Algebraic curves admitting non-collinear Galois points

TL;DR

This paper establishes criteria for algebraic curves to have embeddings with non-collinear Galois points, provides a new example of such curves, and characterizes Fermat curves through this property.

Contribution

It introduces a new criterion for the existence of non-collinear Galois points and offers a novel characterization of Fermat curves based on these points.

Findings

Presented a criterion for birational embeddings with non-collinear Galois points.

Constructed a new example of a plane curve with non-collinear Galois points.

Characterized Fermat curves using non-collinear Galois points.

Abstract

A criterion for the existence of a birational embedding into a projective plane with non-collinear Galois points for algebraic curves is presented. A new example of a plane curve with non-collinear Galois points as an application is described. Furthermore, a new characterization of the Fermat curve in terms of non-collinear Galois points is presented.

Equations

$$x^q+x=y^{q+1}.$$

$$\sum _{\sigma \in G_1}\sigma (P_2)\ne \sum _{\sigma \in G_1}\sigma (P_3).$$

$$k(X{)}^{G_1}=k(f),(f)=\sum _{\sigma \in G_1}\sigma (P_3)-\sum _{\sigma \in G_1}\sigma (P_2)$$

$$k(X{)}^{G_2}=k(g),(g)=\sum _{\tau \in G_2}\tau (P_3)-\sum _{\tau \in G_2}\tau (P_1).$$

$$D:=P_1+\sum _{\sigma \in G_1}\sigma (P_2)=P_2+\sum _{\tau \in G_2}\tau (P_1).$$

$$\phi :X\to {\mathbb{P}}^2;(f:g:1)$$

$$(g/f)$$

$$A_{a}:=\left(\begin{array}[]{ccc}a^{q+1}&0&0\\ 0&a&0\\ 0&0&1\end{array}\right)$$

$$(A_{a^{m}})^{s}=A_{a^{q-1}}=\left(\begin{array}[]{ccc}1&0&0\\ 0&a^{q-1}&0\\ 0&0&1\end{array}\right).$$

$$\sum _{\sigma \in G_i}\sigma (Q_k)=s(q+1)Q_k=\sum _{\tau \in G_j}\tau (Q_k).$$

$${\phi }^{\ast }\overline{P_1{P}_2}=\sum _{Q\in \mathrm{supp}({\phi }^{\ast }\overline{P_1{P}_2})}mQ$$

$${\gamma }^{\ast }{\phi }^{\ast }(\overline{P_1{P}_2})={\phi }^{\ast }(\overline{P_1{P}_2}).$$

$$(f)={\phi }^{\ast }\overline{P_3{P}_2}-{\phi }^{\ast }\overline{P_3{P}_1}.$$

$$(g)={\phi }^{\ast }\overline{P_1{P}_2}-{\phi }^{\ast }\overline{P_1{P}_3}.$$

$$G_{P_{3}}\hookrightarrow{\rm PGL}(3,k);\ \gamma\mapsto\left(\begin{array}[]{ccc}1&0&0\\ 0&1&0\\ 0&0&a(\gamma)\end{array}\right).$$

Full text

The author was partially supported by JSPS KAKENHI Grant Number 19K03438.

1. Introduction

Let $X$ be a (reduced, irreducible) smooth projective curve over an algebraically closed field $k$ of characteristic $p\geq 0$ and let $k(X)$ be its function field. We consider a morphism $\varphi:X\rightarrow\mathbb{P}^{2}$, which is birational onto its image. A point $P\in\mathbb{P}^{2}$ is called a Galois point, if the field extension $k(\varphi(X))/\pi_{P}^{*}k(\mathbb{P}^{1})$ of function fields induced by the projection $\pi_{P}$ from $P$ is a Galois extension. This notion was introduced by Hisao Yoshihara in 1996, to investigate the function fields of algebraic curves ([3, 5]). The associated Galois group is denoted by $G_{P}$, when $P$ is a Galois point. Furthermore, a Galois point $P$ is said to be inner (resp. outer), if $P\in\varphi(X)\setminus{\rm Sing}(\varphi(X))$ (resp. if $P\in\mathbb{P}^{2}\setminus\varphi(X)$). It is a natural and interesting problem to determine the number of Galois points for any curve $X$ and any birational embedding $\varphi:X\rightarrow\mathbb{P}^{2}$. However, this problem is difficult in general.

Until recent years, it was not easy to construct a pair $(X,\varphi)$ such that $\varphi(X)$ admits two Galois points. In 2016, a criterion for the existence of a birational embedding with two Galois points was described by the present author ([1]), and by this criterion, a lot of new examples of plane curves with two Galois points were obtained ([1, 6]). We recall this criterion.

