Reconstruction of one-punctured elliptic curves in positive characteristic by their geometric fundamental groups

Abstract

Our main question is whether the isomorphism class as a scheme of a curve X over an algebraic closure of a finite field can be reconstructed by the etale fundamental group of X. Tamagawa answered this question affirmatively when the genus of X is 0. In this paper, we will discuss the genus 1 case, and prove a similar result when the genus of X is 1, the cardinality of cusps is 1, and the characteristic of X is not equal to 2.

1 Introduction

Let k be a field, $G_{k}$ the absolute Galois group of k, U an algebraic variety over k (i.e. a geometrically connected separated scheme of finite type over k) and $\pi _{1}(U)$ the étale fundamental group of U.

When k is a number field or, more generally, a field finitely generated over the prime field, the following philosophy of anabelian geometry, which is sometimes called the Grothendieck conjecture, was advocated by A.Grothendieck.

When U is an “anabelian variety", the geometry of U can be recovered group-theoretically from $\pi _{1}(U)\twoheadrightarrow G_{k}$.

When k is an algebraically closed field of characteristic 0 and U is a curve (i.e. an integral separated regular scheme of finite type over k and of dimension 1), the isomorphism class of $\pi _{1}(U)$ as a topological group can be recovered group-theoretically from the cardinality of cusps of U and the genus of U. Therefore the isomorphism class of U as a scheme cannot be recovered group-theoretically only from $\pi _{1}(U)$.

When k is an algebraically closed field of characteristic $p > 0$, the isomorphism class of $\pi _{1}(U)$ cannot be recovered from easy invariants such as the cardinality of cusps or the genus. Thus, we can even consider the following problem.

Is the isomorphism class of U as a scheme recovered group-theoretically only from $\pi _{1}(U)$ ?

Regarding this problem, various results are known (cf. [4, 5, 7, 8, 11, 12]). Among others, the following theorem is known.

Theorem 1.1

([11] Theorem 3.5) Let k be an algebraically closed field of characteristic $p > 0$, U a curve over k, $F \subset k$ the algebraic closure of ${\mathbb {F}}_{p}$, $U_{0}$ a curve over F and $X_{0}$ a smooth compactification of $U_{0}$. Assume that the genus of $X_{0}$ is 0. Then

$$\begin{aligned} \pi _{1} (U) \simeq \pi _{1} (U_{0}\times _{F}k) \Leftrightarrow U \simeq U_{0}\times _{F}k \ (as \ a \ scheme) \end{aligned}$$

$\square$

The main result of the present paper is the following generalization of Theorem 1.1.

Theorem 1.2

(Theorem 4.9) Let k be an algebraically closed field of characteristic $p \ne 0 , 2$, U a curve over k, $F \subset k$ the algebraic closure of ${\mathbb {F}}_{p}$, $U_{0}$ a curve over F and $X_{0}$ a smooth compactification of $U_{0}$. Assume that the genus of $X_{0}$ is 1 and that the cardinality of $X_{0}\backslash U_{0}$ is 1. Then

$$\begin{aligned} \pi _{1} (U) \simeq \pi _{1} (U_{0}\times _{F}k) \Leftrightarrow U \simeq U_{0}\times _{F}k \ ( as \ a \ scheme) \end{aligned}$$

In the second section, we will review the reconstruction of various invariants by $\pi _{1}(U)$, which will be used in the later sections.

In the third section, U is assumed to be an open subscheme of an elliptic or hyperelliptic curve. We will prove that linear relations of the images of cusps in ${\mathbb {P}}^{1}$ are encoded in $\pi _{1}(U)$ and a certain closed subgroup $L_{U} \subset \pi _{1}(U)$ (see the third section for the definition of $L_{U}$).

In the fourth section, U is assumed to be a curve of (1,1)-type. At first we will prove that we can apply the main theorem of the third section to certain étale covers of U. Then we will prove that the isomorphism class of U as a scheme is determined only by $\pi _{1}(U)$.

2 The reconstruction of various invariants ([11] §1,§2)

In this section, we will review the reconstruction of various invariants that was shown in [11].

The theorems in the first section are about curves of genus 0 or 1, while the theorems in this section are about curves of arbitrary genus.

Let k be an algebraically closed field of characteristic $p > 0$, U a curve over k (i.e. an integral separated regular scheme of finite type over k and of dimension 1), F the algebraic closure of ${\mathbb {F}}_{p}$ in k.

Definition

(i):

Let $\pi _{1}(U)$ be the étale fundamental group of U, $U_{H}$ the étale cover of U that corresponds to an open subgroup $H \subset \pi _{1}(U)$, $X=U^{cpt}$ the smooth compactification of U, g(X) the genus of X, $S_{U}=X\backslash U$ the complement of U in X, $n_{U}$ the cardinality of $S_{U}$, K the function field of U, $K^{sep}$ a separable closure of K, ${\tilde{K}}$ the maximal Galois extension of K in $K^{sep}$ that is unramified over U, ${\tilde{X}}$ the normalization of X in ${\tilde{K}}$, ${\tilde{S}}_{U}$ the inverse image of $S_{U}$ under ${\tilde{X}} \rightarrow X$, $I_{{\tilde{P}}}$ the inertia subgroup in $\pi _{1}(U)$ associated to ${\tilde{P}} \in {\tilde{S}}_{U}$, $I_{{\tilde{P}}}^{wild}$ the Sylow p-subgroup of $I_{{\tilde{P}}}$, $I_{{\tilde{P}}}^{tame} {\mathop {=}\limits ^{def}}I_{{\tilde{P}}} / I^{wild}_{{\tilde{P}}}$,

$$\begin{aligned}&Sub(\pi _{1}(U)){\mathop {=}\limits ^{def}}\{ H \subset \pi _{1}(U) | \ H \ \mathrm{is\ a\ closed\ subgroup}\},\\&({\mathbb {Q}}/{\mathbb {Z}})' {\mathop {=}\limits ^{def}}\{ a \in {\mathbb {Q}}/{\mathbb {Z}} \ | \ \mathrm{the \ order \ of} \ a \ \mathrm{is \ prime \ to} \ p \} \end{aligned}$$

and

$$\begin{aligned} F_{{\tilde{P}}} {\mathop {=}\limits ^{def}}(I^{tame}_{{\tilde{P}}} \otimes _{{\mathbb {Z}}}({\mathbb {Q}}/{\mathbb {Z}})')\coprod \{ *\} \end{aligned}$$

$(\{ *\}$ means one point set, ${\tilde{P}} \in {\tilde{S}}_{U}$).

(ii):

Suppose ${\mathcal {F}}(U)$ and ${\mathcal {G}}(U)$ are an additional structure on U. We say that ${\mathcal {F}}(U)$ can be recovered group-theoretically from $\pi _{1}(U)$ (resp. $\pi _{1}(U)$ and ${\mathcal {G}}(U)$) if any isomorphism $\pi _{1}(U_{1})\simeq \pi _{1}(U_{2})$ (resp. $\pi _{1}(U_{1})\simeq \pi _{1}(U_{2})$ that induces ${\mathcal {G}}(U_{1}) \simeq {\mathcal {G}}(U_{2})$) induces ${\mathcal {F}}(U_{1})\simeq {\mathcal {F}}(U_{2})$.

Theorem 2.1

([11] §1,§2) From $\pi _{1}(U)$

• $(g(X),n_{U})$ can be recovered group-theoretically.

• When $(g(X),n_{U})\ne (0,0)$, p can be recovered group-theoretically.

• $\pi _{1}(X)$ can be recovered group-theoretically as a quotient group of $\pi _{1}(U)$.

• ${\tilde{S}}_{U}$ can be recovered group-theoretically as a subset of $Sub(\pi _{1}(U))$. More precisely, ${\tilde{S}}_{U}$ can be identified with a subset of $Sub(\pi _{1}(U))$ via ${\tilde{S}}_{U} \rightarrow Sub(\pi _{1}(U))$, ${\tilde{P}}\rightarrow I_{{\tilde{P}}}$, and this subset can be recovered group-theoretically.

• $S_{U}$ can be recovered group-theoretically as a quotient set of ${\tilde{S}}_{U}$.

• For any ${\tilde{P}}\in {\tilde{S}}_{U}$, the field structure of $F_{{\tilde{P}}}$ obtained by identifying $F_{{\tilde{P}}}$ with F (see [11] the argument before Proposition 2.8) can be recovered group-thoretically.

$\square$

Definition

(i):

For any set (resp. group) S and ring R, R[S] stands for the free R-module with basis S (resp. the group ring of S over R).

(ii):

Set $I = I_{{\tilde{P}}}$. Let d be any positive integer. We define $\chi _{I,d}$ as follows

$$\begin{aligned} \chi _{I,d} \ : \ I \twoheadrightarrow I^{tame}/(p^{d}-1)=I^{tame}\otimes _{{\mathbb {Z}}} \frac{1}{p^{d}-1}{\mathbb {Z}}/{\mathbb {Z}} \hookrightarrow F_{{\tilde{P}}}^{\times } \end{aligned}$$

When we identify $F_{{\tilde{P}}}^{\times }$ with $F^{\times }$, $\chi _{I,d}$ coincides with the following character

$$\begin{aligned} I \ni \gamma \mapsto \gamma \left( \pi ^{\frac{1}{p^{d}-1}}\right) /\pi ^{\frac{1}{p^{d}- 1}} \in F^{\times } \end{aligned}$$

where $\pi \in {\mathcal {O}}_{X,P}$ stands for a prime element.

(iii):

Let $M = M_{U}$ be an ${\mathbb {F}}_{p}[\pi _{1}(U)]$-module depending functorially on U. Then for any ${\tilde{P}}\in {\tilde{S}}_{U}$, $i \in {\mathbb {Z}}$ and $d\in {\mathbb {Z}}_{> 0}$, we define $M(\chi _{I_{{\tilde{P}}},d}^{i})$ as the following.

$$\begin{aligned} M(\chi _{I_{{\tilde{P}}},d}^{i}){\mathop {=}\limits ^{def}}\{ x\in M\otimes _{{\mathbb {F}}_{p}}F_{{\tilde{P}}}\ | \ \gamma x = \chi _{I_{{\tilde{P}}},d}^{i}(\gamma )x \ (\gamma \in I_{{\tilde{P}}}) \} \end{aligned}$$

Corollary 2.2

([11] Corollary 2.11) $(M(\chi _{I_{{\tilde{P}}},d}^{i}))_{{\tilde{P}}\in {\tilde{S}}_{U}}$ can be recovered group-theoretically from $\pi _{1}(U)$. $\square$

3 Linear relations of the images of cusps in ${\mathbb {P}}^{1}$

In this section, we will use the same symbols as in the previous sections, and we assume that $p \ne 0,2$ and that X is an elliptic or hyperelliptic curve.

We will prove that linear relations of the images of cusps in ${\mathbb {P}}^{1}$ are encoded in $\pi _{1}(U)$ and a certain closed subgroup $L_{U} \subset \pi _{1}(U)$.

Let $x : X \rightarrow {\mathbb {P}}^{1}$ be a finite morphism of degree 2, $T_{U}{\mathop {=}\limits ^{def}}x(S_{U})$,

$$\begin{aligned} \lambda _{0},\lambda _{\infty },\lambda _{1}, \lambda _{2},\ldots ,\lambda _{m} \in X \end{aligned}$$

the ramified points of x and

$$\begin{aligned} P_{i} {\mathop {=}\limits ^{def}}x(\lambda _{i}) \in {\mathbb {P}}^{1}\ (i = 0, \infty , 1, 2, \ldots , m). \end{aligned}$$

(By Hurwitz’s formula, m is an even number.) In this section, we assume that

$$\begin{aligned}&\lambda _{0},\lambda _{\infty },\lambda _{1},\lambda _{2},\ldots ,\lambda _{m} \in S_{U},\\&S_{U} \backslash \{ \lambda _{0},\lambda _{\infty },\lambda _{1},\lambda _{2},\ldots ,\lambda _{m}\} \ne \emptyset \end{aligned}$$

and

$$\begin{aligned} x^{-1}(T_{U})=S_{U}. \end{aligned}$$

Let

$$\begin{aligned} \mu _{(1,1)},\mu _{(1,2)},\mu _{(2,1)},\ldots ,\mu _{(l,1)}, \mu _{(l,2)} \end{aligned}$$

be the unramified points of x in $S_{U}$ ($\mu _{(i,1)}$ is conjugate with $\mu _{(i,2)}$),

$$\begin{aligned} R_{i} {\mathop {=}\limits ^{def}}x(\mu _{(i,1)})=x( \mu _{(i,2)}) \in T_{U}\ (i = 1,2, \ldots , l) . \end{aligned}$$

Set

$$\begin{aligned}&S_{U,unr} {\mathop {=}\limits ^{def}}\{ \mu _{(1,1)},\mu _{(1,2)},\mu _{(2,1)},\ldots ,\mu _{(l,1)} , \mu _{(l,2)} \},\\&S_{U,ram} {\mathop {=}\limits ^{def}}\{ \lambda _{0},\lambda _{\infty },\lambda _{1}, \lambda _{2},\ldots ,\lambda _{m} \},\\&T_{U,unr} {\mathop {=}\limits ^{def}}\{ R_{1},R_{2},\ldots ,R_{l} \},\\&T_{U,ram} {\mathop {=}\limits ^{def}}\{ P_{0},P_{\infty },P_{1},P_{2},\ldots , P_{m} \}. \end{aligned}$$

Let $I_{{\tilde{\lambda }}} \subset \pi _{1}(U)$ be the inertia group corresponding to ${\tilde{\lambda }} \in {\tilde{X}}$, $I_{{\tilde{\lambda }},{\mathbb {P}}^{1}} \subset \pi _{1}({\mathbb {P}}^{1} \backslash T_{U})$ be the inertia group corresponding to ${\tilde{\lambda }} \in \tilde{{\mathbb {P}}}^{1}$ (Here, $\tilde{{\mathbb {P}}}^{1}$ stands for the integral closure of ${\mathbb {P}}^{1}$ in ${\tilde{K}}$. By definition, ${\tilde{X}}=\tilde{{\mathbb {P}}}^{1}$).

