Algebraic Geometry – Rational Points on Rank 0 Picard Curves for $y^3=x^4+x$

ag.algebraic-geometrydiophantine equationsnt.number-theory

What are the rational solutions to the equation
$$
y^3 = x^4 + x,
$$
in particular, are there any (finite) solutions other than $(x,y)=(0,0)$ and $(-1,0)$?

Context: This is the simplest-looking example of a Picard curve, that is, curve $y^3=P(x)$, where $P(x)$ is a polynomial of degree 4. The rank of its Jacobian is 0, the simplest case possible. Hashimoto and Morrison https://arxiv.org/abs/2002.03291 treated more difficult case of rank $1$. For rank $0$, Jackson Morrow $y^3 = x^4 + x + 2$, and existence of rational points on rank 0 Picard curves presented some Magma code that can prove that there are no rational points (if this is indeed the case). However, the curve $y^3 = x^4 + x$ has some rational points, so deeper analysis seems to be required. Hence the question.

Best Answer

Let $x = \frac{a}{d}$, $y = \frac{b}{d}$, $\gcd(a, d, b) = 1$. $$b^3d = a^4 + ad^3$$ Suppose $p \mid a$ and $p \mid d$ for prime $p$. Then $b$ is not divisible by $p$. $$\nu_p(d) = \nu_p(a^4 + ad^3) = \nu_p(a) + \nu_p(a^3 + d^3) = 4\nu_p(a)$$ The last equality is true because $\nu_p(a) < \nu_p(d)$. Hence, we can make the substitution $a = a_1c$, $d = d_1c^4$; $(a_1, d_1, c)$ are pairwise coprime.

$$b^3d_1 = a_1^4 + a_1d_1^3c^9$$ $d_1 \mid a_1$. But this is only possible if $d_1 = \pm 1$. $(\pm b)^3 = a_1(a_1^3 \pm c^9)$ Whence $a_1^3 \pm c^9$ must be a cube, which contradicts Fermat's theorem.

Related Solutions

Rational Points on Rank 0 Picard Curves – $y^3 = x^4 + x + 2$

The below Magma code determines the size of $J_C(\mathbb{F}_p)$ for various primes $p$, and finally compute the GCD of their orders, which gives you a bound on the size of $J_C(\mathbb{Q})$. For a discussion of this, see Section 4.1 of this paper. To determine $J_C(\mathbb{F}_p)$, you can just evaluate the numerator of the zeta function of the reduction of $C$ mod $p$ (see e.g., Hindry--Silverman Diophantine Geometry Exercise C.4).

The below code does this for primes up to 30 and tells you that the GCD of $\#J_C(\mathbb{F}_p)$ for these primes is 1, and hence $J_C(\mathbb{Q})$ is just the trivial group. The code is general and can work for any equation of this form of rank 0. If you do have some non-trivial torsion in $J_C(\mathbb{Q})$, then you will need to work harder to determine the rational points.

I will also note that you can run this on the online magma calculator.

P2<x,y,z> := ProjectiveSpace(Rationals(),2);
C := Curve(P2,z*y^3 - (x^4 + x*z^3 + 2*z^4));
bad := {2,3};
bool := false; p := 2;
torsOrders := {@@};
while  p le 30 do
  p :=  NextPrime(p);        
  p := p in bad select NextPrime(p) else p;
  Cp := Curve(Reduction(C,p));
torsOrders := torsOrders join  {@ Evaluate(Numerator(ZetaFunction(Cp)),1) @};
end while;
GCD(torsOrders); //1

Number Theory – Finding K=Q(?6)-Rational Points on the Twist of X?(26)

I used Magma to point search on $C/K$ up to a height of $1000$ and it appears that $C(K) = \emptyset$. If that's true, then one can probably use the Mordell-Weil sieve to prove it. Here's a bit more detail.

The curve $C$ has four automorphisms defined over $\mathbb{Q}$ and for one of these (the map $(x,y) \mapsto (1/x,y/x^{3})$), the quotient curve has genus $1$. In particular, there is a map from $\phi : C \to E$, where $E : y^{2} = x^{3} - 651x - 12742$. This curve $E$ has rank $1$ over $K$.

For a finite set $S$ of prime ideals of $\mathcal{O}_{K}$ (all of which are primes of good reduction for $E/K$), one can write down the commutative diagram $$ \require{AMScd} \begin{CD} C(K) @>>> E(K)\\ @VVV @VV{\beta}V\\ \prod_{\mathfrak{p} \in S} C(\mathbb{F}_{\mathfrak{p}}) @>\alpha>> \prod_{\mathfrak{p} \in S} E(\mathbb{F}_{\mathfrak{p}})\\ \end{CD} $$ The horizontal maps in this diagram use the map from $C$ to $E$, while the vertical maps use reduction modulo the prime ideals in $S$. If $P \in C(K)$ is a point, then the image of $\alpha$ and the image of $\beta$ (as subsets of $\prod_{\mathfrak{p} \in S} E(\mathbb{F}_{\mathfrak{p}})$) have a non-trivial intersection. So if one can find a set $S$ of primes for which the image of $\alpha$ and the image of $\beta$ are disjoint, this proves that $C(K)$ is empty.

For more about this, I recommend The Mordell-Weil sieve: Proving non-existence of rational points on curves by Nils Bruin and Michael Stoll.

Best Answer

Related Solutions

Rational Points on Rank 0 Picard Curves – $y^3 = x^4 + x + 2$

Number Theory – Finding K=Q(?6)-Rational Points on the Twist of X?(26)

Related Question