Plane model-fields of definition, fields of definition, the field of moduli of smooth plane curves

TL;DR

This paper constructs a smooth plane curve over the algebraic closure of rationals demonstrating that its field of moduli is not a field of definition, and that fields of definition differ from plane model-fields of definition, highlighting a novel phenomenon.

Contribution

It provides the first known example of a smooth plane curve where the field of moduli is not a field of definition, and fields of definition differ from plane model-fields of definition.

Findings

Field of moduli not a field of definition for the constructed curve

Fields of definition do not coincide with plane model-fields of definition

First example of this phenomenon in the literature

Abstract

Given a smooth plane curve $\overline{C}$ of genus $g\geq 3$ over an algebraically closed field $\overline{k}$, a field $L\subseteq\overline{k}$ is said to be a \emph{plane model-field of definition for $\overline{C}$} if $L$ is a field of definition for $\overline{C}$, i.e. $\exists$ a smooth curve $C'$ defined over $L$ where $C'\times_L\overline{k}\cong \overline{C}$, and such that $C'$ is $L$-isomorphic to a non-singular plane model $F(X,Y,Z)=0$ in $\mathbb{P}^2_{L}$. {In this short note, we construct a smooth plane curve $\overline{C}$ over $\overline{\mathbb{Q}}$, such that the field of moduli of $\overline{C}$ is not a field of definition for $\overline{C}$, and also fields of definition do not coincide with plane model-fields of definition for $\overline{C}$.} As far as we know, this is the first example in the literature with the above property, since this phenomenon does not occur for hyperelliptic curves, replacing plane model-fields of definition with the so-called hyperelliptic model-fields of definition.

Equations

$$S:=\left(\begin{array}[]{ccc}1&0&0\\ 0&\zeta_{3}&0\\ 0&0&\zeta_{3}^{2}\\ \end{array}\right),\,T:=\left(\begin{array}[]{ccc}0&1&0\\ 0&0&1\\ 1&0&0\\ \end{array}\right),\,\,\text{and}\,\,R:=\left(\begin{array}[]{ccc}1&0&0\\ 0&0&1\\ 0&1&0\\ \end{array}\right).$$

$$\varphi$$

$$\psi$$

$$\chi$$

$$c_{{\varphi }^2}$$

$$c_{\varphi \psi }$$

$$c_{{\psi }^2}$$

$$F_{\sqrt{u},\sqrt{v}}(X,Y,Z):=c_{{\varphi }^2}{\varphi }^2-6c_{\varphi \psi }\varphi \psi -18c_{{\psi }^2}{\psi }^2+\chi .$$

$$\frac{c_{{\varphi }^2}}{p^2}(XYZ{)}^2-\frac{6c_{\varphi \psi }}{p}(XYZ)(X^3+\frac{1}{p}{Y}^3+\frac{1}{p^2}{Z}^3)-18c_{{\psi }^2}(X^3+\frac{1}{p}{Y}^3+\frac{1}{p^2}{Z}^3{)}^2+\frac{1}{p}{X}^3{Y}^3+\frac{1}{p^3}(YZ{)}^3+\frac{1}{p^2}{X}^3{Z}^3=0,$$

Full text

In this short note, we construct a smooth plane curve $\overline{C}$ over $\overline{\mathbb{Q}}$, such that the field of moduli of $\overline{C}$ is not a field of definition for $\overline{C}$, and also fields of definition do not coincide with plane model-fields of definition for $\overline{C}$. As far as we know, this is the first example in the literature with the above property, since this phenomenon does not occur for hyperelliptic curves, replacing plane model-fields of definition with the so-called hyperelliptic model-fields of definition.

E. Badr and F. Bars are supported by MTM2016-75980-P

1. Introduction

Consider $F$ the base field for an algebraically closed field $\overline{k}$. Let $F\subseteq L\subseteq\overline{k}$ be fields, given a smooth projective curve $\overline{C}$ over $\overline{k}$, then $\overline{C}$ is defined over $L$ if and only if there is a curve $C^{\prime}$ over $L$ that is $\overline{k}$-isomorphic to $\overline{C}$, i.e. $C^{\prime}\times_{L}\overline{k}\cong\overline{C}$. In such case, $L$ is called a field of definition of $\overline{C}$. We say that $\overline{C}$ is definable over $L$ if there is a curve $C^{\prime}/L$ such that $\overline{C}$ and $C^{\prime}\times_{L}\overline{k}$ are $\overline{k}$-isomorphic.

