Ray class field and ring class field

Question

Let $K$ be quadratic number field and let $O$ be an order of $K$. A modulus $m$ of $K$ is a formal product $m_0\cdot m_\infty$ of finitely many finite primes $m_0$ and finitely many infinite real primes $m_\infty$, with real primes all different from each other. Set $m$ = $[Z_K:O]$. Let $I_{K_\Delta(L/K) }$ be the group of all fractional ideals of $Z_{K}$ prime to $\Delta(L/K)$. By $I_{K_(m) }$ we mean the group of all fractional ideals of $K$ relatively prime to $m_0$. So every prime ideal in $\Delta(L/K)$ is unramified in $L$ and define $P_{K,1(m)}$ to be the subgroup of $I_{K_m}$ generated by principal ideals $\alpha.Z_{K}$, with $\alpha\in Z_{K}$, satisfying $\alpha \equiv 1$ $ \text{mod } m_0 $ and $\sigma(\alpha)>0$ for all infinite primes $\sigma$ of $K$ dividing $m_{\infty}$. Define $P_{K,Z}{(m)}$ to be the subgroup of $I_{K_m}$ generated by the principal ideals $\alpha.Z_{K}$ with $\alpha\in Z_{K}$, satisfying $\alpha \equiv a$ $ \text{mod } m $ for a number $a$ with $gcd(a,m)=1$.

Question: 1.How to prove that $m=[Z_K:O]=1,2,3,4,6$ if and only if $P_{K,1(m)}=P_{K,Z}{(m)}$ so that Ring class field of $O$, $R_i(O)$ and Ray class field of $K$, $R_m(K)$ with modulus $m$ are equal ?

Or 2) can some one suggest me where can I find a proof for this?

Answer 1

Here $Z_k$ is the maximal order and $O=Z[\sqrt{-d}]$. Set $m=[Z_k:O]$, where $\Delta(O)=-4d=m^2 \Delta(K/Q)$.

Eg; for $K=Q(\sqrt{-27})$; $m=6$, Now $K=Q(\sqrt{-3})$ and $O=Z[\sqrt{-27}]$.

$\therefore\Delta(O)=-4\cdot27$ and $\Delta(K/Q)=-3$. Equating, $-4\cdot27=m^2\cdot(-3)$ $\Rightarrow m^2=36$, $m=6$

Answer 2

And continued from above

The cases where m is equal to 1,2,3,4,6 amount to having $ \phi(m) \le 2$, and these are the cases where the global units 1 and -1 "fill up" $(Z/mZ)^*$, and the ray class field equals the ring class field.

courtesy: Professor P Stevenhagen