Welcome to MathOverflow

Question

11

1

It's fairly easy to write down families of elliptic curves over $\mathbb{Q}(t)$ such that almost every (i.e. when the "height" of $t$ is sufficiently large) curve in the family has positive rank over $\mathbb{Q}$. One can do this by constructing the fibration so it has sections /$\mathbb{Q}$ a priori, or by fiddling around with Gauss sums; see papers of Fermigier, Mestre, Arms/Miller/Lozano-Robledo, etc. Are there any known examples of families of genus two curves over $\mathbb{Q}(t)$ such that the Jacobian of almost every curve $C_t$ in the family has provably positive rank? Can one do this while requiring that $\mathrm{Jac}(C_t)$ be $\overline{\mathbb{Q}}$-simple for almost all $t$?

flag	asked Aug 9 2010 at 20:55 David Hansen 8,429●2●18●48

I do not know of a write up of an example, but have you tried the same tricks that are used for elliptic curves? E.g. let $S \to \mathbb{P}^1_{\mathbb{Q}}$ be a rel curve of genus 2, $S$ smooth proj surface, and suppose $\Sigma_1,\Sigma_2$ are sections. Make sure that the line bundle $\Sigma_1-\Sigma_2$ is a non-torsion, e.g. by choosing it so that the reductions modulo different primes in two fibers have coprime order. Presumably, simplicity of the Jacobians should also be easy, at least for a positive proportion of fibers. Not sure how to guarantee simplicity for almost all $t$. – damiano Aug 9 2010 at 22:22

1

$y^2 = x^5 + x + t^2$ the point infinity - (0,t) has infinite order. – Felipe Voloch Aug 9 2010 at 22:39

@Felipe: Very nice, what's the simplest way to see this? Is there Lutz-Nagell for preimages of torsion in the Jacobian? – David Hansen Aug 10 2010 at 0:10

@David: There is no Lutz-Nagell or a particularly easy way to do this. One possibility is to use the Manin map. Another is an algorithm of Poonen's (Computing torsion points on curves) for curves over Q. In his paper he does $y^2=x^5-x+1$ and find that there is no torsion of the form infinity-point except for the ones with y=0. This implies that $y^2=x^5-x+t^2$ will work. A similar calculation can be done for my first example. Like Borcherds said, pretty much anything works. – Felipe Voloch Aug 10 2010 at 2:26

For families with large rank look at Elkies' paper: math.rice.edu/~hassett/conferences/Clay2006/… – Victor Miller Aug 10 2010 at 4:06

show 3 more comments

Richard Borcherds · Accepted Answer · 2010-08-09 22:25:02Z UTC

My guess for some examples is the family of (genus 2) hyperelliptic curves y²=degree 6 poly in x passing through n "randomly chosen" rational points (for n=2,3,4,5, or 6). The family of such curves has dimension 7-n, and I would guess that if a hyperelliptic curve has n "random" rational points on it then the Q rank of its Jacobian is usually at least n-1.

An obvious place to look for explicit examples is
MR1406090 Cassels, J. W. S.; Flynn, E. V. Prolegomena to a middlebrow arithmetic of curves of genus 2. London Mathematical Society Lecture Note Series, 230. Cambridge University Press, Cambridge, 1996. ISBN: 0-521-48370-0

Families of genus 2 curves with positive rank jacobians

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.