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Aug 10 at 18:15 comment added Noam D. Elkies @WillSawin right, I didn't want to complicate matters. (For j=0 or 1728 we must also assume K is not of characteristic 2 or 3, in which case things get even more complicated.)
Aug 9 at 19:27 comment added Will Sawin You of course know, but one should also mention that for curves of $j$ invariant $0$ and $1728$ there are sextic and quartic twists, and unless every element is a square and every element is a cube there will be two non-isomorphic elliptic curves with the same $j$ invariant.
Aug 9 at 18:31 history answered Noam D. Elkies CC BY-SA 4.0