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Feb 18, 2011 at 5:44 comment added Pete L. Clark @Chris: very true.
Feb 8, 2011 at 16:47 comment added Chris Wuthrich You won't need BSD: If the analytic rank of the twist of $E$ corresponding to $K$ were zero, then, by Kolyvagin, the rank of this twist would be zero, too.
Feb 4, 2011 at 14:16 comment added Pete L. Clark @Chris: well, it seems to me that you are imposing one (very mild) additional hypothesis about the $2$-torsion on $E(K)$ and then using some form of BSD to get from the rank to the analytic rank. But notwithstanding those cavils, you have a very nice idea here! I'll think about it and get back to you...
Feb 4, 2011 at 10:52 comment added Chris Wuthrich If the mod-2 representation of the Jacobian $E$ of $C$ is surjective, then there are no $2$-torsion points on $E(K)$ for quadratic $K$. So the rank of $E$ has to grow in every quadratic $K/\mathbb{Q}$ and so must the analytic rank. But that is not possible by non-vanishing results, I believe. Do I miss something ?
Feb 4, 2011 at 1:34 vote accept Eric Larson
Feb 4, 2011 at 1:34
Feb 4, 2011 at 1:03 history answered Pete L. Clark CC BY-SA 2.5