How Many Rational Points Can a Curve Have?

Michael Stoll

Max-Planck-Institut für Mathematik, Bonn Heinrich-Heine-Universität, Düsseldorf

The Problem

Let C be a curve of genus g over ℚ. What can we say about the set of rational points on C, C(ℚ)?

• If g = 0, then either C(ℚ) = ∅, or there is a rational parametrisation ℙ^{1}(ℚ) ⟶ C(ℚ). This is effective. (Legendre, …)

• If g = 1, then either C(ℚ) = ∅, or C/ℚ is an elliptic curve, and C(ℚ) is a finitely generated abelian group. (Mordell) This is not known to be effective, but it is under standard conjectures.

• If g ≥ 2, then C(ℚ) is finite (Faltings), but this is about all one can say.

We will consider the problem of bounding #C(ℚ) in the case g ≥ 2.

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What is Not Known

If the so-called Lang Conjecture on varieties of general type were true, then it would follow that

#C(ℚ) ≤ B(g) ,

the bound depending only on the genus. (Caporaso-Harris-Mazur)

But this conjecture is not known to be true.

So we will be more modest and ask whether perhaps one can have a bound

#C(ℚ) ≤ B(g, r)

depending also on the Mordell-Weil rank r of the Jacobian J of C.

This is also not known. But at least heuristically, for example a generic hyperelliptic curve should not have more than 3 + ⌊gr/(g − 1)⌋ pairs of non-Weierstrass points. (Elkies)

So one might even conjecture that B(g, r) can be taken to grow linearly with r.

But in fact, it is not even known whether B(g, 0) exists!

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What is Known

Since the original problem is too hard, we make it simpler. Let us restrict attention to the set of twists of a given curve C. In this case, there is a result due to Silverman:

#C′(ℚ) ≤ B(C, r) = γ(C)7^{r}

for all twists C′/ℚ of C. This is a bound, but it is exponential in r. Also, γ is effective in principle, but not really explicit. On the other hand, there is now the following result. Theorem (MS). Given C, there is an explicit constant w(C) such that

#C′(ℚ) ≤ w(C) + 2r

for all but an explicitly given finite set of twists C′ of C, under the condition that r < g. This provides a linear bound, but it suffers from the restriction r < g.

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Applications

1. Hyperelliptic Curves. For hyperelliptic curves C and quadratic twists C′ = C_{d}, we have

w(C) = #{Weierstrass points in C(ℚ)} ≤ 2g + 2 .

The result is best possible in some sense, since there is no reason why non-Weierstrass points coming from distinct orbits under the hyperelliptic involution should be dependent in J.

2. Thue Equations. The result can be refined to give the following. Theorem. Let F ∈ ℤ[X, Y] be homogeneous of degree n and without multiple factors. Then for all but finitely many nth-power free h ∈ ℤ, the Thue equation F (x, y) = h has not more than r rational solutions, if the Mordell-Weil rank of the corresponding curve satisfies r ≤ n − 3.

3. A Special Family. For a tenth-power free integer A, let C_{A} be the curve Y^{2} = X^{5} + A of genus 2. Theorem.

1. If r(C_{A}) = 0, then #C_{A}(ℚ) ≤ 4, attained only for A = 1.

2. If r(C_{A}) = 1, then #C_{A}(ℚ) ≤ 7, attained only for A = 18^{2}.

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Chabauty’s Method

Chabauty proved Mordell’s conjecture in the case r < g.

Idea: J(ℚ) sits in a closed subgroup of J(ℚ_{p}) of dimension ≤ r < g and intersects C ⊂ J only in finitely many points.

There is a perfect pairing of ℚ_{p}-vector spaces

Ω^{1}(C/ℚ_{p}) × J(ℚ_{p}) ⊗_{ℤ} ℚ ⟶ ℚ_{p}

given by

⟨ω, [Q − P]⟩ = λ_{ω}([Q − P]) = ∫_{P}^{Q} ω

If P and Q are in the same residue class on C, the integral is given by a converging power series. If ω ≠ 0, it will only have finitely many zeros.

If G ⊂ J(ℚ_{p}) ⊗_{ℤ} ℚ is a proper subspace (e.g., the space generated by J(ℚ) if r < g), then there is a nonzero subspace Λ ⊂ Ω^{1}(C) (of dimension ≥ g − r) such that Λ kills G under the pairing above.

If p is large enough, then λ_{ω} has at most ν + 1 zeros on the residue class of P̄ ∈ C(𝔽_{p}), where ν is the order of vanishing of ω̄ in P̄.

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Refinement

For large p, we get a bound of the form

#C(ℚ) ≤ #{residue classes hit by C(ℚ)} + N(Λ)

where

N(Λ) = ∑_{P̄∈C(𝔽_{p})} min_{ω∈Λ} νP̄(ω̄)

Proposition. If C is hyperelliptic or has (potentially) good reduction at p, then

N(Λ) ≤ 2 codim_{Ω^{1}(C)} Λ ≤ 2r

This improves the trivial bound N(Λ) ≤ 2g − 2.

But one has the problem that there will be many residue classes for p large.

How can we get a bound that does not depend on p?

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New Ideas

Let us consider a hyperelliptic curve C : Y^{2} = F (X) for simplicity.

Instead of looking at a good prime, we take a bad prime that becomes good after a qudratic extension. Say, F (X) = pF_{1}(X), where p does not divide disc F_{1}.

Over ℚ(\sqrt{p}), C is isomorphic to C_{1} : Y^{2} = F_{1}(X) and has therefore good reduction.

Now the nice thing is that the set of residue classes of C_{1}(Q_{p}(\sqrt{p})) that are hit by C(ℚ_{p}) comes from the Weierstrass points in C_{1}(𝔽_{p}) and is therefore bounded in size independently of p.

Each of these residue classes is accounted for by a Weierstrass point in C(ℚ_{p}).

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Applying Chabauty’s Method

We know consider the set X of points in C(ℚ_{p}) that map into the closure of J(ℚ) in J_{1}(ℚ_{p}(\sqrt{p})), where J_{1} is the Jacobian of C_{1}.

This closed subgroup is still of dimension ≤ r. It is therefore killed by a subspace Λ ⊂ Ω^{1}(C_{1}/ℚ_{p}(\sqrt{p})) of codimension at least r.

Chabauty’s method, applied to this situation, then gives

#X ≤ #{Weierstrass points in C(ℚ_{p})} + 2r .

Since X contains all the Weierstrass points in C(ℚ_{p}) (they map to torsion points, hence to zero in J ⊗ ℚ), we get that the number of non-Weierstrass points in C(ℚ) is bounded by N(Λ) ≤ 2r (if p is large enough).

‘Large enough’ here means p > 2r + 5.

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