Quartic points on the Fermat quintic

TL;DR

This paper characterizes algebraic points of degree 4 on the Fermat quintic curve, providing a detailed algebraic description and identifying the unique Galois extension associated with such points.

Contribution

It offers a new algebraic description of degree 4 points on the Fermat quintic and establishes the uniqueness of the Galois extension involved.

Findings

Only one Galois extension of degree 4 arises from these points.

Provides a complete algebraic description of degree 4 points.

Builds on previous geometric and diophantine work to classify points.

Abstract

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 $F_5$.

Equations

$$x^5+y^5+z^5=0.$$

$$a=(0,-1,1),b=(-1,0,1),c=(-1,1,0),w=(\zeta ,{\zeta }^2,1),\overline{w}=({\zeta }^2,\zeta ,1).$$

$$K=\mathbb{Q}(\alpha )with31{\alpha }^4-36{\alpha }^3+26{\alpha }^2-36\alpha +31=0.$$

$$z=1.$$

$$y^4+u{y}^3+(u+2)y^2+uy+1=0withu=\frac{4t^5-1}{t^5+1},$$

$$x=t(y+1).$$

$$t\ne -1.$$

$$x(x+1)(x^2+x+1)=0.$$

$$f=X^4+u{X}^3+(u+2)X^2+uX+1.$$

$${\epsilon }^2=u^2-4u.$$

$$\mathbb{Q}(\epsilon )\subseteq K.$$

$$f=\left(X^2+\frac{u-\epsilon }{2}X+1\right)\left(X^2+\frac{u+\epsilon }{2}X+1\right).$$

$$[\mathbb{Q}(\epsilon ):\mathbb{Q}]=2.$$

$$\Delta =-u^2(u-4{)}^3(3u+4)=5^3\frac{(4t^5-1{)}^2(16t^5+1)}{(t^5+1{)}^6}.$$

$$\Delta is\; a\; square\; in\mathbb{Q}(\epsilon ).$$

$$X^2+\frac{u-\epsilon }{2}X+1and{X}^2+\frac{u+\epsilon }{2}X+1$$

$$\left({\left(\frac{u-\epsilon }{2}\right)}^2-4\right)\left({\left(\frac{u+\epsilon }{2}\right)}^2-4\right)i.e.-(u-4)(3u+4)$$

$$\mathbb{Q}\left(\sqrt{\Delta }\right)=\mathbb{Q}(\epsilon ).$$

$$-u(3u+4)=\frac{(1-4t^5)(16t^5+1)}{(t^5+1{)}^2}.$$

$$Y^2=5(16X^5+1).$$

$$5Z^2=16X^5+1.$$

$$X=\frac{a}{b}.$$

$$b=5c^2.$$

$$v_p(b)\equiv 0\mathrm{mod}2.$$

$$v_2(b)\equiv 0\mathrm{mod}2.$$

$$v_5(b)\equiv 1\mathrm{mod}2.$$

$$16a^5+b^5=d^{2}withd=5^3{c}^{5}Z.$$

$$(a,b,d)=(-1,2,\pm 4).$$

$$Y^2=(1-4X^5)(16X^5+1).$$

Full text

1. Introduction

Let us denote by $F_{5}$ the quintic Fermat curve over $\mathbb{Q}$ given by the equation

[TABLE]

Let $P$ be a point in $F_{5}(\overline{\mathbb{Q}})$. The degree of $P$ is the degree of its field of definition over $\mathbb{Q}$. Write $P=(x,y,z)$ for the projective coordinates of $P$. It is said to be non-trivial if $xyz\neq 0$. Let $\zeta$ be a primitive cubic root of unity and

[TABLE]

It is well known that $F_{5}(\mathbb{Q})=\left\{a,b,c\right\}$. In 1978, Gross and Rohrlich have proved that the only quadratic points of $F_{5}$ are $w$ and $\overline{w}$ [2, th. 5.1]. In 1997, by proving that the group of $\mathbb{Q}$-rational points of the Jacobian of $F_{5}$ is isomorphic to $(\mathbb{Z}/5\mathbb{Z})^{2}$, and by expliciting generators, Klassen and Tzermias have described geometrically all the points of $F_{5}$ whose degrees are less than $6$ [4, th. 1]. I mention that Top and Sall have pushed further this description for points of $F_{5}$ of degrees less than $12$ [5]. In particular, Klassen and Tzermias have established the following statement :

Theorem 1**.**

The points of degree $4$ of $F_{5}$ arise as the intersection of $F_{5}$ with a rational line passing through $a,b$ or $c$.

