Timeline for Another question related to the isogeny theorem for elliptic curves
Current License: CC BY-SA 3.0
8 events
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| Oct 10, 2012 at 20:41 | comment | added | Adam Harris | @Tommaso:Yes, its an interesting model theoretic question too. I think it should be the case that you can recover the root of the class polynomial which corresponds to $j(\mathcal{O})$ (where $\mathcal{O}$ is the order) over the field $\mathbb{Q}(S)$ there $S$ is the set of $j$ values of all CM elliptic curves. | |
| Oct 10, 2012 at 13:34 | comment | added | Tommaso Centeleghe | The problem seems to be that you cannot define j of an ideal class without first regarding it as a lattice in the complex plane C. Right? A posteriori the two embeddings of R in C will tell you that there are two reasonable j-invariants you can attach to each ideal class, they are complex conjugate of each other. Another point about the real j-invariant, is that if you give yourself just the Class polynomial, it is not clear how to get the real root. I think you should first split it in the complex numbers. I find these questions interesting. | |
| Oct 10, 2012 at 13:07 | comment | added | Adam Harris | @Tommaso: I'm not sure what you mean. Aren't the solutions of the class polynomial exactly the j-invariants of the corresponding ideals? In which case - why do you have to mention the j-invariant? I'm not sure why you can't just look for a real solution of the class polynomial? | |
| Oct 10, 2012 at 12:22 | comment | added | Tommaso Centeleghe | I have the feeling that in order to set up your bijection one would have to define what is the j-invariant of an element of the ideal class group H_R, and then make elliptic curves and ideal classes correspond though j. However this requires the choice of an orientation of R\otimes\reals, i.e., of an embedding of R into the complex #s. right? | |
| Oct 10, 2012 at 11:31 | vote | accept | Adam Harris | ||
| Oct 10, 2012 at 8:00 | comment | added | Adam Harris | yes - this was my initial motivation for asking the question, and whether any more can be done when the class number isn't odd | |
| Oct 10, 2012 at 7:43 | comment | added | Tommaso Centeleghe | In case the class group of $R$ is odd, perhaps we can recognize the identity element on the elliptic curve side: it should be the one with real j-invariant, right? Corresponding to the unique lattice isomorphic to its complex conjugate. | |
| Oct 9, 2012 at 23:09 | history | answered | Joe Silverman | CC BY-SA 3.0 |