Set

$$\begin{aligned} Q {\mathop {=}\limits ^{def}}\pi _{1}({\mathbb {P}}^{1} \backslash T_{U})^{ab,p'} \end{aligned}$$

(the maximal pro-prime-to-p abelian quotient of $\pi _{1}({\mathbb {P}}^{1}\backslash T_{U}$)),

$$\begin{aligned} L_{U} {\mathop {=}\limits ^{def}}ker(\pi _{1}(U) \rightarrow \pi _{1}({\mathbb {P}}^{1} \backslash T_{U}) \rightarrow Q) \end{aligned}$$

and

$$\begin{aligned} Q_{U} {\mathop {=}\limits ^{def}}\pi _{1}(U)/L_{U}. \end{aligned}$$

When X is a hyperelliptic curve, x is the unique finite morphism of degree 2 (up to isomorphism of ${\mathbb {P}}^{1}$, see [2] IV Propotition 5.3). When X is an elliptic curve, x is not unique (therefore, $T_{U},$Q,  $L_{U},$$Q_{U},$$S_{U,unr},$$T_{U,unr},$$\lambda _{0},$$P_{0},$$\mu _{(1,1)},$$R_{1}$, etc., depend on the choice of x). In this section, we assume that x is fixed.

Proposition 3.1

$S_{U,ram}, \ S_{U,unr}, \ T_{U}, \ T_{U,ram}, \ T_{U,unr}, \ Q$ and the natural injective map $Q_{U}\hookrightarrow Q$ can be recovered group-theoretically from $\pi _{1}(U)$ and $L_{U}$.

Proof

For each $\lambda \in S_{U}$, we fix ${\tilde{\lambda }} \in {\tilde{S}}_{U}$ above $\lambda$. We define an equivalence relation $\sim$ on $S_{U}$ by saying $\nu \sim \lambda$ if $I_{{\tilde{\nu }}}/(I_{{\tilde{\nu }}} \cap L_{U}) = I_{{\tilde{\lambda }}}/(I_{{\tilde{\lambda }}} \cap L_{U})$ (as subsets of $Q_{U}$). We can identify $T_{U}$ with $S_{U}/\sim$ (see the proof of [11] Lemma 2.1). $S_{U,unr} = \{ \lambda \in S_{U} |$ there exists $\nu \in S_{U} \backslash \{ \lambda \}$ such that $\lambda \sim \nu$ }, $S_{U,unr}$ and $S_{U,ram}$ are recovered from $\pi _{1}(U)$ and $L_{U}$. As $T_{U,ram}$ (resp. $T_{U,unr})$ is the image of $S_{U,ram}$ (resp. $S_{U,unr})$, $T_{U,ram}$ and $T_{U,unr}$ are recovered from $\pi _{1}(U)$ and $L_{U}$.

Via the exact sequence $0\rightarrow Q_{U} \rightarrow Q \rightarrow {\mathbb {Z}}/2{\mathbb {Z}}$, we can regard Q as subset of $\frac{1}{2}Q_{U}$. By G.A.G.A theorems ([1] Exposé 12 , Exposé 13)

$$\begin{aligned} \begin{aligned}&Q\simeq (\oplus _{P\in T_{U}}I^{tame}_{{\tilde{P}},{\mathbb {P}}^{1}})/\varDelta \ , \ I^{tame}_{{\tilde{P}},{\mathbb {P}}^{1}}\simeq \hat{{\mathbb {Z}}}^{p'} \ , \ \varDelta \simeq \hat{{\mathbb {Z}}^{p'}} \\&I_{{\tilde{\lambda }},{\mathbb {P}}^{1}}^{tame}/I_{{\tilde{\lambda }}}^{tame} \simeq {\mathbb {Z}}/2{\mathbb {Z}} \ ( \lambda \in S_{U,ram}) \ , \ I_{{\tilde{\lambda }},{\mathbb {P}}^{1}}^{tame}/I_{{\tilde{\lambda }}}^{tame} = 0 \ ( \lambda \in S_{U,unr}) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&Q_{U} \simeq \left( \left( \oplus _{P \in T_{U,ram}}I^{tame}_{{\tilde{P}},{\mathbb {P}}^{1}}\right) ^{*}+ \left( \sum _{P \in T_{U,unr}} I^{tame}_{{\tilde{P}},{\mathbb {P}}^{1}}\right) \right) \\&\left( \left( \oplus _{\lambda \in T_{U,ram}}I^{tame}_{{\tilde{P}},{\mathbb {P}}^{1}}\right) ^{*} {\mathop {=}\limits ^{def}}ker\left( \left( \oplus _{\lambda \in T_{U,ram}}I^{tame}_{{\tilde{P}},{\mathbb {P}}^{1}}\right) \twoheadrightarrow \oplus {\mathbb {Z}}/2{\mathbb {Z}} {\mathop {\twoheadrightarrow }\limits ^{sum}} {\mathbb {Z}}/2{\mathbb {Z}}\right) \right) \end{aligned} \end{aligned}$$

therefore

$$\begin{aligned} Q \simeq \left( \sum _{P \in T_{U,ram}} \frac{1}{2}I_{{\tilde{P}}}^{tame}\right) + \left( \sum _{P \in T_{U,unr}} I_{{\tilde{P}}}^{tame}\right) \subset \frac{1}{2}Q_{U} \end{aligned}$$

By identifying Q with the right-hand side of this isomorphism, we obtain $Q_{U}\hookrightarrow Q$.

$\square$

We will use the following lemma in the proof of Theorem 3.3.

Lemma 3.2

Let p be an odd prime number, m an even non-negative integer and l a non-negative integer such that $(m,l) \ne (0,0)$.

• (a) For any $a_{1},\ldots , a_{m},b_{1},\ldots , b_{l}\in \{ 0,1,\ldots ,p-1 \}$ , $e_{1},\ldots ,e_{m} , f_{1},\ldots ,f_{l} \in {\mathbb {Z}}_{>0}$ with $p \not \mid (\prod _{i=1}^{m}e_{i})(\prod _{j=1}^{l}f_{j})$ and $\alpha _{1},\ldots ,\alpha _{m},\beta _{1},\ldots ,\beta _{l}\in {\mathbb {Z}}$. Then, there exist $d_{0} ,{\tilde{a}}_{1},\ldots ,{\tilde{a}}_{m},{\tilde{b}}_{1},\ldots ,{\tilde{b}}_{l} \in {\mathbb {Z}}_{> 0}$ such that, for any $d \in {\mathbb {Z}}$ such that $d \ge d_{0}$, we have (i), (ii), (iii)

  $$\begin{aligned} \begin{aligned} \mathrm{(i)}&\ {\tilde{c}} \equiv c \ mod \ p \ \ (c=a_{1},\ldots , a_{m},b_{1},\ldots , b_{l})\\ \mathrm{(ii)}&\ {\tilde{a}}_{i} \equiv \alpha _{i} \ mod \ e_{i}\ ,\ {\tilde{b}}_{j} \equiv \beta _{j} \ mod \ f_{j} \ \ (1\le i\le m\ ,\ 1 \le j \le l) \\ \mathrm{(iii)}&\ \mathrm{For \ all } \ q,t,\delta _{1 },\ldots , \delta _{m},\epsilon _{1},\ldots ,\epsilon _{l}\in {\mathbb {Z}} \ s.t. \ 0 \le q \le \frac{m}{2} \ ,0 \le t \le \frac{m}{2}\ , \\&0 \le \delta _{i} \le {\tilde{a}}_{i}+\frac{p^{d}-1}{2} \ , \ 0 \le \epsilon _{j} \le {\tilde{b}}_{j} \ \mathrm{and }\\&\ \sum _{i} \delta _{i}+\sum _{j} \epsilon _{j}=\frac{p^{d}-1}{2}+s-q+tp^{d}\ ,\\&we \ have \ \prod _{i,j} \left( {\begin{array}{c}{\tilde{a}}_{i}+\frac{p^{d}-1}{2}\\ \delta _{i}\end{array}}\right) \left( {\begin{array}{c}\tilde{b_{j}}\\ \epsilon _{j}\end{array}}\right) \equiv 0 \ { mod }\ p \end{aligned} \end{aligned}$$

• (b) Assume, moreover, that $l \ne 0$ and $(m,l) \ne (0,1)$. Then, for any $a_{1},\ldots , a_{m}$, $b_{1},\ldots , b_{l} \in \{ 0,1,\ldots ,p-1 \}$, there exist $d,{\tilde{a}}_{1},\ldots ,{\tilde{a}}_{m},{\tilde{b}}_{1},\ldots ,{\tilde{b}}_{l} \in {\mathbb {Z}}_{>0}$ which satisfy (i), (iii), (iv), (v), (vi).

  $$\begin{aligned} \begin{aligned} (iv)&\ p^{d} > 4s \ \ \left( s {\mathop {=}\limits ^{def}}\sum _{c=a_{1},\ldots , a_{m},b_{1},\ldots , b_{l}} {\tilde{c}}\right) \\ (v)&\ 2 | \frac{p^{d}-1}{(p^{d}-1,{\tilde{c}})} \ , \ 2 | \frac{p^{d}-1}{(p^{d}-1,s-1)} \ \ (c = a_{1},\ldots , a_{m},b_{1},\ldots , b_{l})\\ (vi)&\ (p^{d}-1,\tilde{b_{1}})=1\\ \end{aligned} \end{aligned}$$

Proof

We take any $u\in {\mathbb {Z}}$ such that

$$\begin{aligned} \begin{aligned}&p^{u}>2\left( \sum _{i}a_{i}+\sum _{j}b_{j}+\frac{m}{2}p+\left( \sum _{i}e_{i} +\sum _{j}f_{j}\right) p\right) -1\\&\quad \left( \iff p^{u}>\left( \sum _{i}a_{i}+\sum _{j}b_{j}+\frac{m}{2}p\right. \right. \\&\quad \left. \left. +\left( \sum _{i}e_{i}+\sum _{j}f_{j}\right) p\right) +\left( \sum _{h=0}^{u-1}\frac{p-1}{2}p^{h}\right) \right) \end{aligned} \end{aligned}$$

and set $d_{0} {\mathop {=}\limits ^{def}}u+3$. We define ${\tilde{a}}_{1},\ldots ,{\tilde{a}}_{m},{\tilde{b}}_{1},\ldots ,{\tilde{b}}_{l}$ to be the unique integers that satisfy (ii) and the following condition.

$$\begin{aligned} \begin{aligned} {\tilde{a}}_{i}&= a_{i}+\frac{p+1}{2}p+\sum _{h=2}^{u}\frac{p-1}{2}p^{h}+A_{i}p \ \ (1\le A_{i} \le e_{i})\\ {\tilde{b}}_{j}&= b_{j}+B_{j}p \ \ (1\le B_{j}\le f_{j}) \end{aligned} \end{aligned}$$

Then for any $d\ge d_{0}$, we have

$$\begin{aligned} \begin{aligned}&s=\sum _{i}a_{i}+\sum _{j}b_{j}+\frac{m}{2}p+\frac{m}{2}p^{u+1}+D \\&\qquad \left( \left( m+l\right) p \le D {\mathop {=}\limits ^{def}}\left( \sum _{i}A_{i}p\right) + \left( \sum _{j}B_{j}p\right) \le \left( \sum _{i}e_{i}+\sum _{j}f_{j}\right) p\right) \\&\qquad {\tilde{a}}_{i}+\frac{p^{d}-1}{2}=a_{i}+\frac{p-1}{2} +\frac{p+1}{2}p^{u+1}+\left( \sum _{h=u+2}^{d-1}\frac{p-1}{2}p^{h}\right) +A_{i}p\\&\qquad \frac{p^{d}-1}{2}+s-q+tp^{d} \\&\quad = \left( \sum _{i}a_{i}+\sum _{j}b_{j}+\frac{m}{2}p+D-q\right) \\&\qquad +\left( \sum _{h=0}^{d-1}\frac{p-1}{2}p^{h}\right) +\frac{m}{2}p^{u+1}+tp^{d} \end{aligned} \end{aligned}$$

Let $\sum _{g}a_{(i,g)}p^{g} \ , \ \sum _{g}b_{(j,g)}p^{g} \ , \ \sum _{g}\delta _{(i,g)}p^{g} \ , \ \sum _{g}\epsilon _{(j,g)}p^{g} \ \ (a_{(i,g)},b_{(j,g)},\delta _{(i,g)},\epsilon _{(j,g)}$$\in \{0,1,\ldots , p-1\})$ be the p-adic expansions of ${\tilde{a}}_{i}+\frac{p^{d}-1}{2}, \ {\tilde{b}}_{j}, \ \delta _{i}, \ \epsilon _{j}$, respectively.

