7
$\begingroup$

Fix a number field $K$.

  1. Is the rank of $J(K)$ unbounded, where $J$ ranges over the Jacobians of all smooth, projective, geometrically connected curves over $K$?

  2. Does there exist an integer $g$ such that the rank of $J(K)$ is unbounded, where $J$ now ranges over the Jacobians of all smooth, projective, geometrically connected curves of genus $g$ over $K$?

I expect (perhaps naively) that the answer to 1. is "yes". Maybe one can write down an explicit family of superelliptic curves and use descent to show that their ranks are not bounded. Or else there may be a construction that, given an elliptic curve $E$ and integer $r$, produces a curve $C$ such that $\mathrm{Jac}(C)$ contains a factor $E^r$ up to isogeny. But I can't make either approach work.

On the other hand, I would be surprised if 2. were known, since it is so famously open in the case $g=1$.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
7
$\begingroup$

The answer to 1. is yes. Take $J_0(N)$ the Jacobian of the modular curve $X_0(N)$ over the rationals. Since all elliptic curves of conductor dividing $N$ are factors of $J_0(N)$ and there are infinitely many isogeny classes of elliptic curves over the rationals with positive rank. My guess, just like yours, is that 2. is open.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Thank you, Felipe. The modularity theorem strikes again. :) $\endgroup$
    – R.P.
    Commented Nov 20, 2012 at 0:18
  • 4
    $\begingroup$ No need to cite modularity here, or even use modular curves at all: just fix an elliptic curve $E$ of positive rank, and let $C$ be a random curve in $E^r$, say a complete intersection for which none of the $r$ $E$-valued coordinates is constant. Then $J(C)$ has rank at least $r$. $\endgroup$
    – Noam D. Elkies
    Commented Nov 20, 2012 at 0:39
  • $\begingroup$ Thanks Noam, but why do you insist that $C$ be a complete intersection? I don't see why that is necessary. $\endgroup$
    – R.P.
    Commented Nov 20, 2012 at 1:02
  • 1
    $\begingroup$ No, it doesn't have to be a complete intersection; that's just a convenient way to construct curves in $E^r$. (I wrote "say a complete intersection", which is not at all insistent...) And yes, unboundedness of $g=1$ ranks would then imply unboundedness for each $g>1$. A similar question is often asked about ranks of simple abelian varieties (or Jacobians) of given dimension to avoid this reduction to genus 1. $\endgroup$
    – Noam D. Elkies
    Commented Nov 20, 2012 at 4:30
  • 3
    $\begingroup$ @Noam Elkies: A belated remark. You not only want the projections to be non-constant, but you'll want them to be "independent". For instance, simply choosing $C$ to be the diagonal $E \subset E^r$ clearly won't do the job. So I'm guessing the condition must be that the images of the $r$ pull-back maps $H^0(E,\Omega_E) \rightarrow H^0(C,\Omega_C)$ must span an $r$-dimensional subspace. Alternatively, someone suggested to me that one could look at curves $C$ in $\prod_{i=1}^r E_i$, where the $E_i$ are non-isogenous elliptic curves of positive rank, with non-trivial projections to each factor. $\endgroup$
    – R.P.
    Commented Nov 21, 2012 at 21:05
3
$\begingroup$

The answer to Question 1 is already contained in the following paper:

ANDRÉ NÉRON, Problèmes arithmétique et géométriques rattachés à la notion de rang d’une courbe algébrique dans un corps, Bulletin de la S. M. F., tome 80 (1952), p. 101-166. http://archive.numdam.org/ARCHIVE/BSMF/BSMF_1952__80_/BSMF_1952__80__101_0/BSMF_1952__80__101_0.pdf

The corollary to Theorem 7 (see page 155 of the article) says that for any given number field $K$, there are infinitely many curves $C$ of genus $g$ such that $\text{Jac}(C)(K)$ has rank at least $3g+5$. (With a bit more work, one can get $3g+6$. And with even more work, for $g=1$, the author gets up to rank at least $11$.)

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .