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Timeline for Isogeny classes of elliptic curves

Current License: CC BY-SA 3.0

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Sep 16, 2012 at 15:39 answer added Ari Shnidman timeline score: 11
Sep 16, 2012 at 14:11 vote accept LMN
Sep 16, 2012 at 13:52 answer added stankewicz timeline score: 16
Sep 16, 2012 at 13:29 comment added Damian Rössler @Will Savin: you mean "countably many isogenous curves"...
Sep 16, 2012 at 13:18 comment added LMN @Damian: That's correct.
Sep 16, 2012 at 10:00 comment added R.P. Dror: you probably mean isogenous, not isomorphic.
Sep 16, 2012 at 8:50 comment added Dror Speiser A curve over a galois extension of $\mathbb{Q}$ that isomorphic to all of its conjugates is called a $\mathbb{Q}$-curve, and by work of Ribet and the recent proof of Serre's conjecture, these are known to be modular (classic $GL_2(\mathbb{A}_\mathbb{Q})$). I think this forces the $j$-invariant, being some integration of a modular form on the upper half plane, to be in either a CM, or a totally real, extension of $\mathbb{Q}$. So any $j$-invariant with field of definition not CM or totally real, should provide a counter example. An expert should verify this, or, more probable, refute this.
Sep 16, 2012 at 7:08 comment added Will Sawin I mean it's clear that if $j$ is transcendental then an automorphism of $\mathbb C$ takes it to any other transcendental $j$ invariant, which is an uncountable set, but there are uncountably many isogenous curves. But there should also be counterexamples for $j$ algebraic.
Sep 16, 2012 at 5:56 comment added Damian Rössler Isogenous is not the same as isomorphic (ie $\simeq$). Are you asking whether in general an elliptic curve over $\bf C$ is isogenous to any of its conjugates by an automorphism of $\bf C$ ? This is very likely not true although I don't have a counterexample at hand.
Sep 16, 2012 at 4:39 history asked LMN CC BY-SA 3.0