The Fermat cubic and quartic curves over cyclic fields

Abstract

First we show that there exist infinitely many distinct cyclic cubic number fields K such that the Fermat cubic $x^3 + y^3 = z^3$ has non-trivial points in K. Second, we show that the Fermat quartic $x^4 + y^4 = z^4$ can have no non-trivial points in any cyclic cubic number field. It remains an open question whether the Fermat quartic has any points in quartic number fields with Galois group of type $\mathbb {Z}/4\mathbb {Z}$ or $A_4$.

1 Introduction

The Fermat cubic

$$\begin{aligned} \mathcal {F}_3(K) : x^3 + y^3 = z^3 \end{aligned}$$

over a field K is an intensively studied equation, and it is well known that the only rational points on the curve $\mathcal {F}_3(\mathbb {Q})$ are the trivial points for which $xyz=0$.

The curve is of course elliptic, with cubic model

$$\begin{aligned} \mathcal {E}_3:\; Y^2 = X^3-432. \end{aligned}$$

It contains points in quadratic fields $K=\mathbb {Q}(\sqrt{d})$ precisely when the K-rank of $\mathcal {E}_3$ is positive. Here, we ask about points on $\mathcal {F}_3$ that are defined over a cubic number field. This seems a rather trivial question, because it is self-evident that $\mathcal {F}_3$ contains non-trivial points in infinitely many non-isomorphic cubic fields. But this self-evidence relates to cubic fields with Galois group $S_3$, for example, cubic fields $\mathbb {Q}(\root 3 \of {a^3+b^3})$ for $a,b \in \mathbb {Z}$. It is less obvious, however, that $\mathcal {F}_3$ has points in infinitely many non-isomorphic cubic fields with cyclic Galois group, and one of the purposes of this note is to prove this assertion, which we do in Sect. 2.

One can ask an analogous question about the Fermat quartic

$$\begin{aligned} \mathcal {F}_4: x^4 + y^4 = z^4. \end{aligned}$$

The curve clearly contains points in cubic fields, by taking the intersection of $\mathcal {F}_4$ with a straight line through a rational point such as (1, 0, 1). But such cubic fields always have Galois group $S_3$. In Sect. 3 we show that this is essentially the only way that cubic points arise on $\mathcal {F}_4$, so that $\mathcal {F}_4$ contains no non-trivial points in any cyclic cubic number field.

Finally, it is easy to see that $\mathcal {F}_4$ contains points in quartic number fields whose Galois groups are of type $\mathbb {Z}/2\mathbb {Z}\times \mathbb {Z}/2\mathbb {Z}$, $D_8$, or $S_4$. But we do not know of any quartic field with Galois group $\mathbb {Z}/4\mathbb {Z}$ or $A_4$ in which $\mathcal {F}_4$ has non-trivial points. We discuss this question in Sect. 4.

2 The Fermat cubic over cyclic cubic fields

Theorem 2.1

There exist infinitely many non-isomorphic cyclic cubic fields K such that the curve

$$\begin{aligned} \mathcal {F}_3 : x^3 + y^3 = z^3 \end{aligned}$$

contains points in K with $xyz \ne 0$.

Proof

Let t be a root of the cubic polynomial

$$\begin{aligned} c^3(1+x^3) - (a x+1)^3 = 0, \qquad a,c \in \mathbb {Q}, \quad a \ne c. \end{aligned}$$

(2.1)

If the cubic is reducible, then there is a rational root $t_0$, with $(c, c t_0, a t_0+1) \in \mathcal {F}_3(\mathbb {Q})$. But this would imply $a=1$ or $c=0,1$. We therefore assume $a \ne c$, $a \ne 1$, $c \ne 0, 1$, so that the cubic is irreducible, and defines a cubic field $K=\mathbb {Q}(\theta )$. In this situation, $(x,y,z)=(c,c \theta ,a \theta +1) \in \mathcal {F}_3(K)$. The condition that K be cyclic is that the discriminant of the cubic be a perfect square, namely:

$$\begin{aligned} -3(a^6 - 2 a^3 c^3 - 2 a^3 -2 c^3 + c^6 + 1) = \square . \end{aligned}$$

(2.2)

Let

$$\begin{aligned} c = a + d, \end{aligned}$$

so that (2.2) becomes

$$\begin{aligned} -27 d^2 a^4 - 6(9 d^3-2) a^3 - 9d(5d^3-2) a^2 - 18 d^2(d^3-1) a - 3(d^3-1)^2 = w^2, \end{aligned}$$

