2. Preliminaries

In this section, we recall the classification [Reference Cardona and Lario2] of twists of the curve $y^2 = x^6+1$ , with an emphasis on the biquadratic case, and some previous results of [Reference Fité and Sutherland5].

For a curve X over a field k, we say that another curve $X'$ over k is a twist of X if $X'$ becomes isomorphic to X over the algebraic closure $\overline {k}$ of k. The set of k-isomorphism classes of X is denoted by $\operatorname {\mathrm {Twist}}(X/k)$ . In what follows, we abbreviate $\operatorname {\mathrm {Aut}}_{\overline {k}}(X):=\operatorname {\mathrm {Aut}}_{\overline {k}}(X_{\overline {k}})$ . It is well known (see, for example, [Reference Silverman6, Theorem X.2.2]) that there is a canonical bijection

From now on, let $X_0/\mathbb {Q}$ denote the genus $2$ curve defined by the equation $ y^2 = x^6+1$ . We recall that $\operatorname {\mathrm {Aut}}_{\overline {\mathbb {Q}}}(X_0)$ is isomorphic to $2D_{12}$ as an abstract group. Following [Reference Cardona and Lario2], we use a group presentation

$$ \begin{align*} A := \langle{U, V, -1 : (-1)^2 = U^2 = V^6 = 1, (UV)^2 = (VU)^2 = -1, -1 \in Z(A)}\rangle, \end{align*} $$

which is isomorphic to $2D_{12}$ as a group. Its automorphism group is isomorphic to $C_2 \times D_{12}$ . We specify the elements of $\operatorname {\mathrm {Aut}}(A)$ as follows: $\iota , \jmath $ are two central involutions, s the noncentral involution, t the automorphism of order $3$ with relation $ts = st^2$ (see [Reference Cardona and Lario2, Section 3]).

Suppose that X is a twist of $X_0$ . Then $\operatorname {\mathrm {Aut}}_{\overline {\mathbb {Q}}}(X)$ is also isomorphic to $2D_{12}$ as a group, so we first consider possible $G_{\mathbb {Q}}$ -module structures on A. Since giving a $G_{\mathbb {Q}}$ -structure on A is equivalent to giving a morphism $G_{\mathbb {Q}} \to \operatorname {\mathrm {Aut}}(A)$ , we concentrate on the latter. Let K be the field of definition of the $G_{\mathbb {Q}}$ -structure on A so that

for a certain subgroup H in $C_2 \times D_{12}$ . Then, a $G_{\mathbb {Q}}$ -structure on A is determined by $K, H$ and a group isomorphism between H and $\operatorname {\mathrm {Gal}}(K/\mathbb {Q})$ . To describe the extension $K/\mathbb {Q}$ , we need to define some subfields of K. We consider subgroups of H, which are

$$ \begin{align*} H \cap \langle \iota, s, t \rangle, \quad H \cap \langle \iota, \jmath, t \rangle, \quad H \cap \langle \jmath, s, t \rangle. \end{align*} $$

They are index $1$ or $2$ , so induce a quadratic or trivial extension of $\mathbb {Q}$ . We denote them by

(2.1) $$ \begin{align} K_1 = \mathbb{Q}(\sqrt{u}), \quad K_2 = \mathbb{Q}(\sqrt{v}), \quad K_3 = \mathbb{Q}(\sqrt{v'}). \end{align} $$

In other words, we define $u, v, v' \in \mathbb {Q}^\times /(\mathbb {Q}^\times )^2$ determined by a $G_{\mathbb {Q}}$ -action on A.

Since we only consider a biquadratic extension $K/\mathbb {Q}$ , the subgroup H is isomorphic to $V_4$ . We recall that there are seven subgroups of $C_2 \times 2D_{12}$ isomorphic to $V_4$ :

(2.2) $$ \begin{align} \langle \iota, \jmath \rangle, \quad \langle \iota, s \rangle, \quad \langle \jmath, s \rangle \quad \langle \iota\jmath, s \rangle, \quad \langle \iota, \jmath s \rangle, \quad \langle \jmath, \iota s \rangle, \quad \langle \iota s, \jmath s \rangle. \end{align} $$

They correspond to the type $V_4^{\bullet }$ for $\bullet \in \{ A, \ldots , G \}$ (see [Reference Cardona and Lario2, Table 3]). To describe a specific isomorphism (in $6=|\operatorname {\mathrm {Aut}}(V_4)|$ -choices) between $\operatorname {\mathrm {Gal}}(K/\mathbb {Q})$ and H, it suffices to give quadratic subfields of K corresponding to the generators of H. For instance, we consider $V_4^A$ , which corresponds to $\langle \iota , \jmath \rangle \leq \operatorname {\mathrm {Aut}}(A)$ . Then an isomorphism between $\operatorname {\mathrm {Gal}}(K/\mathbb {Q})$ and $\langle \iota , \jmath \rangle $ can be described by $K^{\langle \iota \rangle }, K^{\langle \jmath \rangle }$ , which are quadratic extensions of $\mathbb {Q}$ .

We next describe a constraint for our $G_{\mathbb {Q}}$ -module A.

By Lemma 2.1, we only consider $G_{\mathbb {Q}}$ -subgroups of $\operatorname {\mathrm {GL}}_2(\overline {\mathbb {Q}})$ whose underlying groups are isomorphic to $2D_{12}$ . To get a more explicit condition using the parameters in (2.1), we introduce the following definition.

Definition 2.2. If $u,v \in \mathbb {Q}$ satisfy $(u, -3v) = 1 \in \mathrm {Br}_2(\mathbb {Q})$ , then by [Reference Cardona and Lario2, Remark 1],

$$ \begin{align*} x^2 + \frac{3}{v}y^2 = u \end{align*} $$

has solutions in $\mathbb {Q}^\times $ . Once solutions $\alpha ,\beta \in \mathbb {Q}^\times $ are chosen, we define constants

$$ \begin{align*} z = 4 \alpha^3 - 3u\alpha, \quad s = \frac{u^2 - 2\alpha^2u + \alpha z}{3\beta}. \end{align*} $$

Theorem 2.3 [Reference Cardona and Lario2, Theorem 2].

Let A be a $G_{\mathbb {Q}}$ -group with underlying group isomorphic to $2D_{12}$ , K the defining field of the $G_{\mathbb {Q}}$ -module structure on A which is a biquadratic extension of $\mathbb {Q}$ , $K_i$ the subfield of K defined by (2.1) for $i=1, 2, 3$ , and $u, v, v'$ elements in $\mathbb {Q}^\times /(\mathbb {Q}^\times )^2$ also defined by (2.1). Then A can be embedded in $\operatorname {\mathrm {GL}}_2(\overline {\mathbb {Q}})$ if and only if $(u, -3v) = 1 \in \mathrm {Br}_2(\mathbb {Q})$ and $v' \equiv -3v \pmod {\mathbb {Q}^{\times 2}}$ . In this case, $A \cong \langle U, V, -1 \rangle $ , where

$$ \begin{align*} U = \frac{1}{\sqrt{u}} \begin{pmatrix} \alpha & \beta \\ {3\beta}/{v} & -\alpha \end{pmatrix}, \quad V = \frac{\sqrt{-3}}{2} \begin{pmatrix} 1 & {\sqrt{v}}/{3} \\ -{1}/{\sqrt{v}} & 1 \end{pmatrix}, \quad -1 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \end{align*} $$

and the field of definition is $K = \mathbb {Q}(\sqrt {u}, \sqrt {v}, \sqrt {-3})$ .

In particular, the biquadratic extension K should include $\mathbb {Q}(\sqrt {-3})$ . Using these parameters, we can give a concrete defining equation of a twist.

Theorem 2.4 [Reference Cardona and Lario2, Propositions 3 and 4].

For a $G_{\mathbb {Q}}$ -group A embedded in $\operatorname {\mathrm {GL}}_2(\overline {\mathbb {Q}})$ and isomorphic to $2D_{12}$ as an abstract group, there is a twist with defining equation

$$ \begin{align*} y^2 = 27zx^6 - 162svx^5 - 135vzx^4 + 180sv^2x^3 + 45v^2zx^2 - 18sv^3x - v^3z, \end{align*} $$

whose automorphism group is $G_{\mathbb {Q}}$ -isomorphic to A. Furthermore, two twists, whose automorphism group is $G_{\mathbb {Q}}$ -isomorphic to A, differ by a hyperelliptic twist.

Together with a classification of hyperelliptic twists [Reference Cardona and Lario2, Proposition 6], we obtain the complete classification of twists of $X_0$ , whose automorphism group is defined over K.

We note that this terminology works well with [Reference Cardona and Lario2] since the twists of biquadratic type are twists of type $V_4^\bullet $ for $\bullet \in \{ A, B, \ldots , G \}$ in [Reference Cardona and Lario2].

We also have a complete classification of twists of biquadratic type. The classification is given in two steps. First, we specify the $G_{\mathbb {Q}}$ -module structure on $2D_{12}$ by describing the quadratic subfields (2.1). After that, Theorem 2.4 gives a twist whose automorphism group is isomorphic to a given $G_{\mathbb {Q}}$ -module, and any other twists with the same conditions are the hyperelliptic twists of the given twist. By [Reference Cardona and Lario2, Section 7], we give $K_i$ (and hence, a $G_{\mathbb {Q}}$ -module structure on $2D_{12}$ ) for each type $V_4^\bullet $ , $\bullet \in \{ A, B, \ldots , G \}$ . Here is a summary of [Reference Cardona and Lario2, Section 7]:

• • if $(d, -3) = 1 \in \operatorname {\mathrm {Br}}_2(\mathbb {Q})$ , there is the twist of type $V_4^A$ with $u = d, v=1$ ;

• • there is the twist of type $V_4^B$ with $u = 1, v=d$ ;

• • there is the twist of type $V_4^C$ with $u = d, v=-3$ ;

• • if $(d, d) = 1 \in \operatorname {\mathrm {Br}}_2(\mathbb {Q})$ , there is the twist of type $V_4^D$ with $u = d, v=-3u$ ;

• • if $(d, -3d) = 1 \in \operatorname {\mathrm {Br}}_2(\mathbb {Q})$ , there is the twist of type $V_4^E$ with $u = d, v=d$ ;

• • there is no twist of the form $V_4^F$ ;

• • if $(-3d, -3) = 1 \in \operatorname {\mathrm {Br}}_2(\mathbb {Q})$ , there is the twist of type $V_4^G$ with $u = -3, v=d$ .

We recall that $K_1 = \mathbb {Q}(\sqrt {u}), K_2 = \mathbb {Q}(\sqrt {v})$ and $K_3 = \mathbb {Q}(\sqrt {v'})$ .

The following definitions come from [Reference Fité and Sutherland5].

We note that $K, L$ depend on $X'$ and $L_\phi $ further depends on the choice of $\phi $ . In the above example concerning $E_A$ and $E_1$ , there are other choices of $\phi $ making $L_0 = \mathbb {Q}(\zeta _3\!\sqrt [3]{A})$ or $\mathbb {Q}(\zeta _3^2\sqrt [3]{A})$ . However, $K = \mathbb {Q}(\zeta _3)$ and $L = KL_0 = \mathbb {Q}(\zeta _3, \sqrt [3]{A})$ for any choice of $\phi $ . Hence a quadratic, cubic, quartic and sextic twist on an elliptic curve indicates the degree of $L_{\phi }$ , and the word ‘biquadratic’ in the phrase ‘twists of biquadratic type’ means the extension of $K/\mathbb {Q}$ .

