1D701}}^{0}$ is isomorphic to the affine curve defined by $z_{f+1}^{2}=w_{1}(w_{1}^{q-1}-1)^{2}$ , which has genus $0$ and singularities at $w_{1}\in k^{\times }$ , and the complement ${\mathcal{D}}_{\unicode[STIX]{x1D701}}\setminus \mathbf{P}_{\unicode[STIX]{x1D701}}^{0}$ is an open annulus.

Proof. The claim on $\overline{\mathbf{P}}_{\unicode[STIX]{x1D701}}^{0}$ follows from the congruence (4.31) modulo $0+$ . We prove the last assertion. We change a variable as $Z_{1}=1+z_{1}^{\prime }$ with $0<v(z_{1}^{\prime })<1/(8q^{3})$ . Similarly to (4.30), we introduce parameters $\{z_{i}^{\prime }\}_{2\leqslant i\leqslant f+1}$ by $z_{i}^{\prime }+Z^{l_{i}}=z_{i+1}^{\prime }$ for $1\leqslant i\leqslant f$ . Then, by similar computations to those in the proof of Lemma 4.8, we obtain

$$\begin{eqnarray}z_{f+1}^{\prime 2}\equiv Z^{2q-1}\quad \left(\text{mod}~2v(z_{f+1}^{\prime })+\right).\end{eqnarray}$$

By setting $z_{f+2}^{\prime }=Z^{q}/z_{f+1}^{\prime }$ , we obtain

$$\begin{eqnarray}z_{f+2}^{\prime 2}\equiv Z\quad \left(\text{mod}~v(Z)+\right).\end{eqnarray}$$

Then, we can see that all parameters $z_{i}^{\prime }$ for $1\leqslant i\leqslant f+1$ and $Z$ are written as functions of $z_{f+2}^{\prime }$ . Hence, $({\mathcal{D}}_{\unicode[STIX]{x1D701}}\setminus \mathbf{P}_{\unicode[STIX]{x1D701}}^{0})(\mathbf{C})$ is identified with

$$\begin{eqnarray}\{z_{f+2}^{\prime }\in \mathbf{C}\mid 0<v(z_{f+2}^{\prime })<1/(8q^{4})\}.\square\end{eqnarray}$$

2. Elliptic curves  For $\unicode[STIX]{x1D701}^{\prime }\in k^{\times }$ , we choose $c_{2,\unicode[STIX]{x1D701}^{\prime }}\in {\mathcal{O}}_{\mathbf{C}}^{\times }$ such that $\bar{c}_{2,\unicode[STIX]{x1D701}^{\prime }}=\unicode[STIX]{x1D701}^{\prime }$ and

$$\begin{eqnarray}c_{2,\unicode[STIX]{x1D701}^{\prime }}^{4(q-1)}+1+\unicode[STIX]{x1D71B}^{1/(4q^{4})}c_{2,\unicode[STIX]{x1D701}^{\prime }}^{4q-3}=0,\end{eqnarray}$$

and a square root $c_{2,\unicode[STIX]{x1D701}^{\prime }}^{1/2}$ of $c_{2,\unicode[STIX]{x1D701}^{\prime }}$ . Further, we choose $c_{1,\unicode[STIX]{x1D701}^{\prime }}$ such that

$$\begin{eqnarray}c_{1,\unicode[STIX]{x1D701}^{\prime }}^{2}+\unicode[STIX]{x1D71B}^{1/(8q^{4})}c_{2,\unicode[STIX]{x1D701}^{\prime }}^{q}c_{1,\unicode[STIX]{x1D701}^{\prime }}+c_{2,\unicode[STIX]{x1D701}^{\prime }}(c_{2,\unicode[STIX]{x1D701}^{\prime }}^{2(q-1)}+1)={\displaystyle \frac{q}{2}}\unicode[STIX]{x1D71B}^{1/(4q^{4})},\end{eqnarray}$$

and $b_{2,\unicode[STIX]{x1D701}^{\prime }}$ such that $b_{2,\unicode[STIX]{x1D701}^{\prime }}^{3}=\unicode[STIX]{x1D71B}^{1/(4q^{4})}c_{2,\unicode[STIX]{x1D701}^{\prime }}^{4}$ . We put

$$\begin{eqnarray}a_{1,\unicode[STIX]{x1D701}^{\prime }}=\unicode[STIX]{x1D71B}^{1/(8q^{4})}c_{2,\unicode[STIX]{x1D701}^{\prime }}^{q},\qquad b_{1,\unicode[STIX]{x1D701}^{\prime }}=c_{2,\unicode[STIX]{x1D701}^{\prime }}^{(2q-3)/2}b_{2,\unicode[STIX]{x1D701}^{\prime }}.\end{eqnarray}$$

For each $\unicode[STIX]{x1D701}^{\prime }\in k^{\times }$ , we define a subspace ${\mathcal{D}}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }}\subset \mathbf{P}_{\unicode[STIX]{x1D701}}^{0}$ by $v(w_{1}-c_{2,\unicode[STIX]{x1D701}^{\prime }})>0$ . Furthermore, we define $\mathbf{X}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }}\subset {\mathcal{D}}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }}$ by $v(w_{1}-c_{2,\unicode[STIX]{x1D701}^{\prime }})\geqslant 1/(12q^{4})$ . We put

$$\begin{eqnarray}\mathbf{P}_{\unicode[STIX]{x1D701}}=\mathbf{P}_{\unicode[STIX]{x1D701}}^{0}\Big\backslash\mathop{\bigcup }_{\unicode[STIX]{x1D701}^{\prime }\in k^{\times }}{\mathcal{D}}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }}.\end{eqnarray}$$

We take $(\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime })\in k_{2}^{\times }\times k^{\times }$ and compute the reduction of $\mathbf{X}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }}$ . In the following, we omit the subscript $\unicode[STIX]{x1D701}^{\prime }$ of $a_{1,\unicode[STIX]{x1D701}^{\prime }}$ , $b_{1,\unicode[STIX]{x1D701}^{\prime }}$ , $b_{2,\unicode[STIX]{x1D701}^{\prime }}$ $c_{1,\unicode[STIX]{x1D701}^{\prime }}$ and $c_{2,\unicode[STIX]{x1D701}^{\prime }}$ , if there is no confusion. We change variables as

$$\begin{eqnarray}z_{f+1}=a_{1}z+b_{1}w+c_{1},\qquad w_{1}=b_{2}w+c_{2}.\end{eqnarray}$$

By substituting these into (4.31), we acquire

(4.34) $$\begin{eqnarray}a_{1}^{2}(z^{2}+z+w^{3})\equiv 0\quad \left(\text{mod}~{\displaystyle \frac{1}{4q^{4}}}+\right)\end{eqnarray}$$

by the definition of $a_{1}$ , $b_{1}$ , $b_{2}$ , $c_{1}$ and $c_{2}$ .

Proposition 4.10. For each $(\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime })\in k_{2}^{\times }\times k^{\times }$ , the reduction of $\mathbf{X}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }}$ is isomorphic to the affine curve defined by $z^{2}+z=w^{3}$ , and the complement ${\mathcal{D}}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }}\setminus \mathbf{X}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }}$ is an open annulus.

Proof. The first assertion follows from (4.34). We prove the second assertion. We change variables as

$$\begin{eqnarray}z_{f+1}=z^{\prime }+c_{2}^{(2q-3)/2}w^{\prime }+c_{1},\qquad w_{1}=w^{\prime }+c_{2}\end{eqnarray}$$

with $0<v(w^{\prime })<1/(12q^{4})$ . Substituting them into (4.31), we acquire

$$\begin{eqnarray}z^{\prime 2}\equiv c_{2}^{2(q-2)}w^{\prime 3}\quad \left(\text{mod}~2v(z^{\prime })+\right)\end{eqnarray}$$

by the choice of $c_{2}$ . Note that we have

$$\begin{eqnarray}v(z^{\prime })=3v(w^{\prime })/2<{\displaystyle \frac{1}{8q^{4}}}.\end{eqnarray}$$

By setting $z^{\prime \prime }=z^{\prime }/(c_{2}^{q-2}w^{\prime })$ , we obtain

$$\begin{eqnarray}z^{\prime \prime 2}\equiv w^{\prime }\quad \left(\text{mod}~v(w^{\prime })+\right).\end{eqnarray}$$

Then, we can see that $z^{\prime }$ and $w^{\prime }$ are written as functions of $z^{\prime \prime }$ . Hence, $({\mathcal{D}}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }}\setminus \mathbf{X}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }})(\mathbf{C})$ is identified with $\{z^{\prime \prime }\in \mathbf{C}\mid 0<v(z^{\prime \prime })<1/(24q^{4})\}$ .◻

4.5 Stable covering of $\mathbf{X}_{1}(\mathfrak{p}^{3})$

In this subsection, we show the existence of the stable covering of $\mathbf{X}_{1}(\mathfrak{p}^{3})$ over some finite extension of the base field $\widehat{K}^{\text{ur}}$ . See [Reference Coleman and McMurdyCM, Section 2.3] for the notion of semistable coverings. A semistable covering is called stable, if the corresponding semistable model is stable.

In the following, we construct a candidate of a semistable covering of $\mathbf{X}_{1}(\mathfrak{p}^{3})$ over some finite extension. We put

$$\begin{eqnarray}\displaystyle & \displaystyle \mathbf{V}_{1}=\mathbf{W}_{1,1^{\prime }}^{+}\cup \!\mathop{\bigcup }_{2\leqslant i\leqslant 6}\mathbf{W}_{i,1^{\prime }},\qquad \mathbf{V}_{2}=\mathbf{W}_{1,1^{\prime }}^{-}\cup \!\mathop{\bigcup }_{2\leqslant i\leqslant 4}\mathbf{W}_{1,i^{\prime }},\qquad & \displaystyle \nonumber\\ \displaystyle & \displaystyle \mathbf{U}=\mathbf{W}_{1,1^{\prime }}\backslash \!\mathop{\bigcup }_{\unicode[STIX]{x1D701}\in {\mathcal{S}}_{1}}\mathbf{X}_{\unicode[STIX]{x1D701}}. & \displaystyle \nonumber\end{eqnarray}$$

We note that $\mathbf{V}_{1}\supset \mathbf{Y}_{1,2}$ , $\mathbf{V}_{2}\supset \mathbf{Y}_{2,1}$ , $\mathbf{U}\supset \mathbf{Z}_{1,1}$ , $\mathbf{V}_{1}\cap \mathbf{V}_{2}=\emptyset$ , $\mathbf{V}_{1}\cap \mathbf{U}=\mathbf{W}_{1,1^{\prime }}^{+}$ and $\mathbf{V}_{2}\cap \mathbf{U}=\mathbf{W}_{1,1^{\prime }}^{-}$ .

We consider the case where $q$ is even in this paragraph. We set $\hat{{\mathcal{D}}}_{\unicode[STIX]{x1D701}}={\mathcal{D}}_{\unicode[STIX]{x1D701}}\backslash (\bigcup _{\unicode[STIX]{x1D701}^{\prime }\in k^{\times }}\mathbf{X}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }})$ for $\unicode[STIX]{x1D701}\in k_{2}^{\times }$ . Then, $\hat{{\mathcal{D}}}_{\unicode[STIX]{x1D701}}$ contains $\mathbf{P}_{\unicode[STIX]{x1D701}}$ as the underlying affinoid. On the other hand, for $(\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime })\in k_{2}^{\times }\times k^{\times }$ , the space ${\mathcal{D}}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }}$ has the underlying affinoid $\mathbf{X}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }}$ .

