Timeline for Counting Special Rational Points on Cubic Surfaces
Current License: CC BY-SA 3.0
9 events
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| Nov 18, 2013 at 18:05 | comment | added | Daniel Loughran | Right, I'm starting to understand now - sorry for the earlier confusion. It seems like no points can satisfy your condition over the algebraic closure of Q, so your set behaves a lot like the complement of the image of some dominant morphism, e.g. perhaps the complement of a thin set in the sense of Serre. There is not much in the literature on counting points on thin sets, aside from the case of projective space which Serre considers in his book. I don't have much else intelligent to say, good luck with your problem! | |
| Nov 18, 2013 at 11:55 | comment | added | Y. Zhao | @DanielLoughran: Suppose $C$ is a degree 3 rational curve on $x^3+y^3+z^3=3$ through $(1,1,1)$. Then the curve lies on a plane tangent to the surface at a rational point. This observation turns the problem into seeking for rational points on the intersection curve of $x^3+y^3+z^3=3$ and $x^2+y^2+z^2=3$, which is a genus 3 curve. Let $u=x+y$ and $v=z$, we can eliminate $x^2+y^2$ and search for points on a cubic curve $u(9-u^2-3v^2)/2+v^3=3$. A computation on MAGMA shows this curve is isomorphic to the Weierstrass form $y^2=x^3-243x+2430$, whose rank is zero and torsion group trivial. | |
| Nov 18, 2013 at 9:58 | comment | added | Daniel Loughran | Do you have a proof that the point $(1,1,1)$ satisfies your conditions? | |
| Nov 18, 2013 at 9:55 | comment | added | Daniel Loughran | Also I should note that your surfaces have Picard number one, in particular they contain no lines or conics. So your question is really about singular plane cubic curves. | |
| Nov 18, 2013 at 9:54 | comment | added | Daniel Loughran | I think you mean "degree $3$" rather than "genus $3$" here; an irreducible singular plane cubic curve has arithmetic genus one and its normalisation has genus zero. I don't buy your claim that the rational points on such curves have "small height". For example the curve $y^2 = x^3$ may be parametrised by the map $t \mapsto (t^2,t^3)$, so there are many rational points of large height. | |
| Nov 18, 2013 at 7:53 | comment | added | Y. Zhao | @DanielLoughran: Yes,that is what I try to express(of course, all the rational curves are defined over $\mathbb{Q}$). In fact, $(1,1,1)$ is such a point on $x^3+y^3+z^3=3$. As for other points on the surface $x^3+y^3+z^3=3$, I cannot prove they are elements of $\Sigma'$(just because I cannot list all the rational points on some genus 3 curves), but there is some heuristic arguments suggests that rational points on genus 3 curves are of "small" height, so a naive search might give all the rational points on a genus 3 curve, which is key to ascertain whether a point is in $\Sigma'$ or not. | |
| Nov 17, 2013 at 18:53 | comment | added | Daniel Loughran | I don't quite understand your definition of $\Sigma'$, can you elaborate a bit more? Namely the condition that "P′ cannot be a non-singular point on plane rational curves over $\mathbb{Q}$". Do you mean that for every hyperplane $H \subset \mathbb{P}^3$ which contains $P'$ and for which $S \cap H$ is a (possible union of) rational curves, the point $P'$ is always a singular point of $S \cap H$? Can you give some examples of the points which you are interested in? | |
| Nov 17, 2013 at 11:52 | history | edited | Y. Zhao | CC BY-SA 3.0 |
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| Nov 16, 2013 at 14:56 | history | asked | Y. Zhao | CC BY-SA 3.0 |