Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given an homogeneous polynomial $F(X,Y,Z,T)\in \mathbb Q[X,Y,Z,T]$ of degree $>4$, the surface it defines is well-known to be of general type. Suppose, moreover, that this surface doesn't contain any rational or elliptic curve (a fact that it happens "generically", according to a theorem by Geng Xu in Subvarieties of general hypersurfaces in projective space). Then, a conjecture by Bombieri and Lang (see for example the book by Marc Hindry and Joseph H. Silverman "Diophantine Geometry: An Introduction") asserts that this surface should have a finite number of rational points (this is a special case of a strong version of the conjecture) (as pointed out below by Felipe Voloch, one needs to assume the surface is smooth) .

Some questions:

  1. Is it known any example of such a surface which has some rational point, it is smooth, and one can prove there is only a finite number of rational points? I ask to have a rational point in order to prevent a proof by "local considerations": it has no rational points since it has no real points, or $p$-adic points, or "modulo n" points, etc...

  2. A bit less: Is it known explicit examples of such surfaces verifying the hypothesis of the conjecture? The results I find are for "generic" cases.

  3. For example, does there exists some $k$ such that any non-singular hypersurfaces in $\mathbb P^3_{\mathbb Q}$ of degree $>k$ does not contain any rational or elliptic curve?

I understand that Faltings' results on subvarieties of abelian varities cannot be applied here, since there is no trivial map to these varities to an abelian variety (since their albanese variety is trivial, since the dimension of their $H^1$ is zero, by standard results).

share|improve this question
4  
First sentence is not necessarily true unless you assume that the surface is smooth. –  Felipe Voloch Aug 26 '12 at 12:47

1 Answer 1

For question 2, you can look for surfaces of Picard number 1. One way to find such surfaces is to reduce modulo p and show that the reduction has Picard number 1 by computing its L-function. This will only work for odd degree. For even degree, you may need to compute the reduction modulo two primes with Picard number 2 but incompatible lattices so the Picard number of the original surface is 1 (this is a trick due to R. van Luijk). These computations are not easy and there is some work, e.g. by Kedlaya, to make them feasible.

Question 3 is false. You can easily write down smooth surfaces of any degree containing a straight line.

Off the top of my head, I don't have an example for your question 1, which is unconditional. But, if believe the ABC conjecture in its multiple summands variant, then diagonal hypersurfaces give examples (e.g. $x^n+2y^n+3z^n=6w^n, n$ large).

Edit: As Jason points out in the comments, my suggestion for question 2 is not enough. A different idea would be to use the idea for question 1. Namely use the function field ABC (which is a theorem) to show that a "fewnomial" (diagonal equations contain lines) equation has no curves of genus 0 or 1. I haven't worked out the details, though.

share|improve this answer
3  
I'm not sure what you write is correct. Certainly a smooth rational or elliptic curve would give a divisor class on the general type hypersurface that is not commensurable with the hyperplane class. However, these surfaces may contain singular curves whose normalization has geometric genus $0$ or $1$ and whose divisor classes are multiples of the hyperplane class. –  Jason Starr Aug 26 '12 at 13:11
    
@Jason: Good point. Can these be ruled out separately? –  Felipe Voloch Aug 26 '12 at 13:26
    
@Felipe: The best result I know of is the result of Geng Xu cited by the OP. However, as the OP remarks, this only applies to "very general" surfaces, so potentially might fail for <I>every</I> surface defined over a number field. –  Jason Starr Aug 26 '12 at 14:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.