We put

$$\begin{eqnarray}{\mathcal{S}}=\left\{\begin{array}{@{}ll@{}}{\mathcal{S}}_{1}\quad & \text{if }q\text{ is odd},\\ k_{2}^{\times }\times k^{\times }\quad & \text{if }q\text{ is even}.\end{array}\right.\end{eqnarray}$$

Now, we define an admissible covering of $\mathbf{X}_{1}(\mathfrak{p}^{3})$ as

$$\begin{eqnarray}{\mathcal{C}}_{1}(\mathfrak{p}^{3})=\left\{\begin{array}{@{}ll@{}}\{\mathbf{V}_{1},\mathbf{V}_{2},\mathbf{U},\{{\mathcal{D}}_{\unicode[STIX]{x1D701}}\}_{\unicode[STIX]{x1D701}\in {\mathcal{S}}_{1}}\}\quad & \text{if }q\text{ is odd},\\ \{\mathbf{V}_{1},\mathbf{V}_{2},\mathbf{U},\{\hat{{\mathcal{D}}}_{\unicode[STIX]{x1D701}}\}_{\unicode[STIX]{x1D701}\in k_{2}^{\times }},\{{\mathcal{D}}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }}\}_{(\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime })\in {\mathcal{S}}}\}\quad & \text{if }q\text{ is even}.\end{array}\right.\end{eqnarray}$$

In Section 7.2, we show that ${\mathcal{C}}_{1}(\mathfrak{p}^{3})$ is a semistable covering of $\mathbf{X}_{1}(\mathfrak{p}^{3})$ over some finite extension.

6.1 Actions of the Weil group on $\overline{\mathbf{Y}}_{1,2}$ , $\overline{\mathbf{Y}}_{2,1}$ and $\overline{\mathbf{Z}}_{1,1}$

For $\unicode[STIX]{x1D70E}\in W_{K}$ , we put

$$\begin{eqnarray}\unicode[STIX]{x1D706}(\unicode[STIX]{x1D70E})=\overline{\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D71B}^{1/(q^{2}-1)})/\unicode[STIX]{x1D71B}^{1/(q^{2}-1)}}\in k_{2}^{\times }.\end{eqnarray}$$

We note that $\unicode[STIX]{x1D706}$ is not a group homomorphism in general.

Lemma 6.2. Let $\unicode[STIX]{x1D70E}\in W_{K}$ . Then, the element $\unicode[STIX]{x1D70E}$ induces the automorphisms

$$\begin{eqnarray}\displaystyle & & \displaystyle \overline{\mathbf{Y}}_{1,2}\rightarrow \overline{\mathbf{Y}}_{1,2};\qquad (x,y)\mapsto (\unicode[STIX]{x1D706}(\unicode[STIX]{x1D70E})^{q}x^{q^{-r_{\unicode[STIX]{x1D70E}}}},\unicode[STIX]{x1D706}(\unicode[STIX]{x1D70E})^{-1}y^{q^{-r_{\unicode[STIX]{x1D70E}}}}),\nonumber\\ \displaystyle & & \displaystyle \overline{\mathbf{Y}}_{2,1}\rightarrow \overline{\mathbf{Y}}_{2,1};\qquad (x,y)\mapsto (\unicode[STIX]{x1D706}(\unicode[STIX]{x1D70E})^{-1}x^{q^{-r_{\unicode[STIX]{x1D70E}}}},\unicode[STIX]{x1D706}(\unicode[STIX]{x1D70E})^{q}y^{q^{-r_{\unicode[STIX]{x1D70E}}}})\nonumber\end{eqnarray}$$

as schemes over $k$ .

Proof. We prove the claim for $\overline{\mathbf{Y}}_{1,2}$ . We set

$$\begin{eqnarray}\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D71B}^{1/(q^{3}(q^{2}-1))})=\unicode[STIX]{x1D709}\unicode[STIX]{x1D71B}^{1/(q^{3}(q^{2}-1))}\end{eqnarray}$$

with $\unicode[STIX]{x1D709}\in \unicode[STIX]{x1D707}_{q^{3}(q^{2}-1)}(K^{\text{ac}})$ . Let $P\in \mathbf{Y}_{1,2}(\mathbf{C})$ . We have $X_{3}(\unicode[STIX]{x1D70E}(P))=\unicode[STIX]{x1D70E}(X_{3}(P))$ . By applying $\unicode[STIX]{x1D70E}$ to $X_{3}(P)=\unicode[STIX]{x1D71B}^{1/(q^{3}(q^{2}-1))}x_{3}(P)$ , we obtain

$$\begin{eqnarray}x_{3}(\unicode[STIX]{x1D70E}(P))=\unicode[STIX]{x1D709}\unicode[STIX]{x1D70E}(x_{3}(P))\equiv \unicode[STIX]{x1D709}x_{3}(P)^{q^{-r_{\unicode[STIX]{x1D70E}}}}\quad \left(\text{mod}~0+\right).\end{eqnarray}$$

In the same way, we have

$$\begin{eqnarray}x_{1}(\unicode[STIX]{x1D70E}(P))\equiv \unicode[STIX]{x1D709}^{q^{4}}x_{1}(P)^{q^{-r_{\unicode[STIX]{x1D70E}}}}\quad \left(\text{mod}~0+\right).\end{eqnarray}$$

Therefore, we acquire $x^{\unicode[STIX]{x1D70E}}=\bar{\unicode[STIX]{x1D709}}x^{q^{-r_{\unicode[STIX]{x1D70E}}}}$ and $y^{\unicode[STIX]{x1D70E}}=\bar{\unicode[STIX]{x1D709}}^{-q}y^{q^{-r_{\unicode[STIX]{x1D70E}}}}$ by (4.5). Hence, the claim follows from $\bar{\unicode[STIX]{x1D709}}=\unicode[STIX]{x1D706}(\unicode[STIX]{x1D70E})^{q}$ . We can prove the claim for $\overline{\mathbf{Y}}_{2,1}$ similarly.◻

For $\unicode[STIX]{x1D70E}\in W_{K}$ , we put

$$\begin{eqnarray}\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}={\displaystyle \frac{\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D714}_{3})}{\unicode[STIX]{x1D714}_{3}}}\in \unicode[STIX]{x1D707}_{2q^{3}(q-1)}(K^{\text{ac}}).\end{eqnarray}$$

Proof. We use the notation in Section 4.3. Let $P\in \mathbf{Z}_{1,1}(\mathbf{C})$ . Since we set $X_{1}=\unicode[STIX]{x1D714}_{1}^{2q-1}x_{1}$ , $X_{2}=\unicode[STIX]{x1D714}_{1}x_{2}$ and $X_{3}=\unicode[STIX]{x1D714}_{3}x_{3}$ , we have

$$\begin{eqnarray}\displaystyle x_{1}(\unicode[STIX]{x1D70E}(P)) & = & \displaystyle \unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}^{q^{2}(2q-1)}\unicode[STIX]{x1D70E}(x_{1}(P)),\nonumber\\ \displaystyle x_{2}(\unicode[STIX]{x1D70E}(P)) & = & \displaystyle \unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}^{q^{2}}\unicode[STIX]{x1D70E}(x_{2}(P)),\nonumber\\ \displaystyle x_{3}(\unicode[STIX]{x1D70E}(P)) & = & \displaystyle \unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D70E}(x_{3}(P)).\nonumber\end{eqnarray}$$

Hence, we obtain

$$\begin{eqnarray}{\displaystyle \frac{x_{2}(\unicode[STIX]{x1D70E}(P))}{x_{1}(\unicode[STIX]{x1D70E}(P))}}=\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}^{-2q^{2}(q-1)}\unicode[STIX]{x1D70E}\biggl({\displaystyle \frac{x_{2}(P)}{x_{1}(P)}}\biggr)\equiv \unicode[STIX]{x1D70E}\biggl({\displaystyle \frac{x_{2}(P)}{x_{1}(P)}}\biggr)\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{1}+\right).\end{eqnarray}$$

Since we set $x_{2}/x_{1}=-1+\unicode[STIX]{x1D6FE}_{2}(x_{2}^{q}/t)$ , we acquire

$$\begin{eqnarray}t(\unicode[STIX]{x1D70E}(P))\equiv \unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}^{q^{3}}\unicode[STIX]{x1D70E}(t(P))\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{2}+\right).\end{eqnarray}$$

Therefore, we obtain

$$\begin{eqnarray}{\displaystyle \frac{x_{3}(\unicode[STIX]{x1D70E}(P))^{q}}{t(\unicode[STIX]{x1D70E}(P))}}=\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}^{-q(q^{2}-1)}\unicode[STIX]{x1D70E}\biggl({\displaystyle \frac{x_{3}(P)^{q}}{t(P)}}\biggr)\equiv \unicode[STIX]{x1D70E}\biggl({\displaystyle \frac{x_{3}(P)^{q}}{t(P)}}\biggr)\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{2}+\right).\end{eqnarray}$$

Since we set $1+(x_{3}^{q}/t)=\unicode[STIX]{x1D6FE}_{3}Z_{0}$ , we obtain

$$\begin{eqnarray}Z_{0}(\unicode[STIX]{x1D70E}(P))\equiv \unicode[STIX]{x1D70E}(Z_{0}(P))\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{3}+\right).\end{eqnarray}$$

Therefore, we acquire

(6.1) $$\begin{eqnarray}Z(\unicode[STIX]{x1D70E}(P))\equiv \unicode[STIX]{x1D70E}(Z(P))\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{4}+\right)\end{eqnarray}$$

by $Z_{0}+(x_{2}/x_{3})-(x_{3}/x_{1})=\unicode[STIX]{x1D6FE}_{4}(x_{2}/x_{3})Z$ .

The assertion follows from

$$\begin{eqnarray}x_{3}(\unicode[STIX]{x1D70E}(P))=\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D70E}(x_{3}(P))\equiv \unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}x_{3}(P)^{q^{-r_{\unicode[STIX]{x1D70E}}}}\quad \left(\text{mod}~0+\right)\end{eqnarray}$$

and (6.1).◻

6.2 Action of the Weil group on $\overline{\mathbf{X}}_{\unicode[STIX]{x1D702}}$

In this subsection, let $\unicode[STIX]{x1D701}\in \unicode[STIX]{x1D707}_{2(q^{2}-1)}(k^{\text{ac}})$ . Until Lemma 6.8, let $\unicode[STIX]{x1D70E}\in W_{K}$ .

6.2.1 $q$ odd

We assume that $q$ is odd. We use the notation in Section 4.4.1. By (6.1) and $x_{3}(\unicode[STIX]{x1D70E}(P))=\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D70E}(x_{3}(P))$ , we have

(6.2) $$\begin{eqnarray}\displaystyle a_{\bar{\unicode[STIX]{x1D709}}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D701}^{q^{-r_{\unicode[STIX]{x1D70E}}}}}z(\unicode[STIX]{x1D70E}(P))+c_{1,\bar{\unicode[STIX]{x1D709}}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D701}^{q^{-r_{\unicode[STIX]{x1D70E}}}}} & = & \displaystyle Z(\unicode[STIX]{x1D70E}(P))\equiv \unicode[STIX]{x1D70E}(Z(P))\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D70E}(a_{\unicode[STIX]{x1D701}})\unicode[STIX]{x1D70E}(z(P))+\unicode[STIX]{x1D70E}(c_{1,\unicode[STIX]{x1D701}})\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{4}+\right)\end{eqnarray}$$

and

(6.3) $$\begin{eqnarray}\displaystyle b_{\bar{\unicode[STIX]{x1D709}}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D701}^{q^{-r_{\unicode[STIX]{x1D70E}}}}}w(\unicode[STIX]{x1D70E}(P))+c_{2,\bar{\unicode[STIX]{x1D709}}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D701}^{q^{-r_{\unicode[STIX]{x1D70E}}}}} & = & \displaystyle x_{3}(\unicode[STIX]{x1D70E}(P))=\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D70E}(x_{3}(P))\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D70E}(b_{\unicode[STIX]{x1D701}})\unicode[STIX]{x1D70E}(w(P))+\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D70E}(c_{2,\unicode[STIX]{x1D701}})\end{eqnarray}$$

for $P\in \mathbf{X}_{\unicode[STIX]{x1D701}}(\mathbf{C})$ . Note that $c_{1,\bar{\unicode[STIX]{x1D709}}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D701}^{q^{-r_{\unicode[STIX]{x1D70E}}}}}=c_{1,\unicode[STIX]{x1D701}}$ and $c_{2,\bar{\unicode[STIX]{x1D709}}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D701}^{q^{-r_{\unicode[STIX]{x1D70E}}}}}=\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}^{q^{4}}\unicode[STIX]{x1D701}^{q^{-r_{\unicode[STIX]{x1D70E}}}-1}c_{2,\unicode[STIX]{x1D701}}$ . We have

$$\begin{eqnarray}v(\unicode[STIX]{x1D70E}(c_{1,\unicode[STIX]{x1D701}})-c_{1,\unicode[STIX]{x1D701}})\geqslant \unicode[STIX]{x1D716}_{4}\end{eqnarray}$$

by (6.2). We put

$$\begin{eqnarray}\begin{array}{@{}c@{}}a_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}}={\displaystyle \frac{\unicode[STIX]{x1D70E}(a_{\unicode[STIX]{x1D701}})}{\unicode[STIX]{x1D701}^{r_{\unicode[STIX]{x1D70E}}(q^{2}-1)}\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}^{q+1}a_{\unicode[STIX]{x1D701}}}},\qquad b_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}}={\displaystyle \frac{\unicode[STIX]{x1D70E}(c_{1,\unicode[STIX]{x1D701}})-c_{1,\unicode[STIX]{x1D701}}}{\unicode[STIX]{x1D701}^{r_{\unicode[STIX]{x1D70E}}(q^{2}-1)}\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}^{q+1}a_{\unicode[STIX]{x1D701}}}},\\ c_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}}={\displaystyle \frac{\unicode[STIX]{x1D70E}(b_{\unicode[STIX]{x1D701}})}{\unicode[STIX]{x1D701}^{(q^{-r_{\unicode[STIX]{x1D70E}}}-1)(q+3)/2}\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}^{(q+1)/2}b_{\unicode[STIX]{x1D701}}}}.\end{array}\end{eqnarray}$$

Then, we have $a_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}},b_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}},c_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}}\in {\mathcal{O}}_{K^{\text{ac}}}$ . In the following, we omit the subscript $\unicode[STIX]{x1D701}$ of $a_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}}$ , $b_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}}$ and $c_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}}$ .

Proposition 6.4. We have $\bar{a}_{\unicode[STIX]{x1D70E}}\in k^{\times }$ , $\bar{b}_{\unicode[STIX]{x1D70E}}\in k$ and $\bar{a}_{\unicode[STIX]{x1D70E}}=\bar{c}_{\unicode[STIX]{x1D70E}}^{2}$ . Further, $\unicode[STIX]{x1D70E}$ induces the morphism

$$\begin{eqnarray}\overline{\mathbf{X}}_{\unicode[STIX]{x1D701}}\rightarrow \overline{\mathbf{X}}_{\bar{\unicode[STIX]{x1D709}}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D701}^{q^{-r_{\unicode[STIX]{x1D70E}}}}};\qquad (z,w)\mapsto (\bar{a}_{\unicode[STIX]{x1D70E}}z^{q^{-r_{\unicode[STIX]{x1D70E}}}}+\bar{b}_{\unicode[STIX]{x1D70E}},\bar{c}_{\unicode[STIX]{x1D70E}}w^{q^{-r_{\unicode[STIX]{x1D70E}}}}).\end{eqnarray}$$

Proof. We have

$$\begin{eqnarray}v(\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D70E}(c_{2,\unicode[STIX]{x1D701}})-\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}^{q^{4}}\unicode[STIX]{x1D701}^{q^{-r_{\unicode[STIX]{x1D70E}}}-1}c_{2,\unicode[STIX]{x1D701}})\geqslant \unicode[STIX]{x1D716}_{3}\end{eqnarray}$$

by $v(\unicode[STIX]{x1D70E}(c_{1,\unicode[STIX]{x1D701}})-c_{1,\unicode[STIX]{x1D701}})\geqslant \unicode[STIX]{x1D716}_{4}$ . Hence, we have the last assertion by (6.2) and (6.3). By the definition of $a_{\unicode[STIX]{x1D701}}$ , $b_{\unicode[STIX]{x1D701}}$ and $c_{1,\unicode[STIX]{x1D701}}$ , we can check that

$$\begin{eqnarray}\bar{a}_{\unicode[STIX]{x1D70E}}^{q-1}=1,\qquad \bar{b}_{\unicode[STIX]{x1D70E}}^{q}=\bar{b}_{\unicode[STIX]{x1D70E}},\qquad \bar{a}_{\unicode[STIX]{x1D70E}}=\bar{c}_{\unicode[STIX]{x1D70E}}^{2}\end{eqnarray}$$

using $c_{1,\unicode[STIX]{x1D701}}^{q}\equiv -\unicode[STIX]{x1D704}(2-\unicode[STIX]{x1D6FE}_{4}c_{1,\unicode[STIX]{x1D701}})\;(\text{mod}~(q-1)/q^{4})$ .◻

We put $L=K(\unicode[STIX]{x1D71B}^{1/2})$ and $L_{2}=K_{2}(\unicode[STIX]{x1D71B}^{1/2})$ in $K^{\text{ac}}$ . Let $\text{LT}_{L_{2}}$ be the formal ${\mathcal{O}}_{L_{2}}$ -module over ${\mathcal{O}}_{L^{\text{ur}}}$ of dimension $1$ such that

$$\begin{eqnarray}\displaystyle \text{}[\unicode[STIX]{x1D71B}^{1/2}]_{\text{LT}_{L_{2}}}(X) & = & \displaystyle \unicode[STIX]{x1D71B}^{1/2}X-X^{q^{2}},\nonumber\\ \displaystyle \text{}[\unicode[STIX]{x1D701}]_{\text{LT}_{L_{2}}}(X) & = & \displaystyle \unicode[STIX]{x1D701}X\quad \text{for }\unicode[STIX]{x1D701}\in \unicode[STIX]{x1D707}_{q^{2}-1}(L_{2})\cup \{0\}.\nonumber\end{eqnarray}$$

We put $\unicode[STIX]{x1D71B}_{1,L_{2}}=\unicode[STIX]{x1D71B}^{1/(2(q^{2}-1))}$ and take $\unicode[STIX]{x1D71B}_{2,L_{2}}\in {\mathcal{O}}_{K^{\text{ac}}}$ such that $[\unicode[STIX]{x1D71B}^{1/2}]_{\text{LT}_{L_{2}}}(\unicode[STIX]{x1D71B}_{2,L_{2}})=\unicode[STIX]{x1D71B}_{1,L_{2}}$ . Let $\text{Art}_{L_{2}}:L_{2}^{\times }\stackrel{{\sim}}{\rightarrow }W_{L_{2}}^{\text{ab}}$ be the Artin reciprocity map normalized so that the image by $\text{Art}_{L_{2}}$ of a uniformizer is a lift of the geometric Frobenius. We consider the following homomorphism:

$$\begin{eqnarray}\displaystyle & & \displaystyle I_{L_{2}}\rightarrow k_{2}^{\times }\times k_{2};\qquad \unicode[STIX]{x1D70E}\mapsto (\unicode[STIX]{x1D706}_{1}(\unicode[STIX]{x1D70E}),\unicode[STIX]{x1D706}_{2}(\unicode[STIX]{x1D70E}))\nonumber\\ \displaystyle & & \displaystyle \quad =\left(\overline{{\displaystyle \frac{\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D71B}_{1,L_{2}})}{\unicode[STIX]{x1D71B}_{1,L_{2}}}}},\overline{{\displaystyle \frac{\unicode[STIX]{x1D71B}_{1,L_{2}}\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D71B}_{2,L_{2}})-\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D71B}_{1,L_{2}})\unicode[STIX]{x1D71B}_{2,L_{2}}}{\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D71B}_{1,L_{2}})\unicode[STIX]{x1D71B}_{1,L_{2}}}}}\right).\nonumber\end{eqnarray}$$

This map is equal to the composite

$$\begin{eqnarray}I_{L_{2}}\rightarrow {\mathcal{O}}_{L_{2}}^{\times }\rightarrow k_{2}^{\times }\times k_{2},\end{eqnarray}$$

where the first homomorphism is induced from the inverse of $\text{Art}_{L_{2}}$ , and the second homomorphism is given by $a+b\unicode[STIX]{x1D71B}^{1/2}\mapsto (\bar{a},\bar{b}/\bar{a})$ for $a\in \unicode[STIX]{x1D707}_{q^{2}-1}(L_{2})$ and $b\in {\mathcal{O}}_{L_{2}}$ . Then, we rewrite Proposition 6.4 as follows.

Corollary 6.5. Let $\unicode[STIX]{x1D70E}\in I_{L}$ . We put

$$\begin{eqnarray}g_{0}=2\unicode[STIX]{x1D701}^{-(q+1)}(\unicode[STIX]{x1D706}_{2}(\unicode[STIX]{x1D70E})^{q}+\unicode[STIX]{x1D701}^{q^{2}-1}\unicode[STIX]{x1D706}_{2}(\unicode[STIX]{x1D70E}))\in k.\end{eqnarray}$$

Then, $\unicode[STIX]{x1D70E}$ induces the morphism

$$\begin{eqnarray}\overline{\mathbf{X}}_{\unicode[STIX]{x1D701}}\rightarrow \overline{\mathbf{X}}_{\unicode[STIX]{x1D706}_{1}(\unicode[STIX]{x1D70E})^{q+1}\unicode[STIX]{x1D701}};\qquad (z,w)\mapsto (\unicode[STIX]{x1D706}_{1}(\unicode[STIX]{x1D70E})^{-2(q+1)}(z+g_{0}),\unicode[STIX]{x1D706}_{1}(\unicode[STIX]{x1D70E})^{-(q+1)}w).\end{eqnarray}$$

Proof. We can check that $\bar{a}_{\unicode[STIX]{x1D70E}}=\unicode[STIX]{x1D706}_{1}(\unicode[STIX]{x1D70E})^{-2(q+1)}$ and $\bar{c}_{\unicode[STIX]{x1D70E}}=\unicode[STIX]{x1D706}_{1}(\unicode[STIX]{x1D70E})^{-(q+1)}$ easily. We prove that

$$\begin{eqnarray}\bar{b}_{\unicode[STIX]{x1D70E}}=\unicode[STIX]{x1D706}_{1}(\unicode[STIX]{x1D70E})^{-2(q+1)}g_{0}.\end{eqnarray}$$

We simply write $\unicode[STIX]{x1D71B}_{i}$ for $\unicode[STIX]{x1D71B}_{i,L_{2}}$ . We put $\unicode[STIX]{x1D704}=\unicode[STIX]{x1D701}^{q^{2}-1}$ and

$$\begin{eqnarray}C=\unicode[STIX]{x1D71B}_{1}^{(q^{2}-1)/q}\biggl\{\biggl({\displaystyle \frac{\unicode[STIX]{x1D71B}_{2}}{\unicode[STIX]{x1D71B}_{1}}}\biggr)^{q}+\unicode[STIX]{x1D704}\biggl({\displaystyle \frac{\unicode[STIX]{x1D71B}_{2}}{\unicode[STIX]{x1D71B}_{1}}}\biggr)\biggr\}.\end{eqnarray}$$

Then, we have

$$\begin{eqnarray}C^{q}-\unicode[STIX]{x1D704}\unicode[STIX]{x1D6FE}_{1}C\equiv -1\quad \left(\text{mod}\,{\displaystyle \frac{1}{2}}+\right)\end{eqnarray}$$

by $\unicode[STIX]{x1D71B}_{2}^{q^{2}}-\unicode[STIX]{x1D71B}^{1/2}\unicode[STIX]{x1D71B}_{2}=-\unicode[STIX]{x1D71B}_{1}$ . We can easily check the equality

$$\begin{eqnarray}\unicode[STIX]{x1D70E}(C)-C\equiv \unicode[STIX]{x1D71B}^{\unicode[STIX]{x1D716}_{1}}(\unicode[STIX]{x1D706}_{2}(\unicode[STIX]{x1D70E})^{q}+\unicode[STIX]{x1D704}\unicode[STIX]{x1D706}_{2}(\unicode[STIX]{x1D70E}))\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{1}+\right).\end{eqnarray}$$