At first, suppose either that there exist $i\in \{1,2,\ldots ,m\} , g\in \{0,1,\ldots ,u-1\}$ such that $b_{(j,g)} < \epsilon _{(j,g)}$, or that there exist $j\in \{1,2,\ldots ,l\} , g\in \{0,1,\ldots ,u-1\}$ such that $b_{(j,g)} < \epsilon _{(j,g)}$. By Lucas’ theorem ([3]),

$$\begin{aligned} \left( {\begin{array}{c}{\tilde{a}}_{i}+\frac{p^{d}-1}{2}\\ \delta _{i}\end{array}}\right) \equiv 0 \ mod \ p \ \ \mathrm{or} \ \ \left( {\begin{array}{c}{\tilde{b}}_{j}\\ \epsilon _{j}\end{array}}\right) \equiv 0 \ mod \ p \end{aligned}$$

therefore we have (iii).

Next, suppose that $a_{(i,g)} \ge \delta _{(i,g)}$ and $b_{(j,g)}\ge \epsilon _{(j,g)}$ hold for any $i\in \{1,2,\ldots ,m\} , j\in \{1,2,\ldots ,l\} , g\in \{0,1,\ldots ,u-1\}$. Then we have

$$\begin{aligned} p^{u}> & {} \sum _{i}a_{i}+\sum _{j}b_{j}+\frac{m}{2}\left( p-1\right) +D\\\ge & {} \left( \sum _{i}\sum _{g=0}^{u-1}\delta _{\left( i,g\right) } p^{g}\right) +\left( \sum _{j}\sum _{g=0}^{u-1}\epsilon _{\left( j,g\right) }p^{g}\right) \end{aligned}$$

Let $\eta$ be the uth coefficient of the p-adic expansion of $(\sum _{i}\delta _{i})+(\sum _{j}\epsilon _{j})=\frac{p^{d}-1}{2}+s-q+tp^{d}$. Then $\eta$ satisfies $\eta \equiv \sum _{i}\delta _{(i,u)}+\sum _{j}\epsilon _{(j,u)} \ mod \ p$. And we have

$$\begin{aligned} p^{u}> \left( \sum _{i}a_{i}+\sum _{j}b_{j}+\frac{m}{2}p+D-q\right) +\left( \sum _{h=0}^{u-1}\frac{p-1}{2}p^{h}\right) \end{aligned}$$

Then we have $\eta = \frac{p-1}{2}$. Therefore there exists $i\in \{ 1,2,\ldots ,m \}$ such that $\delta _{(i,u)} \ne 0$ or there exists $j \in \{ 1,2,\ldots ,l \}$ such that $\epsilon _{(j,u)} \ne 0$.

On the other hand, any $i \in \{ 1,2\ldots ,m \}$ satisfies

$$\begin{aligned} p^{u}> a_{i}+\frac{p-1}{2}+A_{i}p \end{aligned}$$

Therefore we have $a_{(i,u)}=0$. It is clear that any j satisfies $b_{(j,u)}=0$. By Lucas’s theorem ([3]),

$$\begin{aligned} \left( {\begin{array}{c}{\tilde{a}}_{i}+\frac{p^{d}-1}{2}\\ \delta _{i}\end{array}}\right) \equiv 0 \ mod \ p \ \ or \ \ \left( {\begin{array}{c}{\tilde{b}}_{j}\\ \epsilon _{j}\end{array}}\right) \equiv 0 \ mod \ p \end{aligned}$$

Thus, in both cases, we have (iii). By definition of ${\tilde{a}}_{1},\ldots ,{\tilde{a}}_{m},{\tilde{b}}_{1},\ldots ,{\tilde{b}}_{l}$, we have (i), (ii). This proves (a).

Next, we will prove (b).

• Suppose $b_{1}=0$

We set $f_{1}=1$ and take any $e_{1},\ldots ,e_{m} , f_{2},\ldots ,f_{l} , \alpha _{1},\ldots ,\alpha _{m},\beta _{1},\ldots ,\beta _{l}$ that satisfy $p \not \mid (\prod _{i=1}^{m}e_{i})(\prod _{j=1}^{l}f_{j})$. We apply (a) to them. By the proof of the first half of the lemma, we can take ${\tilde{b}}_{1}=p$. We can take a sufficiently large d that satisfies (v), because $p \ne 2$. Therefore we can take d that satisfies (iv), (v) and (vi).

• Suppose $b_{1}\ne 0$ and $l \equiv 0\ mod \ 2$.

By Dirichlet’s theorem on arithmetic progressions, there exists $N \in {\mathbb {Z}}_{>0}$ such that $b_{1}+p+Np^{2}$ is a prime number. We take

$$\begin{aligned}&f_{1}=1+Np, \ \beta _{1}=b_{1},\\&e_{1}=e_{2}= \cdots =e_{m}=f_{2}=\cdots =f_{l}=2, \\&\alpha _{1}=\cdots =\alpha _{m}=\beta _{2}=\cdots =\beta _{l}=1. \end{aligned}$$

We apply the first half of the lemma to them. By the proof of the first half of the lemma, we can take ${\tilde{b}}_{1}=b_{1}+p+Np^{2}$. Then ${\tilde{b}}_{1}$ is a prime number and ${\tilde{b}}_{1}\ge 1+p+p^{2}$, in particular $(p^{2}-p,{\tilde{b}}_{1})=1$. Any $d\ge d_{0}$ satisfies (v), because ${\tilde{a}}_{1},\ldots ,{\tilde{a}}_{m},{\tilde{b}}_{1},\ldots ,{\tilde{b}}_{l},s-1$ are odd numbers. Thus, if we take sufficiently large d that satisfies (iv), $d,{\tilde{a}}_{1},\ldots ,{\tilde{a}}_{m},{\tilde{b}}_{1},\ldots ,{\tilde{b}}_{l}$ satisfy $(i),(iii)\sim (v)$. If ${\tilde{b}}_{1} \not \mid p^{d}-1$, then we also have (vi). If ${\tilde{b}}_{1} | p^{d}-1$ (i.e. (vi) is not satisfied), we have

$$\begin{aligned} p^{d+1}-1=(p-1)(p^{d}+(p^{d-1}+\cdots +p+1))\equiv (p-1)p^{d} \ mod \ {\tilde{b}}_{1}. \end{aligned}$$

Hence $d+1, {\tilde{a}}_{1}, \ldots ,{\tilde{a}}_{m},{\tilde{b}}_{1},\ldots ,{\tilde{b}}_{l}$ satisfy $(i),(iii)\sim (vi)$.

• Suppose $b_{1}\ne 0$ and $l \equiv 1 \ mod \ 2$.

By assumption, we have $m \ne 0$ or $l\ge 3$. Suppose $l\ge 3$ (resp. $m\ne 0$). By Dirichlet’s theorem on arithmetic progressions, there exists $N \in {\mathbb {Z}}_{>0}$ such that $b_{1}+p+Np^{2}$ is a prime number. We take

$$\begin{aligned}&f_{1}=1+Np, \ \beta _{1}=b_{1}, f_{2}=4, \ \beta _{2}=2, \\&e_{1}=\cdots =e_{m}=f_{3}=\cdots =f_{l}=2,\\&\alpha _{1}=\cdots =\alpha _{m}=\beta _{3}=\cdots =\beta _{l}=1 \end{aligned}$$

(resp.

$$\begin{aligned}&f_{1}=1+Np, \ \beta _{1}=b_{1}, \ e_{1}=4, \ \alpha _{1}=2, \\&e_{2}=\cdots =e_{m}=f_{2}=\cdots =f_{l}=2 , \\&\alpha _{2}=\cdots =\alpha _{m}=\beta _{2}=\cdots =\beta _{l}=1). \end{aligned}$$

We apply the first half of the lemma to them. By the proof of the first half of the lemma, we can take ${\tilde{b}}_{1}=b_{1}+p+Np^{2}$. Then ${\tilde{b}}_{1}$ is a prime number and ${\tilde{b}}_{1}\ge 1+p+p^{2}$, in particular $(p^{3}-p,{\tilde{b}}_{1})=1$. ${\tilde{a}}_{1},\ldots ,{\tilde{a}}_{m},{\tilde{b}}_{1},\ldots ,{\tilde{b}}_{l},s-1$ are odd numbers except ${\tilde{b}}_{2}$ (resp. ${\tilde{a}}_{1})$, and ${\tilde{b}}_{2}$ (resp. ${\tilde{a}}_{1})\equiv 2 \ mod \ 4$. Hence all $d \in 2{\mathbb {Z}}_{>0}$ satisfy (v). Thus, if we take sufficiently large d that satisfies (iv), $d,{\tilde{a}}_{1},\ldots ,{\tilde{a}}_{m},{\tilde{b}}_{1},\ldots ,{\tilde{b}}_{l}$ satisfy $(i),(iii)\sim (v)$. If ${\tilde{b}}_{1} \not \mid p^{d}-1$, then we also have (vi). If ${\tilde{b}}_{1}|p^{d}-1$, then we have

$$\begin{aligned} p^{d+2}-1= & {} (p-1)(p^{d}(p+1)+(p^{d-1}+\cdots +p+1)) \\\equiv & {} (p-1)(p^{d}(p+1)) \ mod \ {\tilde{b}}_{1}. \end{aligned}$$

Hence $d+2, {\tilde{a}}_{1}, \ldots ,{\tilde{a}}_{m}, {\tilde{b}}_{1}, \ldots ,{\tilde{b}}_{l}$ satisfy $(i),(iii)\sim (vi)$.

This proves (b). $\square$

Definition

( [11] §3) Let $\gamma$ be an integer such that $\gamma \ge 1$ , $p \not \mid \gamma$ and $2|\gamma$. We define

$$\begin{aligned} \begin{aligned}&{\tilde{H}}({\mathbb {Z}}/\gamma {\mathbb {Z}}) \\&\quad {\mathop {=}\limits ^{def}}\left\{ (c_{P})_{P\in T_{U}},c_{P}\in {\mathbb {Z}} /\gamma {\mathbb {Z}} \mid ( \ < c_{P}>_{P\in T_{U}} \ = \ {\mathbb {Z}}/\gamma {\mathbb {Z}} \ ) \ and \right. \\&\qquad \quad \left. \left( \sum _{P\in T_{U}} c_{P}=0\right) \right\} \\&\qquad H({\mathbb {Z}}/\gamma {\mathbb {Z}}) {\mathop {=}\limits ^{def}}{\tilde{H}}({\mathbb {Z}}/\gamma {\mathbb {Z}})/({\mathbb {Z}}/\gamma {\mathbb {Z}})^{\times } \end{aligned} \end{aligned}$$

The natural identification Surj$(Q,{\mathbb {Z}}/\gamma {\mathbb {Z}})$$\simeq$${\tilde{H}}({\mathbb {Z}}/\gamma {\mathbb {Z}})$ and the restriction map Hom$(Q,{\mathbb {Z}}/\gamma {\mathbb {Z}})\rightarrow$Hom$(Q_{U},{\mathbb {Z}}/\gamma {\mathbb {Z}})$ yield the following map

$$\begin{aligned} \begin{aligned}&H({\mathbb {Z}}/\gamma {\mathbb {Z}}) \simeq \{ H'\subset \pi _1({\mathbb {P}}^1{\setminus } T_{U}) : open \ subgroup \ \mid \pi _ 1({\mathbb {P}}^{1} {\setminus } T_{U})/H' \simeq {\mathbb {Z}}/\gamma {\mathbb {Z}}\} \\&\quad \rightarrow \left\{ H \subset \pi _{1}(U) : open \ subgroup \ | \right. \\&\quad \left. \left( \ \pi _{1} (U)/H \simeq {\mathbb {Z}}/\gamma {\mathbb {Z}} \ \ \mathrm{or }\ \ \pi _{1}(U)/H \simeq {\mathbb {Z}}/\frac{1}{2} \gamma {\mathbb {Z}}\right) \ and \ L_{U} \subset H \right\} \end{aligned} \end{aligned}$$

Fix closed points $\rho _{0}\ne \rho _{\infty } \in {\mathbb {P}}^{1}$. For each isomorphism $\phi : {\mathbb {P}}^{1} \simeq {\mathbb {P}}^{1}$ with $\phi (\rho _{0})=0,\phi (\rho _{\infty })=\infty$, we obtain a bijection ${\mathbb {P}}^{1}(k)\backslash \{ \rho _{\infty } \} \simeq {\mathbb {P}}^{1}(k)\backslash \{ \infty \}= k$. This bijection does not depend on the choice of $\phi$ up to scalar multiplication. Hence the additive structure on ${\mathbb {P}}^{1}(k)\backslash \{ \rho _{\infty } \}$ that is induced by this bijection does not depend on the choice of $\phi$, and only depends on the choice of $\rho _{0}$ and $\rho _{\infty }$.