3. Twist families of biquadratic type

In this section, we give a concrete parametrisation of twist types $V_4^B$ and $V_4^C$ . Then we prove that L is a subfield of $K(i)$ in both cases.

3.1. Type C

For $V_4^C$ , we have $u = d$ , $v = -3$ . According to Definition 2.2, $ x^2-y^2=d$ has solutions in $\mathbb {Q}^\times $ . We choose (with $d\neq \pm 1$ )

$$ \begin{align*} \alpha=\frac{d+1}{2},\quad\beta=\frac{d-1}{2} \end{align*} $$

so that the defining equation in Theorem 2.4 becomes $y^2 = f_d(x)$ , where

(3.1) $$ \begin{align} f_d(x) =54(d^3 + 1) x^6 & + 324( d^3 - 1) x^5 + 810(d^3 + 1) x^4 + 1080(d^3 - 1) x^3 \notag \\ &+ 810(d^3 + 1) x^2 + 324(d^3 - 1) x + 54( d^3 + 1). \end{align} $$

Let $X_d^C$ be the twist defined by the equation $y^2 = f_d(x)$ . Then, the field K of $X_d^C$ is $\mathbb {Q}(\sqrt {d}, \sqrt {-3})$ since $u = d$ and $v = -3$ . Also, by Theorem 2.4, $\{X_d^C\}$ is the one-parameter twist such that:

• • none of the elements is a hyperelliptic twist of another element;

• • every twist of type $V_4^C$ is a hyperelliptic twist of $\{X_d^C\}$ .

Hence, one can say that this family is a family of twists orthogonal to the hyperelliptic twists. This is the first motivation of our paper.

The polynomial $f_d(x)$ in (3.1) factors into

$$ \begin{align*} &54( (d+1)x^2 + 2(d-1)x + d + 1) \\ &\quad \times ((d^2-d+1) x^4 + 4 (d^2 - 1) x^3 + (6 d^2 + 2 d + 6) x^2 + 4 (d^2 - 1) x + d^2 - d + 1). \end{align*} $$

We further suppose that $d \neq 0$ and set $\delta _C = (1+\sqrt {-3})/2d$ . Then the zeros are

$$ \begin{align*} \frac{-(d-1) \pm 2\sqrt{-d}}{d+1}, \quad -\frac{1-\sqrt{\delta_C}}{1+\sqrt{\delta_C}}, \quad -\frac{1+\sqrt{\delta_C}}{1-\sqrt{\delta_C}}, \quad -\frac{1-\sqrt{\overline{\delta_C}}}{1+\sqrt{\overline{\delta_C}}}, \quad -\frac{1+\sqrt{\overline{\delta_C}}}{1-\sqrt{\overline{\delta_C}}}. \end{align*} $$

Lemma 3.1. Let d be a nonsquare rational number and let $L_d$ be the splitting field of the quartic factor of $f_d(x)$ , that is,

$$ \begin{align*} (d^2-d+1) x^4 + 4 (d^2 - 1) x^3 + (6 d^2 + 2 d + 6) x^2 + 4 (d^2 - 1) x + d^2 - d + 1. \end{align*} $$

Then, $L_d$ is a biquadratic extension of $\mathbb {Q}$ satisfying $L_d \subset K(i)$ when $d \neq \pm 1, \pm 3$ .

Proof. One can easily compute that

$$ \begin{align*} \sqrt{\delta_C \overline{\delta_C}} = \frac{1}{d} \in \mathbb{Q}, \quad \bigg(\sqrt{\delta_C} + \sqrt{\overline{\delta_C}}\bigg)^2 = \delta_C + \overline{\delta_C} + 2\sqrt{\delta_C\overline{\delta_C}} \in \mathbb{Q} \end{align*} $$

and

$$ \begin{align*} \frac{\displaystyle\frac{1 - \sqrt{\delta_C}}{1 + \sqrt{\delta_C}} -1 }{\displaystyle\frac{1 - \sqrt{\delta_C}}{1 + \sqrt{\delta_C}} +1} = -\sqrt{\delta_C}. \end{align*} $$

Hence,

$$ \begin{align*} L_d = \mathbb{Q} \bigg({\frac{1-\sqrt{\delta_C}}{1+\sqrt{\delta_C}}, \frac{1-\sqrt{\overline{\delta_C}}}{1+\sqrt{\overline{\delta_C}}}}\bigg) = \mathbb{Q} \bigg(\frac{1-\sqrt{\delta_C}}{1+\sqrt{\delta_C}}\bigg) = \mathbb{Q}(\sqrt{\delta_C}). \end{align*} $$

We note that each root of the quartic part of $f_d$ is not in $\mathbb {Q}$ because

$$ \begin{align*} \frac{1-\sqrt{\delta_C}}{1 + \sqrt{\delta_C}} = -1 + \frac{2}{1 + \sqrt{\delta_C}} \not\in \mathbb{Q}, \quad \sqrt{\delta_C} \not\in \mathbb{Q}, \quad \delta_C \not\in\mathbb{Q}. \end{align*} $$

Hence, if the quartic part is reducible, then it is a product of two quadratics over $\mathbb {Q}$ . So in this case, $L_d \supset \mathbb {Q}(\sqrt {-3})$ is a biquadratic extension of $\mathbb {Q}$ .