On the other hand, we can check

$$\begin{eqnarray}c_{1,\unicode[STIX]{x1D701}}^{q}\equiv -\unicode[STIX]{x1D704}(2-\unicode[STIX]{x1D6FE}_{4}c_{1,\unicode[STIX]{x1D701}})\quad \left(\text{mod}~{\displaystyle \frac{q-1}{2q^{4}}}+\right)\end{eqnarray}$$

by the definition of $c_{1,\unicode[STIX]{x1D701}}$ . Therefore, the elements $C$ and $c_{1,\unicode[STIX]{x1D701}}^{q^{3}}/(2\unicode[STIX]{x1D704})$ satisfy

$$\begin{eqnarray}x^{q}-\unicode[STIX]{x1D704}\unicode[STIX]{x1D6FE}_{1}x\equiv -1\quad \left(\text{mod}~{\displaystyle \frac{1}{2}}+\right).\end{eqnarray}$$

Hence, we obtain $C\equiv c_{1,\unicode[STIX]{x1D701}}^{q^{3}}/(2\unicode[STIX]{x1D704})\;(\text{mod}~\unicode[STIX]{x1D716}_{1}+)$ . This implies that

$$\begin{eqnarray}(\unicode[STIX]{x1D70E}(c_{1,\unicode[STIX]{x1D701}})-c_{1,\unicode[STIX]{x1D701}})^{q^{3}}\equiv 2\unicode[STIX]{x1D704}(\unicode[STIX]{x1D70E}(C)-C)\quad \left(\text{mod}~\unicode[STIX]{x1D716}_{1}+\right).\end{eqnarray}$$

Therefore, we obtain

$$\begin{eqnarray}\bar{b}_{\unicode[STIX]{x1D70E}}\equiv \bar{b}_{\unicode[STIX]{x1D70E}}^{q^{3}}\equiv \unicode[STIX]{x1D706}_{1}(\unicode[STIX]{x1D70E})^{-2(q+1)}g_{0}\quad \left(\text{mod}~0+\right)\end{eqnarray}$$

by $\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E}}=\unicode[STIX]{x1D706}_{1}(\unicode[STIX]{x1D70E})^{q+1}\;(\text{mod}~0+)$ .◻

6.2.2 $q$ even

We assume that $q$ is even. We use the notation in Section 4.4.2. For $P\in \mathbf{P}^{0}(\mathbf{C})$ , we have

(6.4) $$\begin{eqnarray}w_{1}(\unicode[STIX]{x1D70E}(P))\equiv \unicode[STIX]{x1D70E}(w_{1}(P))\quad \left(\text{mod}~{\displaystyle \frac{1}{4q^{4}}}+\right)\end{eqnarray}$$

by (6.1). We can see that

(6.5) $$\begin{eqnarray}z_{f+1}(\unicode[STIX]{x1D70E}(P))\equiv \unicode[STIX]{x1D70E}(z_{f+1}(P))\quad \left(\text{mod}~{\displaystyle \frac{1}{8q^{4}}}+\right)\end{eqnarray}$$

using (4.31) and (6.4).

Lemma 6.6. The element $\unicode[STIX]{x1D70E}$ induces the morphism

$$\begin{eqnarray}\overline{\mathbf{P}}_{\unicode[STIX]{x1D701}}\rightarrow \overline{\mathbf{P}}_{\bar{\unicode[STIX]{x1D709}}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D701}^{q^{-r_{\unicode[STIX]{x1D70E}}}}};\qquad w_{1}\mapsto w_{1}^{q^{-r_{\unicode[STIX]{x1D70E}}}}.\end{eqnarray}$$

Proof. This follows from Lemma 6.3 and (6.4).◻

We take $\unicode[STIX]{x1D701}^{\prime }\in k^{\times }$ . By (6.4) and (6.5), we have

(6.6) $$\begin{eqnarray}\displaystyle & & \displaystyle a_{1,\unicode[STIX]{x1D701}^{\prime }}z(\unicode[STIX]{x1D70E}(P))+b_{1,\unicode[STIX]{x1D701}^{\prime }}w(\unicode[STIX]{x1D70E}(P))+c_{1,\unicode[STIX]{x1D701}^{\prime }}\nonumber\\ \displaystyle & & \displaystyle \quad \equiv a_{1,\unicode[STIX]{x1D701}^{\prime }}\unicode[STIX]{x1D70E}(z(P))+\unicode[STIX]{x1D70E}(b_{1,\unicode[STIX]{x1D701}^{\prime }})\unicode[STIX]{x1D70E}(w(P))+\unicode[STIX]{x1D70E}(c_{1,\unicode[STIX]{x1D701}^{\prime }})\quad \left(\text{mod}~{\displaystyle \frac{1}{8q^{4}}}+\right)\end{eqnarray}$$

and

(6.7) $$\begin{eqnarray}b_{2,\unicode[STIX]{x1D701}^{\prime }}w(\unicode[STIX]{x1D70E}(P))+c_{2,\unicode[STIX]{x1D701}^{\prime }}\equiv \unicode[STIX]{x1D70E}(b_{2,\unicode[STIX]{x1D701}^{\prime }})\unicode[STIX]{x1D70E}(w(P))+\unicode[STIX]{x1D70E}(c_{2,\unicode[STIX]{x1D701}^{\prime }})\quad \left(\text{mod}~{\displaystyle \frac{1}{4q^{3}}}+\right)\end{eqnarray}$$

using $\unicode[STIX]{x1D70E}(a_{1,\unicode[STIX]{x1D701}^{\prime }})\equiv a_{1,\unicode[STIX]{x1D701}^{\prime }}\;(\text{mod}~1/(8q^{4})+)$ . We put

$$\begin{eqnarray}\displaystyle & \displaystyle a_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}^{\prime }}={\displaystyle \frac{\unicode[STIX]{x1D70E}(b_{2,\unicode[STIX]{x1D701}^{\prime }})}{b_{2,\unicode[STIX]{x1D701}^{\prime }}}},\qquad b_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}^{\prime }}={\displaystyle \frac{\unicode[STIX]{x1D70E}(b_{1,\unicode[STIX]{x1D701}^{\prime }})b_{2,\unicode[STIX]{x1D701}^{\prime }}-b_{1,\unicode[STIX]{x1D701}^{\prime }}\unicode[STIX]{x1D70E}(b_{2,\unicode[STIX]{x1D701}^{\prime }})}{a_{1,\unicode[STIX]{x1D701}^{\prime }}b_{2,\unicode[STIX]{x1D701}^{\prime }}}}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle b_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}^{\prime }}^{\prime }={\displaystyle \frac{\unicode[STIX]{x1D70E}(c_{2,\unicode[STIX]{x1D701}^{\prime }})-c_{2,\unicode[STIX]{x1D701}^{\prime }}}{b_{2,\unicode[STIX]{x1D701}^{\prime }}}},\qquad c_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}^{\prime }}={\displaystyle \frac{\unicode[STIX]{x1D70E}(c_{1,\unicode[STIX]{x1D701}^{\prime }})-c_{1,\unicode[STIX]{x1D701}^{\prime }}-b_{1,\unicode[STIX]{x1D701}^{\prime }}b_{2,\unicode[STIX]{x1D701}^{\prime }}^{-1}(\unicode[STIX]{x1D70E}(c_{2,\unicode[STIX]{x1D701}^{\prime }})-c_{2,\unicode[STIX]{x1D701}^{\prime }})}{a_{1,\unicode[STIX]{x1D701}^{\prime }}}}. & \displaystyle \nonumber\end{eqnarray}$$

In the following, we omit the subscript $\unicode[STIX]{x1D701}^{\prime }$ of $a_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}^{\prime }}$ , $b_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}^{\prime }}$ , $b_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}^{\prime }}^{\prime }$ and $c_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D701}^{\prime }}$ . We note that $v(a_{\unicode[STIX]{x1D70E}})=0$ . We have $v(b_{\unicode[STIX]{x1D70E}}^{\prime })\geqslant 0$ by (6.7). This implies that $v(b_{\unicode[STIX]{x1D70E}})\geqslant 0$ . By (6.6) and (6.7), we obtain $v(c_{\unicode[STIX]{x1D70E}})\geqslant 0$ using $v(b_{\unicode[STIX]{x1D70E}})\geqslant 0$ .

Proposition 6.7. The element $\unicode[STIX]{x1D70E}$ induces the morphism

$$\begin{eqnarray}\overline{\mathbf{X}}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }}\rightarrow \overline{\mathbf{X}}_{\bar{\unicode[STIX]{x1D709}}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D701}^{q^{-r_{\unicode[STIX]{x1D70E}}}},\unicode[STIX]{x1D701}^{\prime }};\qquad (z,w)\mapsto (z^{q^{-r_{\unicode[STIX]{x1D70E}}}}+\bar{b}_{\unicode[STIX]{x1D70E}}w^{q^{-r_{\unicode[STIX]{x1D70E}}}}+\bar{c}_{\unicode[STIX]{x1D70E}},\bar{a}_{\unicode[STIX]{x1D70E}}w^{q^{-r_{\unicode[STIX]{x1D70E}}}}+\bar{b}_{\unicode[STIX]{x1D70E}}^{\prime }).\end{eqnarray}$$

Proof. This follows from (6.6) and (6.7).◻

In the following, we simplify the description of $\bar{a}_{\unicode[STIX]{x1D70E}}$ , $\bar{b}_{\unicode[STIX]{x1D70E}}$ , $\bar{b}_{\unicode[STIX]{x1D70E}}^{\prime }$ and $\bar{c}_{\unicode[STIX]{x1D70E}}$ . Let $\tilde{\unicode[STIX]{x1D701}}^{\prime }\in \unicode[STIX]{x1D707}_{q-1}(K)$ be the lift of $\unicode[STIX]{x1D701}^{\prime }$ . We put

$$\begin{eqnarray}h_{\unicode[STIX]{x1D701}^{\prime }}(x)=x^{4}-\unicode[STIX]{x1D71B}^{1/4}\tilde{\unicode[STIX]{x1D701}}^{\prime 4}x-\tilde{\unicode[STIX]{x1D701}}^{\prime 4}.\end{eqnarray}$$

Lemma 6.8. There is a root $\unicode[STIX]{x1D6FF}_{1}$ of $h_{\unicode[STIX]{x1D701}^{\prime }}(x)=0$ such that

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{1}\equiv c_{2,\unicode[STIX]{x1D701}^{\prime }}^{q^{4}}+{\displaystyle \frac{q}{2}}\unicode[STIX]{x1D71B}^{1/4}\tilde{\unicode[STIX]{x1D701}}^{\prime 2}\quad \left(\text{mod}~{\displaystyle \frac{1}{4}}+\right).\end{eqnarray}$$

Proof. We put

$$\begin{eqnarray}h(x)=x^{4(q-1)}+1+\unicode[STIX]{x1D71B}^{1/4}x^{4q-3}.\end{eqnarray}$$

By the definition of $c_{2,\unicode[STIX]{x1D701}^{\prime }}$ , we have $h(c_{2,\unicode[STIX]{x1D701}^{\prime }}^{q^{4}})\equiv 0\;(\text{mod}~1)$ . Hence, we have a root $c_{2}^{\prime }$ of $h$ such that $c_{2}^{\prime }\equiv c_{2,\unicode[STIX]{x1D701}^{\prime }}^{q^{4}}\;(\text{mod}~3/4)$ by Newton’s method. We can check that

$$\begin{eqnarray}c_{2}^{\prime }\equiv \tilde{\unicode[STIX]{x1D701}}^{\prime }+\unicode[STIX]{x1D71B}^{1/16}\tilde{\unicode[STIX]{x1D701}}^{\prime 5/4}\quad \left(\text{mod}~{\displaystyle \frac{1}{16}}+\right).\end{eqnarray}$$

