Martin's Blog

Siegel's theorem for curves of genus 0

Posted by Martin Orr on Friday, 28 October 2011 at 12:35

Last time we proved Siegel's theorem on the finiteness of integer points on affine curves of genus at least 1. The theorem applies also to curves of genus 0 with at least 3 points at infinity. I shall give a simple proof that deduces this from the higher genus case, then another proof using Baker's theorem from transcendental number theory which gives an effective bound on the heights of the points.

Theorem. Let K be a number field and S a finite set of places of K. Let X be an affine K-curve of genus 0 such that there are at least 3 \bar{K}-points in the projective closure of X which are not in X. Then X has finitely many S-integer points.

The condition that there should be at least 3 points at infinity is necessary: the affine line is a genus 0 curve with 1 point at infinity and infinitely many integer points, and the curve x^2 - Dy^2 = 1 for D a non-square positive integer has 2 points at infinity and infinitely many integer points.

The unit equation

The unit equation is a special case of Siegel's theorem for genus 0. We give a simple proof that it has finitely many solutions, using Siegel's theorem for higher genus. Then we show that the general Siegel's theorem for any curve of genus 0 can be reduced to the unit equation.

Theorem. Let K be a number field and S a finite set of places of K. Let A, B, C \in K - \{0\}.
Then the equation Ax + By = C has finitely many solutions in S-units x, y.

Proof 1. The group \mathcal{O}_S^\times/\mathcal{O}_S^{\times 3} is finite (by Dirichlet's unit theorem) so any S-units has px_0^3 for x_0 \in \mathcal{O}_S^\times and p from a fixed finite set. Hence there are finitely many equations of the form A'x_0^3 + B'y_0^3 = C which give all the solutions to the unit equation.

But each equation A'x_0^3 + B'y_0^3 = C defines an affine curve of genus 1, so by Siegel's theorem it has finitely many integer solutions.

To see that this is a special case of Siegel's theorem for genus 0, observe that solutions to the unit equation are the same as S-integer points on the affine curve  Ax + By = C,  \,  xz = 1,  \,  yt = 1. Projection onto the x-coordinate gives an isomorphism with \mathbb{A}^1 - \{0, C/A \} = \mathbb{P}^1 - \{ 0, C/A, \infty \} so this is a genus 0 curve with three points at infinity.

Any genus 0 curve is isomorphic to \mathbb{P}^1 minus a finite set of points, so if it has at least 3 points at infinity then it has an injective morphism into \mathbb{P}^1 - \{0, \lambda, \infty \} for some \lambda \in K. This morphism is given by polynomials with coefficients in K; by multiplying them by a constant we may suppose that the coefficients are in \mathcal{O}_S and so the morphism takes S-integral points to S-integral points. Then the finiteness of the solutions to the unit equation implies the finiteness of S-integer points on the original curve.

Baker's method

Baker's theorem gives an effective bound on how closely it is possible to approximate zero by a linear combination of logarithms. We can use this to prove the finiteness of solutions to the unit equation, and get an effective bound on their height.

Baker's Theorem. Let K be a number field and \alpha_1, \ldots, \alpha_n \in K. Then there is an effective constant c(\alpha), depending only on \alpha_1, \ldots, \alpha_n and on the completion K_v, such that for any integers b_1, \ldots, b_n with B = \max(4, |b_1|, \ldots, |b_n|), either  \prod \alpha_i^{b_i} = 1 \quad \text{or} \quad \left\lvert \prod \alpha_i^{b_i} - 1 \right\rvert_v > B^{-c(\alpha)}.

(This statement of the theorem contains no logarithms, but it is equivalent to a statement that \sum b_i \log \alpha_i is either zero or bounded away from zero.)

To apply this to the unit equation, we start with Dirichlet's unit theorem: \mathcal{O}_S^\times is isomorphic to the direct product of a finite group \mu and a free abelian group; let \epsilon_1, \dotsc, \epsilon_r be a basis for the latter.

Given x = \zeta \epsilon_1^{b_1} \ldots \epsilon_r^{b_r} \in \mathcal{O}_S^\times, define b(x) = \max(|b_1|, \ldots, |b_n|). This defines a discrete norm on \mathcal{O}_S^\times/\mu. We can define another such norm by  n(x) = \max_{v \in S} \log \lvert x \rvert_v. To see that the latter is a norm, we need to prove that it is positive for all x \neq 0. This follows from the fact that \sum_{v \in S} \log \lvert x \rvert_v = 0 for x \in \mathcal{O}_S^\times. (We need to enlarge S here to contain all the archimedean places. Then for an S-unit, we have \log \lvert x \rvert_v = 0 for v \not\in S and the product formula gives that the sum of the logs of the absolute values is 0.)

Any two discrete norms on a finitely generated free abelian group are equivalent, so there exists a constant c such that  n(x) \geq c b(x) \text{ for all } x \in \mathcal{O}_S^\times. Or in other words, for any x \in \mathcal{O}_S^\times, there is a valuation v \in S for which \log \lvert x \rvert_v \geq cb(x).

Since S is finite, it will suffice to prove that for each v \in S, Ax + By = C has finitely many solutions in S-units satisfying b(x) \geq b(y) and \log \lvert x \rvert_v \geq cb(x).

So given such a solution, write x = \zeta \epsilon_1^{b_1} \cdots \epsilon_r^{b_r} and y = \zeta' \epsilon'_1^{b_1} \cdots \epsilon'_r^{b_r}, and \alpha = -B\zeta' / A\zeta.

Then  \left\lvert 1 - \alpha \epsilon_1^{b'_1 - b_1} \cdots \epsilon_r^{b'_r - b_r} \right\rvert_v
 = \left\rvert 1 + \frac{By}{Ax} \right\rvert_v = \left\lvert \frac{C}{Ax} \right\rvert_v
 < c_1 c_2^{-b(x)} for some constants c_1 > 0, c_2 > 1 depending only on A, B, C, v.

But by Baker's theorem, the LHS of this inequality is greater than b(x)^{-c_3}. So we have b(x)^{-c_3} < c_1 c_2^{-b(x)} and b(x) is bounded. The constants c_1, c_2, c_3 are effectively computable, and hence we can get an effective bound for b(x).

Tags maths, number-theory

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Comments

  1. Barinder Banwait said on Sunday, 06 November 2011 at 19:17 :
    1. Small typo: In the paragraph before "The Unit Equation", I think you want your curve to be x^2 - Dy^2 = 1.

    2. Why is b(x) a norm, that is, why is it non-negative? So far as I can see, all of the b_is could be negative. Don't we need to bound b(x) from below as well?

  2. Martin Orr said on Monday, 07 November 2011 at 12:36 :

    You area right on both counts. Both in the definition of b(x) and in B in Baker's theorem, the b_is need to be replaced by |b_i|.

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