Annales Mathématiques Blaise Pascal

Alain Kraus

ANNALES MATHÉMATIQUES
BLAISE PASCAL
Alain Kraus
Quartic points on the Fermat quintic
Volume 25, no 1 (2018), p. 199-205.
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Annales mathématiques Blaise Pascal 25, 199-205 (2018)
Quartic points on the Fermat quintic
Alain Kraus
Abstract
We study the algebraic points of degree 4 overQ on the Fermat curveF5/Q of equation x5+y5+z5 = 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 Q of degree 4
that arises as the field of definition of a non-trivial point of F5.
Points quartiques sur la quintique de Fermat
Résumé
Nous étudions les points algébriques de degré 4 sur Q de la courbe de Fermat F5/Q d’équation
x5 + y5 + z5 = 0. Klassen et Tzermias ont donné en 1997 une description géométrique de ces points. En
utilisant leur résultat et le travail de Bruin portant sur les équations diophantiennes de signature (5, 5, 2),
nous donnons une description algébrique de ces points. Nous prouvons en particulier qu’il existe une
unique extension galoisienne de Q de degré 4 qui apparaît comme le corps de définition d’un point non
trivial de F5.
1. Introduction
Let us denote by F5 the quintic Fermat curve over Q given by the equation
x5 + y5 + z5 = 0.
Let P be a point in F5(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, ζ2, 1), w = (ζ2, ζ, 1).
It is well known that F5(Q) = {a, b, c}. In 1978, Gross and Rohrlich have proved that
the only quadratic points of F5 are w and w [2, Theorem 5.1]. In 1997, by proving that
the group of Q-rational points of the Jacobian of F5 is isomorphic to (Z/5Z)2, and by
expliciting generators, Klassen and Tzermias have described geometrically all the points
of F5 whose degrees are less than 6 in [4, Theorem 1]. I mention that Top and Sall have
I thank D. Bernardi for his remarks during the writing of this paper, as well as the referee for his suggestions.
Keywords: Fermat quintic, number fields, rational points.
2010 Mathematics Subject Classification: 11D41, 11G30.
199
A. Kraus
pushed further this description for points of F5 of degrees less than 12 in [5]. In particular,
Klassen and Tzermias have proved that F5 has no cubic points and they have established
the following statement:
Theorem 1.1. The points of degree 4 of F5 arise as the intersection of F5 with a rational
line passing through exactly one of points a, b, c.
Using this result, and Bruin’s work about the diophantine equations 16x5 + y5 = z2
and 4x5 + y5 = z2 [1, 3], we propose in this paper to give an algebraic description of the
non-trivial quartic points of F5.
2. Statement of the results
Let K be a number field of degree 4 over Q.
Theorem 2.1. Suppose that F5(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.
(2) One has
K = Q(α) with 31α4 − 36α3 + 26α2 − 36α + 31 = 0. (2.1)
The extension K/Q is cyclic. Up to Galois conjugation and permutation,
(2, 2α,−α − 1) is the only non-trivial point in F5(K).
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 F5(K) is contained in {w,w}.
All that follows is devoted to the proof of Theorem 2.1.
3. Preliminary results
Let P = (x, y, z) ∈ F5(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
z = 1. (3.1)
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Quartic points on the Fermat quintic
Lemma 3.1. One has K = Q(y). There exists t ∈ Q, t , −1, such that
y4 + uy3 + (u + 2)y2 + uy + 1 = 0 with u =
4t5 − 1
t5 + 1
, (3.2)
x = t(y + 1). (3.3)
Proof. The equation of the tangent line to F5 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
x = t(y + 1).
In particular, one has K = Q(y). Furthermore, one has
t , −1. (3.4)
Indeed, if t = −1, the equalities x + y + 1 = 0 and x5 + y5 + 1 = 0 imply
x(x + 1)(x2 + x + 1) = 0.
Since P is non-trivial, one has x(x + 1) , 0, so x2 + 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 x5 + y5 + 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.
Lemma 3.2.
(1) One has |G | ∈ {4, 8}.
(2) Suppose that |G | = 4. One of the two following conditions is satisfied:
5(16t5 + 1) is a square in Q. (3.5)
(1 − 4t5)(16t5 + 1) is a square in Q. (3.6)
Proof. Let us denote
f = X4 + uX3 + (u + 2)X2 + uX + 1
in Q[X]. One has f (y) = 0 (Lemma 3.1). Let ε ∈ Q such that
ε2 = u2 − 4u.
The element y + 1y is a root of the polynomial X
2 + uX + u. So we have the inclusion
Q(ε) ⊆ K . (3.7)
201
A. Kraus
Moreover, we have the equality
f =
(
X2 +
u − ε
2
X + 1
) (
X2 +
u + ε
2
X + 1
)
. (3.8)
Since K = Q(y) and [K : Q] = 4, we have
[Q(ε) : Q] = 2. (3.9)
From (3.8), we deduce that the roots of f belong to at most two quadratic extensions of
Q(ε). The equality (3.9) then implies |G | ≤ 8. Since 4 divides |G |, this proves the first
assertion.
Henceforth let us suppose |G | = 4, i.e. the extension K/Q is Galois. Let ∆ be the
discriminant of f . One has the equalities
∆ = −u2(u − 4)3(3u + 4) = 53
(4t5 − 1)2(16t5 + 1)
(t5 + 1)6
. (3.10)
Let us prove that
∆ is a square in Q(ε). (3.11)
From (3.8) and our assumption, the roots of the polynomials
X2 +
u − ε
2
X + 1 and X2 +
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((u − ε
2
)2
− 4
) ((u + ε
2
)2
− 4
)
i.e. − (u − 4)(3u + 4)
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
Q
(√
∆
)
= Q(ε).
