Quartic points on the Fermat quintic

In this paper, we study the algebraic points of degree $4$ over $\mathbb{Q}$ on the Fermat curve $F_5/\mathbb{Q}$ of equation $x^5+y^5+z^5=0$. A geometrical description of these points has been given in 1997 by Klassen and Tzermias. Using their result, as well as Bruin's work about diophantine equations of signature $(5,5,2)$, we give here an algebraic description of these points. In particular, we prove there is only one Galois extension of $\mathbb{Q}$ of degree $4$ that arises as the field of definition of a non-trivial point of $F_5$.


Introduction
Let us denote by F 5 the quintic Fermat curve over Q given by the equation Let P be a point in F 5 (Q). The degree of P is the degree of its field of definition over Q. Write P = (x, y, z) for the projective coordinates of P. It is said to be non-trivial if xyz 0. Let ζ be a primitive cubic root of unity and a = (0, −1, 1), b = (−1, 0, 1), c = (−1, 1, 0) w = (ζ, ζ 2 , 1), w = (ζ 2 , ζ, 1).
It is well known that F 5 (Q) = {a, b, c}. In 1978, Gross and Rohrlich have proved that the only quadratic points of F 5 are w and w [2, Theorem 5.1]. In 1997, by proving that the group of Q-rational points of the Jacobian of F 5 is isomorphic to (Z/5Z) 2 , and by expliciting generators, Klassen and Tzermias have described geometrically all the points of F 5 whose degrees are less than 6 in [4,Theorem 1]. I mention that Top and Sall have pushed further this description for points of F 5 of degrees less than 12 in [5]. In particular, Klassen and Tzermias have proved that F 5 has no cubic points and they have established the following statement: Theorem 1.1. The points of degree 4 of F 5 arise as the intersection of F 5 with a rational line passing through exactly one of points a, b, c.
Using this result, and Bruin's work about the diophantine equations 16x 5 + y 5 = z 2 and 4x 5 + y 5 = z 2 [1,3], we propose in this paper to give an algebraic description of the non-trivial quartic points of F 5 .

Statement of the results
Let K be a number field of degree 4 over Q.
Theorem 2.1. Suppose that F 5 (K) has a non-trivial point of degree 4. One of the following conditions is satisfied: (1) the Galois closure of K is a dihedral extension of Q of degree 8.
As a direct consequence of [2, Theorem 5.1] and the previous Theorem, we obtain: Corollary 2.2. Suppose that K does not satisfy one of the two conditions above. The set of non-trivial points of All that follows is devoted to the proof of Theorem 2.1.

Preliminary results
Let P = (x, y, z) ∈ F 5 (K) be a non-trivial point of degree 4. By permuting x, y, z if necessary, we can suppose that P belongs to a Q-rational line L passing through a = (0, −1, 1) (Theorem 1.1). Moreover, P being non-trivial we shall assume Lemma 3.1. One has K = Q(y). There exists t ∈ Q, t −1, such that Proof. The equation of the tangent line to F 5 at the point a is Y + Z = 0. Since x 0, it is distinct from L. According to (3.1), it follows there exists t ∈ Q such that In particular, one has K = Q(y). Furthermore, one has Indeed, if t = −1, the equalities x + y + 1 = 0 and x 5 + y 5 + 1 = 0 imply Since P is non-trivial, one has x(x + 1) 0, so x 2 + x + 1 = 0. This leads to P = w or P = w, which contradicts the fact that P is not a quadratic point, and proves (3.4). From the equalities (3.3) and x 5 + y 5 + 1 = 0, as well as the condition y −1, we then deduce the Lemma.
Let G be the Galois group of the Galois closure of K over Q. Let us denote by |G| the order of G.
(2) Suppose that |G| = 4. One of the two following conditions is satisfied: Proof. Let us denote . One has f (y) = 0 (Lemma 3.1). Let ε ∈ Q such that The element y + 1 y is a root of the polynomial X 2 + uX + u. So we have the inclusion Moreover, we have the equality Let us prove that ∆ is a square in Q(ε).
(3.11) From (3.8) and our assumption, the roots of the polynomials X 2 + u − ε 2 X + 1 and X 2 + u + ε 2 X + 1 belong to K, which is a quadratic extension of Q(ε) ((3.7) and (3.9)). Therefore, the product of their discriminants must be a square in Q(ε). The first equality of (3.10) then implies (3.11). Suppose that the condition (3.5) is not satisfied. From the second equality of (3.10), we deduce that ∆ in not a square in Q. It follows from (3.11) that we have Therefore, ∆(u 2 − 4u) is a square in Q, in other words, such is the case for −u(3u + 4). One has the equality This implies the condition (3.6) and proves the Lemma. 4. The curve C 1 /Q Let us denote by C 1 /Q the curve, of genus 2, given by the equation We obtain 5Z 2 = 16X 5 + 1. (4.1) Let a and b be coprime integers, with b ∈ N, such that Let us prove there exists c ∈ N such that b = 5c 2 .

(4.2)
For every prime number p, let v p be the p-adic valuation over Q. If p is a prime number dividing b, distinct from 2, 5, one has Moreover, one has v 2 (X) < 0 (5 is not a square modulo 8), so 4 − 5v 2 (b) = 2v 2 (Z).
We obtain X = −1/2, which is not the abscissa of a point of C 1 (Q), hence the result.
Let us denote by C 2 /Q the curve, of genus 4, given by the equation Proposition 5.1. One has Proof. Let (X, Y ) be a point of C 2 (Q). Let a and b be coprime integers such that We obtain the equality is the square of an integer. Moreover, b 5 − 4a 5 and 16a 5 + b 5 are coprime apart from 2 and 5. So, changing Suppose b 5 − 4a 5 ∈ {2d 2 , 10d 2 }. In this case, b must be even, therefore v 2 (2d 2 ) = 2, which is not.
We obtain X = 0 or X = −1/2, which leads to the announced points in the statement. Suppose b 5 − 4a 5 = 5d 2 . It follows from (5.1) that there exists c ∈ N such that 16a 5 + b 5 = 5c 2 . Since a and b are coprime, 5 does not divide ab. We then directly verify that the two equalities b 5 − 4a 5 = 5d 2 and 16a 5 + b 5 = 5c 2 do not have simultaneously any solutions modulo 25, hence the result.

End of the proof of Theorem 2.1
The group G is isomorphic to a subgroup of the symmetric group S 4 and one has |G| = 4 or |G| = 8 (Lemma 3.2). In case |G| = 8, G is isomorphic to a 2-Sylow subgroup of S 4 , that is dihedral.
Suppose |G| = 4 and let us prove the assertion 2 of the Theorem. First, we directly verify that the extension K/Q defined by the condition (2.1) is cyclic of degree 4, and that the point (2, 2α, −α − 1) belongs to F 5 (K).
Conversely, from the Proposition 4.1, the condition (3.5) of the Lemma 3.2 is not satisfied. The condition (3.6) and the Proposition 5.1 imply that t = 0 or t = −1/2. The case t = 0 is excluded because P is non-trivial. With the condition (3.2), we obtain u = − 36 31 .
Thus, necessarily y is a root of the polynomial 31X 4 − 36X 3 + 26X 2 − 36X + 31, in other words y is a conjugate over Q of α. The equality (3.3), x = − y + 1 2 then implies the result.