Otherwise, the discriminant of the quartic is ${2^{16}\cdot 3^2 \cdot d^6}/{(d^2-d+1)^6}$ and the resolvent of the quartic is

$$ \begin{align*} &x^3 + \frac{-6d^2 - 2d - 6}{d^2 - d + 1} \cdot x^2 + \frac{12d^4 + 8d^3 - 44d^2 + 8d + 12}{d^4 - 2d^3 + 3d^2 - 2d + 1} \cdot x \\ &\qquad+ \frac{-8d^4 - 16d^3 + 104d^2 - 16d - 8}{d^4 - 2d^3 + 3d^2 - 2d + 1}\\ &\quad=\bigg(x + \frac{-2d^2 - 10d - 2}{d^2 - d + 1}\bigg) \cdot (x - 2) \cdot \bigg(x + \frac{-2d^2 + 6d - 2}{d^2 - d + 1}\bigg). \end{align*} $$

Hence, $L_d$ is again a biquadratic extension of $\mathbb {Q}$ , since the discriminant is square and the resolvent is reducible.

We can say that $L_d = \mathbb {Q}(\sqrt {-3}, \sqrt {t})$ for some $t \in \mathbb {Q}^\times /(\mathbb {Q}^\times )^2$ since $\sqrt {-3} \in L_d$ . Then, $K(\sqrt {\delta _C}) = \mathbb {Q}(\sqrt {-3}, \sqrt {d}, \sqrt {t})$ should be a triquadratic extension containing $\mathbb {Q}(\sqrt {\delta _C})$ or $\mathbb {Q}(\sqrt {\delta _C})$ itself. However,

$$ \begin{align*} \sqrt{d}\sqrt{\delta_C} = \sqrt{ \frac{1 + \sqrt{-3}}{2} } = \frac{i}{\frac{1+\sqrt{-3}}{2}} \in K(\sqrt{\delta_C}), \end{align*} $$

which implies that $i \in K(\sqrt {\delta _C})$ . Hence, when $d \neq \pm 3$ , $K(\sqrt {\delta _C}) = \mathbb {Q}(\sqrt {-3}, i, \sqrt {d} )$ and $\mathbb {Q}(\sqrt {\delta _C}) \subset K(i)$ .

3.2. Type B

For $V_4^B$ , we have $u=1$ and $v=d$ . According to Definition 2.2, $x^2+({3}/{d})y^2=1$ has solutions in $\mathbb {Q}^\times $ . We choose (with $d\neq \pm 3$ )

$$ \begin{align*} \alpha = \frac{d-3}{d+3}, \quad \beta = \frac{2d}{d+3}, \end{align*} $$

so that the defining equation in Theorem 2.4 becomes of the form $y^2 = f_d(x)$ , where

$$ \begin{align*} f_d(x) &= 27 \frac{(d - 3)(d^2 - 42d + 9)}{(d + 3)^3} \bigg( x^2 + \frac{{4d}}{{d - 3}}x - \frac{{1}}{{3}}d \bigg) \\ &\quad \times \bigg( x^4 + \frac{{32d^2 - 96d}}{{d^2 - 42d + 9}}x^3 + \frac{{-\frac{{14}}{{3}}d^3 + 68d^2 - 42d}}{{d^2 - 42d + 9}}x^2 + \frac{{-\frac{{32}}{{3}}d^3 + 32d^2}}{{d^2 - 42d + 9}}x + \frac{{1}}{{9}}d^2 \bigg). \end{align*} $$

As before, we denote by $X_d^B$ the twist $y^2 = f_d(x)$ . Then the family $\{X_d^B\}$ has the same properties as the family $\{X_d^C\}$ discussed in the previous section.

The zeros of the quadratic factors of $f_d$ are

$$ \begin{align*} \frac{-6d \pm (d+3)\sqrt{3d}}{3(d-3)}, \end{align*} $$

and the zeros of the quartic factor are

$$ \begin{align*} -\frac{1}{2}\Bigg({\delta_B \pm \sqrt{\delta_B^2 + \frac{4d}{3}} }\Bigg), \quad -\frac{1}{2}\Bigg({\overline{\delta_B} \pm \sqrt{\overline{\delta_B}^2 + \frac{4d}{3}} }\Bigg), \end{align*} $$

where

$$ \begin{align*} \delta_B := \frac{-16d^2 + 48d + 2(d+3)^2 \sqrt{d}}{d^2-42d+9}, \quad \overline{\delta_B} := \frac{-16d^2 + 48d - 2(d+3)^2 \sqrt{d}}{d^2-42d+9}. \end{align*} $$

The defining fields of zeros of the quadratic and the quartic are

$$ \begin{align*} \mathbb{Q}(\sqrt{3d}), \quad \mathbb{Q}\Bigg(\sqrt{d}, \sqrt{\delta_B^2 + \frac{4d}{3}}, \sqrt{\overline{\delta_B}^2 + \frac{4d}{3}}\Bigg). \end{align*} $$

The latter is denoted by $L_d$ , which is the splitting field of the quartic part of $f_d$ .

Lemma 3.2. In the above settings, $L_d = \mathbb {Q}(\sqrt {d}, \sqrt {3})$ when $d\neq \pm 3$ is a nonsquare rational number.