Definition 1.1**.**

The field of moduli of a smooth projective curve $\overline{C}$ defined over $\overline{k}$, denoted by $K_{\overline{C}}$, is the intersection of all fields of definition of $\overline{C}$.

It becomes very natural to ask when the field of moduli of a smooth projective curve $\overline{C}$ is also a field of definition. A necessary and sufficient condition (Weil’s cocycle criterion of descent) for the field of moduli to be a field of definition was provided by Weil [12]. If $\operatorname{Aut}(\overline{C})$ is trivial, then this condition becomes trivially true and so the field of moduli needs to be a field of definition. It is also quite well known that a smooth curve $\overline{C}$ of genus $g=0$ or $1$ can be defined over its field of moduli, where $g$ is the geometric genus of $\overline{C}$. However, if $g>1$ and $\operatorname{Aut}(\overline{C})$ is non-trivial, then Weil’s conditions are difficult to be checked and so there is no guarantee that the field of moduli is a field of definition for $\overline{C}$. This was first pointed out by Earle [4] and Shimura [11]. More precisely, in page 177 of [11], the first examples not definable over their field of moduli are introduced, which are hyperelliptic curves over $\mathbb{C}$ with two automorphisms. There are also examples of non-hyperelliptic curves not definable over their field of moduli given in [2, 5]. B. Huggins [6] studied this problem for hyperelliptic curves over a field $\overline{k}$ of characteristic $p\neq 2$, proving that a hyperelliptic curve $\overline{C}$ of genus $g\geq 2$ with hyperelliptic involution $\iota$ can be defined over $K_{\overline{C}}$ when $\operatorname{Aut}(\overline{C})/\langle\iota\rangle$ is not cyclic or is cyclic of order divisible by $p$.

On the other hand, one may define fields of definition of models of the same concrete type for a smooth projective curve $\overline{C}$. For example, if $\overline{C}$ is hyperelliptic, a field $M$ is called a hyperelliptic model-field of definition for $\overline{C}$ if $M$, as a field of definition for $\overline{C}$, satisfies that $\overline{C}$ is $M$-isomorphic to a hyperelliptic model of the form $y^{2}=f(x)$, for some polynomial $f(x)$ of degree $2g+1$ or $2g+2$.

By the work of Mestre [10], Huggins [6, 5], Lercier-Ritzenthaler [7], Lercier-Ritzenthaler-Sijsling [8] and Lombardo-Lorenzo in [9], one gets fair-enough characterizations for the interrelations between the three fields; the field of moduli, fields of definition and hyperelliptic model-fields of definition. For instance, if $\overline{C}$ is hyperelliptic, then there are always two of these fields, which are equal. Summing up, one obtains the next table issued from Lercier-Ritzenthaler-Sijsling [8], where $k=F$ is a perfect field of characteristic $char(F)\neq 2$:

[TABLE]

By tamely cyclic, we mean that the group is cyclic of order not divisible by the $char(F)$.

Now, consider a smooth plane curve $\overline{C}$, i.e. $\overline{C}$ viewed as a smooth curve over $\overline{k}$ admits a non-singular plane model defined by an equation of the form $F(X,Y,Z)=0$ in $\mathbb{P}^{2}_{\overline{k}}$, where $F(X,Y,Z)$ is a homogenous polynomial of degree $d\geq 4$ over $\overline{k}$ with $g=\frac{1}{2}(d-1)(d-2)\geq 3$. Similarly, we define a so-called plane model-fields of definition for $C$:

Definition 1.2**.**

Given a smooth plane curve $\overline{C}$ over $\overline{k}$, a subfield $M\subset\overline{k}$ is said to be a plane model-field of definition for $C$ if and only if the following conditions holds

1. (i)

$M$ is a field of definition for $\overline{C}$. 2. (ii)

$\exists$ a smooth curve $C^{\prime}$ defined over $M$, which is $\overline{k}$-isomorphic to $\overline{C}$, and $M$-isomorphic to a non-singular plane model $F(X,Y,Z)=0$, for some homogenous polynomial $F(X,Y,Z)\in M[X,Y,Z]$ of degree $d\geq 3$.

In this short note, we start with a smooth plane curve $\overline{C}$ over $\overline{\mathbb{Q}}$ where the field of moduli is not a field of definition by the work of B. Huggins in [5]. Next, we go further, following the techniques developed in [1], to construct a twist of $\overline{C}$, for which there is a field of definition for $\overline{C}$, which is not a plane model-field of definition.