We define a parameter $s$ with $v(s)\geqslant 1/16$ by $x=\tilde{\unicode[STIX]{x1D701}}^{\prime }+s$ . Then, we have

$$\begin{eqnarray}\displaystyle h(\tilde{\unicode[STIX]{x1D701}}^{\prime }+s) & \equiv & \displaystyle \tilde{\unicode[STIX]{x1D701}}^{\prime -4}s^{4}+\left(\frac{q}{2}-1\right)\tilde{\unicode[STIX]{x1D701}}^{\prime -8}s^{8}+\unicode[STIX]{x1D71B}^{1/4}(\tilde{\unicode[STIX]{x1D701}}^{\prime }+s)\nonumber\\ \displaystyle & \equiv & \displaystyle \tilde{\unicode[STIX]{x1D701}}^{\prime -4}h_{\unicode[STIX]{x1D701}^{\prime }}(x)+\left({\displaystyle \frac{q}{2}}-1\right)\tilde{\unicode[STIX]{x1D701}}^{\prime -8}s^{8}+\unicode[STIX]{x1D71B}^{1/4}s^{4}\tilde{\unicode[STIX]{x1D701}}^{\prime -3}\quad \left(\text{mod}~{\displaystyle \frac{1}{2}}+\right).\nonumber\end{eqnarray}$$

This implies that

$$\begin{eqnarray}h_{\unicode[STIX]{x1D701}^{\prime }}(c_{2}^{\prime })\equiv {\displaystyle \frac{q}{2}}\unicode[STIX]{x1D71B}^{1/2}\tilde{\unicode[STIX]{x1D701}}^{\prime 6}\quad \left(\text{mod}\,{\displaystyle \frac{1}{2}}+\right).\end{eqnarray}$$

Therefore, we have a root $\unicode[STIX]{x1D6FF}_{1}$ of $h_{\unicode[STIX]{x1D701}^{\prime }}(x)=0$ such that

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{1}\equiv c_{2}^{\prime }+{\displaystyle \frac{q}{2}}\unicode[STIX]{x1D71B}^{1/4}\tilde{\unicode[STIX]{x1D701}}^{\prime 2}\quad \left(\text{mod}~{\displaystyle \frac{1}{4}}+\right)\end{eqnarray}$$

by Newton’s method. ◻

By the definition of $b_{2,\unicode[STIX]{x1D701}^{\prime }}$ , we have

$$\begin{eqnarray}b_{2,\unicode[STIX]{x1D701}^{\prime }}^{3q^{4}}\unicode[STIX]{x1D71B}^{-1/4}\equiv \tilde{\unicode[STIX]{x1D701}}^{\prime 4}\quad \left(\text{mod}~0+\right).\end{eqnarray}$$

Let $\unicode[STIX]{x1D701}^{\prime \prime }$ be the element of $\unicode[STIX]{x1D707}_{3(q-1)}(K^{\text{ur}})$ satisfying $\unicode[STIX]{x1D701}^{\prime \prime }\equiv b_{2,\unicode[STIX]{x1D701}^{\prime }}^{q^{4}}\unicode[STIX]{x1D71B}^{-1/12}\;(\text{mod}~0+)$ . Note that $\unicode[STIX]{x1D701}^{\prime \prime 3}=\tilde{\unicode[STIX]{x1D701}}^{\prime 4}$ . We take $\unicode[STIX]{x1D6FF}_{1}$ as in Lemma 6.8 and put $\unicode[STIX]{x1D6FF}=\unicode[STIX]{x1D6FF}_{1}/(\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/12})$ . Then, we have

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}^{4}-\unicode[STIX]{x1D6FF}={\displaystyle \frac{1}{\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/3}}}.\end{eqnarray}$$

Note that $v(\unicode[STIX]{x1D6FF})=-1/12$ . We take $\unicode[STIX]{x1D701}_{3}\in \unicode[STIX]{x1D707}_{3}(K^{\text{ur}})$ such that $\unicode[STIX]{x1D701}_{3}\neq 1$ , and put

$$\begin{eqnarray}h_{\unicode[STIX]{x1D6FF}_{1}}(x)=x^{2}-(1+2\unicode[STIX]{x1D701}_{3})\unicode[STIX]{x1D71B}^{1/4}\unicode[STIX]{x1D6FF}_{1}^{2q}x-\unicode[STIX]{x1D71B}^{1/4}\unicode[STIX]{x1D6FF}_{1}^{4q-1}(1+2\unicode[STIX]{x1D71B}^{1/4}\unicode[STIX]{x1D6FF}_{1}).\end{eqnarray}$$

Lemma 6.9. There is a root $\unicode[STIX]{x1D703}_{1}$ of $h_{\unicode[STIX]{x1D6FF}_{1}}(x)=0$ such that

$$\begin{eqnarray}\unicode[STIX]{x1D703}_{1}\equiv c_{1,\unicode[STIX]{x1D701}^{\prime }}^{2q^{4}}\quad \left(\text{mod}~\frac{1}{4}+\right).\end{eqnarray}$$

Proof. By the definition of $c_{1,\unicode[STIX]{x1D701}^{\prime }}$ and $c_{2,\unicode[STIX]{x1D701}^{\prime }}$ , we have $h_{\unicode[STIX]{x1D6FF}_{1}}(c_{1,\unicode[STIX]{x1D701}^{\prime }}^{2q^{4}})\equiv 0\;(\text{mod}~1/2+)$ . Hence, we can show the claim using Newton’s method.◻

We take $\unicode[STIX]{x1D703}_{1}$ as in Lemma 6.9 and put

$$\begin{eqnarray}\unicode[STIX]{x1D703}={\displaystyle \frac{\unicode[STIX]{x1D703}_{1}}{\unicode[STIX]{x1D71B}^{1/4}\unicode[STIX]{x1D6FF}_{1}^{2q}}}-\unicode[STIX]{x1D701}_{3}.\end{eqnarray}$$

Then, we have $\unicode[STIX]{x1D703}^{2}-\unicode[STIX]{x1D703}=\unicode[STIX]{x1D6FF}^{3}$ . Note that $v(\unicode[STIX]{x1D703})=-1/8$ . Let $\unicode[STIX]{x1D70E}\in W_{K}$ in this paragraph. We put

$$\begin{eqnarray}\unicode[STIX]{x1D701}_{3,\unicode[STIX]{x1D70E}}={\displaystyle \frac{\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/3})}{\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/3}}}.\end{eqnarray}$$

We take $\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}\in \unicode[STIX]{x1D707}_{3}(K^{\text{ur}})\cup \{0\}$ such that $\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FF})\equiv \unicode[STIX]{x1D701}_{3,\unicode[STIX]{x1D70E}}^{-1}(\unicode[STIX]{x1D6FF}+\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}})\;(\text{mod}~5/6)$ . Then, we have

(6.8) $$\begin{eqnarray}\begin{array}{@{}c@{}}\displaystyle (\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D703})-\unicode[STIX]{x1D703}+\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}^{2}\unicode[STIX]{x1D6FF})^{2}\equiv \unicode[STIX]{x1D70E}(\unicode[STIX]{x1D703})-\unicode[STIX]{x1D703}+\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}^{2}\unicode[STIX]{x1D6FF}+\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}^{3},\\ \displaystyle (\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D703})-\unicode[STIX]{x1D703}+\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}^{2}\unicode[STIX]{x1D6FF}+\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}^{3})^{2}\equiv \unicode[STIX]{x1D70E}(\unicode[STIX]{x1D703})-\unicode[STIX]{x1D703}+\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}^{2}\unicode[STIX]{x1D6FF}\quad \left(\text{mod}~0+\right).\end{array}\end{eqnarray}$$

By these equations, we can take $\unicode[STIX]{x1D707}_{\unicode[STIX]{x1D70E}}\in \unicode[STIX]{x1D707}_{3}(K^{\text{ur}})\cup \{0\}$ such that

$$\begin{eqnarray}\unicode[STIX]{x1D707}_{\unicode[STIX]{x1D70E}}\equiv \unicode[STIX]{x1D70E}(\unicode[STIX]{x1D703})-\unicode[STIX]{x1D703}+\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}^{2}\unicode[STIX]{x1D6FF}+\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}^{3}+\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D701}_{3})-\unicode[STIX]{x1D701}_{3}\quad \left(\text{mod}~0+\right).\end{eqnarray}$$

Then, we have $\unicode[STIX]{x1D707}_{\unicode[STIX]{x1D70E}}^{2}+\unicode[STIX]{x1D707}_{\unicode[STIX]{x1D70E}}\equiv \unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}^{3}\;(\text{mod}~1)$ by (6.8) and $\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}},\unicode[STIX]{x1D707}_{\unicode[STIX]{x1D70E}}\in \unicode[STIX]{x1D707}_{3}(K^{\text{ur}})\cup \{0\}$ .

Lemma 6.10.

1. (1) Let $\unicode[STIX]{x1D70E}\in W_{K}$ . Then, we have

   $$\begin{eqnarray}a_{\unicode[STIX]{x1D70E}}\equiv \unicode[STIX]{x1D701}_{3,\unicode[STIX]{x1D70E}},\qquad b_{\unicode[STIX]{x1D70E}}\equiv \unicode[STIX]{x1D701}_{3,\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}^{2},\qquad b_{\unicode[STIX]{x1D70E}}^{\prime }\equiv \unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}},\qquad c_{\unicode[STIX]{x1D70E}}\equiv \unicode[STIX]{x1D707}_{\unicode[STIX]{x1D70E}}\quad \left(\text{mod}~0+\right).\end{eqnarray}$$

2. (2) Let $\unicode[STIX]{x1D70E}\in W_{K}$ . Then we have $\bar{a}_{\unicode[STIX]{x1D70E}}\in \mathbb{F}_{4}^{\times }$ and $\bar{b}_{\unicode[STIX]{x1D70E}},\bar{b}_{\unicode[STIX]{x1D70E}}^{\prime },\bar{c}_{\unicode[STIX]{x1D70E}}\in \mathbb{F}_{4}$ . Further, $\bar{a}_{\unicode[STIX]{x1D70E}}\bar{b}_{\unicode[STIX]{x1D70E}}^{2}=\bar{b}_{\unicode[STIX]{x1D70E}}^{\prime }$ and $\bar{b}_{\unicode[STIX]{x1D70E}}^{3}=\bar{c}_{\unicode[STIX]{x1D70E}}^{2}+\bar{c}_{\unicode[STIX]{x1D70E}}$ hold.

Proof. By the definition of $b_{2,\unicode[STIX]{x1D701}^{\prime }}$ , we have

$$\begin{eqnarray}a_{\unicode[STIX]{x1D70E}}^{4q^{4}}\equiv {\displaystyle \frac{\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D701}^{\prime \prime 4}\unicode[STIX]{x1D71B}^{1/3})}{\unicode[STIX]{x1D701}^{\prime \prime 4}\unicode[STIX]{x1D71B}^{1/3}}}\quad \left(\text{mod}~0+\right).\end{eqnarray}$$

Hence, we have $\bar{a}_{\unicode[STIX]{x1D70E}}^{4q^{4}}=\bar{\unicode[STIX]{x1D701}}_{3,\unicode[STIX]{x1D70E}}\in \mathbb{F}_{4}^{\times }$ . This implies that $\bar{a}_{\unicode[STIX]{x1D70E}}=\bar{\unicode[STIX]{x1D701}}_{3,\unicode[STIX]{x1D70E}}\in \mathbb{F}_{4}^{\times }$ .

