Timeline for Rational points à la Chabauty-Coleman
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| May 28, 2013 at 1:20 | comment | added | Maarten Derickx | A small remark. I just read the paper on Bjorn Poonen his website and in his article and he talks about "Guess" and "Question" and not about conjecture. I have the feeling he does this on purpose and that this is meant to indicate that he is not really convinced yet that the answer to his question will be positive. | |
| Mar 31, 2011 at 13:30 | vote | accept | Barinder Banwait | ||
| Mar 28, 2011 at 21:31 | comment | added | Barinder Banwait | Thanks for that, it was just what I needed. In the toy example, $J$ is indeed simple, so there (assuming the conjecture) $r' = 2$, and the Chabauty-Coleman method will not work. | |
| Mar 28, 2011 at 16:28 | comment | added | Tzanko Matev | There is a conjecture due to Bjorn Poonen (www-math.mit.edu/~poonen/papers/leopoldt.pdf) which implies that if $J$ is simple, then $r'=r$. So, if you believe it you will have to pick $J$ to be a product of two elliptic curves, with ranks 2 and 0 respectively.For any Jacobian of this type $r'$ will be 1. | |
| Mar 26, 2011 at 13:03 | comment | added | Chris Wuthrich | If you know how to compute the $p$-adic logarithm on the formal group for your $J$, then you could just check by computing with sufficient precision if the image of the generators of $J(\mathbb{Q})$ are independent in the Lie algebra. | |
| Mar 26, 2011 at 2:54 | answer | added | Felipe Voloch | timeline score: 7 | |
| Mar 25, 2011 at 23:52 | history | asked | Barinder Banwait | CC BY-SA 2.5 |