Let $K$ be an imaginary quadratic field and suppose that $E/K$ has complex multiplication by $\mathcal{O}_K$. Let $\psi$ be the Hecke character associated with $E$ and $\mathfrak{f}$ its conductor (i.e it is the integral ideal of $\mathcal{O}_K$ coinciding exactly with the primes of bad reduction of $E$). Let $\mathfrak{a} \triangleleft \mathcal{O}_K$ and $K(\mathfrak{a})$ denote the ray class field of $K$ modulo $\mathfrak{a}$.

Given that $\mathfrak{a}$ is prime to $6\mathfrak{f}$ and $E[\mathfrak{af}] \subseteq E(K(\mathfrak{af}))$ and that $\mathrm{Gal}(K(E[\mathfrak{a}])/K) \hookrightarrow (\mathcal{O}_K/\mathfrak{a})^\times$, show that in fact $\mathrm{Gal}(K(E[\mathfrak{a}]/K) \cong (\mathcal{O}_K/\mathfrak{a})^\times$.

I have been struggling to prove this for a long while now. It is effectively the content of Corollary 5.20 of Rubin's paper found here . I don't particularly understand the assertions regarding the kernel and isomorphism that follow the first paragraph in the proof. Does anyone have any ideas? Help with the other assertions of Corollary 5.20 (iii - v) would also be greatly appreciated.


P.S I have crossposted this from MathSE as I have yet to receive an answer there.


Given your assumption that $E/K$ has CM by $\mathcal{O}_K$, it follows that $K$ has class numer $1$. In particular, there are isomorphisms $\mathrm{Gal}(E[\mathfrak{a}])/K\cong\mathrm{Gal}(K(\mathfrak{a})/K)$ as well as $\mathrm{Gal}(K(\mathfrak{a})/K)\cong(\mathcal{O}_K/\mathfrak{a})^\times$, the first coming from the main theorem of Complex Multiplication stating that adding $\mathfrak{a}$-torsion points together with $j(\mathcal{O}_K)$ gives you the ray class field; and the second coming from Class Field Theory, which explicitely describes the difference between ray and Hilbert class fields. The inclusion in your statement needs therefore be an isomorphism.

  • $\begingroup$ Thanks for the quick reply! This answer makes sense to me but I'm trying to argue directly from Rubin's proof (and the rest of his paper). In particular, he doesn't show that $K(E[\mathfrak{a}]) = K(\mathfrak{a})$. All that he has proved is that $E[\mathfrak{af}] \subseteq K(\mathfrak{af})$. It is from this stage that I am trying to understand the proof. Do you have any ideas given these assumptions? $\endgroup$ – Alexandre Daoud Apr 1 '17 at 17:48
  • $\begingroup$ I think that the different points in Rubin's corollary aren't logically connected. For instance, the first and the second point have different assumptions. If you care about point (ii), just concentrate ln the second item in the proof. $\endgroup$ – Filippo Alberto Edoardo Apr 2 '17 at 9:32

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