Timeline for Isogeny classes of elliptic curves
Current License: CC BY-SA 3.0
10 events
when toggle format | what | by | license | comment | |
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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 |