Acknowledgments

We would like to thank Elisa Lorenzo and Christophe Ritzenthaler for bringing this problem to our attention, as a consequence of our discussion with them in BGSMath-Barcelona Graduate School in March 2017.

2. The example

Consider the Hessian group of order $18$, denoted by $\operatorname{Hess}_{18}$, which is $\operatorname{PGL}_{3}(\overline{\mathbb{Q}})$-conjugate to the group generated by

[TABLE]

First, we reproduce an example, by B. Huggins in [5, Chp. 7, §2], of a smooth $\overline{\mathbb{Q}}$-plane curve of genus $10$ not definable over its field of moduli, and with full automorphism groups $\operatorname{Hess}_{18}$.

Definition 2.1**.**

A quaternion extension of a field $K$ is a Galois extension $K^{\prime}/K$ such that $\operatorname{Gal}(K^{\prime}/K)$ is isomorphic to the quaternion group of order $8$.

Definition 2.2**.**

([5, Lemma 7.2.3]) A field $K$ is of level $2$ if $-1$ is not a square in $K$, but it is a sum of two squares in $K$.

Lemma 2.3**.**

([5, Lemma 7.2.3]) Let $K$ be a field of level $2$. Then, for $u,v\in K^{*}\setminus(K^{*})^{2}$ such that $uv\notin(K^{*})^{2}$, $K(\sqrt{u},\sqrt{v})$ is embeddable into a quaternion extension of $K$ if and only if $-u$ is a norm from $K(\sqrt{-v})$ to $K$ (i.e. $-u=x^{2}+vy^{2}$ for some $x,y\in K$).

For instance, the field $K:=\mathbb{Q}(\zeta_{3})$ is of level $2$, since $(\zeta_{3}^{2})^{2}+\zeta_{3}^{2}=-1$ and $\sqrt{-1}\notin K$. It is easily shown that $\pm 2$ are not norms from $K(\sqrt{-13})$ to $K$. So neither $K(\sqrt{2},\sqrt{13})$ nor $K(\sqrt{-2},\sqrt{13})$ are embeddable into a quaternion extension of $K$.

Now fix $K$ to be the field $\mathbb{Q}(\zeta_{3})$, and define the following:

[TABLE]

Suppose that $u,v\in\mathbb{Q}^{*}$, such that $L:=K(\sqrt{u},\sqrt{v})$ is a $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ extension of $K$ that can not be embedded into a quaternion extension of $K$. Let

[TABLE]

Fix an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$ containing $L$ as above.

Theorem 2.4**.**

(B. Huggins, [5, Lemma 7.2.5 and Proposition 7.2.6]) Following the above notations, let

[TABLE]

Then the equation $F_{\sqrt{u},\sqrt{v}}(X,Y,Z)=0$ such that $F_{\sqrt{u},\sqrt{v}}(X,1,1)$ is square free, defines a smooth $\overline{\mathbb{Q}}$-plane curve $\overline{C}$ over $\overline{\mathbb{Q}}$, with automorphism group $\operatorname{Hess}_{18}$. The field of moduli $K_{\overline{C}}$ is $K=\mathbb{Q}(\zeta_{3})$, but it is not a field of definition.

Remark 2.5**.**

The condition that $F_{\sqrt{u},\sqrt{v}}(X,1,1)$ is square free is possible. For example, with $u=2$ and $v=13$, the resultant of $F_{\sqrt{2},\sqrt{13}}(X,1,1)$ and $\frac{\partial F}{\partial X}(X,1,1)$ is not zero.

Lemma 2.6**.**

Let $\overline{C}$ be a smooth curve defined over an algebraically closed field $\overline{k}$, with $F=k$ and $k$ perfect. An $\overline{k}$-isomorphism $\phi:\overline{C^{\prime}}\rightarrow\overline{C}$ does not change the field of moduli or fields of definition, that is both $\overline{C}$ and $\overline{C^{\prime}}$ have the same fields of moduli and fields of definitions.

Proof.