Proof. The quartic part of $f_d$ is

$$ \begin{align*} &x^4 + \frac{32d^2 - 96d}{d^2 - 42d + 9} x^3 + \frac{-\frac{14}{3}d^3 + 68d^2 - 42d}{d^2 - 42d + 9} x^2 + \frac{-\frac{32}{3}d^3 + 32d^2}{d^2 - 42d + 9} x + \frac{1}{9}d^2. \end{align*} $$

Since

$$ \begin{align*} \sqrt{\delta_B^2 + \frac{4d}{3}} \sqrt{\overline{\delta_B}^2 + \frac{4d}{3}} = \frac{16}{3} \frac{d(d+3)^2}{(d^2 - 42 d + 9)} \in \mathbb{Q}, \end{align*} $$

we have

$$ \begin{align*} L_d = \mathbb{Q}\Bigg(\sqrt{d}, \sqrt{\delta_B^2 + \frac{4d}{3}}, \sqrt{\overline{\delta_B}^2 + \frac{4d}{3}}\Bigg) = \mathbb{Q}\Bigg({\sqrt{\delta_B^2 + \frac{4d}{3}}}\Bigg). \end{align*} $$

However,

$$ \begin{align*} \Bigg({\sqrt{\delta_B^2 + \frac{4d}{3}} + \sqrt{\overline{\delta_B}^2 + \frac{4d}{3}}}\Bigg)^2 &= \delta_B^2 + \overline{\delta_B}^2 +\frac{8d}{3} + \sqrt{\delta_B^2 + \frac{4d}{3}} \sqrt{\overline{\delta_B}^2 + \frac{4d}{3}} \in \mathbb{Q}, \end{align*} $$

which means that

$$ \begin{align*} \mathbb{Q}\Bigg({\sqrt{\delta_B^2 + \frac{4d}{3}} + \sqrt{\overline{\delta_B}^2 + \frac{4d}{3}}}\Bigg) \subset \mathbb{Q} \Bigg({\sqrt{\delta_B^2 + \frac{4d}{3}} }\Bigg) \end{align*} $$

is a quadratic subfield. Since

$$ \begin{align*} \Bigg({\sqrt{\delta_B^2 + \frac{4d}{3}} + \sqrt{\overline{\delta_B}^2 + \frac{4d}{3}}}\Bigg)^2 = \Bigg({\frac{8}{3} \frac{(d-3)(d+3)}{d^2 - 42d + 9}\sqrt{3d}}\Bigg)^2, \end{align*} $$

we can conclude that $L_d \supset \mathbb {Q}(\sqrt {d}, \sqrt {3})$ .

Next, we claim that each zero of the quartic is not rational. Suppose that

$$ \begin{align*} \delta_B\pm\sqrt{\delta_B^2+\frac{4d}{3}}\in\mathbb{Q}. \end{align*} $$

Then there is a rational number a such that

$$ \begin{align*} \sqrt{\delta_B^2+\frac{4d}{3}}=a\mp\frac{2(d+3)^2}{d^2-42d+9}\sqrt{d}. \end{align*} $$

Taking the squares of both sides,

$$ \begin{align*} \frac{16d(d+3)^2(d^2+30d+9)}{3(d^2-42d+9)^2} & \mp\frac{64(d-3)d(d+3)^2}{(d^2-42d+9)^2}\sqrt{d} \\ =a^2+\frac{4d(d+3)^4}{(d^2-42d+9)^2} & \mp\frac{4a(d+3)^2}{(d^2-42d+9)}\sqrt{d}. \end{align*} $$

Comparing the coefficients of $\sqrt {d}$ ,

$$ \begin{align*} a=\frac{16d(d-3)}{d^2-42d+9}. \end{align*} $$

Substituting this yields

$$ \begin{align*} \frac{16d(d+3)^2(d^2+30d+9)}{3(d^2-42d+9)^2}=\frac{16^2d^2(d-3)^2}{(d^2-42d+9)^2}+\frac{4d(d+3)^4}{(d^2-42d+9)^2}. \end{align*} $$

This is equivalent to

$$ \begin{align*} \tfrac{4}{3} (d^2 - 42d + 9)^2 = 0, \end{align*} $$

which does not have a solution in $\mathbb {Q}$ . To deal with the other case:

$$ \begin{align*} \overline{\delta}_B\pm\sqrt{\overline{\delta}_B^2+\frac{4d}{3}}\in\mathbb{Q}, \end{align*} $$

it suffices to replace $\sqrt {d}$ by $-\sqrt {d}$ in the above computation. Therefore, we arrive at the same conclusion. Hence, if the quartic part of f is reducible over $\mathbb {Q}$ , it is a product of two quadratics, which means that $L_d$ is a biquadratic extension of $\mathbb {Q}$ .

Suppose that the quartic part is irreducible over $\mathbb {Q}$ . Its discriminant is

$$ \begin{align*} \frac{2^{16}}{3^4} \frac{d^6 (d + 3)^{12}}{(d^2 - 42d + 9)^{6}}, \end{align*} $$

which is a square, and its resolvent is

$$ \begin{align*} \bigg(x + \frac{\frac{{-2}}{{3}}d^3 - 36d^2 - 6d}{d^2 - 42d + 9} \bigg) \cdot \bigg(x + \frac{{2}}{{3}}d \bigg) \cdot \bigg(x + \frac{\frac{{14}}{{3}}d^3 - 4d^2 + 42d }{d^2 - 42d + 9}\bigg), \end{align*} $$

which is reducible. Hence, $L_d$ is also a biquadratic extension of $\mathbb {Q}$ . Consequently, we have $L_d = \mathbb {Q}(\sqrt {d}, \sqrt {3})$ in both cases.

3.3. Computation of the number field L

Proposition 3.3. Let $X_d : y^2 = f_d(x)$ be a twist of type $V_4^{\bullet }$ with $\bullet \in \{ A, B, \ldots , G \}$ and $K = \mathbb {Q}(\sqrt {d}, \sqrt {-3})$ . Suppose that $f_d(x)$ is decomposed in $K(i)$ . Then, L is $K(i)$ or K.

