Timeline for Is there an infinite family of primes $q_{1},q_{2},...$ so that the rank of $E(\mathbb{Q}(\sqrt{-q_{i}}))$ equals that of $E(\mathbb{Q})$?
Current License: CC BY-SA 3.0
7 events
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| Jul 16, 2016 at 9:33 | vote | accept | The Thin Whistler | ||
| Jul 15, 2016 at 9:10 | history | edited | The Thin Whistler | CC BY-SA 3.0 |
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| Jul 14, 2016 at 13:51 | answer | added | The Thin Whistler | timeline score: 3 | |
| Jul 13, 2016 at 11:54 | comment | added | Pasten | One can obtain what you want if the elements of $Q$ are only required to be squarefree (rather than primes). This follows from results in MR1482805 (say) and the analytic rank 0 case of BSD. If you insist that the elements of $Q$ be primes, then the necessary analytic result can be obtained under GRH by a slight modification of MR1106677 (perhaps it is known unconditionally, but I don't know). | |
| Jul 13, 2016 at 11:46 | comment | added | Chris Wuthrich | Not too keen to look for references, but this should be provable. Use some analytic results on the non-vanishing of twists of the $L$-function of $E$ by quadratic characters. There should be a large density of those. (2) then is a congruence condition that can be imposed, too. The result by Kato implies then that the rank of $E$ does not grow in the corresponding quadratic extensions. | |
| Jul 13, 2016 at 11:00 | history | edited | The Thin Whistler | CC BY-SA 3.0 |
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| Jul 13, 2016 at 10:50 | history | asked | The Thin Whistler | CC BY-SA 3.0 |