A field $L\subseteq\overline{k}$ is a field of definition for $\overline{C}$ if and only if there exists a smooth curve $C^{\prime\prime}$ over $L$, such that $C^{\prime\prime}\times_{L}\overline{k}$ is $\overline{k}$-isomorphic to $\overline{C}$ through some $\psi:C^{\prime\prime}\times_{L}\overline{k}\rightarrow\overline{C}$. Hence $\phi^{-1}\circ\psi:C^{\prime\prime}\times_{L}\overline{k}$ is a $\overline{k}$-isomorphism, and $L$ is a field of definition for $\overline{C^{\prime}}$. The converse is true by a similar discussion. Consequently, the field of moduli for $\overline{C}$ and $\overline{C^{\prime}}$ coincides, being the intersection of all fields of definition. ∎

Corollary 2.7**.**

Consider a smooth $\overline{\mathbb{Q}}$-plane curve $\overline{C}$ defined by an equation of the form

[TABLE]

where $p\in\mathbb{Q}$, in particular $\overline{C}$ admits $\mathbb{Q}(\sqrt{u},\sqrt{v},\zeta_{3})$ as a plane model-field of definition for $\overline{C}$. Then $\operatorname{Aut}(\overline{C})$ is isomorphic to $\operatorname{Hess}_{18}$. Moreover, the field of moduli $K_{\overline{C}}$ is $K=\mathbb{Q}(\zeta_{3})$, but it is not a field of definition.

Proof.

Since $\overline{C}$ is $\mathbb{Q}(\sqrt[3]{p})$-isomorphic to $F_{\sqrt{u},\sqrt{v}}(X,Y,Z)=0$ through a change of variables of the shape $\phi=\operatorname{diag}(1,1/\sqrt[3]{p},1/\sqrt[3]{p^{2}})$, therefore they have conjugate automorphism groups. Moreover, fields of definition and the field of moduli of both curves are the same by Lemma 2.6. Consequently, the field of moduli $K_{\overline{C}}$ is $K=\mathbb{Q}(\zeta_{3})$, but it is not a field of definition, using Theorem 2.4. ∎

Theorem 2.8**.**

Consider the family $\mathcal{C}_{p}$ of smooth plane curves over the plane model-field of definition $L=\mathbb{Q}(\sqrt{u},\sqrt{v},\zeta_{3})$ given by an equation of the form

[TABLE]

where $p$ is a prime integer such that $p\equiv 3$ or $5$ mod $7$. Given a smooth plane curve $C$ over $L$ in $\mathcal{C}_{p}$, then there exists a twist $C^{\prime}$ of $C$ over $L$ which does not have $L$ as a plane model-field of definition. Moreover, the field of moduli of $C^{\prime}$ is $\mathbb{Q}(\zeta_{3})$, and is not a field of definition for $C^{\prime}$.

Proof.

Consider the Galois extension $M^{\prime}/L$ with $M^{\prime}=L(cos(2\pi/7),\sqrt[3]{p})$, where all the automorphisms of $\overline{C}:=C\times_{L}\overline{\mathbb{Q}}$ are defined. Let $\sigma$ be a generator of the cyclic Galois group $\text{Gal}(L(cos(2\pi/7))/L)$. We define a 1-cocycle on $\text{Gal}(M^{\prime}/L)\cong\text{Gal}(L(cos(2\pi/7))/L)\times\text{Gal}(L(\sqrt[3]{p})/L)$ to $\text{Aut}(\overline{C})$ by mapping $(\sigma,id)\mapsto[Y:Z:pX]$ and $(id,\tau)\mapsto id$. This defines an element of $\text{H}^{1}(\text{Gal}(M^{\prime}/L),\text{Aut}(\overline{C}))$, coming from the inflation of an element in $\text{H}^{1}(\text{Gal}(L(\cos(2\pi/7))/L),\text{Aut}(\overline{C}))^{Gal(M^{\prime}/L(cos(2\pi/7)))})$.

This $1$-cocycle is trivial if and only if $p$ is a norm of an element of $L(\cos(2\pi/7)$ over $L$. However, this is not the case, since $\mathbb{Q}(\cos(2\pi/7))$ and $L$ are disjoint with $[L:\mathbb{Q}]$ and $[\mathbb{Q}(\cos(2\pi/7)):\mathbb{Q}]$ coprime, and moreover $p$ is not a norm of an element of $\mathbb{Q}(\cos(2\pi/7))$ over $\mathbb{Q}$ being inert by our assumption. Consequently, the twist $C^{\prime}$ is not $L$-isomorphic to a non-singular plane model in $\mathbb{P}^{2}_{L}$ by [1, Theorem 4.1]. That is, $L$ is not a plane model-field of definition for $C^{\prime}$. The last sentence in the theorem follows by Lemma 2.6 and Corollary 2.7. ∎

Remark 2.9**.**

By our work in [1], we know that a non-singular plane model of $C^{\prime}$ exists over at least a degree degree $3$ extension of $L$.