Therefore, ∆(u2 − 4u) is a square in Q, in other words, such is the case for −u(3u + 4).
One has the equality
−u(3u + 4) =
(1 − 4t5)(16t5 + 1)
(t5 + 1)2
.
This implies the condition (3.6) and proves the Lemma. 
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Quartic points on the Fermat quintic
4. The curve C1/Q
Let us denote by C1/Q the curve, of genus 2, given by the equation
Y2 = 5(16X5 + 1).
Proposition 4.1. The set C1(Q) is empty.
Proof. Suppose there exists a point (X,Y ) ∈ C1(Q). Let Z = Y5 . We obtain
5Z2 = 16X5 + 1. (4.1)
Let a and b be coprime integers, with b ∈ N, such that
X =
a
b
.
Let us prove there exists c ∈ N such that
b = 5c2. (4.2)
For every prime number p, let vp be the p-adic valuation over Q. If p is a prime number
dividing b, distinct from 2, 5, one has
2vp(Z) = −5vp(b),
consequently
vp(b) ≡ 0 mod 2. (4.3)
Moreover, one has v2(X) < 0 (5 is not a square modulo 8), so
4 − 5v2(b) = 2v2(Z).
In particular, one has
v2(b) ≡ 0 mod 2. (4.4)
Let us verify the congruence
v5(b) ≡ 1 mod 2. (4.5)
One has v5(X) ≤ 0. Suppose v5(X) = 0. In this case, one has X5 ≡ ±1,±7 mod 25.
The equality (4.1) implies X5 ≡ −1 mod 25 and Z2 ≡ 2 mod 5, which leads to a
contradiction. Therefore, we have 1 + 2v5(Z) = −5v5(b), which proves (4.5).
The conditions (4.3), (4.4) and (4.5) then imply (4.2).
We deduce from (4.1) and (4.2) the equality
16a5 + b5 = d2 with d = 53c5Z .
One has ab , 0. From the informations given in the Appendix of [3], this implies
(a, b, d) = (−1, 2,±4).
We obtain X = −1/2, which is not the abscissa of a point of C1(Q), hence the result. 
203
A. Kraus
5. The curve C2/Q
Let us denote by C2/Q the curve, of genus 4, given by the equation
Y2 = (1 − 4X5)(16X5 + 1).
Proposition 5.1. One has
C2(Q) = {(0,±1), (−1/2,±3/4)}.
Proof. Let (X,Y ) be a point of C2(Q). Let a and b be coprime integers such that
X =
a
b
.
We obtain the equality
(Y b5)2 = (b5 − 4a5)(16a5 + b5). (5.1)
Therefore, (b5 − 4a5)(16a5 + b5) is the square of an integer. Moreover, b5 − 4a5 and
16a5 + b5 are coprime apart from 2 and 5. So, changing (a, b) by (−a,−b) if necessary,
there exists d ∈ N such that
b5 − 4a5 ∈ {d2, 2d2, 5d2, 10d2}.
Suppose b5 − 4a5 ∈ {2d2, 10d2}. In this case, b must be even, therefore v2(2d2) = 2,
which is not.
Suppose b5 − 4a5 = d2. One has b , 0. It then comes from [3] that
a = 0 or (a, b, d) = (−1, 2,±6).
We obtain X = 0 or X = −1/2, which leads to the announced points in the statement.
Suppose b5 − 4a5 = 5d2. It follows from (5.1) that there exists c ∈ N such that
16a5 + b5 = 5c2. Since a and b are coprime, 5 does not divide ab. We then directly verify
that the two equalities b5 − 4a5 = 5d2 and 16a5 + b5 = 5c2 do not have simultaneously
any solutions modulo 25, hence the result. 
6. End of the proof of Theorem 2.1
The group G is isomorphic to a subgroup of the symmetric group S4 and one has |G | = 4
or |G | = 8 (Lemma 3.2). In case |G | = 8, G is isomorphic to a 2-Sylow subgroup of S4,
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 F5(K).
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Quartic points on the Fermat quintic
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 31X4 − 36X3 + 26X2 − 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.
References
[1] Nils Bruin. Chabauty methods and covering techniques applied to generalised Fermat
equations. PhD thesis, Leiden University, Niederland, 1999.
[2] Benedict H. Gross and David E. Rohrlich. Some results on the Mordell-Weil group
of the Jacobian of the Fermat curve. Invent. Math., 44(3):201–224, 1978.
[3] Wilfrid Ivorra. Sur les équations xp + 2βyp = z2 et xp + 2βyp = 2z2. Acta Arith.,
108(4):327–338, 2003.
[4] Matthew Klassen and Pavlos Tzermias. Algebraic points of low degree on the Fermat
quintic. Acta Arith., 82(4):393–401, 1997.
[5] Thiéyacine Top and Oumar Sall. Points algébriques de degrés au plus 12 sur la
quintique de Fermat. Acta Arith., 169(4):385–395, 2015.
Alain Kraus
Université de Paris VI
Institut de Mathématiques de Jussieu
4 place Jussieu
75005 Paris, France
alain.kraus@imj-prg.fr
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