Timeline for Elliptic Curves with CM and Class Field Theory
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 30, 2019 at 19:34 | comment | added | Rdrr | I don't follow why $K(j(E_2))\supseteq K(j(E_1))$ implies there is an isogeny $E_2\to E_1$. Using the fact that $E_2$ is isogenous to at most one twist of $E_1$, I can replace, $E_1$ with a twist over $\mathbb{Q}(j(E_1))$, such that we will still have the containment of ring class fields but $E_2$ wont be isogenous to $E_1$. | |
| Oct 28, 2014 at 14:36 | comment | added | Damien Robert | Sorry, I missed your comment! It comes from the fact that all isogenies of degree $m$ starting from $E_1$ will give you an elliptic curve $E_3$ with endomorphism ring of conductor a divisor of $m$, so in particular $\mathrm{End}(E_3) \supset \mathcal{O}$. This means that $E_3$ is rational over $F$, so the isogeny is rational over $F$. In particular the Galois action is of the form $\lambda \mathrm{Id}$ on the $m$-torsion; but since we have a rational point $\lambda=1$. | |
| Apr 24, 2014 at 18:06 | comment | added | Pete L. Clark | Upon further thought, I am having some trouble filling in the details of this answer in the general case. I see that if $E_2$ has full $N$-torsion over a field $F$ containin the CM field, then (i) $F$ contains the $N$th roots of unity and (ii) $E_1$ has at least one $F$-rational point of order $N$. This means that the mod $N$ Galois rep on $E_1$ has the shape $[[1 * ][0 1]]$. That last $*$ means that we get full $N$-torsion via a cyclic extension of degree dividing $N$. If $N$ is unramified in the CM field, this forces the torsion to be rational over $F$. But if $N$ is ramified....? | |
| Apr 15, 2014 at 18:02 | history | edited | Pete L. Clark | CC BY-SA 3.0 |
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| Apr 15, 2014 at 11:50 | history | answered | Damien Robert | CC BY-SA 3.0 |