(2.3)

say. When $d=\frac{2}{7}$, Eq. (2.3) has solution $(a,w)=(37/7,117/49)$. Accordingly, the curve (2.3) with $d=\frac{2}{7}$, namely

$$\begin{aligned} -3 (86436 a^4 - 421204 a^3 - 189924 a^2 - 56280 a + 112225) = \square , \end{aligned}$$

represents an elliptic quartic curve. A cubic model is

$$\begin{aligned} Y^2 + Y = X^3 - 78438 X - 128334357, \end{aligned}$$

with rank 1, and generator of infinite order $P=(365049,220559930)$. The multiples nP of P pull back to infinitely many solutions $(a_n,c_n)$, $n=1,2,3 , \ldots {}$, of (2.2). We claim that the corresponding cubic polynomials at (2.1) (which are irreducible by earlier remarks) give infinitely many non-isomorphic cubic fields. $\square$

Lemma 2.2

Let $\mathbb {Q}(u)$ be the cubic field defined by $u^3+A u+B=0$, and write $\Delta ^2=(-4A^3-27B^2)/9 \ne 0$. Let $\mathbb {Q}(v)$ be the cubic field defined by $v^3+R v+S=0$. Then $\mathbb {Q}(u)=\mathbb {Q}(v)$ if and only if there exists a rational number f such that

$$\begin{aligned} (B^2 R^3-A^3 S^2) + \Delta ^2 f (-\Delta ^2 f^2+A R)(-\Delta ^2 f^3 +A R f+3S) = 0. \end{aligned}$$

(2.4)

Proof

If $\mathbb {Q}(u)=\mathbb {Q}(v)$, then $v=f u^2+g u+h$, say, for some $f,g,h \in \mathbb {Q}$, and the minimal polynomial of v is

$$\begin{aligned}&v^3+(2Af-3h)v^2+(3h^2+g^2A-4h A f+A^2 f^2+3g B f)v \\&\quad -\,h^3-h g^2A+2h^2A f-h A^2 f^2+g^3B-3h g B f+g A B f^2-B^2 f^3 = 0. \end{aligned}$$

Thus $h=\frac{2A f}{3}$, and

$$\begin{aligned} R=A g^2+3B f g-\frac{A^2}{3} f^2, \quad S=B g^3-\frac{2}{3}A^2 f g^2-A B f^2 g-\frac{2A^3+27B^2}{27} f^3. \end{aligned}$$

(2.5)

Eliminating g gives (2.4).

Conversely, suppose that (2.4) holds.

If $A=0$, then $3\Delta ^4 f^6-9 \Delta ^2 S f^3-R^3=0$, so that $S=(3\Delta ^4 f^6-R^3)/(9 \Delta ^2 f^3)$. This gives $v=f u^2 +R/(3 B f) u$, so that $\mathbb {Q}(v) \subseteq \mathbb {Q}(u)$, and thus $\mathbb {Q}(u)=\mathbb {Q}(v)$.

If $A \ne 0$, then as a quadratic in S, Eq. (2.4) has discriminant a rational square, implying

$$\begin{aligned} B^2 (-3\Delta ^2 f^2 + A R)^2 (-3\Delta ^2 f^2 + 4 A R) = \square . \end{aligned}$$

Thus either $-3\Delta ^2 f^2 + A R = 0$ or $-3\Delta ^2 f^2 + 4 A R = f^2 T^2$, say. The first instance leads to

$$\begin{aligned} R=\frac{3\Delta ^2}{A} f^2, \qquad S=\frac{3 \Delta ^4}{A^3} f^3, \end{aligned}$$

with

$$\begin{aligned} v = \left( u^2 + \frac{3(D-B)}{2A} u + \frac{2A}{3} \right) f. \end{aligned}$$

Thus $\mathbb {Q}(v) \subseteq \mathbb {Q}(u)$ and $\mathbb {Q}(u)=\mathbb {Q}(v)$.

The second instance leads to

$$\begin{aligned} R=\frac{3\Delta ^2+T^2}{4A} f^2, \quad S=\frac{-3\Delta ^4+9 B\Delta ^2 T+3\Delta ^2 T^2-B T^3}{8A^3} f^3, \end{aligned}$$

and

$$\begin{aligned} v = \left( u^2 - \frac{3B+T}{2A} u + \frac{2A}{3} \right) f. \end{aligned}$$

Thus $\mathbb {Q}(u) \subseteq \mathbb {Q}(v)$, so that $\mathbb {Q}(u)=\mathbb {Q}(v)$. The Lemma is proved. $\square$

We complete the proof of Theorem 2.1 by induction. Suppose pullbacks of multiples nP of P have generated k distinct cubic fields $K_i$ given by

$$\begin{aligned} x^3+A_i x+B_i=0, \quad i=1 , \ldots , k, \end{aligned}$$

for $k \ge 1$ (where, necessarily, $-4A_i^3-27B_i^2$ is a non-zero square). The cubic field K corresponding to $(a_n,c_n)$ is defined by

$$\begin{aligned} c_n^3(1+x^3)-(a_n x+1)^3 = 0, \end{aligned}$$

or equivalently, under $x = (a_n^2+c_n \phi )/(-a_n^3+c_n^3)$, by

$$\begin{aligned} \phi ^3 - 3 a_n c_n \phi + a_n^6-a_n^3-2a_n^3c_n^3-c_n^3+c_n^6 = 0, \quad c_n=a_n+\frac{2}{7}. \end{aligned}$$