Fact 1**.**

Let $G_{1}$ and $G_{2}$ be finite subgroups of ${\rm Aut}(X)$ and let $P_{1}$ and $P_{2}$ be different points of $X$. Then, three conditions

• (a)

$X/{G_{i}}\cong\mathbb{P}^{1}$* for $i=1,2$,*

• (b)

$G_{1}\cap G_{2}=\{1\}$, and

• (c)

$P_{1}+\sum_{\sigma\in G_{1}}\sigma(P_{2})=P_{2}+\sum_{\tau\in G_{2}}\tau(P_{1})$* in ${\rm Div}(X)$*

are satisfied, if and only if there exists a birational embedding $\varphi:X\rightarrow\mathbb{P}^{2}$ of degree $|G_{1}|+1$ such that $\varphi(P_{1})$ and $\varphi(P_{2})$ are different inner Galois points for $\varphi(X)$ and the associated Galois group $G_{\varphi(P_{i})}$ coincides with $G_{i}$ for $i=1,2$.

It is a natural problem to find a condition for the existence of non-collinear Galois points (see also [6]). This problem is solved as follows.

Theorem 1**.**

Let $G_{1}$, $G_{2}$ and $G_{3}\subset{\rm Aut}(X)$ be finite subgroups, and let $P_{1}$, $P_{2}$ and $P_{3}$ be different points of $X$. Then, four conditions

• (a)

$X/{G_{i}}\cong\mathbb{P}^{1}$* for $i=1,2,3$,*

• (b)

$G_{i}\cap G_{j}=\{1\}$* for any $i,j$ with $i\neq j$,*

• (c)

$P_{i}+\sum_{\sigma\in G_{i}}\sigma(P_{j})=P_{j}+\sum_{\tau\in G_{j}}\tau(P_{i})$* for any $i,j$ with $i\neq j$, and*

• (d)

$G_{i}P_{j}\neq G_{i}P_{k}$* for any $i,j,k$ with $\{i,j,k\}=\{1,2,3\}$*

are satisfied, if and only if there exists a birational embedding $\varphi:X\rightarrow\mathbb{P}^{2}$ of degree $|G_{1}|+1$ such that $\varphi(P_{1})$, $\varphi(P_{2})$ and $\varphi(P_{3})$ are non-collinear inner Galois points for $\varphi(X)$ and $G_{\varphi(P_{i})}=G_{i}$ for $i=1,2,3$.

Theorem 2**.**

Let $G_{1}$, $G_{2}$ and $G_{3}\subset{\rm Aut}(X)$ be finite subgroups, and let $Q_{1}$, $Q_{2}$ and $Q_{3}$ be different points of $X$. Then, four conditions

• (a)

$X/{G_{i}}\cong\mathbb{P}^{1}$* for $i=1,2,3$,*

• (b)

$G_{i}\cap G_{j}=\{1\}$* for any $i,j$ with $i\neq j$,*

• (c’)

$\sum_{\sigma\in G_{i}}\sigma(Q_{k})=\sum_{\tau\in G_{j}}\tau(Q_{k})$* for any $i,j,k$ with $\{i,j,k\}=\{1,2,3\}$, and*

• (d’)

$G_{i}Q_{j}\neq G_{i}Q_{k}$* for any $i,j,k$ with $\{i,j,k\}=\{1,2,3\}$*

are satisfied, if and only if there exists a birational embedding $\varphi:X\rightarrow\mathbb{P}^{2}$ of degree $|G_{1}|$ and non-collinear outer Galois points $P_{1},P_{2}$ and $P_{3}$ exist for $\varphi(X)$ such that $G_{P_{i}}=G_{i}$ and $\overline{P_{i}P_{j}}\ni\varphi(Q_{k})$ for any $i,j,k$ with $\{i,j,k\}=\{1,2,3\}$, where $\overline{P_{i}P_{j}}$ is the line passing through $P_{i}$ and $P_{j}$.

As an application, a new example of a plane curve with non-collinear outer Galois points is constructed as follows.

Theorem 3**.**

Let $p>0$, $q$ be a power of $p$, and let $X\subset\mathbb{P}^{2}$ be the Hermitian curve, which is (the projective closure of) the curve given by

[TABLE]

If a positive integer $s$ divides $q-1$, then there exists a plane model of $X$ of degree $s(q+1)$ admitting non-collinear outer Galois points $P_{1},P_{2}$ and $P_{3}$.

A next task is to classify plane curves with non-collinear Galois points. We consider the group $G:=\langle G_{P_{1}},G_{P_{2}},G_{P_{3}}\rangle\subset{\rm Aut}(X)$ for non-collinear outer Galois points $P_{1},P_{2}$ and $P_{3}$. The case where the orbit $GQ$ of $Q$ is included in $\varphi^{-1}(\bigcup_{i\neq j}\overline{P_{i}P_{j}})$ for any $Q\in\varphi^{-1}(\bigcup_{i\neq j}\overline{P_{i}P_{j}})$ is determined as follows.