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 $F_{5}$.

2. Statement of the results

Let $K$ be a number field of degree $4$ over $\mathbb{Q}$.

Theorem 2**.**

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 $\mathbb{Q}$ of degree $8$.

2) One has

[TABLE]

The extension $K/\mathbb{Q}$ is cyclic. Up to Galois conjugation and permutation, $(2,2\alpha,-\alpha-1)$ is the only non-trivial point in $F_{5}(K)$.

As a direct consequence of [2, th. 5.1] and the previous theorem, we obtain :

Corollary 1**.**

Suppose that $K$ does not satisfy one of the two conditions above. The set of non-trivial points of $F_{5}(K)$ is contained in $\left\{w,\overline{w}\right\}$.

All that follows is devoted to the proof of theorem 2.

3. Preliminary results

Let $P=(x,y,z)\in F_{5}(K)$ be a non-trivial point of degree $4$. By permuting $x,y,z$ if necessary, we can suppose that $P$ belongs to a $\mathbb{Q}$-rational line $\mathcal{L}$ passing through $a=(0,-1,1)$ (th. 1). Moreover, $P$ being non-trivial we shall assume

[TABLE]

Lemma 1**.**

One has $K=\mathbb{Q}(y)$. There exists $t\in\mathbb{Q}$, $t\neq-1$, such that

[TABLE]

[TABLE]

Proof.

The equation of the tangent line to $F_{5}$ at the point $a$ is $Y+Z=0$. Since $x\neq 0$, it is distinct from $\mathcal{L}$. According to (3.1), it follows there exists $t\in\mathbb{Q}$ such that

[TABLE]

In particular, one has $K=\mathbb{Q}(y)$. Furthermore, one has

[TABLE]

Indeed, if $t=-1$, the equalities $x+y+1=0$ and $x^{5}+y^{5}+1=0$ imply

[TABLE]

Since $P$ is non-trivial, one has $x(x+1)\neq 0$, so $x^{2}+x+1=0$. This leads to $P=w$ or $P=\overline{w}$, 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 $y\neq-1$, we then deduce the lemma. ∎

Let $G$ be the Galois group of the Galois closure of $K$ over $\mathbb{Q}$. Let us denote by $|G|$ the order of $G$.

Lemma 2**.**

1) One has $|G|\in\left\{4,8\right\}$.

2) Suppose that $|G|=4$. One of the two following conditions is satisfied :

• (1)

$5(16t^{5}+1)$* is a square in $\mathbb{Q}$.*

• (2)

$(1-4t^{5})(16t^{5}+1)$* is a square in $\mathbb{Q}$.*

Proof.

Let us denote in $\mathbb{Q}[X]$

[TABLE]

One has $f(y)=0$ (lemma 1). Let $\varepsilon\in\overline{\mathbb{Q}}$ such that

[TABLE]

The element $y+\frac{1}{y}$ is a root of the polynomial $X^{2}+uX+u$. So we have the inclusion

[TABLE]

Moreover, we have the equality

[TABLE]

Since $K=\mathbb{Q}(y)$ and $[K:\mathbb{Q}]=4$, we have

[TABLE]

From (3.6), we deduce that the roots of $f$ belong to at most two quadratic extensions of $\mathbb{Q}(\varepsilon)$. The equality (3.7) then implies $|G|\leq 8$. Since $4$ divides $|G|$, this proves the first assertion.

Henceforth let us suppose $|G|=4$ i.e. the extension $K/\mathbb{Q}$ is Galois. Let $\Delta$ be the discriminant of $f$. One has the equalities

[TABLE]

Let us prove that

[TABLE]

From (3.6) and our assumption, the roots of the polynomials

[TABLE]

belong to $K$, which is a quadratic extension of $\mathbb{Q}(\varepsilon)$ ((3.5) and (3.7)). Therefore, the product of their discriminants

[TABLE]

must be a square in $\mathbb{Q}(\varepsilon)$. The first equality of (3.8) then implies (3.9).