This is the cubic field defined by

$$\begin{aligned} X^3+R X+S=0, \end{aligned}$$

where

$$\begin{aligned} R = -147 t(t+1), \quad S = -335 - 981 t - 909 t^2 - 542 t^3 + 72 t^4 , \end{aligned}$$

with $t=\frac{7}{2} a_n$; and necessarily $-4R^3-27S^2=\square$. By Lemma 2.2, this field K is isomorphic to the field $K_i$ if and only if there exists a rational number f such that (2.4) holds for $(A,B,\Delta )=(A_i,B_i,\Delta _i)$ with $\Delta _i^2=(-4A_i^3-27B_i^2)/9$.

Write (2.4) in the form

$$\begin{aligned} Q(A_i,B_i,R,S) : \Delta _i^2 (2\Delta _i^2 f^3-2 A_i R f-3 S)^2 + B_i^2 (4R^3+27S^2) = 0. \end{aligned}$$

For each $i=1 , \ldots ,k$, this equation represents a curve $\mathcal {C}_i$ of degree 6 in f and degree 8 in t in the f–t space.

Lemma 2.3

For any $A_i, B_i \in \mathbb {Q}$ with $A_i B_i (-4A_i^3-27B_i^2) \ne 0$, the genus of the curve $\mathcal {C}_i$ is equal to 3.

Proof

For any $A_i$, $B_i$, the curve $\mathcal {C}_i$ has only one singularity at infinity. This singular point, expressible in homogeneous coordinates as $(1,\,0,\,0)$, is a cusp, with singularity multiplicity count of 4 (see, for example, Abhyankar [1, Lecture 19]). Finite singularities of $\mathcal {C}_i$ occur where $Q(A_i,B_i,R,S)=0$ and $\partial Q/\partial f=\partial Q/\partial t=0$; that is, where

$$\begin{aligned} \Delta _i^2 (2\Delta _i^2 f^3-2 A_i R f-3 S)^2 + B_i^2 (4R^3+27S^2)&= 0, \nonumber \\ (2 \Delta _i^2 f^3-2 A_i R f-3 S)(3\Delta _i^2 f^2 - A_i R)&= 0, \nonumber \\ 2 \Delta _i^2 (2 \Delta _i^2 f^3\!-\!2 A_i R f\!-\!3 S)\frac{\partial }{\partial t}(-2 A_i R f-3 S) + B_i^2 \frac{\partial }{\partial t} (4R^3+27S^2)&= 0. \end{aligned}$$

(2.6)

Case (i): $2 \Delta _i^2 f^3 - 2 A_i R f - 3 S = 0$. Then

$$\begin{aligned} 4R^3+27S^2=0, \qquad \frac{\partial }{\partial t} (4R^3+27S^2) = 0; \end{aligned}$$

and since

$$\begin{aligned} 4R^3+27S^2 = 27(1+3t+3t^2)^2(112225-16080t-15504t^2-9824t^3+576t^4), \end{aligned}$$

we have

$$\begin{aligned} 1+3t+3t^2 = 0, \qquad 2 \Delta _i^2 f^3 - 2 A_i R f - 3 S = 0. \end{aligned}$$

The discriminant of the latter cubic is $26244 \Delta _i^2 (16A^3 R^3-243 \Delta _i^2 S^2)$, which for $1+3t+3t^2 = 0$ reduces to $-2^4 3^{11} 7^6 B_i^2 \Delta _i^2 \ne 0$. Thus for arbitrary $A_i$, $B_i$, we obtain precisely three singular points (ordinary double points) at each root of $1+3t+3t^2 = 0$.

Case (ii): $3 \Delta _i^2 f^2 - A_i R = 0$. From the first equation in (2.6), we have

$$\begin{aligned} 6 A_i \Delta _i^2 R S f = \Delta _i^2 R^3+3 A_i^3 S^2, \end{aligned}$$

and eliminating f from the last two equations gives

$$\begin{aligned} \Delta _i^2 R^3 - 3 A_i^3 S^2 = 0, \end{aligned}$$

(2.7)

with

$$\begin{aligned} f = (\Delta _i^2 R^3+3 A_i^3 S^2)/ (6 A_i \Delta ^2 R S) = R^2/(3 A_i S). \end{aligned}$$

All three equations in (2.6) are now satisfied. The discriminant of (2.7) with respect to t cannot vanish, so that we acquire precisely eight singular points (ordinary double points) of $\mathcal {C}_i$ at the eight distinct roots of (2.7).