Theorem 4**.**

Let $\varphi:X\rightarrow\mathbb{P}^{2}$ be a birational embedding of degree $d\geq 3$ and let $C=\varphi(X)$. Then, the following conditions are equivalent.

• (a)

There exist non-collinear Galois points $P_{1},P_{2}$ and $P_{3}\in\mathbb{P}^{2}\setminus C$ such that $GQ\subset\varphi^{-1}(\bigcup_{i\neq j}\overline{P_{i}P_{j}})$ for any $Q\in\varphi^{-1}(\bigcup_{i\neq j}\overline{P_{i}P_{j}})$, where $G=\langle G_{P_{1}},G_{P_{2}},G_{P_{3}}\rangle$.

• (b)

$p=0$* or $d$ is prime to $p$, and $C$ is projectively equivalent to the Fermat curve $X^{d}+Y^{d}+Z^{d}=0$.*

2. Proof of Theorems 1 and 2

Proof of Theorem 1.

We consider the if-part. According to Fact 1, conditions (a), (b) and (c) are satisfied. Since points $\varphi(P_{1}),\varphi(P_{2})$ and $\varphi(P_{3})$ are not collinear, condition (d) is satisfied.

We consider the only-if part. By condition (d),

[TABLE]

Then, by condition (a), there exists a function $f\in k(X)\setminus k$ such that

[TABLE]

(see also [4, III.7.1, III.7.2, III.8.2]). Similarly, there exists $g\in k(X)\setminus k$ such that

[TABLE]

Considering condition (c), we take a divisor

[TABLE]

Then, $f,g\in\mathcal{L}(D)$ and the sublinear system of $|D|$ corresponding to a linear space $\langle f,g,1\rangle$ is base-point-free. Using condition (b), the induced morphism

[TABLE]

is birational onto its image, and points $\varphi(P_{1})=(0:1:0)$ and $\varphi(P_{2})=(1:0:0)$ are inner Galois points for $\varphi(X)$ such that $G_{\varphi(P_{1})}=G_{1}$ and $G_{\varphi(P_{2})}=G_{2}$ (see [1, Proofs of Proposition 1 and of Theorem 1]). Furthermore, $\varphi(P_{3})=(0:0:1)$. Using condition (c),

[TABLE]

Then, the subfield $k(g/f)$ induced by the projection from $P_{3}$ coincides with $k(X)^{G_{3}}$. Therefore, this point $\varphi(P_{3})$ is inner Galois with $G_{\varphi(P_{3})}=G_{3}$. ∎

The proof of Theorem 2 is very similar.

3. A new example

Let $X\subset\mathbb{P}^{2}$ be the Hermitian curve of degree $q+1$. The set of all $\mathbb{F}_{q^{2}}$-rational points of $X$ is denoted by $X(\mathbb{F}_{q^{2}})$. See [2] for properties of the Hermitian curve.

Proof of Theorem 3.

Let $Q_{1}=(1:0:0)$ and $Q_{2}=(0:0:1)$, and let $Q_{3}\in X(\mathbb{F}_{q^{2}})$ with $Q_{3}\not\in\overline{Q_{1}Q_{2}}=\{Y=0\}$. Then, the matrix

[TABLE]

acts on $X$ and fixes $Q_{1}$ and $Q_{2}$, where $a\in\mathbb{F}_{q^{2}}\setminus\{0\}$. Let $sm=q-1$ and let $G_{3}\subset{\rm Aut}(X)$ be the cyclic group of order $s(q+1)$ consisting of all $A_{a^{m}}$. Note that each element of $G_{3}\setminus\{1\}$ does not fix $Q_{3}$. Considering the Sylow $p$-group of ${\rm Aut}(X)$ fixing $Q_{1}$, it follows that there exists an automorphism $\Phi\in{\rm Aut}(X)$ such that $\Phi(Q_{1})=Q_{1}$, $\Phi(Q_{2})=Q_{3}$ and $\Phi(Q_{3})=Q_{2}$. Then, the group $\Phi G_{3}\Phi^{-1}$ fixes points $Q_{1}$ and $Q_{3}$, and each element of this group different from identity does not fix $Q_{2}$. Therefore, for each pair $(Q_{i},Q_{j})$, there exists a cyclic group $G_{k}$ of order $s(q+1)$ such that $G_{k}$ fixes points $Q_{i}$ and $Q_{j}$ and each element of $G_{k}\setminus\{1\}$ does not fix $Q_{k}$. We would like to show that conditions (a), (b), (c’) and (d’) in Theorem 2 are satisfied for groups $G_{1},G_{2}$ and $G_{3}$.

