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 $L$. 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?

Answer 1

The question was asked and answered by the same user on MSE.

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.

For more on this, see: Andrew Sutherland's 18.785 Number theory I Fall 2017 Lecture #21 (11/20/2017) at MIT.edu. Specifically, Example 21.6.