The well-known Plücker formula for the genus of a curve now gives

$$\begin{aligned} {\text{ genus }}(\mathcal {C}_i) = (8-1)(8-2)/2 - 4 - 3 - 3 - 8 = 3. \end{aligned}$$

Remark: the above lemma is in accord with Magma, which tells us that the genus of $\mathcal {C}_i$ considered as a curve over the function field $\mathbb {Q}(A,B)$ has genus 3.

It follows that the curves $\mathcal {C}_i$, $i=1 , \ldots ,k$, can possess only finitely many rational points $(f',t')$. Thus n may be chosen so that $t=\frac{7}{2} a_n$ avoids the finite set of $t'$, and gives rise to a $(k+1)$th cubic field thereby completing the induction. $\square$

3 Cubic points on the Fermat quartic

Klassen and Tzermias [4], inter alia, have shown that there are no cubic points on the Fermat quintic

$$\begin{aligned} \mathcal {F}_5: x^5+y^5=z^5. \end{aligned}$$

However, we are not aware of theorems in the literature related to cubic points on the Fermat quartic, possibly because of the obvious existence of such by taking the residual intersection of a straight line through one of the trivial rational points. We show here that there are no non-trivial points on $\mathcal {F}_4$ in any cyclic cubic number field. The argument is elementary, and will show that any non-trivial point in $\mathcal {F}_4(K)$, where K is a cubic field, lies on a rational straight line containing one of the four rational points $(\pm 1,0,1)$, $(0,\pm 1,1)$ in $\mathcal {F}_4(\mathbb {Q})$. It follows, without loss of generality, that the coordinates satisfy $z=x+t y$, say, $t \in \mathbb {Q}$. This line meets $\mathcal {F}_4$ at (1, 0, 1) and the cubic point $(\theta ,1,\theta +t)$, where

$$\begin{aligned} 4 t \theta ^3 +6t^2 \theta ^2 +4 t^3 \theta + (t^4-1) = 0. \end{aligned}$$

But the discriminant of this cubic field equals $-16 t^2 (t^8+27)<0$, so cannot be a perfect square; thus, $\mathbb {Q}(\theta )$ cannot be a cyclic extension.

Our argument is modelled on Bremner [2], which is itself based on Cassels [3]. In this latter paper, Cassels considered quartic curves of type $\mathcal {C}:F(x^2,y^2,z^2)=0$ where F is a homogeneous rational quadratic form. Suppose that P is a point on $\mathcal {C}$ whose coordinates generate a cubic extension K of the rationals. For each $i=1,2,3$, it is shown that P induces rational points $P_i$ respectively on the quartic curves $\mathcal {D}_i$ of genus 1 defined by

$$\begin{aligned} \mathcal {D}_1:\; F(X,y^2,z^2)=0, \qquad \mathcal {D}_2:\; F(x^2,Y,z^2)=0, \qquad \mathcal {D}_3:\; F(x^2,y^2,Z)=0. \end{aligned}$$

Further, for any triple of such rational points $\{P_1,P_2,P_3\}$, a construction is then given which leads to three homogeneous equations in the coefficients A, B, C, D of a polynomial $g(x)=A x^3+B x^2 + C x+D$ defining the cubic field. Integer solutions A, B, C, D of this system for which g(x) is irreducible thus determine the cubic fields that can arise from the given triple. The construction is effective, and used in Bremner [2] to study the curves $x^4+y^4=D z^4$, $D \in \mathbb {Z}$ (where the extra symmetry gave additional information). Here, the curves $\mathcal {D}_i$, $i=1,2,3$, corresponding to the Fermat quartic $\mathcal {F}_4$ are all elliptic quartics of rational rank 0, so that there are only finitely many triples $\{P_1,P_2,P_3\}$ to consider.

Theorem 3.1

Let K be a cubic number field. If (x, y, z) is a non-trivial point in $\mathcal {F}_4(K)$, then x, y, z are collinear, and in particular K is a field with Galois group $S_3$.

Proof

Suppose that $P=(x_1,y_1,z_1) \in \mathcal {F}_4(K)$, where K is a cubic field generated by $x_1$, $y_1$, $z_1$. Denote by T the triple of points $\{x_i,y_i,z_i\}$, $i=1,2,3$, the three conjugates of P. From the Pythagorean parametrization, there exist $l,m \in K$ such that

$$\begin{aligned} x_1^2 : y_1^2 : z_1^2 = l^2-m^2 : 2 l m : l^2+m^2, \qquad l:m = x_1^2+z_1^2 : y_1^2. \end{aligned}$$

There exist $A,B,C,D \in \mathbb {Z}$ such that

$$\begin{aligned} G(l,m) = A l^3+B l^2 m +C l m^2 +D m^3 = 0, \end{aligned}$$

and when G(l, m) is irreducible, a root determines the cubic field K.