Suppose that the condition (1) is not satisfied. From second equality of (3.8), we deduce that $\Delta$ in not a square in $\mathbb{Q}$. It follows from (3.9) that we have

[TABLE]

Therefore, $\Delta(u^{2}-4u)$ is a square in $\mathbb{Q}$, in other words, such is the case for $-u(3u+4)$. One has the equality

[TABLE]

This implies the condition (2) and proves the lemma. ∎

4. The curve $C_{1}/\mathbb{Q}$

Let us denote by $C_{1}/\mathbb{Q}$ the curve, of genus $2$, given by the equation

[TABLE]

Proposition 1**.**

The set $C_{1}(\mathbb{Q})$ is empty.

Proof.

Suppose there exists a point $(X,Y)\in C_{1}(\mathbb{Q})$. Let $Z=\frac{Y}{5}$. We obtain

[TABLE]

Let $a$ and $b$ be coprime integers, with $b\in\mathbb{N}$, such that

[TABLE]

Let us prove there exists $c\in\mathbb{N}$ such that

[TABLE]

For every prime number $p$, let $v_{p}$ be the $p$-adic valuation over $\mathbb{Q}$.

If $p$ is a prime number dividing $b$, distinct from $2,5$, one has $2v_{p}(Z)=-5v_{p}(b)$, consequently

[TABLE]

Moreover, one has $v_{2}(X)<0$ ($5$ is not a square modulo $8$), so $4-5v_{2}(b)=2v_{2}(Z)$. In particular, one has

[TABLE]

Let us verify the congruence

[TABLE]

One has $v_{5}(X)\leq 0$. Suppose $v_{5}(X)=0$. In this case, one has $X^{5}\equiv\pm 1,\pm 7\mod 25$. The equality (4.1) implies $X^{5}\equiv-1\mod 25$ and $Z^{2}\equiv 2\mod 5$, which leads to a contradiction. Therefore, we have $1+2v_{5}(Z)=-5v_{5}(b)$, 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

[TABLE]

One has $ab\neq 0$. From the informations given in the Appendix of [3], this implies

[TABLE]

We obtain $X=-1/2$, which is not the abscissa of a point of $C_{1}(\mathbb{Q})$, hence the result. ∎

5. The curve $C_{2}/\mathbb{Q}$

Let us denote by $C_{2}/\mathbb{Q}$ the curve, of genus $4$, given by the equation

[TABLE]

Proposition 2**.**

One has

[TABLE]

Proof.

Let $(X,Y)$ be a point of $C_{2}(\mathbb{Q})$. Let $a$ and $b$ be coprime integers such that

[TABLE]

We obtain the equality

[TABLE]

Therefore, $(b^{5}-4a^{5})(16a^{5}+b^{5})$ 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 $(a,b)$ by $(-a,-b)$ if necessary, there exists $d\in\mathbb{N}$ such that

[TABLE]

Suppose $b^{5}-4a^{5}\in\left\{2d^{2},10d^{2}\right\}.$ In this case, $b$ must be even, therefore $v_{2}(2d^{2})=2$, which is not.

Suppose $b^{5}-4a^{5}=d^{2}$. One has $b\neq 0$. It then comes from [3] that

[TABLE]

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 $c\in\mathbb{N}$ 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. ∎

6. End of the proof of Theorem 2

The group $G$ is isomorphic to a subgroup of the symmetric group $\mathbb{S}_{4}$ and one has $|G|=4$ or $|G|=8$ (lemma 2). In case $|G|=8$, $G$ is isomorphic to a $2$-Sylow subgroup of $\mathbb{S}_{4}$, that is dihedral.

Suppose $|G|=4$ and let us prove the assertion 2 of the theorem.

First, we directly verify that the extension $K/\mathbb{Q}$ defined by the condition (2.1) is cyclic of degree $4$, and that the point $(2,2\alpha,-\alpha-1)$ belongs to $F_{5}(K)$.

Conversely, from the proposition 1, the condition (1) of the lemma 2 is not satisfied. The condition (2) and the proposition 2 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

[TABLE]

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 $\mathbb{Q}$ of $\alpha$. The equality (3.3),

[TABLE]

then implies the result.

Acknowledgments. I thank D. Bernardi for his remarks during the writing of this paper.