Theorem 3.3

Let $A(U,P_{0},P_{\infty })$ be the following.

$$\begin{aligned} \begin{aligned}&A(U,P_{0},P_{\infty }){\mathop {=}\limits ^{def}}\\&\quad \left\{ (a_{P})_{P\in T_{U} \backslash \{ P_{0},P_{\infty }\}} \ | \ a_{P}\in {\mathbb {F}}_{p}, \sum _{P \in T_{U} \backslash \{ P_{0},P_{\infty }\}}a_{P}P=P_{0} \right\} \\&\quad (with\ respect\ to\ the\ additive\ structure\ associated\ with\ P_{0}\ \mathrm{and }\ P_{\infty }) \end{aligned} \end{aligned}$$

Then $A(U,P_{0},P_{\infty })$ can be recovered group-theoretically from $\pi _{1}(U)$, $L_{U}$, $P_{0}$ and $P_{\infty }$.

Proof

Step 1. Construct a suitable cover $U_{H}$ over U defined by $\{ a_{P} \}_{P}$.

We define $a_{1}, \ldots ,a_{m}, b_{1}, \ldots ,b_{l} \in \{0,1, \ldots ,p-1\}$ by $a_{i} \ mod \ p =a_{P_{i}} , b_{j} \ mod \ p = a_{R_{j}}\ (i=1,\ldots ,m \ , \ j=1,\ldots ,l)$, and apply Lemma 3.2 (b) to them. Then we obtain ${\tilde{a}}_{P_{i}}{\mathop {=}\limits ^{def}}{\tilde{a}}_{i} , \ {\tilde{a}}_{R_{j}}{\mathop {=}\limits ^{def}}{\tilde{b}}_{j} , \ d$ that satisfy (i), (iii), (iv), (v),  (vi) of Lemma 3.2. Let H (resp. $H'$) be the open subgroup of $\pi _{1}(U)$ (resp. $\pi _{1}({\mathbb {P}}^{1} \backslash T_{U})$) associated with $(c_{P})_{P \in T_{U}} \in H({\mathbb {Z}}/(p^{d}-1){\mathbb {Z}})$, where

$$\begin{aligned}&c_{P_{\infty }}=1, \\&c_{P_{0}}=s-1 {\mathop {=}\limits ^{def}}\sum _{P \in T_{U} \backslash \{ P_{0} , P_{\infty }\}} {\tilde{a}}_{P}-1,\\&c_{P}=-{\tilde{a}}_{P} \ (P \ne P_{0},P_{\infty }). \end{aligned}$$

Set

$$\begin{aligned}&X_{H}{\mathop {=}\limits ^{def}}(U_{H})^{cpt},\\&(\mathbb {P}^{1})_{H'} {\mathop {=}\limits ^{def}}((\mathbb {P}^{1}\backslash T_{U})_{H'})^{cpt}, \\&\phi : X_{H}\rightarrow X,\\&\psi :X_{H} \rightarrow (\mathbb {P}^{1})_{H'}. \end{aligned}$$

By Lemma 3.2 (vi), we have $(p^d-1,{{\tilde{a}}}_{R_1})=1$. Then $R_{1}$ is totally ramified in $({\mathbb {P}}^1)_{H'}\rightarrow {\mathbb {P}}^1$. On the other hand, by definition, $R_{1}$ is unramified in $X\rightarrow {\mathbb {P}}^1$. Hence the above commutative diagram is a cartesian product on generic points. In particular, the degree of $X_H\rightarrow X$ is $p^d-1$, that is the degree of $({\mathbb {P}}^1)_{H'}\rightarrow {\mathbb {P}}^1$.

For $r \in {\mathbb {Z}}_{\ge 0}$, let ${\mathcal {H}}_{r}$ be $H^{0}(X_{H},\varOmega _{X_{H}})((\chi _{{\tilde{\mu }}_{(1,1)} ,d}^{-{\tilde{a}}_{R_{1}}})^{p^{r}})$.

Step 2. Prove that whether the action of the Cartier operator C on $\sum _{r} {\mathcal {H}}_{r}$ is nilpotent or not can be determined group-theoretically.

Then by Theorem 2.1 and Corollary 2.2, whether $(\pi _{1}(X_{H})^{ab}/p)(\chi _{{\tilde{\mu }}_{(1,1)} , d}^{-{\tilde{a}}_{R_{1}}})=0$ holds or not can be determined group-theoretically (here, ${\tilde{\mu }}_{(1,1)} \in {\tilde{X}}$ is a point above $\mu _{(1,1)}$). By Artin-Schreier theory,

$$\begin{aligned} \begin{aligned}&(\pi _{1}(X_{H})^{ab}/p)^{*} {\mathop {=}\limits ^{def}}Hom(\pi _{1}(X_{H})^{ab}/p , {\mathbb {F}}_{p}) = Hom_{cont}(\pi _{1}(X_{H}) , {\mathbb {F}}_{p}) \\&\quad = H^{1}_{et}(X_{H},{\mathbb {F}}_{p}) = H^{1}(X_{H},{\mathcal {O}}_{X_{H}})[F-1] \end{aligned} \end{aligned}$$

This, together with [9] Proposition 9, implies

$$\begin{aligned} \begin{aligned}&(\pi _{1}(X_{H})^{ab}/p)(\chi _{{\tilde{\mu }}_{(1,1)} , d}^{-{\tilde{a}}_{R_{1}}})=0 \\&\quad \Leftrightarrow (\pi _{1}(X_{H})^{ab}/p)^{*}((\chi _{{\tilde{\mu }}_{(1,1)} , d}^{-{\tilde{a}}_{R_{1}}})^{-1})=0 \\&\quad \Leftrightarrow The\ Frobenius\ F\ on \sum _{r}H^{1}(X_{H},{\mathcal {O}}_{X_{H}})((\chi _{{\tilde{\mu }}_{(1,1)} , d}^{-{\tilde{a}}_{R_{1}}})^{-p^{r}}) \ is\ nilpotent \\&\quad \Leftrightarrow The\ Cartier\ operator\ C \ on \sum _{r} {\mathcal {H}}_{r} \ is\ nilpotent \end{aligned} \end{aligned}$$

Step 3. Construct a suitable k-basis of ${\mathcal {H}}_{0}$.

By fixing a suitable coordinate choice of ${\mathbb {P}}^{1}$, Let

$$\begin{aligned}&B {\mathop {=}\limits ^{def}}k[x,x^{-1},(x-P_{1})^{-1},\ldots ,(x-P_{m})^{-1}, (x-R_{1})^{-1},\ldots ,(x-R_{l})^{-1}][z]\\&\quad /\langle z^{2}-x(x-P_{1})\ldots (x-P_{m})\rangle , \end{aligned}$$

then we can write $U=SpecB$. Set

$$\begin{aligned} B_{H}{\mathop {=}\limits ^{def}}B[y]/\langle y^{p^{d}-1}-x^{s-1}\prod _{P \in T_{U} \backslash \{ P_{0},P_{\infty } \}}(x-P)^{-{\tilde{a}}_{P}}\rangle , \end{aligned}$$

then we can write $U_{H}=SpecB_{H}$. Because

$$\begin{aligned} \varOmega _{{\mathbb {P}}^{1}\backslash T_{U}}={\mathcal {O}}_{{\mathbb {P}}^{1}\backslash T_{U}}(dx)={\mathcal {O}}_{{\mathbb {P}}^{1}\backslash T_{U}}(dx/x) \end{aligned}$$

and ${\mathbb {P}}^{1}\backslash T_{U} \leftarrow U_{H}$ is étale, we have $\varOmega _{U_{H}}={\mathcal {O}}_{U_{H}}(dx/x)$. By the above coordinate and the definition of $\chi _{{\tilde{\mu }}_{(1,1)} , d}^{-{\tilde{a}}_{R_{1}}}$, we have

$$\begin{aligned} \chi _{{\tilde{\mu }}_{(1,1)} , d}^{-{\tilde{a}}_{R_{1}}}(\gamma ) = \gamma (y)/y\ (\gamma \in I_{{\tilde{\mu }}_{(1,1)}}). \end{aligned}$$

Then Lemma 3.2(vi) (i.e. $(p^{d}-1,{\tilde{a}}_{R_{1}}) = 1$) implies that we have

$$\begin{aligned} \varGamma (U_{H},\varOmega _{U_{H}})(\chi _{{\tilde{\mu }}_{(1,1)},d}^{-{\tilde{a}}_{R_{1}}}) = By(dx/x). \end{aligned}$$

Let $f\in B$ and set $\omega = fy(\frac{dx}{x})\in \varGamma (U_{H},\varOmega _{U_{H}})(\chi _{{\tilde{\mu }}_{(1,1)},d}^{ -{\tilde{a}}_{R_{1}}})$. We will consider a necessary and sufficient condition for $\omega \in \varGamma (X_{H},\varOmega _{X_{H}})(\chi _{{\tilde{\mu }}_{(1,1)},d}^{ -{\tilde{a}}_{R_{1}}})$. This can be checked at each $\nu \in X_H\setminus U_H$. Let $t_{\nu }$ be a prime element of ${\mathcal {O}}_{X_{H},\nu }$.

• Suppose $\phi (\nu ) = \lambda _{\infty }$

The ramification index of $\psi (\nu )$ over $P_{\infty }$ is $p^{d}-1$. The ramification index of $\phi (\nu )=\lambda _{\infty }$ over $P_{\infty }$ is 2. By Abhyankar’s lemma, the ramification index of $\nu$ over $\lambda _{\infty }$ is $(p^{d}-1)/2$ and $\nu$ is unramified over $\psi (\nu )$. By $(dx/dt_{\nu })=-x^{2}(dx^{-1}/dt_{\nu })$ and $ord_{\nu }(dx^{-1}/dt_{\nu })=p^{d}-2$, we have $ord_{\nu }(dx/dt_{\nu })=-p^{d}$, and

$$\begin{aligned} \begin{aligned} ord_{\nu }\left( fy\frac{dx}{dt_{\nu }}x^{-1}\right)&=\frac{p^{d}-1}{2}ord_{\lambda _{\infty }}(f)+1-p^{d}+(p^{d}-1)\\&=\frac{p^{d}-1}{2}ord_{\lambda _{\infty }}(f) \end{aligned} \end{aligned}$$

therefore

$$\begin{aligned} \begin{aligned} \omega&= \left( fy\frac{dx}{dt_{\nu }}x^{-1}\right) dt_{\nu } \in \varOmega _{X_{H},\nu } \\&\Leftrightarrow \ ord_{\nu }\left( fy\frac{dx}{dt_{\nu }}x^{-1}\right) \ge 0 \\&\Leftrightarrow \ ord_{\lambda _{\infty }}(f) \ge 0 \end{aligned} \end{aligned}$$

• Suppose $\phi (\nu )=\lambda _{0}$

Set $e_{P_{0}}{\mathop {=}\limits ^{def}}(p^{d}-1)/(p^{d}-1,s-1)$ which is the ramification index of $\psi (\nu )$ over $P_{0}$. By Lemma 3.2 (v), we have $2|e_{P_{0}}$. By the same argument as above, the ramification index of $\nu$ over $\lambda _{0}$ is $e_{P_{0}}/2$ and that $\nu$ is unramified over $\psi (\nu )$. Then

$$\begin{aligned} \begin{aligned} ord_{\nu }\left( fy\frac{dx}{dt_{\nu }}x^{-1}\right)&= \frac{e_{P_{0}}}{2}ord_{\lambda _{0}}(f)+\frac{(s-1) e_{P_{0}}}{p^{d}-1}+(e_{P_{0}}-1)-e_{P_{0}}\\&= \frac{e_{P_{0}}}{2}\left( ord_{\lambda _{0}}(f)+\frac{2\left( (s-1) -(p^{d}-1,s-1)\right) }{p^{d}-1}\right) \end{aligned} \end{aligned}$$

