The seminal theorem of Faltings confirms Mordell's conjecture: that is, curves of genus at least 2 have at most finitely many rational points. The proof of Faltings' theorem is not effective, meaning there is no way to estimate the number of rational points on a given curve.

What is the conjectured truth for these curves? In some cases, a curve with a large genus actually have no non-trivial rational points (such as Fermat curves). Do there exist algebraic curves with genus at least 2 and arbitrarily many rational points?

  • 2
    $\begingroup$ Seems to be a duplicate of mathoverflow.net/questions/103327/… $\endgroup$ Mar 25, 2014 at 14:03
  • $\begingroup$ Perhaps not exactly a duplicate: here the genus is not fixed, if I understand correctly (but my english is poor). However I am completely ignorant: what is the genus of $y^m=(x-r_1)(x-r_2)\cdots(x-r_n)$ ? $\endgroup$
    – user46855
    Mar 25, 2014 at 14:34
  • 7
    $\begingroup$ Of course you need to fix the genus, for example taking $y^2=(x-1)(x-2)...(x-2g-2)$ you get a curve of genus $g$ with at least $2g+2$ rational points. $\endgroup$ Mar 25, 2014 at 14:47
  • 1
    $\begingroup$ The number of rational points on a given curve can actually be estimated, similarly to the bounds for the number of solutions in Roth's theorem. Ineffectivity means that there is no current procedure of determining the set of rational points on a given curve (or the set of solutions in Roth's theorem). It is, however, known that an effective version of the ABC conjecture would supply such a procedure. $\endgroup$ Mar 25, 2014 at 16:49
  • 1
    $\begingroup$ If $f(x)$ is the unique polynomial of degree $2g+2$ whose values on $x=1,\dots, 2g+3$ are your favorite sequence of perfect squares, then $y^2=f(x)$, if smooth, is a curve of genus $g$ with $4g+6$ rational points. For a generic sequence of perfect squares it is smooth. The equation for making $f(2g+4)$ a perfect square is a quadratic form, hence satisfies the Hasse principle, so showing it has solutions $p$-adically for all $p$ could give a $4g+8$ lower bound. $\endgroup$
    – Will Sawin
    Mar 25, 2014 at 16:54

1 Answer 1


Apparently it is a conjecture of Lang that there is a bound depending on the genus only. Check out Lucia Caporaso's paper (I am sure there are more recent references).

  • 6
    $\begingroup$ To be really precise: it is a "non-trivial" consequence of a conjecture of Lang that, for all integers $d$ and $g$, there exists a real number $c(d,g)$ depending only on $d$ and $g$, such that if $X$ is a curve of genus $g$ over a number field $K$ of degree $d$ (over $\mathbf Q$), then $\# X(K) \leq c(d,g)$. This is proven by Caporaso, Harris and Mazur using the "correlation theorem" for $d=1$, and by others (Pacelli and Abramovich?) for $d>1$. $\endgroup$ Mar 25, 2014 at 16:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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