Step I: We work with the elliptic quartic curve $\mathcal {D}_1:\;X^2+y^4=z^4$.

For fixed i, the four quantities $x_i^2$, $y_i^2$, $z_i^2$, $y_iz_i$, are linearly dependent in the cubic field K, so that there exist $r,s,t \in \mathbb {Q}$ such that

$$\begin{aligned} x_i^2 = r y_i^2 +s z_i^2 +t y_i z_i, \quad i=1,2,3. \end{aligned}$$

Therefore the conic

$$\begin{aligned} X = r y^2+s z^2+t y z \end{aligned}$$

meets $\mathcal {D}_1$ in the three points $(x_i^2,y_i,z_i)$ and in $\nu _1(T)$, say, a rational point on $\mathcal {D}_1$, which is perforce equal to $(\pm 1, 0, 1)$, $(0,\pm 1,1)$.

• Suppose $\nu _1(T)=(1,0,1)$. Thus $s=1$. We have

  $$\begin{aligned} (X-r y^2-z^2)^2 - t^2 y^2 z^2 = 0, \end{aligned}$$

  so that the homogeneous quartic in l, m

  $$\begin{aligned} \left( l^2-m^2 - 2 l m r - (l^2+m^2) \right) ^2 - 2 l m (l^2+m^2) t^2 \end{aligned}$$

  has $m=0$ as a root, corresponding to $\nu _1(T)$, and G(l, m) as a residual cubic factor. Thus there exists $q \in \mathbb {Q}$ such that

  $$\begin{aligned} \left( l^2-m^2-2 l m r-(l^2+m^2) \right) ^2 - 2 l m (l^2+m^2) t^2 = 2 q m G(l,m), \end{aligned}$$

  equivalently,

  $$\begin{aligned} -t^2 l^3+2 r^2 l^2 m +(4r-t^2)l m^2+2 m^3 = q (A l^3+B l^2 m+C l m^2+D m^3). \end{aligned}$$

  Eliminating q, r, t gives

  $$\begin{aligned} A^2-2 A C+C^2-4 B D = 0. \end{aligned}$$

  (3.1)

  (Remark: it should be noted that, before elimination, t only occurs to the second power, so it is possible that values of (A, B, C, D) satisfying (3.1) pull back to a value of t satisfying only $t^2 \in \mathbb {Q}$).

• Suppose $\nu _1(T)=(-1,0,1)$. Then $s=-1$, and arguing as above, there exists $q \in \mathbb {Q}$ with

  $$\begin{aligned} \left( l^2-m^2-2 l m r+(l^2+m^2) \right) ^2 - 2 l m (l^2+m^2) t^2 = 2 q l G(l,m). \end{aligned}$$

  Equivalently,

  $$\begin{aligned} 2 l^3 +(-4r-t^2) l^2 m+2 r^2 l m^2 - t^2 m^3 = q (A l^3+B l^2 m+C l m^2+D m^3), \end{aligned}$$

  and eliminating q, r, t gives

  $$\begin{aligned} B^2-2 B D+D^2 - 4 A C = 0. \end{aligned}$$

  (3.2)

• Suppose $\nu _1(T)=(0,\pm 1,1)$, thus $r+s \pm t=0$. We have

  $$\begin{aligned} (X-r y^2-s z^2)^2 - (r+s)^2 y^2 z^2 = 0, \end{aligned}$$

  and, as above, there exists $q \in \mathbb {Q}$ with

  $$\begin{aligned} \left( l^2-m^2 - 2 l m r - (l^2+m^2) s \right) ^2 - 2 l m (l^2+m^2) (r+s)^2 = q(l-m) G(l,m), \end{aligned}$$

  from which

  $$\begin{aligned}&(1-s)^2 l^3 + (1-4r-2r^2-2s-s^2) l^2 m + (-1-4r+2r^2-2s+s^2) l m^2&\\&\quad -\, (1+s)^2 m^3 = q (A l^3 + B l^2 m + C l m^2 + D m^3). \end{aligned}$$

  Eliminating q, r, s gives

  $$\begin{aligned}&A^4 + 4A^3B + 6A^2B^2 + 4AB^3 + B^4 - 12A^3C - 20A^2BC - 4AB^2C + 4B^3C \nonumber \\&+\, 38A^2C^2 - 20ABC^2 + 6B^2C^2 - 12AC^3 + 4BC^3 + C^4 + 100A^3D - 68A^2BD \nonumber \\&+\, 76AB^2D - 12B^3D - 84A^2CD + 24ABCD - 20B^2CD + 76AC^2D - 4BC^2D \nonumber \\&+\, 4C^3D + 198A^2D^2 - 84ABD^2 + 38B^2D^2 - 68ACD^2 - 20BCD^2 + 6C^2D^2 \nonumber \\&+\, 100AD^3 - 12BD^3 + 4CD^3 + D^4 = 0. \end{aligned}$$

  (3.3)

Step II: Now we work with the elliptic quartic curve $\mathcal {D}_2 : x^4+Y^2=z^4$.