By the definition of $a_{1,\unicode[STIX]{x1D701}^{\prime }}$ and $b_{1,\unicode[STIX]{x1D701}^{\prime }}$ , we have

$$\begin{eqnarray}\displaystyle b_{\unicode[STIX]{x1D70E}}^{2q^{4}} & \equiv & \displaystyle {\displaystyle \frac{\unicode[STIX]{x1D70E}(b_{2,\unicode[STIX]{x1D701}^{\prime }}^{2q^{4}})(\unicode[STIX]{x1D70E}(c_{2,\unicode[STIX]{x1D701}^{\prime }}^{q^{4}(2q-3)})-c_{2,\unicode[STIX]{x1D701}^{\prime }}^{q^{4}(2q-3)})}{\unicode[STIX]{x1D71B}^{1/4}c_{2,\unicode[STIX]{x1D701}^{\prime }}^{2q^{5}}}}\equiv {\displaystyle \frac{a_{\unicode[STIX]{x1D70E}}^{2q^{4}}b_{2,\unicode[STIX]{x1D701}^{\prime }}^{2q^{4}}(\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FF}_{1}^{2q-3})-\unicode[STIX]{x1D6FF}_{1}^{2q-3})}{\unicode[STIX]{x1D71B}^{1/4}\unicode[STIX]{x1D6FF}_{1}^{2q}}}\nonumber\\ \displaystyle & \equiv & \displaystyle {\displaystyle \frac{\unicode[STIX]{x1D701}_{3,\unicode[STIX]{x1D70E}}^{2}\unicode[STIX]{x1D701}^{\prime \prime 2}(\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FF}_{1})-\unicode[STIX]{x1D6FF}_{1})}{\unicode[STIX]{x1D71B}^{1/12}\unicode[STIX]{x1D6FF}_{1}^{4}}}\equiv \unicode[STIX]{x1D701}_{3,\unicode[STIX]{x1D70E}}^{2}\biggl({\displaystyle \frac{\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/12})}{\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/12}}}\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FF})-\unicode[STIX]{x1D6FF}\biggr)\nonumber\\ \displaystyle & \equiv & \displaystyle \unicode[STIX]{x1D701}_{3,\unicode[STIX]{x1D70E}}^{2}\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}\quad \left(\text{mod}~0+\right),\nonumber\end{eqnarray}$$

where we use Lemma 6.8 in the second congruence, $b_{2,\unicode[STIX]{x1D701}^{\prime }}^{q^{4}}/\unicode[STIX]{x1D71B}^{1/12}\equiv \unicode[STIX]{x1D701}^{\prime \prime }\;(\text{mod}~0+)$ in the third congruence, $\unicode[STIX]{x1D6FF}_{1}^{4}=\tilde{\unicode[STIX]{x1D701}}^{\prime 4}\;(\text{mod}~1/4)$ and $\unicode[STIX]{x1D701}^{\prime \prime 3}=\tilde{\unicode[STIX]{x1D701}}^{\prime 4}$ in the fourth congruence, and $\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/12})/(\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/12})\equiv \unicode[STIX]{x1D701}_{3,\unicode[STIX]{x1D70E}}\;(\text{mod}~0+)$ in the last congruence. Hence, we obtain $\bar{b}_{\unicode[STIX]{x1D70E}}=\bar{\unicode[STIX]{x1D701}}_{3,\unicode[STIX]{x1D70E}}\bar{\unicode[STIX]{x1D708}}_{\unicode[STIX]{x1D70E}}^{2}\in \mathbb{F}_{4}$ .

By Lemma 6.8 and $b_{2,\unicode[STIX]{x1D701}^{\prime }}^{q^{4}}/\unicode[STIX]{x1D71B}^{1/12}\equiv \unicode[STIX]{x1D701}^{\prime \prime }\;(\text{mod}~0+)$ , we have

$$\begin{eqnarray}b_{\unicode[STIX]{x1D70E}}^{\prime q^{4}}\equiv {\displaystyle \frac{\unicode[STIX]{x1D70E}(c_{2,\unicode[STIX]{x1D701}^{\prime }}^{q^{4}})-c_{2,\unicode[STIX]{x1D701}^{\prime }}^{q^{4}}}{b_{2,\unicode[STIX]{x1D701}^{\prime }}^{q^{4}}}}\equiv {\displaystyle \frac{\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FF}_{1})-\unicode[STIX]{x1D6FF}_{1}}{\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/12}}}={\displaystyle \frac{\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/12})}{\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/12}}}\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FF})-\unicode[STIX]{x1D6FF}\equiv \unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}\quad \left(\text{mod}~0+\right).\end{eqnarray}$$

Hence, we obtain $\bar{b}_{\unicode[STIX]{x1D70E}}^{\prime }=\bar{\unicode[STIX]{x1D708}}_{\unicode[STIX]{x1D70E}}\in \mathbb{F}_{4}$ .

By Lemmas 6.8 and 6.9 and the definition of $a_{1,\unicode[STIX]{x1D701}^{\prime }}$ , we have

$$\begin{eqnarray}\displaystyle c_{\unicode[STIX]{x1D70E}}^{2q^{4}} & \equiv & \displaystyle {\displaystyle \frac{\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D703}_{1})-\unicode[STIX]{x1D703}_{1}-\unicode[STIX]{x1D6FF}_{1}^{2q-3}(\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FF}_{1}^{2})-\unicode[STIX]{x1D6FF}_{1}^{2})}{\unicode[STIX]{x1D71B}^{1/4}\unicode[STIX]{x1D6FF}_{1}^{2q}}}\nonumber\\ \displaystyle & \equiv & \displaystyle {\displaystyle \frac{\unicode[STIX]{x1D6FF}_{1}^{-2q}\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FF}_{1}^{2q}\unicode[STIX]{x1D71B}^{1/4}(\unicode[STIX]{x1D703}+\unicode[STIX]{x1D701}_{3}))-\unicode[STIX]{x1D71B}^{1/4}(\unicode[STIX]{x1D703}+\unicode[STIX]{x1D701}_{3})-\unicode[STIX]{x1D6FF}_{1}^{-3}(\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FF}_{1})-\unicode[STIX]{x1D6FF}_{1})^{2}}{\unicode[STIX]{x1D71B}^{1/4}}}\nonumber\\ \displaystyle & \equiv & \displaystyle {\displaystyle \frac{\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D71B}^{1/4}(\unicode[STIX]{x1D703}+\unicode[STIX]{x1D701}_{3}))-\unicode[STIX]{x1D71B}^{1/4}(\unicode[STIX]{x1D703}+\unicode[STIX]{x1D701}_{3})-\unicode[STIX]{x1D71B}^{1/12}\unicode[STIX]{x1D6FF}(\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D71B}^{1/12}\unicode[STIX]{x1D6FF})-\unicode[STIX]{x1D71B}^{1/12}\unicode[STIX]{x1D6FF})^{2}}{\unicode[STIX]{x1D71B}^{1/4}}}\nonumber\\ \displaystyle & \equiv & \displaystyle \unicode[STIX]{x1D70E}(\unicode[STIX]{x1D703})-\unicode[STIX]{x1D703}+\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}^{2}\unicode[STIX]{x1D6FF}+\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D701}_{3})-\unicode[STIX]{x1D701}_{3}\quad \left(\text{mod}~0+\right),\nonumber\end{eqnarray}$$

where we use $\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FF}_{1})\equiv \unicode[STIX]{x1D6FF}_{1}\;(\text{mod}~1/4)$ in the second congruence, and $\unicode[STIX]{x1D6FF}_{1}^{4}=\tilde{\unicode[STIX]{x1D701}}^{\prime 4}\;(\text{mod}~1/4)$ in the third congruence. Then, we have $\bar{c}_{\unicode[STIX]{x1D70E}}^{2q^{4}}\in \mathbb{F}_{4}$ by (6.8). Hence, we have $\bar{c}_{\unicode[STIX]{x1D70E}}\in \mathbb{F}_{4}$ and $c_{\unicode[STIX]{x1D70E}}\equiv \unicode[STIX]{x1D707}_{\unicode[STIX]{x1D70E}}\;(\text{mod}~0+)$ , again by (6.8).

By the above calculations, we can easily check that $\bar{a}_{\unicode[STIX]{x1D70E}}\bar{b}_{\unicode[STIX]{x1D70E}}^{2}=\bar{b}_{\unicode[STIX]{x1D70E}}^{\prime }$ and $\bar{b}_{\unicode[STIX]{x1D70E}}^{3}=\bar{c}_{\unicode[STIX]{x1D70E}}^{2}+\bar{c}_{\unicode[STIX]{x1D70E}}$ .◻

Lemma 6.11. The field $K(\unicode[STIX]{x1D701}_{3},\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/3},\unicode[STIX]{x1D703})$ is a Galois extension over $K$ .

Proof. Let $\unicode[STIX]{x1D70E}\in W_{K}$ . It suffices to show that $\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D703})\in K(\unicode[STIX]{x1D701}_{3},\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/3},\unicode[STIX]{x1D703})$ . We put

$$\begin{eqnarray}\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D70E}}=\unicode[STIX]{x1D703}+\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}^{2}\unicode[STIX]{x1D6FF}+\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}^{3}+\unicode[STIX]{x1D707}_{\unicode[STIX]{x1D70E}}+\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D701}_{3})-\unicode[STIX]{x1D701}_{3}.\end{eqnarray}$$

Then, we have $\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D70E}}^{2}-\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D70E}}\equiv \unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FF})^{3}\;(\text{mod}~2/3)$ . Hence, we can find $\unicode[STIX]{x1D703}^{\prime }$ such that $\unicode[STIX]{x1D703}^{\prime 2}-\unicode[STIX]{x1D703}^{\prime }=\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FF})^{3}$ and $\unicode[STIX]{x1D703}^{\prime }\equiv \unicode[STIX]{x1D703}_{\unicode[STIX]{x1D70E}}\;(\text{mod}~2/3)$ . By the choice of $\unicode[STIX]{x1D707}_{\unicode[STIX]{x1D70E}}$ , we have $\unicode[STIX]{x1D703}^{\prime }=\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D703})\;(\text{mod}~0+)$ . Hence, we obtain $\unicode[STIX]{x1D703}^{\prime }=\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D703})$ .

We take $\unicode[STIX]{x1D70E}^{\prime }\in W_{K}$ such that $\unicode[STIX]{x1D70E}^{\prime }(\unicode[STIX]{x1D703})\neq \unicode[STIX]{x1D70E}(\unicode[STIX]{x1D703})$ . We can define $\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D70E}^{\prime }}$ as above, and have $\unicode[STIX]{x1D70E}^{\prime }(\unicode[STIX]{x1D703})\equiv \unicode[STIX]{x1D703}_{\unicode[STIX]{x1D70E}^{\prime }}\;(\text{mod}~2/3)$ . If $\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}=\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}^{\prime }}$ , then we have $\unicode[STIX]{x1D701}_{3,\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FF})\equiv \unicode[STIX]{x1D701}_{3,\unicode[STIX]{x1D70E}^{\prime }}\unicode[STIX]{x1D70E}^{\prime }(\unicode[STIX]{x1D6FF})\;(\text{mod}~5/6)$ , which implies that $\unicode[STIX]{x1D701}_{3,\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FF})=\unicode[STIX]{x1D701}_{3,\unicode[STIX]{x1D70E}^{\prime }}\unicode[STIX]{x1D70E}^{\prime }(\unicode[STIX]{x1D6FF})$ because both sides are roots of

$$\begin{eqnarray}x^{4}-x-{\displaystyle \frac{1}{\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/3}}}=0.\end{eqnarray}$$

Hence, if $\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FF})^{3}\neq \unicode[STIX]{x1D70E}^{\prime }(\unicode[STIX]{x1D6FF})^{3}$ , we have $\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}}\neq \unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70E}^{\prime }}$ , which implies that

$$\begin{eqnarray}\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D703})\equiv \unicode[STIX]{x1D703}_{\unicode[STIX]{x1D70E}}\not \equiv \unicode[STIX]{x1D703}_{\unicode[STIX]{x1D70E}^{\prime }}\equiv \unicode[STIX]{x1D70E}^{\prime }(\unicode[STIX]{x1D703})\quad \left(\text{mod}~0+\right).\end{eqnarray}$$

