3
$\begingroup$

Suppose $F(x,y,z)$ is a homogeneous polynomial over $\mathbb{Q}$, where $C:F(x,y,z)=0$ is a curve of genus $g\geq 2$.

Question: Faltings proved that $C$ has finite many rational points. Suppose that there is no rational points on $C$. Is there any effective version of Mordell conjecture, i.e., $$\vert F(x,y,z)\vert\gg_{F,g} g(H(x,y,z)),(x,y,z)\in\mathbb{Z}^3$$ where $H$ is naive height and $g(x)$ tends to infinity as $x$ tends to infinity?

$\endgroup$
  • 3
    $\begingroup$ This statement is not what is normally understood as effective Mordell. It is false. Take $F=x^4+y^4-z^4$ and $x=z,y=1$. $\endgroup$ – Felipe Voloch Oct 25 '14 at 11:07
  • $\begingroup$ @FelipeVoloch: Thanks for your correction. If there is no rational point on C, is this statement false as well? $\endgroup$ – Y. Zhao Oct 25 '14 at 12:25
  • 1
    $\begingroup$ Your new question is equivalent to whether $F(x,y,z)=c$ has finitely many integral points for all $c$. Your new hypothesis precludes this surface from containing a line, but it may contain a conic and thus infinitely many integral points. $\endgroup$ – Felipe Voloch Oct 25 '14 at 15:27
1
$\begingroup$

Have a look at the work of Gael Remond; see e.g.

http://plms.oxfordjournals.org/content/101/3/759.abstract

In that paper Remond proves an explicit upper bound on the number of rational points on a curve. You won't be able to do better than that I think with the current state-of-affairs.

Added in edit: The above is not relevant to your question, but it does address but people usually refer to as Effective Mordell. As Felipe Voloch points out, what is usually meant by "Effective Mordell" is an explicit/effective bound on the height of a rational point on a smooth projective curve $X$ over a number field $K$ in terms of data associated to $X$ and $K$. The work of Gael Remond (cited above) is concerned with the number of points on $X$, and gives upper bounds for this number. If you are willing to believe that the Shafarevich conjecture can be made effective (see another paper of Gael Remond) then you can also prove an effective version of the Mordell conjecture. Anyway, this doesn't answer your question, but it might be good to know.

$\endgroup$
  • $\begingroup$ Thanks for your answer. Once I heard that rational points cannot lie very close to curves whose genus>1. Does abc conjecture give a conjectural lower bound? $\endgroup$ – Y. Zhao Oct 25 '14 at 10:30
  • 1
    $\begingroup$ @zy_ : Neglecting the restriction on the genus, this is similar to asking for e.g. lower bounds on non-vanishing quantities of the form $|y^2 - x^3|$, $x,y \in \mathbb{Z}$. Baker's theory of logarithmic linear forms does yield a lower bound here of the shape $(\log{|x|})^{\kappa}$ for a small positive $\kappa > 0$, while the best possible statement $|y^2 - x^3| \gg_{\varepsilon} |x|^{\frac{1}{2}-\varepsilon}$ is a consequence of the ABC conjecture. Your general problem is a question about integral point on surfaces, and not much is known here (there has been some progress by Corvaja, Zannier). $\endgroup$ – Vesselin Dimitrov Oct 26 '14 at 0:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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