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 in the text, 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.

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

(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 $\psi$ divides $\frak{g}$ and that $\varphi$ has type $(1,0)$.

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) and 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

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

are proved simultaneously!

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!

The proofs of Corollary 1.7 is fine. I had misunderstood his proof. Le me just repeat de Shalit's argument.

(i) We have $[F[E[{\frak{m}}]]:F]\leq \# \mathcal{O}_{K}/\frak{m}$ and $K({\frak{m}\frak{g}})=F[E[{{\frak{m}}}]]\cdot K(\frak{g})$

The key observation 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 the Groessencharacter associated to $E/F$ divides $\frak{g}$). From the previous observation, it follows from class field theory that

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

(iii) The result now follows by combining (i), (ii) and the Hasse diagram which appears below Corollary 1.7.

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 in the text, 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.

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

(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 $\psi$ divides $\frak{g}$ and that $\varphi$ has type $(1,0)$. 

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) and 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

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

are proved simultaneously!