If $\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FF})^{3}=\unicode[STIX]{x1D70E}^{\prime }(\unicode[STIX]{x1D6FF})^{3}$ , we have $\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D703})\not \equiv \unicode[STIX]{x1D70E}^{\prime }(\unicode[STIX]{x1D703})\;(\text{mod}~0+)$ . Therefore, we have

$$\begin{eqnarray}v(\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D703})-\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D70E}})>v(\unicode[STIX]{x1D70E}^{\prime }(\unicode[STIX]{x1D703})-\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D70E}}).\end{eqnarray}$$

Then, we obtain

$$\begin{eqnarray}\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D703})\in K(\unicode[STIX]{x1D703}_{\unicode[STIX]{x1D70E}})\subset K(\unicode[STIX]{x1D701}_{3},\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/3},\unicode[STIX]{x1D703})\end{eqnarray}$$

by Krasner’s lemma. ◻

Let $E$ be the elliptic curve over $k^{\text{ac}}$ defined by $z^{2}+z=w^{3}$ . We put

$$\begin{eqnarray}Q=\left\{g(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE})=\left(\begin{array}{@{}ccc@{}}\unicode[STIX]{x1D6FC} & \unicode[STIX]{x1D6FD} & \unicode[STIX]{x1D6FE}\\ & \unicode[STIX]{x1D6FC}^{2} & \unicode[STIX]{x1D6FD}^{2}\\ & & \unicode[STIX]{x1D6FC}\end{array}\right)\in \text{GL}_{3}(\mathbb{F}_{4})~\Bigg|~\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FE}^{2}+\unicode[STIX]{x1D6FC}^{2}\unicode[STIX]{x1D6FE}=\unicode[STIX]{x1D6FD}^{3}\right\}.\end{eqnarray}$$

We note that $|Q|=24$ and $Q$ is isomorphic to $\text{SL}_{2}(\mathbb{F}_{3})$ (cf. [Reference SerreSe, 8.5. Exercices 2]). Let $Q\rtimes \mathbb{Z}$ be a semidirect product, where $r\in \mathbb{Z}$ acts on $Q$ by $g(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE})\mapsto g(\unicode[STIX]{x1D6FC}^{q^{r}},\unicode[STIX]{x1D6FD}^{q^{r}},\unicode[STIX]{x1D6FE}^{q^{r}})$ . Then, $Q\rtimes \mathbb{Z}$ acts faithfully on $E$ as a scheme over $k$ , where $(g(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE}),r)\in Q\rtimes \mathbb{Z}$ acts on $E$ by

$$\begin{eqnarray}(z,w)\mapsto (z^{q^{-r}}+\unicode[STIX]{x1D6FC}^{-1}(\unicode[STIX]{x1D6FD}w^{q^{-r}}+\unicode[STIX]{x1D6FE}),\unicode[STIX]{x1D6FC}(w^{q^{-r}}+(\unicode[STIX]{x1D6FC}^{-1}\unicode[STIX]{x1D6FD})^{2}))\end{eqnarray}$$

for $k^{\text{ac}}$ -valued points.

Proposition 6.12. The element $\unicode[STIX]{x1D70E}\in W_{K}$ sends $\overline{\mathbf{X}}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }}$ to $\overline{\mathbf{X}}_{\bar{\unicode[STIX]{x1D709}}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D701}^{q^{-r_{\unicode[STIX]{x1D70E}}}},\unicode[STIX]{x1D701}^{\prime }}$ . We identify $\overline{\mathbf{X}}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }}$ with $\overline{\mathbf{X}}_{\bar{\unicode[STIX]{x1D709}}_{\unicode[STIX]{x1D70E}}\unicode[STIX]{x1D701}^{q^{-r_{\unicode[STIX]{x1D70E}}}},\unicode[STIX]{x1D701}^{\prime }}$ by $(z,w)\mapsto (z,w)$ . Then, the action of $W_{K}$ gives a homomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D6E9}_{\unicode[STIX]{x1D701}^{\prime }}:W_{K}\rightarrow Q\rtimes \mathbb{Z}\subset \text{Aut}_{k}(\overline{\mathbf{X}}_{\unicode[STIX]{x1D701},\unicode[STIX]{x1D701}^{\prime }});\qquad \unicode[STIX]{x1D70E}\mapsto (g(\bar{\unicode[STIX]{x1D701}}_{3,\unicode[STIX]{x1D70E}},\bar{\unicode[STIX]{x1D701}}_{3,\unicode[STIX]{x1D70E}}^{2}\bar{\unicode[STIX]{x1D708}}_{\unicode[STIX]{x1D70E}}^{2},\bar{\unicode[STIX]{x1D701}}_{3,\unicode[STIX]{x1D70E}}\bar{\unicode[STIX]{x1D707}}_{\unicode[STIX]{x1D70E}}),r_{\unicode[STIX]{x1D70E}}).\end{eqnarray}$$

Proof. This follows from Proposition 6.7 and Lemma 6.10. ◻

Proposition 6.13. The homomorphism $\unicode[STIX]{x1D6E9}_{\unicode[STIX]{x1D701}^{\prime }}$ factors through $W(K^{\text{ur}}(\unicode[STIX]{x1D71B}^{1/3},\unicode[STIX]{x1D703})/K)$ and gives an isomorphism $W(K^{\text{ur}}(\unicode[STIX]{x1D71B}^{1/3},\unicode[STIX]{x1D703})/K)\simeq Q\rtimes \mathbb{Z}$ .

Proof. By Lemma 6.10(1), the homomorphism $\unicode[STIX]{x1D6E9}_{\unicode[STIX]{x1D701}^{\prime }}$ factors through $W(K^{\text{ur}}(\unicode[STIX]{x1D71B}^{1/3},\unicode[STIX]{x1D703})/K)$ and induces an injective homomorphism

$$\begin{eqnarray}W(K^{\text{ur}}(\unicode[STIX]{x1D71B}^{1/3},\unicode[STIX]{x1D703})/K)\rightarrow Q\rtimes \mathbb{Z}.\end{eqnarray}$$

To prove the surjectivity, it suffices to show that $\unicode[STIX]{x1D6E9}_{\unicode[STIX]{x1D701}^{\prime }}$ sends $I_{K}$ onto $Q$ . Let $g=g(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE})\in Q$ . We take $\unicode[STIX]{x1D701}_{\unicode[STIX]{x1D6FC}}\in \unicode[STIX]{x1D707}_{3}(K^{\text{ur}})$ , $\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D6FD}},\unicode[STIX]{x1D707}_{\unicode[STIX]{x1D6FE}}\in \unicode[STIX]{x1D707}_{3}(K^{\text{ur}})\cup \{0\}$ such that $\bar{\unicode[STIX]{x1D701}}_{\unicode[STIX]{x1D6FC}}=\unicode[STIX]{x1D6FC}$ , $\bar{\unicode[STIX]{x1D708}}_{\unicode[STIX]{x1D6FD}}=\unicode[STIX]{x1D6FC}^{-1}\unicode[STIX]{x1D6FD}$ and $\bar{\unicode[STIX]{x1D707}}_{\unicode[STIX]{x1D6FE}}=\unicode[STIX]{x1D6FC}^{-1}\unicode[STIX]{x1D6FE}$ . We put $\unicode[STIX]{x1D6FF}_{g}=\unicode[STIX]{x1D701}_{\unicode[STIX]{x1D6FC}}^{-1}(\unicode[STIX]{x1D6FF}+\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D6FD}})$ and $\unicode[STIX]{x1D703}_{g}=\unicode[STIX]{x1D703}+\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D6FD}}^{2}\unicode[STIX]{x1D6FF}+\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D6FD}}^{3}+\unicode[STIX]{x1D707}_{\unicode[STIX]{x1D6FE}}$ . Then, we have

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{g}^{4}-\unicode[STIX]{x1D6FF}_{g}\equiv {\displaystyle \frac{1}{\unicode[STIX]{x1D701}_{\unicode[STIX]{x1D6FC}}\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/3}}}\quad \left(\text{mod}~{\displaystyle \frac{5}{6}}\right).\end{eqnarray}$$

Hence, we can find $\unicode[STIX]{x1D6FF}_{g}^{\prime }$ such that $\unicode[STIX]{x1D6FF}_{g}^{\prime 4}-\unicode[STIX]{x1D6FF}_{g}^{\prime }=1/(\unicode[STIX]{x1D701}_{\unicode[STIX]{x1D6FC}}\unicode[STIX]{x1D701}^{\prime \prime }\unicode[STIX]{x1D71B}^{1/3})$ and $\unicode[STIX]{x1D6FF}_{g}^{\prime }\equiv \unicode[STIX]{x1D6FF}_{g}$ $(\text{mod}~5/6)$ . Further, we have $\unicode[STIX]{x1D703}_{g}^{2}-\unicode[STIX]{x1D703}_{g}\equiv \unicode[STIX]{x1D6FF}_{g}^{\prime 3}\;(\text{mod}~2/3)$ . Hence, we can find $\unicode[STIX]{x1D703}_{g}^{\prime }$ such that $\unicode[STIX]{x1D703}_{g}^{\prime 2}-\unicode[STIX]{x1D703}_{g}^{\prime }=\unicode[STIX]{x1D6FF}_{g}^{\prime 3}$ and $\unicode[STIX]{x1D703}_{g}^{\prime }\equiv \unicode[STIX]{x1D703}_{g}\;(\text{mod}~2/3)$ . Then, $\unicode[STIX]{x1D71B}^{1/3}\mapsto \unicode[STIX]{x1D701}_{\unicode[STIX]{x1D6FC}}\unicode[STIX]{x1D71B}^{1/3}$ , and $\unicode[STIX]{x1D703}\mapsto \unicode[STIX]{x1D703}_{g}^{\prime }$ gives an element of $I_{K}$ , whose image by $\unicode[STIX]{x1D6E9}_{\unicode[STIX]{x1D701}^{\prime }}$ is $g$ .◻

7.1 Cohomology of reductions

Let $X_{\text{DL}}$ be the smooth compactification of the affine curve over $k$ defined by $X^{q}-X=Y^{q+1}$ . The curve $X_{\text{DL}}$ is also the smooth compactification of the Deligne–Lusztig curve $x^{q}y-xy^{q}=1$ for $\text{SL}_{2}(\mathbb{F}_{q})$ . Then, $a\in k$ acts on $X_{\text{DL}}$ by

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{a}:(X,Y)\mapsto (X+a,Y).\end{eqnarray}$$

On the other hand, $\unicode[STIX]{x1D701}\in k_{2}^{\times }$ acts on $X_{\text{DL}}$ by

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D701}}:(X,Y)\mapsto (\unicode[STIX]{x1D701}^{q+1}X,\unicode[STIX]{x1D701}Y).\end{eqnarray}$$

By these actions, we consider $H^{1}(X_{\text{DL}},\overline{\mathbb{Q}}_{\ell })$ as a $\overline{\mathbb{Q}}_{\ell }[k\times k_{2}^{\times }]$ -module.

Lemma 7.1. We have an isomorphism

$$\begin{eqnarray}H^{1}(X_{\text{DL}},\overline{\mathbb{Q}}_{\ell })\simeq \bigoplus _{\unicode[STIX]{x1D713}\in k^{\vee }\backslash \{1\}}\bigoplus _{\unicode[STIX]{x1D712}\in \unicode[STIX]{x1D707}_{q+1}(k_{2})^{\vee }\backslash \{1\}}\unicode[STIX]{x1D713}\otimes \unicode[STIX]{x1D712}\end{eqnarray}$$

as $\overline{\mathbb{Q}}_{\ell }[k\times \unicode[STIX]{x1D707}_{q+1}(k_{2})]$ -modules.