By Lemma 3.2 (iv), we have

$$\begin{aligned} p^{d}-1 \ge 2s > 2((s-1)-(s-1,p^{d}-1))\ge 0. \end{aligned}$$

Therefore

$$\begin{aligned} \begin{aligned} \omega&= \left( fy\frac{dx}{dt_{\nu }}x^{-1}\right) dt_{\nu } \in \varOmega _{X_{H},\nu }\\&\Leftrightarrow \ ord_{\nu }\left( fy\frac{dx}{dt_{\nu }}x^{-1}\right) \ge 0 \\&\Leftrightarrow \ ord_{\lambda _{0}}(f) \ge -\frac{2((s-1)-(p^{d}-1,s-1))}{p^{d}-1}\\&\Leftrightarrow \ ord_{\lambda _{0}}(f) \ge 0 \end{aligned} \end{aligned}$$

• Suppose $\phi (\nu )=\lambda _{i} \ (i= 1,2,\ldots ,m )$

Set $e_{P_{i}}{\mathop {=}\limits ^{def}}((p^{d}-1)/(p^{d}-1,{\tilde{a}}_{P_{i}}))$, this is the ramification index of $\psi (\nu )$ over $P_{i}$. By Lemma 3.2 (v), we have $2|e_{P_{i}}$. By the same argument as above, the ramification index of $\nu$ over $\lambda _{i}$ is $e_{P_{i}}/2$ and $\nu$ is unramified over $\psi (\nu )$. Then

$$\begin{aligned} \begin{aligned} ord_{\nu }\left( fy\frac{dx}{dt_{\nu }}x^{-1}\right)&= \frac{e_{P_{i}}}{2}ord_{\lambda _{i}}(f)-\frac{{\tilde{a}}_{P_{i} }e_{P_{i}}}{p^{d}-1}+(e_{P_{i}}-1)\\&= \frac{e_{P_{i}}}{2}\left( ord_{\lambda _{i}}(f)+2\frac{(p^{d}-1) -({\tilde{a}}_{P_{i}}+(p^{d}-1,{\tilde{a}}_{P_{i}}))}{p^{d}-1}\right) \end{aligned} \end{aligned}$$

By definition,

$$\begin{aligned} 2>2((p^{d}-1)-({\tilde{a}}_{P_{i}}+(p^{d}-1,{\tilde{a}}_{P_{i}})))/(p^{d}-1) \end{aligned}$$

is clear. By Lemma 3.2 (iv), we have

$$\begin{aligned} p^{d}-1 \ge 4s > 2({\tilde{a}}_{P_{i}}+(p^{d}-1,{\tilde{a}}_{P_{i}})), \end{aligned}$$

hence

$$\begin{aligned} 2((p^{d}-1)-({\tilde{a}}_{P_{i}}+(p^{d}-1,{\tilde{a}}_{P_{i}})))/(p^{d}-1)>1. \end{aligned}$$

Therefore

$$\begin{aligned} \begin{aligned} \omega&= \left( fy\frac{dx}{dt_{\nu }}x^{-1}\right) dt_{\nu } \in \varOmega _{X_{H},\nu }\\&\Leftrightarrow \ ord_{\nu }\left( fy\frac{dx}{dt_{\nu }}x^{-1}\right) \ge 0 \\&\Leftrightarrow \ ord_{\lambda _{i}}(f) \ge -2\frac{(p^{d}-1)-({\tilde{a}}_{P_{i}}+(p^{d}-1, {\tilde{a}}_{P_{i}}))}{p^{d}-1})\\&\Leftrightarrow \ ord_{\lambda _{i}}(f) \ge -1 \end{aligned} \end{aligned}$$

• Suppose $\phi (\nu )=\mu _{(i,j)} \ (i= 1,2,\ldots ,l \ , \ j =1,2 )$

Set $e_{R_{i}}{\mathop {=}\limits ^{def}}((p^{d}-1)/(p^{d}-1,{\tilde{a}}_{R_{i}}))$, which is the ramification index of $\psi (\nu )$ over $R_{i}$. $\mu _{(i,j)}$ is unramified over $R_{i}$. Thus the ramification index of $\nu$ over $\mu _{(i,j)}$ is $e_{R_{i}}$ and $\nu$ is unramified over $\psi (\nu )$. Then

$$\begin{aligned} \begin{aligned} ord_{\nu }\left( fy\frac{dx}{dt_{\nu }}x^{-1}\right)&= e_{R_{i}}ord_{\mu _{(i,j)}}(f)-\frac{{\tilde{a}}_{R_{i}} e_{R_{i}}}{p^{d}-1}+(e_{R_{i}}-1)\\&= e_{R_{i}}\left( ord_{\mu _{(i,j)}}(f)-\frac{{\tilde{a}}_{R{i}} +(p^{d}-1,{\tilde{a}}_{R_{i}})}{p^{d}-1}+1\right) \end{aligned} \end{aligned}$$

By Lemma 3.2 (iv), we have

$$\begin{aligned} p^{d}-1>{\tilde{a}}_{R{i}}+(p^{d}-1,{\tilde{a}}_{R_{i}})>0. \end{aligned}$$

Therefore

$$\begin{aligned} \begin{aligned} \omega&= \left( fy\frac{dx}{dt_{\nu }}x^{-1}\right) dt_{\nu } \in \varOmega _{X_{H},\nu }\\&\Leftrightarrow \ ord_{\nu }\left( fy\frac{dx}{dt_{\nu }}x^{-1}\right) \ge 0 \\&\Leftrightarrow \ ord_{\mu _{(i,j)}}(f) \ge \frac{{\tilde{a}}_{R{i}}+(p^{d}-1,{\tilde{a}}_{R_{i}})}{p^{d}-1}-1\\&\Leftrightarrow \ ord_{\mu _{(i,j)}}(f) \ge 0 \end{aligned} \end{aligned}$$

Set $D {\mathop {=}\limits ^{def}}\lambda _{1}+\lambda _{2}+\cdots +\lambda _{m} \in Div(X)$. By the above computation,

$$\begin{aligned} \omega \in \varGamma (X_{H}, \varOmega _{X_{H}}) \Leftrightarrow f \in \varGamma (X,{\mathscr {L}}(D)) \end{aligned}$$

Let $K_{X}$ be the canonical divisor of X. By Hurwitz’s formula, we have $g{\mathop {=}\limits ^{def}}g(X)=m/2$, and $deg(K_{X})=m-2$. Thus by the Riemann-Roch theorem, we have $dim_{k}\varGamma (X,{\mathscr {L}}(D))=g+1$. The valuations of

$$\begin{aligned} 1,(x/z),(x^{2}/z),\ldots ,(x^{g}/z) \in \varGamma (X,{\mathscr {L}}(D)) \end{aligned}$$

at $\lambda _{0}$ are mutually different, hence these functions are linearly independent over k. Then we have

$$\begin{aligned} \varGamma (X,{\mathscr {L}}(D)) = \langle 1,(x/z),(x^{2}/z),\ldots ,(x^{g}/z)\rangle . \end{aligned}$$

Step 4. Calculate the action of C on the k-basis of ${\mathcal {H}}_{0}$ in Step 3.

By Lemma 3.2 (iv) (which implies $p^{d}-1>s-1$) and the following formula

$$\begin{aligned} C^{d}\left( x^{j}y^{\alpha p^{d}}z^{\beta p^{d}}\frac{dx}{x}\right) = \left\{ \begin{array}{ll} x^{j/p^{d}}y^{\alpha }z^{\beta }\frac{dx}{x} &{}\quad (j \in p^{d}{\mathbb {Z}}) \\ 0 &{} \quad (j \in {\mathbb {Z}} \backslash p^{d}{\mathbb {Z}}) \end{array}\right. \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned}&C^{d}\left( y\frac{dx}{x}\right) \\&\quad = C^{d}\left( x^{1-s}(x-P_{1})^{{\tilde{a}}_{P_{1}}}\ldots (x-P_{m})^{{\tilde{a}}_{m}} (x-R_{1})^{{\tilde{a}}_{R_{1}}}\ldots (x-R_{l})^{{\tilde{a}}_{R_{l}}}y^{p^{d}}\frac{dx}{x}\right) \\&\quad = -(a_{P_{1}}P_{1}+\cdots +a_{P_{m}}P_{m}+a_{R_{1}}R_{1}+\cdots +a_{R_{l}}R_{l})y\frac{dx}{x} \end{aligned} \end{aligned}$$

(1)

On the other hand, for any $q \in \{ 1,2\ldots ,g \}$, we have

$$\begin{aligned} \begin{aligned}&C^{d}\left( \frac{x^{q}}{z}y\frac{dx}{x}\right) \\&\quad = C^{d}\left( x^{\left( q+1-s+\frac{p^{d}-1}{2}\right) }(x-P_{1})^{\left( {\tilde{a}}_{P_{1}} +\frac{p^{d}-1}{2}\right) }\ldots \right. \\&\qquad \left. (x-P_{m})^{\left( {\tilde{a}}_{P_{m}}+\frac{p^{d}-1}{2}\right) } (x-R_{1})^{{\tilde{a}}_{R_{1}}}\ldots (x-R_{l})^{{\tilde{a}}_{R_{l}}}\left( \frac{y}{z} \right) ^{p^{d}}\frac{dx}{x}\right) \\&\quad = \sum _{t} \sum _{(\delta _{1},\ldots ,\delta _{m},\epsilon _{1},\ldots ,\epsilon _{l})} \left( \left( \prod _{i} \left( {\begin{array}{c}{\tilde{a}}_{P_{1}}+\frac{p^{d}-1}{2}\\ \delta _{i}\end{array}}\right) (-P_{i})^{\left( {\tilde{a}}_{P_{i}}+ \frac{p^{d}-1}{2}-\delta _{i}\right) }\right) \right. \\&\qquad \left. \left( \prod _{j}\left( {\begin{array}{c}{\tilde{a}}_{R_{j}}\\ \epsilon _{j}\end{array}}\right) (-R_{j})^{({\tilde{a}}_{R_{j}}-\epsilon _{j})}\right) x^{t+1}\frac{y}{z}\frac{dx}{x}\right) \end{aligned} \end{aligned}$$

In this formula, t runs over all the integers that satisfy

$$\begin{aligned} q+1-s+((p^{d}-1)/2) \le (t+1)p^{d} \le q+1+(m+1)((p^{d}-1)/2) \end{aligned}$$

(hence $(m/2)-1\ge t \ge 0$ by Lemma 3.2 (iv)). $\delta _{1},\ldots ,\delta _{m},\epsilon _{1},\ldots ,\epsilon _{l}$ run over all the non-negative integers that satisfy

$$\begin{aligned} \sum _{i} \delta _{i}+\sum _{j} \epsilon _{j}=\frac{p^{d}-1}{2}+s-q+tp^{d}. \end{aligned}$$

By Lemma 3.2 (iii), for any $q\in \{ 1,2,\ldots ,g \}$, we have

$$\begin{aligned} C^{d}\left( \frac{x^{q}}{z}y\frac{dx}{x}\right) = 0 \end{aligned}$$

(2)

Step 5. End of proof.

By the fact that $C(y^{p^{n}}\frac{dx}{x})\in {\mathcal {H}}_{n-1}$ ($n \in \{ 1,2,\ldots , d \}$) and the formulas of (1) and (2), we have that $C^{d}=0$ if and only if C is nilpotent. Thus, $a_{P_{1}}P_{1}+\cdots +a_{P_{m}}P_{m}+a_{R_{1}}R_{1}+\cdots a_{R_{l}}R_{l}=0$ holds if and only if the Cartier operator C on $\sum _{r}{\mathcal {H}}_{r}$ is nilpotent. Therefore whether

$$\begin{aligned} a_{P_{1}}P_{1}+\cdots +a_{P_{m}}P_{m}+a_{R_{1}}R_{1}+\cdots a_{R_{l}}R_{l}=0 \end{aligned}$$

holds or not can be determined group-theoretically from $\pi _{1}(U)$ and $L_{U}$. $\square$

4 Reconstruction of curves of (1,1)-type by their fundamental groups

In this section, we consider curves of (1,1)-type, which are one-punctured elliptic curves (We are considering that the unique cusp is the identity element of the elliptic curve) . We will first prove that the linear relations of the images of m-torsion points in ${\mathbb {P}}^{1}$ are determined by the fundamental group (Corollary 4.8). Then we will use this corollary, and prove that the isomorphism class (as a scheme) of such a curve is determined by the fundamental group (Theorem 4.9). We will use the same symbols as in the previous sections for elliptic curves and their open subschemes. Let E be a (complete) elliptic curve over k.

Proposition 4.1

Let ${\mathcal {O}}\in E(k)$. Let $x,x'$ be finite morphisms $E \rightarrow {\mathbb {P}}^{1}$ of degree 2 that are ramified at ${\mathcal {O}}$. Then there exists an isomorphism $\phi : {\mathbb {P}}^{1} \simeq {\mathbb {P}}^{1}$ that satisfies $x=\phi \circ x'$.

