# Ray class field and ring class field

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 $L$. Define $P_{K,1(m)}$ to be the subgroup of $I_{K_m}$ generated by principal ideals $\alpha\cdot Z_{K}$, 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 $\alpha\cdot Z_{K}$ 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 I can find a proof for this?

Basically, these cases of $$m$$ are where the Euler totient function $$\varphi(m)$$ is $$\leq 2$$ and the relatively prime numbers to $$m$$ are $$\equiv \pm 1$$ modulo $$m$$. The ideals $$\mathfrak{i} \in P_{K,1}(m)$$ are principal ideals equal to $$(\alpha)$$ for $$\alpha \in Z_K$$, with $$\alpha \equiv 1$$ (mod $$m$$). When $$(\alpha) \equiv 1$$ (mod $$m$$), then, as $$(-\alpha) = (\alpha)$$, we must include the case $$\alpha \equiv -1$$ as well, and thus $$P_{K,1}(m) = P_{K,Z}(m)$$.
In more detail. For example, if $$m=4$$ and $$\alpha = 5$$, then $$-5 \in (\alpha)$$. Reset $$\alpha$$, now for $$\alpha =-5$$, $$(\alpha) \in P_{K,1}$$, but with $$\alpha \equiv -1$$ (mod 4) as we have $$-5 \equiv -1$$ (mod 4). Thus, one can see why for $$\alpha \equiv 1$$ (mod 4) the case $$\alpha \equiv -1$$ (mod 4) has to be included as well.