Proof. By Lemma 2.1, every isomorphism from $X_0$ to $X_d$ can be represented by a linear fractional transformation on x. More precisely, an isomorphism is given by

$$ \begin{align*} x \to \frac{mx + n}{px + q}, \quad y \to \frac{(mq-pn)y}{(px+q)^3} \end{align*} $$

with $mq-pn \neq 0$ . Let $\gamma _i$ denote the zeros of $f_d$ . Since the zeros of $x^6+1$ are $\pm i, \pm i\zeta _3, \pm i\zeta _3^2$ , at least one of the linear fractional transformations that satisfy

$$ \begin{align*} \gamma_{k_1} \to i, \quad \gamma_{k_2} \to -i, \quad \gamma_{k_3} \to i\zeta_3 \end{align*} $$

for $k_i \in \{ 1, \ldots , 6 \}$ is the isomorphism from $C_1$ to $C_d$ , which is defined over (possibly a subfield of) $\mathbb {Q}(i, \zeta _3, \gamma _1, \ldots , \gamma _6)$ .

Let $L_{\phi }$ be the defining field of this isomorphism so that $L_{\phi } \subset \mathbb {Q}(i, \zeta _3, \gamma _1, \ldots , \gamma _6)$ . By assumption, $\mathbb {Q}(\gamma _1, \ldots , \gamma _6)$ and also $\mathbb {Q}(i, \zeta _3, \gamma _1, \ldots , \gamma _6)$ are subfields of $K(i)$ . Therefore,

$$ \begin{align*} L = \left\{ \begin{array}{@{}ll} K & \textrm{if } L_{\phi} \subset K, \\ K(i) & \textrm{otherwise}, \end{array} \right. \end{align*} $$

since $L_{\phi }K = L$ by Lemma 2.8.

Corollary 3.4. Let $X_d$ be a twist of type $V_4^B$ or $V_4^C$ . Then, L is $K(i)$ or K.

Proof. This is a direct consequence of Lemmas 3.1, 3.2 and Proposition 3.3.

4. Proof of the main theorem

4.1. Proof for type $B, C$

Let E be an elliptic curve. Then, its L-function is a product of L-factors, $L_p(E/\mathbb {Q}, s) = (1 - a_p(E) p^{-s} + p^{1 - 2s})^{-1}$ , where E has good reduction at p. Here, $a_p(E)$ is the trace of the Frobenius which is in the interval $[-2\sqrt {p}, 2\sqrt {p}]$ . Similarly,

$$ \begin{align*} L_p(X/\mathbb{Q}, s) = (1 + a_{p, 1}(X)p^{-s} + a_{p, 2}(X)p^{-2s} + a_{p, 1}(X)p^{1-3s} + p^{2-4s} )^{-1}, \end{align*} $$

when X is a curve of genus $2$ and has good reduction at p. For an elliptic curve E and a genus $2$ curve X, we denote by $E^{(D)}$ and $X^{(D)}$ the respective hyperelliptic twists given by the field $\mathbb {Q}(\sqrt {D})$ . It is also well known that

$$ \begin{align*} L_p(E^{(D)}/\mathbb{Q}, s) = (1 - a_p(E) \chi_D(p) p^{-s} + p^{1 - 2s})^{-1}, \end{align*} $$

where $\chi _D$ is the quadratic character attached to the field extension $\mathbb {Q}(\sqrt {D})/\mathbb {Q}$ .

Table 1

The L-factors of $J_d$ corresponding to $I(p)$ .

Let $X_d^B$ and $X_d^C$ be the twists of $X_0$ of biquadratic type $B, C$ studied in the previous section. We denote by $J_0, J_d^B, J_d^C$ their Jacobians, by $X_0^{(D)}, X_d^{B, (D)}, X_d^{C, (D)}$ the hyperelliptic twists and by $J_0^{(D)}, J_d^{B, (D)}, J_d^{C, (D)}$ the Jacobians of the hyperelliptic twists. Finally, we denote by $E_0$ the elliptic curve over $\mathbb {Q}$ defined by the equation $y^2 = x^3 + 1$ .

We also recall some notation from [Reference Fité and Sutherland5, Section 4]. Let $M = \mathbb {Q}(\sqrt {-3})$ . The definition of $L, K$ which depends on the choice of the twist X of $X_0$ is given by Definition 2.7. For a number field F, the residue degree at p in $F/\mathbb {Q}$ is denoted by $f_F(p)$ . We define $ I(p) = I(p, C) := (f_L(p), f_K(p), f_M(p))$ .

Proposition 4.1. Let $X_d^{\bullet }$ be a twist of $X_0$ with $\bullet \in \{ B, C \}$ , and let p be a prime greater than $3$ , where $X_d^{\bullet }$ have good reduction at p. Table 1 gives the L-factors for $J_d^{\bullet }$ (with $a_p = a_{p}(E_0)$ ).

Proof. This is an application of [Reference Fité and Sutherland5, Proposition 4.9] in our cases, but we note that [Reference Fité and Sutherland5] uses a normalisation $a_{p, i}(J)/p^{{i}/{2}}$ and $a_1(E)(p) = -a_p(E_0)$ . (For example, the L-factor of an elliptic curve is $1 + a_1(E)(p)T + T^2$ in [Reference Fité and Sutherland5, page 555].)

By Corollary 3.4, $L = K(i)$ or $L = K$ . By [Reference Fité and Sutherland5, Proposition 4.9], the possible values of $I(p)$ are

$$ \begin{align*} (1, 1, 1), \quad (2, 1, 1), \quad (2, 2, 1), \quad (4, 2, 1), \quad (2,2,2), \quad (4, 2, 2), \end{align*} $$

and $I(p)$ determines $a_{p, 1}$ and $a_{p, 2}$ . When $L=K(i)$ , which is a triquadratic field, the first entry of $I(p)$ cannot be $4$ . This gives the first three columns of Table 1 and the last one can be easily computed.