Proof

Set $P{\mathop {=}\limits ^{def}}x({\mathcal {O}}), \ P'{\mathop {=}\limits ^{def}}x'({\mathcal {O}})$. When we think of $P,P'$ as elements of $Div({\mathbb {P}}^{1})$, we have ${\mathscr {L}}(P)\simeq {\mathscr {L}}(P')\simeq {\mathscr {O}}(1)$. By definition, we have $x^{*}({\mathscr {L}}(P))={\mathscr {L}}(2{\mathcal {O}})=x'^{*}({\mathscr {L}}(P'))$. Then both x and $x'$ correspond to a linear system that is a subset of $|{\mathscr {L}}(2{\mathcal {O}})|$ of dimension 1.

By the Riemann–Roch theorem, we have $dim|{\mathscr {L}}(2{\mathcal {O}})|=1$. Thus both x and $x'$ correspond to $|{\mathscr {L}}(2{\mathcal {O}})|$ . By [2] II Remark 7.8.1, they are equivalent up to an isomorphism of ${\mathbb {P}}^{1}$. $\square$

Let ${\mathcal {O}}\in E(k)$ and consider the group structure on E defined by the elliptic curve $(E,{\mathcal {O}})$. Let $A \subset E(k)$ be a finite subset that includes E[2] and satisfies $-A=A$. Then, by Proposition 4.1, a finite morphism $x : E\rightarrow {\mathbb {P}}^{1}$ of degree 2 that is ramified at ${\mathcal {O}}$ is unique up to an automorphism of ${\mathbb {P}}^{1}$, hence the subgroup $L_{E\backslash A}\subset \pi _{1}(E\backslash A)$ with respect to such x depend only on ${\mathcal {O}}$. So, we sometimes write $L_{E\backslash A, {\mathcal {O}}}$ for $L_{E\backslash A}$. For any elliptic curve that has additive structure with respect to ${\mathcal {O}}$, we fix a finite morphisms $x:E\rightarrow {\mathbb {P}}^{1}$ of degree 2 that is ramified at ${\mathcal {O}}$ from now on

Lemma 4.2

For any $m\in {\mathbb {Z}}_{>0}$, the open subgroup $\pi _{1}(E\backslash E[m]) \subset \pi _{1}(E\backslash {\mathcal {O}})$ that corresponds to the multiplication-by-m map $[m] : E\backslash E[m]\rightarrow E \backslash {\mathcal {O}}$ can be recovered group-theoretically from $\pi _{1}(E \backslash {\mathcal {O}})$.

Proof

By Theorem 2.1, the natural morphism $\pi _{1}(E\backslash {\mathcal {O}})\rightarrow \pi _{1}(E) \rightarrow \pi _{1}(E)/m$, hence its kernel $\pi _{1}(E\backslash E[m])$, can be recovered from $\pi _{1}(E\backslash {\mathcal {O}})$. $\square$

Theorem 4.3

(Tamagawa) For any positive even integer m, $(L_{E \backslash E[m],{\mathcal {P}}})_{P\in E[m]}$ can be recovered group-theoretically from $\pi _{1}(E \backslash {\mathcal {O}} )$.

We will need some definitions and lemmas for the proof of Theorem 4.3.

Definition

Let N be a group, M a left N-module. Set

$$\begin{aligned}&M^{N} {\mathop {=}\limits ^{def}}\{ m\in M | \ \forall g\in N, gm=m \},\\&M_{N} {\mathop {=}\limits ^{def}}M/\langle gm-m \ | \ g\in N, m\in M\rangle , \\&M^{\vee } {\mathop {=}\limits ^{def}}Hom_{{\mathbb {Z}}}(M,{\mathbb {Q}}/{\mathbb {Z}}) \end{aligned}$$

(the action of N on ${\mathbb {Q}}/{\mathbb {Z}}$ is trivial).

Lemma 4.4

Let k be an algebraically closed field of characteristic $p\ge 0$, l a prime that is not p, X and Y curves over k, $X \rightarrow Y$ a finite morphism over k, U (resp. V) a non-empty open subscheme of X (resp. Y). Suppose that $X\rightarrow Y$ restricts to a Galois cover $U \rightarrow V$. Let G be the Galois group of $U \rightarrow V$. Then we get a natural isomorphism

$$\begin{aligned} ((\pi _{1}(X)^{ab,l})_{G})\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l}\simeq (\pi _{1}(Y)^{ab,l})\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l} \end{aligned}$$

Proof

Applying [6] Corollary 7.2.5 (Hochschild–Serre spectral sequence) to the natural exact sequence $1 \rightarrow \pi _{1}(U) \rightarrow \pi _{1}(V) \rightarrow G \rightarrow 1$, we get an exact sequence

$$\begin{aligned} 0 \rightarrow H^{1}(G,{\mathbb {Q}}/{\mathbb {Z}})\rightarrow H^{1}(\pi _{1}(V),{\mathbb {Q}}/{\mathbb {Z}}) \rightarrow H^{1}(\pi _{1}(U),{\mathbb {Q}}/{\mathbb {Z}})^{G}\rightarrow H^{2}(G,{\mathbb {Q}}/{\mathbb {Z}}) \end{aligned}$$

By the general property of homological algebra $H^{1}(N,{\mathbb {Q}}/{\mathbb {Z}})\simeq Hom(N^{ab},{\mathbb {Q}}/{\mathbb {Z}})$ and [6] Theorem 2.9.6(Pontryagin duality), We get an exact sequence

$$\begin{aligned} 0 \leftarrow G^{ab} \leftarrow \pi _{1}(V)^{ab} \leftarrow (\pi _{1}(U)^{ab})_{G} \leftarrow H^{2}(G,{\mathbb {Q}}/{\mathbb {Z}})^{\vee } \end{aligned}$$

Take the l-Sylow subgroups and the tensor products with ${\mathbb {Q}}_{l}$, we have

$$\begin{aligned} \begin{aligned}&0 \leftarrow G^{ab,l}\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l} \leftarrow (\pi _{1}(V)^{ab,l})\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l} \\&\quad \leftarrow ((\pi _{1}(U)^{ab,l})_{G})\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l} \leftarrow H^{2}(G,{\mathbb {Q}}_{l}/{\mathbb {Z}}_{l})^{\vee }\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l} \end{aligned} \end{aligned}$$

Since $G^{ab,l}\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l}$ and $H^{2}(G,{\mathbb {Q}}_{l}/{\mathbb {Z}}_{l})^{\vee }\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l}$ are torsion ${\mathbb {Q}}_{l}$ vector spaces, they are trivial. Then we have

$$\begin{aligned} ((\pi _{1}(U)^{ab,l})_{G})\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l}\simeq (\pi _{1}(V)^{ab,l})\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l} \end{aligned}$$

By the general theory of étale fundamental groups (cf, [1] Exposé V, corollaire 2.4), the kernel of $(\pi _{1}(V)^{ab,l})\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l} \rightarrow (\pi _{1}(Y)^{ab,l})\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l}$ is $A {\mathop {=}\limits ^{def}}(\varSigma _{P\in Y\backslash V}I_{P})\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l}$ and the kernel of $(\pi _{1}(U)^{ab,l})_{G}\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l} \rightarrow (\pi _{1}(X)^{ab,l})_{G}\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l}$ is $B {\mathop {=}\limits ^{def}}$ (the image of $(\varSigma _{P\in X\backslash U}I_{P})\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l}$ in $(\pi _{1}(U)^{ab,l})_{G}\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l}$) (Here, for $P\in Y\backslash V$ (resp. $P\in X \backslash U$), $I_{P}$ stands for the image of the inertia subgroup at P in $\pi _{1}(V)^{ab,p'}$ (resp. $\pi _{1}(U)^{ab,p'}$)). Observe that the isomorphism $((\pi _{1}(U)^{ab,l})_{G})\otimes _{{\mathbb {Z}}_{l}} {\mathbb {Q}}_{l}\xrightarrow {\sim } (\pi _{1}(V)^{ab,l})\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l}$ sends A onto B. Therefore we have

$$\begin{aligned} ((\pi _{1}(X)^{ab,l})_{G})\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l}\simeq (\pi _{1}(Y)^{ab,l})\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l} \end{aligned}$$

$\square$

Definition

Let M be an abelian group equipped with a ${\mathbb {Z}}/2{\mathbb {Z}}$-action. We define $M^{+}{\mathop {=}\limits ^{def}}M^{{\mathbb {Z}}/2{\mathbb {Z}}}$, $M^{-} {\mathop {=}\limits ^{def}}\{ a \in M | \tau a = -a \}$, where $\tau$ is the unique generator of ${\mathbb {Z}}/2{\mathbb {Z}}$.

Let m be an even positive integer. The Galois group of $E\backslash E[m]$$\rightarrow$${\mathbb {P}}^{1}\backslash T_{E\backslash E[m]}$ acts on $E\backslash E[m]$ and is isomorphic to ${\mathbb {Z}}/2{\mathbb {Z}}$.

Lemma 4.5

$$\begin{aligned} (\pi _{1}(E\backslash E[m])^{ab,p'})^{-} = ker(\pi _{1}(E\backslash E[m])^{ab,p'}\rightarrow \pi _{1}({\mathbb {P}}^{1}\backslash T_{E\backslash E[m]})^{ab,p'}) \end{aligned}$$

Proof

By G.A.G.A Theorems, $\pi _{1}({\mathbb {P}}^{1}\backslash T_{E\backslash E[m]})^{ab, p'}$ is a free $\hat{{\mathbb {Z}}}^{p'}$-module. It is clear that

$$\begin{aligned} (\pi _{1}(E\backslash E[m])^{ab,p'})^{-} \subset ker(\pi _{1}(E\backslash E[m])^{ab,p'}\rightarrow \pi _{1}({\mathbb {P}}^{1}\backslash T_{E\backslash E[m]})^{ab,p'}). \end{aligned}$$

Thus, we have a natural surjective morphism

$$\begin{aligned} \pi _{1}(E\backslash E[m])^{ab,p'}/(\pi _{1}(E\backslash E[m])^{ab,p'})^{-} \twoheadrightarrow R, \end{aligned}$$

where $R {\mathop {=}\limits ^{def}}Im(\pi _{1}(E\backslash E[m])^{ab,p'}\rightarrow \pi _{1}({\mathbb {P}}^{1}\backslash T_{E\backslash E[m]})^{ab,p'})$. At first we will prove that R is a free $\hat{{\mathbb {Z}}}^{p'}$-module. We have a short exact sequence

$$\begin{aligned} 1 \rightarrow R \rightarrow \pi _{1}({\mathbb {P}}^{1}\backslash T_{E\backslash E[m]})^{ab,p'} \rightarrow {\mathbb {Z}}/2{\mathbb {Z}} \rightarrow 1 \end{aligned}$$

Because R and $\pi _{1}({\mathbb {P}}^{1}\backslash T_{E\backslash E[m]})^{ab,p'}$ are profinite abelian groups, we have $R^{2'} \simeq \pi _{1}({\mathbb {P}}^{1}\backslash T_{E\backslash E[m]})^{ab,p',2'}$ and

$$\begin{aligned} 1 \rightarrow R^{2} \rightarrow \pi _{1}({\mathbb {P}}^{1}\backslash T_{E\backslash E[m]})^{ab,2} \rightarrow {\mathbb {Z}}/ 2{\mathbb {Z}} \rightarrow 1 \end{aligned}$$

(here, $R^{2}$ stands for the Sylow 2-subgroup of R). This exact sequence is a sequence of ${\mathbb {Z}}_{2}$-modules and ${\mathbb {Z}}_{2}$ is a PID, therefore $R^{2}$ is a free ${\mathbb {Z}}_{2}$-module and $rank_{{\mathbb {Z}}_{2}}(R^{2})=rank_{{\mathbb {Z}}_{2}}(\pi _{1}({\mathbb {P}}^{1} \backslash T_{E\backslash E[m]})^{ab,2})$. Thus R is a free $\hat{{\mathbb {Z}}}^{p'}$-module and $rank_{\hat{{\mathbb {Z}}}^{p'}}(R)=rank_{\hat{{\mathbb {Z}}}^{p'}} (\pi _{1}({\mathbb {P}}^{1}\backslash T_{E\backslash E[m]})^{ab,p'})$.

