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I'm using the same setup as Corollary 1.7 on p. 44 of de Shalit manuscript (Iwasawa theory of elliptic curves with complex multiplication).

I think there is a mistake in his Corollary 1.7 and I'm wondering if it is possible to fix it. So let $E/F$ be an elliptic curve defined over a number field $F$ such that

(1) $E$ has complex multiplication by $\mathcal{O}_K$,

(2) $F(E_{tor})$ is abelian over $K$ (this is a strong assumption, since a priori we only know that $F(E_{tor})$ is abelian over $F$).

In particular, it follows from (2) that $K(1)\subseteq F$, where $K(1)$ stands for the Hilbert class field of $K$. Let $\mathfrak{f}\subseteq\mathcal{O}_F$ be the conductor of the Groessencharacter associated to $E/F$ (which depends just on the $F$-isogeny class of $E$).

For an integral ideal $\mathfrak{m}\subseteq\mathcal{O}_K$, we let

(a) $F[\mathfrak{m}]=F(E[\mathfrak{m}])$

(b) $F(\mathfrak{m})=F(x(P):P\in E[\mathfrak{m}])$

It is easy to see that $[L[\mathfrak{m}]:L(\mathfrak{m})]\leq 2$ and that we have an injection

$\theta_{\mathfrak{m}}:Gal(F[\mathfrak{m}]/F)\hookrightarrow (\mathcal{O}_K/\mathfrak{m})^{\times}$.

[added: Most interesting example: If $F=K(1)$ and if $\mathfrak{m}$ is such that $1\in\mu_K$ is the only root of unity $\zeta\in\mu_K$ such that $\zeta\equiv 1\pmod{\mathfrak{m}}$, then $Gal(F(\mathfrak{m})/F)\simeq (\mathcal{O}_K/\mathfrak{m})^{\times}/\mu_K$. In particular, if $\mu_K=\{\pm 1\}$ then $im(\theta_{\mathfrak{m}})$ has at most index $2$.]

Let $\mathfrak{g}\subseteq\mathcal{O}_K$ be another ideal and let us assume that $(\mathfrak{f}\cap\mathcal{O}_K) |\mathfrak{g}$ and that $(\mathfrak{m},\mathfrak{g})=1$.

---------(added on 6th of April 2017: in fact de Shalit's assumptions on the ideals $\mathfrak{g}$ and $\mathfrak{m}$ are more restrictive than what I initially wrote above)--------.

Then de Shalit claims the following:

(c) $\theta_{\mathfrak{m}}$ is surjective

(d) $F[\mathfrak{m}]$ is linearly disjoint from $F[\mathfrak{g}]$ over $F$.

In general, I don't see why $\theta_{\mathfrak{m}}$ is surjective, since in one of his computations, he forgets the contribution coming from the roots of unity of $K$. Moreover, if we take into account this omission, it seems to me that we only obtain that $F(\mathfrak{m})$ and $F(\mathfrak{g})$ are linearly disjoint over $F$.

Q1 Are (c) and (d) still true ?

There seems also to be something wrong with his proposition 1.6 (his proof of Corollary 1.7 is partly based on it), since it seems to me that one could assume from the outset that $F$ contains the coordinates of the points $E[\mathfrak{m}]$ without changing the assumptions (1) and (2) above (but may be his proposition 1.6 is correct if $F=K(1)$ which is the most interesting case).

Here is one related question to the previous paragraph (still assuming that (1) and (2) hold true) which I don't know the answer:

Q2 Let us assume that $F=K(1)$. Then we know from CM theory that $F(\mathfrak{m})$ corresponds to the ray class field of $K$ of modulus $\mathfrak{m}$. Let us assume that $[F[\mathfrak{m}]:F(\mathfrak{m})]=2$ (which I expect to be the "generic" case). What is the conductor of the abelian extension $F[\mathfrak{m}]/K$ (note in particular that this conductor must be divisible by $\mathfrak{m}$) ?

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  • $\begingroup$ Corollary 5.20.ii in Rubin's part in the Cetraro notes also says that $\theta_{\mathfrak m}$ is surjective if $\mathfrak m$ is coprime to $6\mathfrak f$, where $\mathfrak f$ is the conductor of the cooresponding Hecke character. Maybe the proof there helps. $\endgroup$ Commented Jul 19, 2016 at 19:58
  • $\begingroup$ Hi Chris thanks for the reference, I'll look at it tomorrow. Have a look also at the proof of (iv) of Lemma 21 of his paper "congruences for special values of L-functions...", where he takes into account the roots of unity in $K$ (which is assumed to have class number one, so included in the set up of my question). $\endgroup$ Commented Jul 19, 2016 at 20:14
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    $\begingroup$ Hi Chris, so I just had a look at Rubin's argument and he assumes from the outset the $E$ can be defined over K, so this means that $K$ has class number one. His proof does not seem to generalize to the more general setup of de Shalit. In fact, I suspect de Shalit statement to be false. $\endgroup$ Commented Jul 20, 2016 at 10:19

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The proof of Corollary 1.7 is fine. I had misunderstood his proof. His proof uses in a crucial way his assumption (ii) which appears on the top of p. 41. As is explained on p. 41, this assumption implies that the Groessencharacter $\psi$ associated to E/F (a Groessencharacter on F) comes from a Groessencharacter $\varphi$ on K of type $(1,0)$.

Le me just repeat de Shalit's argument in a hopefully slightly more detailed way:

(1) We have $[F[E[{\frak{m}}]]:F]\leq \# (\mathcal{O}_{K}/\frak{m})^{\times}$, $K({\frak{g}})=F(E[{\frak{g}}])$ (from his Proposition 1.6) and $K({\frak{m}\frak{g}})=F[E[{{\frak{m}}}]]\cdot K({\frak{g}})=K(E[{\frak{m}}{\frak{g}}])$ (again from his Proposition 1.6)

(2) The key observation now is that the set of roots of unity of $F[\frak{g}]$ which are congruent to $1$ modulo $\frak{g}$ is reduced to $\{1\}$. This comes from our assumption that the conductor of $\varphi$ divides $\frak{g}$ and that $\varphi$ has type $(1,0)$ (I had missed the type $(1,0)$ assumption initially).

From (2), it follows that

(3) $[ K( {\frak{m}} {\frak{g}} ) : K ({\frak{g}}) ]=\# (\mathcal{O}_{K}/\frak{m})^{\times}$.

The result now follows by incorporating (1) and (3) into the Hasse diagram which appears below his Corollary 1.7. Note that

(a) the linear disjointness of $F(E[\frak{g}])$ and $F(E[\frak{m}])$ over $F$

and the equality

(b) $[F[E[{\frak{m}}]]:F]=\# (\mathcal{O}_{K}/\frak{m})^{\times}$

are proved simultaneously!

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