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Fix a number field $K$.

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

flag	asked Nov 19 at 21:10 René Pannekoek 1,212●3●13

Felipe Voloch · Accepted Answer · 2012-11-19 21:53:03Z UTC

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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.

answered Nov 19 at 21:53

Felipe Voloch
16.7k●2●36●61

Thank you, Felipe. The modularity theorem strikes again. :) – René Pannekoek Nov 20 at 0:18

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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 $r$. – Noam D. Elkies Nov 20 at 0:39

Thanks Noam, but why do you insist that $C$ be a complete intersection? I don't see why that is necessary. – René Pannekoek Nov 20 at 1:02

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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. – Noam D. Elkies Nov 20 at 4:30

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@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. – René Pannekoek Nov 21 at 21:05

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Are ranks of Jacobians over number fields unbounded?

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