Let T be the torsion subgroup of $(\pi _{1}(E\backslash E[m])^{ab,p'})_{{\mathbb {Z}}/2{\mathbb {Z}}}$. By Lemma 4.4, we have $(\pi _{1}(E\backslash E[m])^{ab,l})_{{\mathbb {Z}}/2{\mathbb {Z}}}\otimes _{{\mathbb {Z}}_{l}} {\mathbb {Q}}_{l} \simeq \pi _{1}({\mathbb {P}}^{1}\backslash T_{E\backslash E[m]})^{ab,l}\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l}$ for any prime number l that is not p. From this, we deduce $(\pi _{1}(E\backslash E[m])^{ab,p'})_{{\mathbb {Z}}/2{\mathbb {Z}}}/T \simeq R$. By an easy computation, we have

$$\begin{aligned} 2((\pi _{1}(E\backslash E[m])^{ab,p'})^{-}) \subset (\tau -1)(\pi _{1}(E\backslash E[m])^{ab,p'}) \subset (\pi _{1}(E\backslash E[m])^{ab,p'})^{-} \end{aligned}$$

and that $\pi _{1}(E\backslash E[m])^{ab,l}/(\pi _{1}(E\backslash E[m])^{ab,l})^{-}$ is torsion free. Thus we have

$$\begin{aligned} \pi _{1}(E\backslash E[m])^{ab,p'}/(\pi _{1}(E\backslash E[m])^{ab,p'})^{-} \simeq (\pi _{1}(E\backslash E[m])^{ab,p'})_{{\mathbb {Z}}/2{\mathbb {Z}}}/ T\simeq R. \end{aligned}$$

$\square$

Lemma 4.6

$$\begin{aligned} (\pi _{1}(E\backslash E[m])^{ab,p'})^{E[m]} \subset (\pi _{1}(E\backslash E[m])^{ab,p'})^{-} \end{aligned}$$

Proof

$[m] : E \backslash E[m] \rightarrow E \backslash \{ {\mathcal {O}} \}$ is a Galois cover with Galois group E[m] (when p|m, $[m] : E \rightarrow E$ is decomposed uniquely as $[m]=[m]' \circ \phi$, where $[m]' : E' \rightarrow E$ (resp. $\phi : E \rightarrow E'$) is a separable (resp. purely inseparable) isogeny of eliptic curves, and we consider $[m]' : E' \rightarrow E$ instead of $[m] : E \rightarrow E$). $E\backslash E[2] \rightarrow {\mathbb {P}}^{1}\backslash \{ 0, 1, \lambda , \infty \}$ is a Galois cover with Galois group ${\mathbb {Z}}/2{\mathbb {Z}}$. $[m] : E \rightarrow E$ is the unique maximal abelian cover whose Galois group is killed by m. Then $E\backslash E[2m]\xrightarrow { [m] } E\backslash E[2] \rightarrow {\mathbb {P}}^{1}\backslash \{ 0, 1, \lambda , \infty \}$ is a Galois cover with Galois group $G{\mathop {=}\limits ^{def}}E[m]\rtimes {\mathbb {Z}}/2{\mathbb {Z}}$. By Lemma 4.4, we have

$$\begin{aligned} (\pi _{1}(E\backslash E[m])^{ab,l})_{G}\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l}\simeq (\pi _{1}( {\mathbb {P}}^{1} \backslash \{ \infty \})^{ab,l})\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l} = 0 \end{aligned}$$

(for each $l \ne p$). Because G is a finite group and ${\mathbb {Q}}_{l}$ is a field of characteristic 0, then we have

$$\begin{aligned} (\pi _{1}(E\backslash E[m])^{ab,l})^{G}\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l} \simeq (\pi _{1}(E\backslash E[m])^{ab,l})_{G}\otimes _{{\mathbb {Z}}_{l}}{\mathbb {Q}}_{l} = 0. \end{aligned}$$

As $(\pi _{1}(E\backslash E[m])^{ab,l})^{G}$ is a free ${\mathbb {Z}}_{l}$-module, then

$$\begin{aligned} ((\pi _{1}(E\backslash E[m])^{ab,l})^{E[m]})^{+} = (\pi _{1}(E\backslash E[m])^{ab,l})^{G} = 0. \end{aligned}$$

Therefore

$$\begin{aligned} (\pi _{1}(E\backslash E[m])^{ab,l})^{E[m]} \subset (\pi _{1}(E\backslash E[m])^{ab,l})^{-}, \end{aligned}$$

hence

$$\begin{aligned} (\pi _{1}(E\backslash E[m])^{ab,p'})^{E[m]} \subset (\pi _{1}(E\backslash E[m])^{ab,p'})^{-}. \end{aligned}$$

$\square$

Let W be the sum of all inertia subgroups in $\pi _{1}(E\backslash E[m])^{ab,p'}$. By G.A.G.A theorems, W is isomorphic to $(\oplus _{P\in E[m]}\hat{{\mathbb {Z}}}^{p'})/\varDelta (\hat{{\mathbb {Z}}}^{p'})$, where $\hat{{\mathbb {Z}}}^{p'}$ at each $P \in E[m]$ corresponds to the inertia subgroup at P and $\varDelta (\hat{{\mathbb {Z}}}^{p'})$ stands for the diagonal. W is closed under the action of the Galois group of

$$\begin{aligned} E\backslash E[2m] \xrightarrow { [m] } E\backslash E[2] \rightarrow {\mathbb {P}}^{1}\backslash \{ 0, 1, \lambda , \infty \}. \end{aligned}$$

Lemma 4.7

$$\begin{aligned} \#(\pi _{1}(E\backslash E[m])^{ab,p'})^{-}/(W^{-}\oplus (\pi _{1}(E\backslash E[m])^{ab,p'})^{E[m]}) < \infty \end{aligned}$$

Proof

At first, we prove

$$\begin{aligned} \#(\pi _{1}(E\backslash E[m])^{ab,p'})/(W\oplus (\pi _{1}(E\backslash E[m])^{ab,p'})^{E[m]})<\infty . \end{aligned}$$

We consider the following diagram

Where C is the cokernel of $(\pi _{1}(E\backslash E[m])^{ab,p'})^{E[m]} \rightarrow (\pi _{1}(E)^{ab,p'})^{E[m]}$. Note that $\pi _{1}(E)$ is abelian.

The two horizontal sequences are exact. C is a subgroup of $H^{1}(E[m],W).$E[m] acts transitively on E[m], hence we have $W^{E[m]}=1$. Let P be an element of E[m]. We have the following commutative diagram.

This implies that E[m] acts trivially on $\pi _{1}(E)^{p'}$, hence we have

$$\begin{aligned} (\pi _{1}(E)^{ab,p'})^{E[m]}=\pi _{1}(E)^{ab,p'}. \end{aligned}$$

By chasing the diagram, we have

$$\begin{aligned} W\cap (\pi _{1}(E\backslash E[m])^{ab,p'})^{E[m]}=W^{E[m]} = 1 \end{aligned}$$

(in $\pi _{1}(E\backslash E[m])^{ab,p'})$ and

$$\begin{aligned} \pi _{1}(E\backslash E[m])^{ab,p'}/(W\oplus (\pi _{1}(E\backslash E[m])^{ab,p'})^{E[m]})\simeq C. \end{aligned}$$

By the general property of homological algebra, we have

$$\begin{aligned} (\# (E[m]))\cdot H^{1}(E[m],W)=1. \end{aligned}$$

Thus we have

$$\begin{aligned} (\#(E[m]))\cdot ((\pi _{1}(E\backslash E[m])^{ab,p'})/(W\oplus (\pi _{1}(E\backslash E[m])^{ab,p'})^{E[m]}))=0. \end{aligned}$$

As $\pi _{1}(E\backslash E[m])^{ab,p'}$ is a finitely generated $\hat{{\mathbb {Z}}}^{p'}$-module, this shows

$$\begin{aligned} \#(\pi _{1}(E\backslash E[m])^{ab,p'})/(W\oplus (\pi _{1}(E\backslash E[m])^{ab,p'})^{E[m]})) <\infty . \end{aligned}$$

By definition,

$$\begin{aligned} \begin{aligned}&((\pi _{1}(E\backslash E[m])^{ab,p'})^{-})\cap (W\oplus ((\pi _{1}(E\backslash E[m]))^{ab,p'})^{E[m]})\\&\quad =W^{-}\oplus ((\pi _{1}(E\backslash E[m]))^{ab,p'})^{E[m]}. \end{aligned} \end{aligned}$$

Then we have a natural injective homomorphism

$$\begin{aligned} \begin{aligned}&(\pi _{1}(E\backslash E[m])^{ab,p'})^{-}/(W^{-}\oplus (\pi _{1}(E\backslash E[m])^{ab,p'})^{E[m]}) \\&\quad \hookrightarrow (\pi _{1}(E\backslash E[m])^{ab,p'})/(W\oplus (\pi _{1}(E\backslash E[m])^{ab,p'})^{E[m]}). \end{aligned} \end{aligned}$$

Thus,

$$\begin{aligned} \#(\pi _{1}(E\backslash E[m])^{ab,p'})^{-}/(W^{-}\oplus (\pi _{1}(E\backslash E[m])^{ab,p'})^{E[m]})<\infty . \end{aligned}$$

$\square$

Proof of Theorem 4.3

By Theorem 2.1 and Lemma 4.2, $\pi _{1}(E\backslash E[m])^{ab,p'}$ can be recovered from $\pi _{1}(E\backslash {\mathcal {O}})$. Then if

$$\begin{aligned} ker(\pi _{1}(E\backslash E[m])^{ab,p'}\rightarrow \pi _{1}({\mathbb {P}}^{1}\backslash T_{E\backslash E[m]})^{ab,p'}) \end{aligned}$$

could be recovered, $L_{E\backslash E[m]}$ could be recovered. By Lemma 4.7 and the fact that R (see the proof of Lemma 4.5) is torsion free, we have

$$\begin{aligned}&ker(\pi _{1}(E\backslash E[m])^{ab,p'}\rightarrow \pi _{1}({\mathbb {P}}^{1}\backslash T_{E\backslash E[m]})^{ab,p'})\\&\quad =\{ a \in \pi _{1}(E\backslash E[m])^{ab,p'} | \ \mathrm{for\ some}\ { n} \in {\mathbb {N}},\ na \in W^{-}\\&\qquad \quad \oplus (\pi _{1}(E\backslash E[m])^{ab,p'})^{E[m]} \}. \end{aligned}$$

It is clear that the action of E[m] on $\pi _{1}(E\backslash E[m])^{ab,p'}$ can be recovered from $\pi _{1}(E\backslash {\mathcal {O}})$, hence $(\pi _{1}(E\backslash E[m])^{ab,p'})^{E[m]}$ can be recovered from $\pi _{1}(E\backslash {\mathcal {O}})$. Recall that W is isomorphic to $(\oplus _{P\in E[m]}\hat{{\mathbb {Z}}}^{p'})/\varDelta (\hat{{\mathbb {Z}}}^{p'})$. Let $pr_{P}$ be a projection map $\oplus _{P\in E[m]}\hat{{\mathbb {Z}}}^{p'} \rightarrow \hat{{\mathbb {Z}}}^{p'}$ at P and $i_{P}$ an isomorphism

Then

$$\begin{aligned} W^{-} = \langle i_{P}(a)-i_{-P}(a) | a\in \varDelta (\hat{{\mathbb {Z}}}^{p'}), P \in E[m]\rangle . \end{aligned}$$

By Theorem 2.1, E[m] can be recovered as (a quotient of) the set of inertia subgroups from $\pi _{1}(E\backslash E[m])$. Then W, the additive structure on E[m] with identity element ${\mathcal {P}}$ (cf. the proof of Theorem 4.9 below) and the action of E[m] on W can be recovered from $\pi _{1}(E\backslash {\mathcal {O}})$. Therefore $W^{-}$ can be recovered from $\pi _{1}(E\backslash {\mathcal {O}})$. Hence $L_{E\backslash E[m]}$ can be recovered from $\pi _{1}(E\backslash {\mathcal {O}})$. $\square$

Corollary 4.8

For any even integer $m>2$, $A(E\backslash E[m], x(\lambda _{0}),x(\lambda _{\infty }))$ can be recovered group-theoretically from $\pi _{1}(E \backslash {\mathcal {O}}$). (The definition of $A(U,P_{0},P_{\infty })$ is in the statement of Theorem 3.3.)

Proof

This is established by Theorems 3.3 and 4.3. $\square$

Recall that F means the algebraic closure of ${\mathbb {F}}_{p}$ in k.

Theorem 4.9

Let U be a curve over k. Supose that there exists a curve $E'$ over F that satisfies $E\simeq E'\times _{F}k$. Then the following equivalence holds.

$$\begin{aligned} \pi _{1}(U)\simeq \pi _{1}(E\backslash {\mathcal {O}}) \Leftrightarrow U \simeq E\backslash {\mathcal {O}}\ (\mathrm{as\ schemes}) \end{aligned}$$

Proof

$(\Leftarrow )$ is clear. Thus it is sufficient to show ($\Rightarrow$). Fix an isomorphisms $\pi _{1}(E\backslash {\mathcal {O}})\simeq \pi _{1}(U)$.

By Theorem 2.1, the genus of $X {\mathop {=}\limits ^{def}}\ U^{cpt}$ is 1, and $\#(X\backslash U)=1$. Set $X\backslash U=\{ {\mathcal {O}}' \}$. We consider the additive structure on E (resp. X) defined by the elliptic curve $(E,{\mathcal {O}})$ (resp. $(X,{\mathcal {O}}')$). Let m be an even integer such that $E[2]\subsetneq E[m]$. Then the isomorphism $\pi _{1}(E\backslash {\mathcal {O}}) \simeq \pi _{1}(U)$ induces an isomorphism $\pi _{1}(E\backslash E[m])\simeq \pi _{1}(X\backslash X[m])$ by Lemma 4.2, which induces a bijection $\phi : E[m] \simeq X[m]$ by Theorem 2.1. We may consider a unique translation of X that sends $\phi ({\mathcal {O}})$ to ${\mathcal {O}}'$, and assume $\phi ({\mathcal {O}})={\mathcal {O}}'$. By using the group isomorphisms

$$\begin{aligned} E[m]\simeq Aut((E\backslash E[m])/(E\backslash {\mathcal {O}})) \ (Q \mapsto (R \mapsto R+Q) ) \end{aligned}$$

and

$$\begin{aligned} X[m]\simeq Aut((X\backslash X[m])/(X\backslash \mathcal {O'}), \end{aligned}$$

we see that $\phi$ is a group isomorphism.

By taking suitable closed immersions to ${\mathbb {P}}^{2}$, we may assume that X is defined by $y^{2}=x(x-1)(x-\lambda )$, ${\mathcal {O}}'=\infty$, E is defined by $y^{2}=x(x-1)(x-\lambda _{E})$, ${\mathcal {O}}=\infty$, $\phi ((\lambda _{E} , 0))= (\lambda , 0)$ and $\phi ((i,0)) = (i,0) \ \ (i=0,1)$.

For any $P \in k \simeq {\mathbb {P}}^{1}(k)\backslash \{ \infty \}$, let $\alpha (P)$ (resp.$\beta (P))$ be a point of E (resp.X) above P. For any P except $0 , 1 , \lambda _{E}$ (resp. $\lambda )$, there exist two points above P, but we choose $\alpha$ and $\beta$ that satisfy $\phi (E[m]\cap Im(\alpha ))=X[m]\cap Im(\beta )$.

Set $P , Q , P' , Q' \in {\mathbb {P}}^{1}(k)\backslash \{ \infty \}$. Suppose $\alpha (P),\alpha (Q),\alpha (P+Q)\in E[m], \phi (\alpha (P))=\beta (P'), \ \phi (\alpha (Q))=\beta (Q')$. By the equation

$$\begin{aligned} x(\alpha (P+Q))-x(\alpha (P))-x(\alpha (Q))=0 \end{aligned}$$

and Corollaly 4.8, we have

$$\begin{aligned} x(\phi (\alpha (P+Q)))-x(\beta (P'))-x(\beta (Q'))=0. \end{aligned}$$

Thus,

$$\begin{aligned} \begin{aligned}&\alpha (P),\alpha (Q),\alpha (P+Q)\in E[m]\ , \ \phi (\alpha (P))=\beta (P') \ , \ \phi (\alpha (Q))=\beta (Q') \\&\quad \Rightarrow \phi (\alpha (P+Q))=\beta (P'+Q') \end{aligned} \end{aligned}$$

By [10] Theorem 1.16 (Addition theorem), for any $a , b \in {\mathbb {F}}_{p}$ ($b\ne 0$),

$$\begin{aligned} \begin{aligned}&x(\alpha (aP)+\alpha (aP+b))+x(\alpha (aP)-\alpha (aP+b))\\&\quad = \frac{4}{b^{2}}a^{3}P^{3}+\left( \frac{6}{b}-\frac{4\lambda _{E}}{b^{2}}-\frac{4}{b^{2}}\right) a^{2}P^{2}\\&\qquad +\left( 2-\frac{4}{b}-\frac{4\lambda _{E}}{b}+\frac{4\lambda _{E}}{b^{2}}\right) aP +\frac{2\lambda _{E}}{b} \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&x(\beta (aP')+\beta (aP'+b))+x(\beta (aP')-\beta (aP'+b))\\&\quad =\frac{4}{b^{2}}a^{3}P'^{3}+\left( \frac{6}{b}-\frac{4\lambda }{b^{2}} -\frac{4}{b^{2}}\right) a^{2}P'^{2}\\&\qquad +\left( 2-\frac{4}{b} -\frac{4\lambda }{b}+\frac{4\lambda }{b^{2}}\right) aP'+\frac{2\lambda }{b} \end{aligned} \end{aligned}$$

Suppose $a=\pm 1 , \ b=1$. Then

$$\begin{aligned} \begin{aligned}&(x(\alpha (P)+\alpha (P+1))+x(\alpha (P)-\alpha (P+1))\\&\qquad + x(\alpha (-P)+\alpha (-P+1))+x(\alpha (-P)-\alpha (-P+1)))\\&\quad =-8\lambda _{E}P^{2}+4P^{2}+4\lambda _{E} \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&x(\beta (P')+\beta (P'+1))+x(\beta (P')-\beta (P'+1))\\&\qquad + x(\beta (-P')+\beta (-P'+1))+x(\beta (-P')-\beta (-P'+1))\\&\quad =-8\lambda P'^{2}+4P'^{2}+4\lambda \end{aligned} \end{aligned}$$

Therefore, by Corollary 4.8,

$$\begin{aligned} \begin{aligned}&\alpha (\pm P),\alpha (\pm P+1),\alpha (P^{2}),\alpha (\lambda _{E}P^{2})\in E[m], \\&\quad \phi (\alpha (P))=\beta (P') \ , \ \phi (\alpha (P^{2}))=\beta (P'^{2})\\&\quad \Rightarrow \ \phi (\alpha (\lambda _{E}P^{2}))=\beta (\lambda P'^{2}) \end{aligned} \end{aligned}$$

(3)

Suppose $a=1 , \ b=\pm 1$. Then

$$\begin{aligned} \begin{aligned}&x(\alpha (P)+\alpha (P+1))+x(\alpha (P)-\alpha (P+1))\\&\qquad - x(\alpha (P)+\alpha (P-1))-x(\alpha (P)-\alpha (P-1))\\&\quad =12P^{2}-8\lambda _{E}P-8P +4\lambda _{E} \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&x(\beta (P')+\beta (P'+1))+x(\beta (P')-\beta (P'+1))\\&\qquad - x(\beta (P')+\beta (P'-1))-x(\beta (P')-\beta (P'-1))\\&\quad =12P'^{2}-8\lambda P'-8P'+4\lambda \end{aligned} \end{aligned}$$

Therefore, by Corollary 4.8, when $p\ne 3$,

$$\begin{aligned} \begin{aligned}&\alpha (P),\alpha (P\pm 1),\alpha (\lambda _{E}P),\alpha (P^{2}) \in E[m] \\&\quad \phi (\alpha (P))=\beta (P') \ , \ \phi (\alpha (\lambda _{E}P))=\beta (\lambda _{E}P')\\&\quad \Rightarrow \phi (\alpha (P^{2}))=\beta (P'^{2}) \end{aligned} \end{aligned}$$

(4)

When $p=3$,

$$\begin{aligned} \begin{aligned}&\alpha (P),\alpha (P\pm 1),\alpha (\lambda _{E}P) \in E[m] \ , \ \phi (\alpha (P))=\beta (P')\\&\quad \Rightarrow \phi (\alpha (\lambda _{E}P))=\beta (\lambda P') \end{aligned} \end{aligned}$$

(5)

By using [10] Theorem 1.16 (Addition theorem) again, we have

$$\begin{aligned} \begin{aligned}&x(\alpha (\lambda _{E})+\alpha (\lambda _{E}+1))=\lambda _{E}^{2}\\&x(\beta (\lambda )+\beta (\lambda +1))=\lambda ^{2} \end{aligned} \end{aligned}$$

Therefore, by Corollary 4.8,

$$\begin{aligned} \alpha (\lambda _{E}+1),\alpha (\lambda _{E}^{2}) \in E[m] \Rightarrow \ \phi (\alpha (\lambda _{E}^{2}))=\beta (\lambda ^{2}) \end{aligned}$$

(6)

Let f be the minimal polynomial of $\lambda _{E}$ over ${\mathbb {F}}_{p}$. We take m such that $\alpha (-\lambda _{E}),\alpha (\lambda _{E}- 1),\alpha (\pm \lambda _{E}+1), \alpha (\pm \lambda _{E}^{2}),\alpha (\lambda _{E}^{2}- 1),\alpha (\pm \lambda _{E}^{2}+1), \alpha (\pm \lambda _{E}^{3}),\ldots , \alpha (\pm \lambda _{E}^{degf-1}),\alpha (\lambda _{E}^{degf-1}- 1),\alpha (\pm \lambda _{E}^{degf-1}+1),\alpha (\lambda _{E}^{degf})\in E[m]$. We will prove $\phi (\alpha (\lambda _{E}^{i}))=\beta (\lambda ^{i}) \ \ (i=0,1,\ldots , degf)$ by induction.

Suppose $p=3$.

By (5), for any $i=1,2,\ldots , degf-1$,

$$\begin{aligned} \phi (\alpha (\lambda _{E}^{i}))=\beta (\lambda ^{i})\Rightarrow \phi (\alpha (\lambda _{E}^{i+1}))=\beta (\lambda ^{i+1}) \ \end{aligned}$$

Thus, by induction, we have $\phi (\alpha (\lambda _{E}^{i}))=\beta (\lambda ^{i}) \ \ (i=0,1,\ldots , degf)$.

Suppose $p\ne 3$.

By (3), for any $i=1,2,\ldots , degf-1$,

$$\begin{aligned} \begin{aligned}&i \equiv 0 \ mod \ 2, \ \phi (\alpha (\lambda _{E}^{i/2}))=\beta (\lambda ^{i/2}), \ \phi (\alpha (\lambda _{E}^{i}))=\beta (\lambda ^{i}) \\&\quad \Rightarrow \phi (\alpha (\lambda _{E}^{i+1}))=\beta (\lambda ^{i+1}) \end{aligned} \end{aligned}$$

By (4),

$$\begin{aligned} \begin{aligned}&i \equiv 1 \ mod \ 2, \ i \ne 1,\\&\quad \phi (\alpha (\lambda _{E}^{(i+1)/2}))=\beta (\lambda ^{(i+1)/2}), \ \phi (\alpha (\lambda _{E}^{(i+3)/2}))=\beta (\lambda ^{(i+3)/2}) \\&\quad \Rightarrow \phi (\alpha (\lambda _{E}^{i+1}))=\beta (\lambda ^{i+1}) \end{aligned} \end{aligned}$$

By (6),

$$\begin{aligned} i=1 \Rightarrow \phi (\alpha (\lambda _{E}^{i+1}))=\beta (\lambda ^{i+1}) \end{aligned}$$

Thus, by induction, we have $\phi (\alpha (\lambda _{E}^{i}))=\beta (\lambda ^{i}) \ \ (i=0,1,\ldots , degf)$.

By Corollary 4.8, we conclude $f(\lambda )=0$. Therefore there exists an isomorphism $\varphi : k \simeq k$ that satisfies $\varphi (\lambda _{E})=\lambda$. Thus,

$$\begin{aligned} E\backslash \{ {\mathcal {O}} \} \simeq (E\backslash \{ {\mathcal {O}} \})\times _{k,\varphi }k \simeq U \end{aligned}$$

$\square$

Corollary 4.10

Supose that there exists a curve $E'$ over F that satisfies $E\simeq E'\times _{F}k$. Let $S\subset E(k)$ be a finite set that is not empty and U a curve over k. Then the following implication holds.

$$\begin{aligned} \pi _{1}(U)\simeq \pi _{1}(E\backslash S) \Rightarrow U^{cpt} \simeq E \ (\mathrm{as\ schemes}) \end{aligned}$$

Proof

Fix $P \in S$. By Theorem 2.1, the isomorphism $\pi _{1}(U)\simeq \pi _{1}(E\backslash S)$ induces an isomorphism $\pi _{1}(U^{cpt}\backslash P')\simeq \pi _{1}(E\backslash {\mathcal {P}})$ for some $P' \in (U^{cpt} \backslash U)(k)$. By applying Theorem 4.9 to the latter isomorphism, we obtain $U^{cpt}\backslash P' \simeq E\backslash P$, hence $U^{cpt}\simeq E$. $\square$

Remark 4.11

We established the results of Sect. 3 for hyperelliptic curves. However, in order to generalize the results of Sect.  4 to hyperelliptic curves, there remain (at least) two difficulties for the present. First, we need to generalize Theorem 4.3 to recover $L_{U} \subset \pi _{1}(U)$. Second, we need to generalize the $\lambda$ -invariant and the Addition Theorem ([10] Theorem 1.16) of elliptic curves in the case of hyperelliptic curves.