Note that

[TABLE]

Let $G_{3}^{\prime}\subset G_{3}$ be a subgroup consisting of all $A_{a^{q-1}}$. Since $k(X)^{G_{3}}\subset k(X)^{G_{3}^{\prime}}=k(x)$, by Lüroth’s theorem, $X/G_{3}$ is rational. Condition (a) is satisfied. Since $G_{1}$ fixes $Q_{2}$ and the set $G_{2}\setminus\{1\}$ does not contain an element fixing $Q_{2}$, $G_{1}\cap G_{2}=\{1\}$. Condition (b) is satisfied. For any $i,j,k$ with $\{i,j,k\}=\{1,2,3\}$,

[TABLE]

Condition (c’) is satisfied. Since $G_{i}Q_{j}=\{Q_{j}\}\neq\{Q_{k}\}=G_{i}Q_{k}$, condition (d’) is satisfied. ∎

4. A characterization of the Fermat curve

Proof of Theorem 4.

(a) $\Rightarrow$ (b). Let $Q\in\varphi^{-1}(\overline{P_{1}P_{2}})$. By the definition of outer Galois points, $G_{P_{1}}Q\subset\varphi^{-1}(\overline{P_{1}P_{2}})$, $G_{P_{2}}Q\subset\varphi^{-1}(\overline{P_{1}P_{2}})$ and $G_{P_{3}}Q\subset\varphi^{-1}(\overline{P_{3}\varphi(Q)})$. If $\gamma(Q)\in\varphi^{-1}(\overline{P_{2}P_{3}})$ for some $\gamma\in G_{P_{3}}$, then $\varphi(\gamma(Q))\in\overline{P_{3}\varphi(Q)}\cap\overline{P_{2}P_{3}}=\{P_{3}\}$. This is a contradiction. Therefore, condition (a) implies that $G_{P_{3}}Q\subset\varphi^{-1}(\overline{P_{1}P_{2}})$. It follows that $\gamma\in G_{P_{3}}$ induces a bijection of ${\rm supp}(\varphi^{*}\overline{P_{1}P_{2}})$. Since $G_{P_{1}}$ acts on ${\rm supp}(\varphi^{*}\overline{P_{1}P_{2}})$ transitively,

[TABLE]

for some integer $m\geq 1$. Therefore,

[TABLE]

Let $D:=\varphi^{*}\overline{P_{3}P_{1}}$. We take a function $f\in k(X)$ with $k(f)=k(X)^{G_{3}}$ such that

[TABLE]

Similarly, we can take a function $g\in k(X)^{G_{1}}$ such that

[TABLE]

Since $\overline{P_{1}P_{2}}$ does not pass through $P_{3}$, $g\not\in\langle f,1\rangle\subset\mathcal{L}(D)$. It follows from the condition $\gamma^{*}\varphi^{*}(\overline{P_{1}P_{2}})=\varphi^{*}(\overline{P_{1}P_{2}})$ that $\gamma^{*}g=a(\gamma)g$ for some $a(\gamma)\in k$. Therefore, a linear subspace $\langle f,1,g\rangle\subset\mathcal{L}(D)$ is invariant under the action of $\gamma\in G_{P_{3}}$. Since $\varphi$ is represented by $(f:1:g)$, there exists an injective homomorphism

[TABLE]

It follows that $d$ is prime to $p$, and the map $G_{P_{3}}\rightarrow k\setminus\{0\}$; $\gamma\mapsto a(\gamma)$ is an injective homomorphism. This implies that $G_{P_{3}}$ is a cyclic group, and $C$ is invariant under the linear transformation $(X:Y:Z)\mapsto(X:Y:\zeta Z)$, where $\zeta$ is a primitive $d$-th root of unity. Similarly, $G_{P_{1}}$ is generated by the automorphism given by the linear transformation $(X:Y:Z)\mapsto(\zeta X:Y:Z)$. It follows that $C$ is defined by $X^{d}+Y^{d}+Z^{d}=0$.

(b) $\Rightarrow$ (a). This is derived from the fact that groups $G_{P_{1}}$, $G_{P_{2}}$ and $G_{P_{3}}$ fix all points on the lines $\{X=0\}$, $\{Y=0\}$ and $\{Z=0\}$ respectively for the Fermat curve, where $P_{1}=(1:0:0)$, $P_{2}=(0:1:0)$ and $P_{3}=(0:0:1)$. ∎

Acknowledgements

The author is grateful to Doctor Kazuki Higashine for helpful discussions.