Timeline for Do we need the Weber function to generate ray class fields of imaginary quadratic fields of class number one?
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| Jul 28, 2021 at 21:49 | comment | added | Rdrr | I don't think this is true. Even if you have an elliptic curve $E$ for which $K(E[\mathfrak{m}]) =K(h(E[\mathfrak{m}]))$, one can twist $E$ by a quadratic field $K(\sqrt{d}) \not\subseteq K(h(E[\mathfrak{m}]))$ to get a curve $E_d$. Now, $h$ doesn't depend on which twist you pick, so $K(h(E[\mathfrak{m}])) =K(h(E_d[\mathfrak{m}]))$; but $K(E_d[\mathfrak{m}]) \neq K(h(E_d[\mathfrak{m}]))$. In fact, there are infinitely many such $d$. | |
| Oct 31, 2018 at 9:08 | history | edited | yourstruly2095 | CC BY-SA 4.0 |
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| Oct 31, 2018 at 9:01 | history | edited | yourstruly2095 | CC BY-SA 4.0 |
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| Oct 31, 2018 at 8:56 | history | edited | yourstruly2095 | CC BY-SA 4.0 |
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| Oct 30, 2018 at 18:10 | comment | added | Jesse Silliman | I think the question is not asking about the general case, but the particular case of class number 1. In that case, it is known that $K(E^{tors}) = K(h(E^{tors})) = K^{ab}$. The question is whether an analogous equality holds for the $\mathfrak{m}$-torsion, i.e. what is the relationship between $K(E[\mathfrak{m}])$ and $K(h(E[\mathfrak{m}]))$. You say in your last sentence that these fields are equal -- why is this true? I have not seen this stronger statement in standard texts on this. | |
| Oct 30, 2018 at 17:17 | history | edited | yourstruly2095 | CC BY-SA 4.0 |
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| Oct 30, 2018 at 16:59 | comment | added | yourstruly2095 | Yes, indeed — thanks for pointing this out! You are of course right that there is only one isomorphism class when $K$ has class number one. In this case, the maximal abelian extension of $K$ is in fact generated by the torsion of an elliptic curve $E$ with CM by $O_K$. In general, adjoining the torsion of an elliptic curve $E$ with CM by $\mathcal{O}_K$ generates abelian extensions of the Hilbert class field $K(j(E))$ which are not necessarily abelian over $K$. The role of the Weber function $h$ (as defined geometrically as in my comment) is to select out subfields which are abelian over $K$. | |
| Oct 29, 2018 at 19:59 | comment | added | pierre de fermat | Doesn't the $\mathscr O_K$-linear Galois representation of $G_K$ on $E[\mathfrak{m}]$ depend intrinsically on $E$? I thought that in the class number $1$ case, there was just one isomorphism class of elliptic curve $E$ with CM by $\mathscr O_K$. | |
| Oct 28, 2018 at 21:20 | review | First posts | |||
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| Oct 28, 2018 at 21:17 | history | answered | yourstruly2095 | CC BY-SA 4.0 |