There exist $r,s,t \in \mathbb {Q}$ such that

$$\begin{aligned} y_i^2 = r x_i^2 + s z_i^2 + t x_i z_i, \quad i=1,2,3, \end{aligned}$$

so that the conic

$$\begin{aligned} Y = r x^2+s z^2+t x z \end{aligned}$$

meets $\mathcal {D}_2$ in the three points $(x_i,y_i^2,z_i)$ and in $\nu _2(T)$, a rational point on $\mathcal {D}_2$, which is therefore equal to $(0,\pm 1,1)$, $(\pm 1, 0, 1)$.

• Suppose $\nu _2(T)=(0,1,1)$, so that $s=1$. We have

  $$\begin{aligned} (Y-r x^2-z^2)^2 - t^2 x^2 z^2 = 0, \end{aligned}$$

  so that

  $$\begin{aligned} \left( 2 l m - (l^2-m^2) r - (l^2+m^2) \right) ^2 - (l^2-m^2)(l^2+m^2) t^2 \end{aligned}$$

  has $l=m$ as a root, corresponding to $\nu _2(T)$, and G(l, m) as the residual cubic. Thus there exists $q \in \mathbb {Q}$ such that

  $$\begin{aligned} \left( 2 l m - (l^2-m^2) r - (l^2+m^2) \right) ^2 - (l^2-m^2)(l^2+m^2) t^2 = q(l-m) G(l,m) \end{aligned}$$

  from which

  $$\begin{aligned}&((1+r)^2-t^2) l^3 + (-3-2r+r^2-t^2) l^2 m + (3-2r-r^2-t^2) l m^2 \\&+ (-1+2r-r^2-t^2) m^3 = q (A l^3+B l^2 m+C l m^2 +D m^3). \end{aligned}$$

  Eliminating q, r, t gives

  $$\begin{aligned} A^2 - B^2 + 2 A C - C^2 - 4 A D + 2 B D + D^2 = 0. \end{aligned}$$

  (3.4)

• Suppose $\nu _2(T)=(0,-1,1)$. Then $s=-1$ leading to

  $$\begin{aligned} ((r-1)^2-t^2) l^3&+ (3-2r-r^2+t^2) l^2 m + (3+2r-r^2-t^2) l m^2 \\&+ (1+2r+r^2+t^2) m^3 = q (A l^3+B l^2 m+C l m^2+D m^3). \end{aligned}$$

  Eliminating q, r, t gives

  $$\begin{aligned} A^2 - B^2 + 2 A C - C^2 + 4 A D + 2 B D + D^2 = 0. \end{aligned}$$

  (3.5)

• Suppose $\nu _2(T)=(\pm 1,0,1)$. Then $r+s \pm t=0$ leading to

  $$\begin{aligned} 2(r + s) l^3 + (-2+r^2-s^2) l^2 m - 2(r - s) l m^2 - (r^2+s^2) m^3 = q G(l,m). \end{aligned}$$

  Eliminating q, r, s gives

  $$\begin{aligned} A^4 + 2 A^2 C^2 + C^4 - 4 A^2 B D - 4 B C^2 D + 8 A C D^2 = 0. \end{aligned}$$

  (3.6)

Step III: Finally, we work with the elliptic quartic curve $\mathcal {D}_3 : x^4+y^4=Z^2$.

There exist $r,s,t \in \mathbb {Q}$ such that

$$\begin{aligned} z_i^2 = r x_i^2 + s y_i^2 + t x_i y_i, \end{aligned}$$

so that the conic

$$\begin{aligned} Z = r x^2+s y^2+t x y \end{aligned}$$

meets $\mathcal {D}_3$ in the three points $(x_i,y_i,z_i^2)$ and in $\nu _3(T)$, a rational point on $\mathcal {D}_3$, which can only be $(0,\pm 1,1)$, $(\pm 1, 0, 1)$.

• Suppose $\nu _3(T)=(0,\pm 1,1)$. Thus $s=1$. We have

  $$\begin{aligned} (Z-r x^2-y^2)^2 - t^2 x^2 y^2 = 0, \end{aligned}$$

  and arguing as above, there exists $q \in \mathbb {Q}$ such that

  $$\begin{aligned} \left( l^2+m^2 - (l^2-m^2) r - 2 l m \right) ^2 - 2 l m (l^2-m^2) t^2 = q (l-m) G(l,m), \end{aligned}$$

  or, equivalently,

  $$\begin{aligned}&(r-1)^2 l^3 + (-3+2r+r^2-2t^2) l^2 m + (3+2r-r^2-2t^2) l m^2 - (r+1)^2 m^3 \\&\quad = q G(l,m). \end{aligned}$$

