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Jan 5, 2022 at 5:25 answer added Harun Kir timeline score: 3
Nov 16, 2018 at 16:03 vote accept Dror Speiser
Apr 13, 2017 at 12:58 history edited CommunityBot
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Feb 8, 2012 at 22:03 answer added Yuri Zarhin timeline score: 14
Aug 11, 2010 at 0:42 comment added JSE By the way, the strategy of Gouvea, Mazur, Rubin, Silverberg for generating lots of quadratic twists of high rank is to find a hyperelliptic curve C on E^r (whose Jacobian thus has E^r as an isogeny quotient.) Then the presence of lots of points on C over quadratic fields (plus some care about linear independence under specialization) shows that E acquires rank at least r over lots of quadratic fields.
Aug 10, 2010 at 21:43 answer added JSE timeline score: 7
Aug 10, 2010 at 16:42 answer added Francesco Polizzi timeline score: 2
Aug 10, 2010 at 13:50 comment added David Hansen A nice example due to Jacobi: the Jacobian of $y^2=x(1-x)(1+ax)(1+bx)(1-abx)$ is a product of elliptic curves.
Aug 10, 2010 at 11:53 answer added David Lehavi timeline score: 3
Aug 10, 2010 at 8:00 answer added Dan Petersen timeline score: 9
Aug 10, 2010 at 7:12 answer added damiano timeline score: 8
Aug 9, 2010 at 23:28 history asked Dror Speiser CC BY-SA 2.5