Timeline for Are ranks of Jacobians over number fields unbounded?
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8 events
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| Nov 21, 2012 at 21:05 | comment | added | R.P. | @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. | |
| Nov 20, 2012 at 4:30 | comment | added | Noam D. Elkies | 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. | |
| Nov 20, 2012 at 1:07 | comment | added | stankewicz | Perhaps a simple-minded comment, but it seems then that Elkies' example shows how the unboundedness of ranks of elliptic curves implies (2) for any specific genus $g$. | |
| Nov 20, 2012 at 1:02 | comment | added | R.P. | Thanks Noam, but why do you insist that $C$ be a complete intersection? I don't see why that is necessary. | |
| Nov 20, 2012 at 0:39 | comment | added | Noam D. Elkies | 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 $r$. | |
| Nov 20, 2012 at 0:18 | vote | accept | R.P. | ||
| Nov 20, 2012 at 0:18 | comment | added | R.P. | Thank you, Felipe. The modularity theorem strikes again. :) | |
| Nov 19, 2012 at 21:53 | history | answered | Felipe Voloch | CC BY-SA 3.0 |