When $L = K$ , the only possible value of $I(p)$ is one of $(1,1,1)$ , $(2, 2, 1)$ and $(2, 2, 2)$ . The other results are not changed.

We note that each entry of the fourth column in Table 1 can be factorised, namely $(1 - a_pp^{-s} + p^{1-2s})^2$ , $(1 + a_pp^{-s} + p^{1-2s})^2$ , $(1 - a_pp^{-s} + p^{1-2s})(1 + a_pp^{-s} + p^{1-2s})$ and $(1 + p^{1-2s})^2$ .

Suppose that $L = K(i)$ . Since there is no rational prime whose residue degree is $4$ in a biquadratic extension, the conditions on $I(p)$ are:

• • $I(p) = (1, 1, 1)$ if and only if p totally splits in L;

• • $I(p) = (2, 1, 1)$ if and only if p is inert in $\mathbb {Q}(i)$ and totally splits in K;

• • $I(p) = (2, 2, 1)$ if and only if p is inert in $\mathbb {Q}(\sqrt {d})$ and splits in $\mathbb {Q}(\sqrt {-3})$ ;

• • $I(p) = (2, 2, 2)$ if and only if p is inert in $\mathbb {Q}(\sqrt {-3})$ .

Let us define

$$ \begin{align*} &S_1 = \{ p : I(p) = (1, 1, 1) \}, \quad S_2 = \{ p : I(p) = (2, 1, 1) \}, \\ & S_3 = \{ p : I(p) = (2, 2, 1) \}, \quad S_4 = \{ p : I(p) = (2, 2, 2) \}. \end{align*} $$

It follows that half of the primes are in $S_4$ , $\tfrac 14$ in $S_3$ and $\tfrac 18$ in each of $S_2, S_1$ .

When $L = K$ , a rational prime is an element of $S_1$ , $S_3$ or $S_4$ . Also in this case:

• • $p \in S_1$ if and only if p totally splits in $L=K$ ;

• • $p \in S_3$ if and only if p is inert in $\mathbb {Q}(\sqrt {d})$ and splits in $\mathbb {Q}(\sqrt {-3})$ ;

• • $p \in S_4$ if and only if p is inert in $\mathbb {Q}(\sqrt {-3})$ .

Hence, in this case, half of the primes are in $S_4$ and $\tfrac 14$ in each of $S_3, S_1$ .

Proof of Theorem 1.1.

We first consider the case $L = K(i)$ . By Proposition 4.1,

$$ \begin{align*} L(J_d^{\bullet}/\mathbb{Q}, s)^{-1} &\sim \prod_{p \in S_1} ({ 1 - a_pp^{-s} + p^{1-2s} })^2 \prod_{p \in S_2} ({ 1 + a_pp^{-s} + p^{1-2s} })^2\\ &\quad \times \prod_{p \in S_3} (1 - a_pp^{-s} + p^{1-2s})(1 + a_p p^{-s} + p^{1-2s}) \prod_{p \in S_4} (1 + p^{1-2s})^2, \end{align*} $$

where $\bullet \in \{ B, C \}$ . Here, $\sim $ means that the quantities are the same at unramified primes. We consider L-functions of $E_0^{(-1)} \times E_0^{(-d)}$ . Note that $p\equiv 2\pmod {3}$ if and only if p is inert in $\mathbb {Q}(\sqrt {-3})$ . Since $a_p = 0$ when $p \equiv 2 \pmod {3}$ ,

$$ \begin{align*} L(E_0^{(D)}/\mathbb{Q}, s)^{-1} &= \prod_p (1 - a_p\chi_{D}(p)p^{-s} + p^{1 - 2s} ) \\ &= \prod_{p \in S_1} (1 - a_p\chi_{D}(p)p^{-s} + p^{1 - 2s} ) \prod_{p \in S_2} (1 - a_p\chi_{D}(p)p^{-s} + p^{1 - 2s} )\\ &\quad \times \prod_{p \in S_3} (1 - a_p\chi_{D}(p)p^{-s} + p^{1 - 2s} ) \prod_{p \in S_4} (1 + p^{1 - 2s} ). \end{align*} $$

We have $\chi _{-1}(p) = \chi _{-d}(p) = 1$ for a prime in $S_1$ and $\chi _{-1}(p) = -1, \chi _{-d}(p) = 1$ for a prime in $S_2$ . In $S_3$ , there are two subclasses of primes:

$$ \begin{align*} S_{3, 1} := \{ p \in S_3 : p \textrm{ splits in } \mathbb{Q}(i) \}, \quad S_{3, 2} := \{ p \in S_3 : p \textrm{ is inert in } \mathbb{Q}(i) \}. \end{align*} $$

Since $p \in S_3$ if and only if $\chi _d(p) = -1$ , we have $\chi _{-d}(p)=\chi _{-1}(p)\chi _d(p)=-\chi _{-1}(p)$ . Hence, for a prime $p \in S_3$ , p is in $S_{3, 1}$ if and only if $\chi _{-1}(p) = 1$ and p is in $S_{3, 2}$ if and only if $\chi _{-d}(p) = 1$ . Consequently, $L(E_0^{(-1)}/\mathbb {Q}, s)^{-1}$ is given by

$$ \begin{align*} \prod_{p \in S_1 \cup S_{3, 1}} (1 - a_pp^{-s} + p^{1 - 2s} ) \prod_{p \in S_2 \cup S_{3, 2}} (1 + a_pp^{-s} + p^{1 - 2s} ) \prod_{p \in S_4} (1 + p^{1 - 2s} ) \end{align*} $$

and $L(E_0^{(-d)}/\mathbb {Q}, s)^{-1}$ is given by

$$ \begin{align*} \prod_{p \in S_1 \cup S_{3, 2}} (1 - a_pp^{-s} + p^{1 - 2s} ) \prod_{p \in S_2 \cup S_{3, 1}} (1 + a_pp^{-s} + p^{1 - 2s} ) \prod_{p \in S_4} (1 + p^{1 - 2s} ). \end{align*} $$

Therefore, $L(J_d^{\bullet }/\mathbb {Q}, s) \sim L(E_0^{(-1)}/\mathbb {Q}, s)L(E_0^{(-d)}/\mathbb {Q}, s),$ which means that $J_d^{\bullet }$ is isogenous to $E_0^{(-1)} \times E_{0}^{(-d)}$ over $\mathbb {Q}$ . When $L = K$ , the same argument shows that $J_d^\bullet $ is isogenous to $E_0 \times E_0^{(d)}$ .

Let $\rho _{J, \ell }$ be the $\ell $ -adic Galois representation attached to the abelian variety J. It is well known that $\rho _{J^{(D)}, \ell } \sim \rho _{J, \ell } \otimes \chi _D$ and $J_1$ is isogenous to $J_2$ if and only if $\rho _{J_1} \sim \rho _{J_2}$ (see [Reference Silverman6, Section III.7]). Hence, for $\bullet \in \{ B, C \}$ ,

$$ \begin{align*} \rho_{J_d^{\bullet, (D)}, \ell} & \sim \rho_{J_d^\bullet, \ell} \otimes \chi_D \sim \rho_{E_0^{(-1)} \times E_0^{(-d)}} \otimes \chi_D = ({\rho_{E_0^{(-1)}} \oplus \rho_{E_0^{(-d)}} }) \otimes \chi_D \\ & = ({ \rho_{E_0^{(-1)}} \otimes \chi_D }) \oplus ({ \rho_{E_0^{(-d)}} \otimes \chi_D }) = \rho_{E_0^{(-D)}} \oplus \rho_{E_0^{(-dD)}}. \end{align*} $$

This proves the result.

We give two remarks on Theorem 1.1. Some values of d are naturally excluded in the family $\{X_d^B\}$ or $\{X_d^C\}$ since they do not define a twist of biquadratic type. Also, Theorem 1.1 shows that two curves in the family $\{X_d^B\}$ (or $\{X_d^C\}$ ), where $d \in \mathbb {Q}^\times /(\mathbb {Q}^\times )^2$ , are not isogenous.

4.2. Remarks for other types

In this section, we give some remarks on the other types $A, D, E$ and G. Compared with the types B and C, there are two main obstacles.

1. (i) The Brauer group condition becomes nontrivial.

2. (ii) For each integer d, we have to find a solution $x,y\in \mathbb {Q}$ of

   (4.1) $$ \begin{align} x^2+\frac{3y^2}{v(d)}=u(d). \end{align} $$
   In practice, we should express x and y as rational functions in d.

Note that when we deal with types B and C, we have easy solutions for obstacle (ii).

In some cases, we may restrict d to make obstacle (i) simpler. However, even in this case, obstacle (ii) may remain nontrivial. In the following examples, we numerically verify the expected nonsimplicity for some small d by finding explicit solutions $(\alpha , \beta )$ of (4.1).

Example 4.2. The Brauer group condition $(d, -3) = 1 \in \operatorname {\mathrm {Br}}_2(\mathbb {Q})$ for type A is equivalent to $d \equiv 1 \pmod {3}$ when d is prime. In this case, $(\alpha , \beta )$ is a solution of

$$ \begin{align*} x^2 + 3y^2 = d. \end{align*} $$

For example, we have $(\alpha , \beta ) = (2, 1)$ when $d = 7$ . With the calculations of previous sections, we find the list of $(\alpha , \beta )$ and $f_d$ for primes $d < 70$ shown in Table 2.

Table 2

The solutions $(\alpha,\beta)$ associated to d and the resulting $f_d$ for type A.

By computing the discriminant of each quadratic divisor of $f_d$ , we have $L_d = \mathbb {Q}(\sqrt {3d})$ , which is a subfield of $K(i)$ . By Proposition 3.3, $L = K$ or $L = K(i)$ . Hence, the proof of Proposition 4.1, and Theorem 1.1 also works. Consequently, $J_d^A$ is not simple for primes $d < 70$ satisfying the Brauer group condition.

Example 4.3. In the case of type D and prime d, the Brauer group condition gives $d \equiv 1 \pmod {4}$ . Since numerical computation is too complicated, we omit a detailed description as in the previous example. For such primes $d < 70$ , the factorisation of $f_d(x)$ over $\mathbb {Q}(i)$ shows that $L_d$ is a quartic extension of $\mathbb {Q}$ whose unique intermediate field is $\mathbb {Q}(i)$ . By the proof of Proposition 3.3, $L_{\phi }$ is a subfield of $L_d(i) = L_d$ . Hence, L is a subfield of $L_dK$ with $[L:K] \leq 2$ . Therefore, we can conclude that $L = K$ or $K(i)$ . This implies that $J_d^D$ is not simple for primes $d < 70$ .

Example 4.4. For type E, the Brauer group condition makes the prime $d \equiv 1 \pmod {12}$ . Analogous computation shows that $f_d$ has three quadratic factors over $\mathbb {Q}(\sqrt {3})$ and $L_d$ is a quartic extension whose quadratic subfield is $\mathbb {Q}(\sqrt {3})$ . By the same argument as for type D, we have checked that $J_d^E$ is not simple for primes $d < 190$ .

Example 4.5. In the case of type G, a simple computation shows that $d = -p$ with $p \equiv 1 \pmod {3}$ satisfies the Brauer group condition. This case may require more effort to find explicit $\alpha , \beta $ .