# Salberger: Arithmetic Bezout and Rational Points of Bounded Height

We take the height function

If is a closed subvariety over , an open subvariety also over , we have the counting function

Already the case of smooth cubic surfaces, away from the lines, this conjecture would imply linear growth, but it was only known recently for quadratic growth; therefore this is hard enough, looking for asymptotic formulas is often asking for too much.

Recently, there has been work which also works in low dimension (curves and surfaces), due to Heath-Brown (2002); this also has applications to other problems (e.g. Waring's problem).

Theorem. [Heath-Brown] Let be an absolutely irreducible curve of degree ; then

The implicit constant does not depend on ; the results of Faltings have no uniformity. We have applications to surfaces. This result is also best possible, taking the Veronese embedding of . There was a result of Bombieri-Pila which proved . Broberg treats arbitrary number fields (and the case ).

This theorem relies on the following result, which is Theorem 14 in the paper of Heath-Brown.

Theorem. [Theorem 14] Let be an absolutely irreducible projective -variety of dimension defined by forms of degree . Let , be given. Then there exists a -hypersurface such that:

1. ;
2. All -points on of height lie on ;
3. We have

4. The irreducible components of have degrees bounded in terms of .

Remark. Heath-Brown treats only the case . Taking , the result of Heath-Brown is an immediate consequence of this theorem if one applies Bezout's theorem in the plane. The case is due to Broberg, and with arbitrary .

If is smooth, then you may replace by in (iii) and (iv).

Lemma. [Colliot-Thélène] Let be a smooth projective surface. Then there exists at most curves of degree on .

This implies that a cubic has only finitely many lines, a quartic has only finitely many conics, and so on. This is best possible, for one might have infinitely many such curves, for example, infinitely many conics on a cubic surface. Removing these curves, we still have a surface, and we get:

Theorem. [Heath-Brown] Let be a smooth projective surface and let be the complement of all curves of degree . Then

This is the best known result if . To do this, apply Theorem 14 by cutting with an auxiliary hyperplane; the same implicit constant applies everywhere, the is the maximum number of irreducible components. We considering for example the Veronese embedding of the projective plane to see that we would expect .

Theorem. [S] Let be a smooth absolutely irreducible projective -variety of dimension defined by forms of degree . Let , be given. Then there exists a -hypersurface such that:

1. ;
2. All -points on of height lie on ;
3. We have

For the moment, it is not clear how to use this theorem to deduce .

Proof. Let be monomials in which form a basis of

where . Let represent the -points on of height . We need to show that the rank of the matrix

is . This is trivial if ; otherwise, we must look at all sub -determinants. Since these points are of bounded height, each term in the determinant is bounded by , so one has a bound on the archimedean height of the determinant. With more points, the determinants are divisible by high powers of prime numbers (by the Weil conjectures, points must coincide); under certain circumstances, these divisibilities contradict the bounds on the determinant. We use for example that

This gives the result.

Remark. The theorem is also true for surfaces with at most rational double points. Already, the theorem is not known for elliptic singularities.

We apply this theorem to count -points on when . Then . By the adjunction formula (and Bezout),

All -points on irreducible, not absolutely irreducible components are singular. Therefore it suffices to count smooth -points on absolutely irreducible components of . Let be absolutely irreducible components of of degree , then

where . This is

If you throw out curves of smallest degree, this is smaller than .

But we still must deal with curves of high degree, e.g. the case when is irreducible.

Lemma. Let be an absolutely irreducible degree , and a prime . Then:

1. The number of -points on of height which specializes to a given smooth -point on is .
2. The number of -points on with smooth specialization at is .

Corollary. We have

If we could replace by , then we would have get . It might still be useful to find something like . Using arithmetic Bezout, bounding the heights of the subvarieties (due to Faltings), this might succeed. It would also be better to work systematically with all primes , some savings might arise.

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