Proof. As $\overline{\mathbb{Q}}_{\ell }[k\times \unicode[STIX]{x1D707}_{q+1}(k_{2})]$ -modules, we have the short exact sequence

(7.1) $$\begin{eqnarray}0\rightarrow \bigoplus _{\unicode[STIX]{x1D713}\in k^{\vee }}\unicode[STIX]{x1D713}\rightarrow H_{c}^{1}(X_{\text{DL}}\backslash X_{\text{DL}}(k),\overline{\mathbb{Q}}_{\ell })\rightarrow H^{1}(X_{\text{DL}},\overline{\mathbb{Q}}_{\ell })\rightarrow 0.\end{eqnarray}$$

Let ${\mathcal{L}}_{\unicode[STIX]{x1D713}}$ denote the Artin–Schreier $\overline{\mathbb{Q}}_{\ell }$ -sheaf associated to $\unicode[STIX]{x1D713}\in k^{\vee }$ . Let ${\mathcal{K}}_{\unicode[STIX]{x1D712}}$ denote the Kummer $\overline{\mathbb{Q}}_{\ell }$ -sheaf associated to $\unicode[STIX]{x1D712}\in \unicode[STIX]{x1D707}_{q+1}(k_{2})^{\vee }$ . Since

$$\begin{eqnarray}X_{\text{DL}}\backslash X_{\text{DL}}(k)\rightarrow \mathbb{G}_{m};\qquad (X,Y)\mapsto Y^{q+1}\end{eqnarray}$$

is a finite etale Galois covering with a Galois group $k\times \unicode[STIX]{x1D707}_{q+1}(k_{2})$ , we have the isomorphism

(7.2) $$\begin{eqnarray}H_{c}^{1}(X_{\text{DL}}\backslash X_{\text{DL}}(k),\overline{\mathbb{Q}}_{\ell })\simeq \bigoplus _{\unicode[STIX]{x1D713}\in k^{\vee }}\bigoplus _{\unicode[STIX]{x1D712}\in \unicode[STIX]{x1D707}_{q+1}(k_{2})^{\vee }}H_{c}^{1}(\mathbb{G}_{m},{\mathcal{L}}_{\unicode[STIX]{x1D713}}\otimes {\mathcal{K}}_{\unicode[STIX]{x1D712}})\end{eqnarray}$$

as $\overline{\mathbb{Q}}_{\ell }[k\times \unicode[STIX]{x1D707}_{q+1}(k_{2})]$ -modules. Note that we have

$$\begin{eqnarray}\dim H_{c}^{1}(\mathbb{G}_{m},{\mathcal{L}}_{\unicode[STIX]{x1D713}}\otimes {\mathcal{K}}_{\unicode[STIX]{x1D712}})=1\end{eqnarray}$$

if $\unicode[STIX]{x1D713}\neq 1$ by the Grothendieck–Ogg–Shafarevich formula (cf. [Reference GrothendieckSGA5, Exposé X, Théorème 7.1]). Clearly, if $\unicode[STIX]{x1D712}\neq 1$ , we have $H_{c}^{1}(\mathbb{G}_{m},{\mathcal{K}}_{\unicode[STIX]{x1D712}})=0$ and $H_{c}^{1}(\mathbb{G}_{m},{\mathcal{L}}_{\unicode[STIX]{x1D713}})\simeq \unicode[STIX]{x1D713}$ . Hence, we acquire the isomorphism

(7.3) $$\begin{eqnarray}\displaystyle & & \displaystyle \bigoplus _{\unicode[STIX]{x1D713}\in k^{\vee }}\bigoplus _{\unicode[STIX]{x1D712}\in \unicode[STIX]{x1D707}_{q+1}(k_{2})^{\vee }}H_{c}^{1}(\mathbb{G}_{m},{\mathcal{L}}_{\unicode[STIX]{x1D713}}\otimes {\mathcal{K}}_{\unicode[STIX]{x1D712}})\nonumber\\ \displaystyle & & \displaystyle \quad \simeq \bigoplus _{\unicode[STIX]{x1D713}\in k^{\vee }\setminus \{1\}}\bigoplus _{\unicode[STIX]{x1D712}\in \unicode[STIX]{x1D707}_{q+1}(k_{2})^{\vee }\setminus \{1\}}H_{c}^{1}(\mathbb{G}_{m},{\mathcal{L}}_{\unicode[STIX]{x1D713}}\otimes {\mathcal{K}}_{\unicode[STIX]{x1D712}})\oplus \bigoplus _{\unicode[STIX]{x1D713}\in k^{\vee }}\unicode[STIX]{x1D713}\end{eqnarray}$$

as $\overline{\mathbb{Q}}_{\ell }[k\times \unicode[STIX]{x1D707}_{q+1}(k_{2})]$ -modules. By (7.1), (7.2) and (7.3), the required assertion follows.◻

For a character $\unicode[STIX]{x1D713}\in k^{\vee }$ and an element $\unicode[STIX]{x1D701}\in k^{\times }$ , we denote by $\unicode[STIX]{x1D713}_{\unicode[STIX]{x1D701}}$ the character $x\mapsto \unicode[STIX]{x1D713}(\unicode[STIX]{x1D701}x)$ . We consider a character group $(k^{\times })^{\vee }$ as a subgroup of $(k_{2}^{\times })^{\vee }$ by $\text{Nr}_{k_{2}/k}^{\vee }$ .

Lemma 7.2. We have an isomorphism

$$\begin{eqnarray}H^{1}(X_{\text{DL}},\overline{\mathbb{Q}}_{\ell })\simeq \bigoplus _{\tilde{\unicode[STIX]{x1D712}}\in (k_{2}^{\times })^{\vee }\backslash (k^{\times })^{\vee }}\tilde{\unicode[STIX]{x1D712}}\end{eqnarray}$$

as $\overline{\mathbb{Q}}_{\ell }[k_{2}^{\times }]$ -modules.

Proof. By Lemma 7.1, we take a basis

$$\begin{eqnarray}\{e_{\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}}\}_{\unicode[STIX]{x1D713}\in k^{\vee }\backslash \{1\},\,\unicode[STIX]{x1D712}\in \unicode[STIX]{x1D707}_{q+1}(k_{2})^{\vee }\backslash \{1\}}\end{eqnarray}$$

of $H^{1}(X_{\text{DL}},\overline{\mathbb{Q}}_{\ell })$ over $\overline{\mathbb{Q}}_{\ell }$ such that $k\times \unicode[STIX]{x1D707}_{q+1}(k_{2})$ acts on $e_{\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}}$ by $\unicode[STIX]{x1D713}\otimes \unicode[STIX]{x1D712}$ . For $\unicode[STIX]{x1D701}\in k_{2}^{\times }$ and $a\in k$ , we have

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D701}}\circ \unicode[STIX]{x1D6FC}_{a}\circ \unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D701}}^{-1}=\unicode[STIX]{x1D6FC}_{\unicode[STIX]{x1D701}^{q+1}a}\end{eqnarray}$$

in $\text{Aut}_{k_{2}}(X_{\text{DL}})$ . Hence, $\unicode[STIX]{x1D701}\in k_{2}^{\times }$ acts on $H^{1}(X_{\text{DL}},\overline{\mathbb{Q}}_{\ell })$ by

$$\begin{eqnarray}e_{\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}}\mapsto c_{\unicode[STIX]{x1D713},\unicode[STIX]{x1D712},\unicode[STIX]{x1D701}}e_{\unicode[STIX]{x1D713}_{\unicode[STIX]{x1D701}^{-(q+1)}},\unicode[STIX]{x1D712}}\end{eqnarray}$$

with some constant $c_{\unicode[STIX]{x1D713},\unicode[STIX]{x1D712},\unicode[STIX]{x1D701}}\in \overline{\mathbb{Q}}_{\ell }^{\times }$ . Therefore, we acquire an isomorphism

$$\begin{eqnarray}H^{1}(X_{\text{DL}},\overline{\mathbb{Q}}_{\ell })\simeq \bigoplus _{\unicode[STIX]{x1D712}\in \unicode[STIX]{x1D707}_{q+1}(k_{2})^{\vee }\backslash \{1\}}\text{Ind}_{\unicode[STIX]{x1D707}_{q+1}(k_{2})}^{k_{2}^{\times }}(\unicode[STIX]{x1D712})\end{eqnarray}$$

as $\overline{\mathbb{Q}}_{\ell }[k_{2}^{\times }]$ -modules. Hence, the required assertion follows.◻

Proposition 7.3. We have isomorphisms

$$\begin{eqnarray}\displaystyle & \displaystyle H^{1}(\overline{\mathbf{Y}}_{1,2}^{\text{c}},\overline{\mathbb{Q}}_{\ell })\simeq \bigoplus _{\tilde{\unicode[STIX]{x1D712}}\in (k_{2}^{\times })^{\vee }\backslash (k^{\times })^{\vee }}(\tilde{\unicode[STIX]{x1D712}}\circ \unicode[STIX]{x1D706})\otimes (\tilde{\unicode[STIX]{x1D712}}^{q}\circ \unicode[STIX]{x1D705}_{1}), & \displaystyle \nonumber\\ \displaystyle & \displaystyle H^{1}(\overline{\mathbf{Y}}_{2,1}^{\text{c}},\overline{\mathbb{Q}}_{\ell })\simeq \bigoplus _{\tilde{\unicode[STIX]{x1D712}}\in (k_{2}^{\times })^{\vee }\backslash (k^{\times })^{\vee }}(\tilde{\unicode[STIX]{x1D712}}\circ \unicode[STIX]{x1D706})\otimes (\tilde{\unicode[STIX]{x1D712}}\circ \unicode[STIX]{x1D705}_{1}) & \displaystyle \nonumber\end{eqnarray}$$

as $(I_{K}\times {\mathcal{O}}_{D}^{\times })$ -representations over $\overline{\mathbb{Q}}_{\ell }$ .

Let $X_{\text{AS}}$ be the smooth compactification of the affine curve $X_{\text{AS}}^{\prime }$ over $k$ defined by $z^{q}-z=w^{2}$ . Let $a\in k$ act on $X_{\text{AS}}$ by

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{a}:(z,w)\mapsto (z+a,w).\end{eqnarray}$$

By this action, we consider $H^{1}(X_{\text{AS}},\overline{\mathbb{Q}}_{\ell })$ as a $\overline{\mathbb{Q}}_{\ell }[k]$ -module. On the other hand, let $b\in \unicode[STIX]{x1D707}_{2(q-1)}(k^{\text{ac}})$ act on $X_{\text{AS}}$ by

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}_{b}:(z,w)\mapsto (b^{2}z,bw).\end{eqnarray}$$

Proof. We take $a\in F$ such that $a^{q}-a=1/s^{2}$ . Then, $sa^{(q-1)/2}$ is a uniformizer of $F$ . Let $v_{F}$ be the normalized valuation of $F$ . For $\unicode[STIX]{x1D70E}\in G$ and an integer $i$ , the condition

$$\begin{eqnarray}v_{F}(\unicode[STIX]{x1D70E}(sa^{(q-1)/2})-sa^{(q-1)/2})\geqslant i\end{eqnarray}$$

is equivalent to the condition

$$\begin{eqnarray}v_{F}(\unicode[STIX]{x1D70E}(a)-a)\geqslant i-3.\end{eqnarray}$$

Hence, the claim follows. ◻

For a character $\unicode[STIX]{x1D713}\in k^{\vee }$ and $x\in k^{\times }$ , we write $\unicode[STIX]{x1D713}_{x}\in k^{\vee }$ for the character $y\mapsto \unicode[STIX]{x1D713}(xy)$ . We set

$$\begin{eqnarray}V=\bigoplus _{\unicode[STIX]{x1D713}\in k^{\vee }\backslash \{1\}}\unicode[STIX]{x1D713}\end{eqnarray}$$

as $\overline{\mathbb{Q}}_{\ell }[k]$ -modules. Let $\{e_{\unicode[STIX]{x1D713}}\}_{\unicode[STIX]{x1D713}\in k^{\vee }\backslash \{1\}}$ be the standard basis of $V$ .