  Eliminating q, r, t gives

  $$\begin{aligned} A^2 + 2 A B + B^2 - 2 A C - 2 B C + C^2 + (14 A - 2 B + 2 C) D + D^2 = 0. \end{aligned}$$

  (3.7)

• Suppose $\nu _3(T)=(\pm 1,0,1)$. Then $r=1$, leading to

  $$\begin{aligned} -t^2 l^3 +2 s^2 l^2 m +(-4s+t^2) l m^2 +2 m^3 = q G(l,m), \end{aligned}$$

  and on eliminating q, s, t,

  $$\begin{aligned} A^2 + 2 A C + C^2 - 4 B D = 0. \end{aligned}$$

  (3.8)

We summarize the above three steps as follows. If there exists a point in $\mathcal {F}_4(K)$ for a cubic field K, then necessarily there exist integers A, B, C, D that satisfy one of the conditions (3.1), (3.2), (3.3); one of the conditions (3.4), (3.5), (3.6); and one of the conditions (3.7), (3.8). This gives eighteen possible sets of three homogeneous equations to be satisfied by A, B, C, D. It is straightforward to check in turn each of these systems for solutions, noting that we can discard solutions (A, B, C, D) for which G(l, m) is reducible (so, in particular, when $A D = 0$). Only the following four systems give solutions.

System 1: Equations (3.3), (3.4), (3.7). The system reduces to either

$$\begin{aligned} A = (C+D)(C-3D)/(4D), \qquad B = -(C+D)^2/(4D), \end{aligned}$$

in which case G(l, m) is reducible, or

$$\begin{aligned} 3A-B-C+3D=0, \quad 4 B^2-4 B C+C^2+18 C D-27 D^2=0. \end{aligned}$$

This is parametrized by

$$\begin{aligned} A:B:C:D = 1 : 1-2U-2U^2 : 2+2U-U^2 : -U^2. \end{aligned}$$

The pullback gives

$$\begin{aligned} (1-U^2)^2 X^4 + Y^4 = Z^4, \end{aligned}$$

where

$$\begin{aligned} (X,Y,Z) = \left( \frac{(l-m)(-m+l U)}{1-U^2}, \frac{(l+m)(l+m U)}{1+U}, l^2+m^2 \right) . \end{aligned}$$

Accordingly, we get a point on $\mathcal {F}_4$ precisely when $1-U^2=\pm \square$.

(i) Put $U=(1-V^2)/(1+V^2)$, $V \in \mathbb {Q}$, to give

$$\begin{aligned} A:B:C:D = (1+V^2)^2 : -3+6V^2+V^4 : 3+6V^2-V^4 : -(1-V^2)^2. \end{aligned}$$

Setting $\theta =l/m$, we obtain

$$\begin{aligned} (X,Y,Z) = \left( \frac{(1-\theta )(1-\theta +(1+\theta )V^2)}{2V}, \frac{(1+\theta )(1+\theta +(-1+\theta )V^2)}{2}, 1+\theta ^2 \right) \end{aligned}$$

with

$$\begin{aligned} X^4 + Y^4 = Z^4 \end{aligned}$$

in the field $\mathbb {Q}(\theta )$ where

$$\begin{aligned} -(1-V^2)^2 + (3+6V^2-V^4) \theta + (-3+6V^2+V^4) \theta ^2 + (1+V^2)^2 \theta ^3 = 0. \end{aligned}$$

Note that (X,Y, Z) lies on the line $Z=V X+Y$ through the point (0, 1, 1).

(ii) Putting $U=(1+V^2)/(1-V^2)$ gives

$$\begin{aligned} A: B : C : D = (1-V^2)^2 : -3-6V^2+V^4 : 3-6V^2-V^4 : -(1+V^2)^2, \end{aligned}$$

and we have simply recovered the reciprocal polynomial to that of (i).

System 2: Equations (3.3), (3.5), (3.8). The system has the solution $(A,B,C,D)=(1,-1,-7,-9)$. With $K=\mathbb {Q}(u)$, $u^3-u^2-7u-9=0$, this corresponds to

$$\begin{aligned} \left( 2\sqrt{2}(1+u) \right) ^4 + (u^2-3)^4 = (u^2+1)^4, \end{aligned}$$

so does not give a point of $\mathcal {F}_4(K)$.

System 3: Equations (3.1), (3.6), (3.8). The system reduces to

$$\begin{aligned} C=0, \qquad A^2 = 4 B D, \end{aligned}$$

with parametrization

$$\begin{aligned} A:B:C:D = 2a : a^2 : 0 : 1. \end{aligned}$$

The pullback corresponds to

$$\begin{aligned} (l^2+a l m)^4 +\frac{1}{a^2} (-a l m+m^2)^4 = (l^2+m^2)^4, \end{aligned}$$

so gives a point in $\mathcal {F}_4(K)$ precisely when $a=\pm \square$. By replacing l / m by $-l/m$ if necessary, we need only consider the case $a=b^2$. This gives the following point of $\mathcal {F}_4(K)$:

$$\begin{aligned} (X,Y,Z)=( b^2 \theta +\theta ^2, \; \frac{1}{b}-b \theta , \; 1+\theta ^2), \end{aligned}$$

in the field $K=\mathbb {Q}(\theta )$, $2b^2 \theta ^3 + b^4 \theta ^2 + 1=0$. Note that (X, Y, Z) lies on the line $Z=X+b Y$ through the point (1, 0, 1).

System 4: Equations (3.2), (3.6), (3.7). The system has the solution $(A,B,C,D) = (4,-9,4,-1)$. With $K=\mathbb {Q}(u)$, $4u^3-9u^2+4u-1=0$, this corresponds to

$$\begin{aligned} (1-u^2)^4 + \big ( (1-u)(1-2u) \sqrt{2} \big )^4 = (1-4u+u^2)^4, \end{aligned}$$

so does not give a point of $\mathcal {F}_4(K)$.

In conclusion, a cubic point has collinear coordinates, and the theorem follows from the remarks at the beginning of this section. $\square$

4 Remarks on the Fermat quartic

It is of interest to ask an analogous question about the Fermat quartic

$$\begin{aligned} \mathcal {F}_4: x^4 + y^4 = z^4 \end{aligned}$$

over cyclic quartic extensions. Clearly $\mathcal {F}_4$ contains points in quartic number fields K with Galois group isomorphic to $\mathbb {Z}/2\mathbb {Z}\times \mathbb {Z}/2\mathbb {Z}$, for example $K=\mathbb {Q}(\sqrt{3},\sqrt{5})$ with

$$\begin{aligned} \big ( \sqrt{3} \big )^4 + 2^4 = \big ( \sqrt{5} \big )^4. \end{aligned}$$

Similarly, there are quartic fields K with Galois group $D_8$ in which $\mathcal {F}_4$ has non-trivial points. For example, $K=\mathbb {Q}(\theta )$, $\theta ^4=17$, with

$$\begin{aligned} 1^4 + 2^4 = \theta ^4. \end{aligned}$$

Finally, there are quartic fields K with Galois group $S_4$ in which $\mathcal {F}_4$ has non-trivial points. For example, $K=\mathbb {Q}(\phi )$, $\phi ^4-16\phi ^3+24\phi ^2-40\phi +16=0$, with

$$\begin{aligned} (2-\phi )^4+(1-\phi )^4 = (1+\phi )^4. \end{aligned}$$

Moreover, it is straightforward to construct infinitely many such fields, with Galois groups of type $\mathbb {Z}/2\mathbb {Z}\times \mathbb {Z}/2\mathbb {Z}$, $D_8$ or $S_4$.

But do there exist quartic fields K with Galois group of type either $\mathbb {Z}/4\mathbb {Z}$ or $A_4$, in which $\mathcal {F}_4$ contains non-trivial points? We have been unable to find any such examples.

A non-trivial point on $\mathcal {F}_4$ over K necessarily implies that the K-rank of the two elliptic quartics

$$\begin{aligned} E_1:\; X^4+Y^2=1, \qquad E_2:\; X^4+1=Z^2 \end{aligned}$$

is positive. Even examples satisfying this weaker requirement seem hard to find. For the case of cyclic Galois group, we know of two such non-isomorphic fields K:

• 1. $K=\mathbb {Q}(\theta )$, $\theta ^4-30\theta ^2+180=0$. Here, $E_1(K)$ has rank 1, and $E_2(K)$ has rank 2.

• 2. $K=\mathbb {Q}(\theta )$, $\theta ^4-65\theta ^2+260=0$, with $E_1(K)$ and $E_2(K)$ both of rank 2.

For the case of alternating Galois group, we know of the following five examples.

• 1. $K=\mathbb {Q}(\theta )$, $\theta ^4+3\theta ^2+7\theta +4=0$; ranks 2, 2;

• 2. $K=\mathbb {Q}(\theta )$, $\theta ^4-10\theta ^2+4\theta +6=0$; ranks 1, 1;

• 3. $K=\mathbb {Q}(\theta )$, $\theta ^4-12\theta ^2+4\theta +19=0$; ranks 1, 1;

• 4. $K=\mathbb {Q}(\theta )$, $\theta ^4+3\theta ^2+17\theta +15=0$; ranks 2, 2;

• 5. $K=\mathbb {Q}(\theta )$, $\theta ^4-21\theta ^2+25\theta +4=0$; ranks 2, 2.

In none of these instances does it appear, however, that $\mathcal {F}_4(K)$ contains non-trivial points.