A type of principal ideal theorem of class field theory for ramified primes

Question

Let $K$ be a number field and $\mathcal{O}_K$ be its ring of integers. Also let $p$ be a prime number, $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$ and $\zeta_{m}$ be a primitive mth root of unity. I have two questions:

1. When we work over $‎\mathbb{Q}$, the (finite) abelian extensions of $‎\mathbb{Q}$ which are unramified outside prime $p$ are subfields of $L=‎\mathbb{Q}(\zeta_{p^{r}})$ for some integer $r$. In fact, the ray class field of $‎\mathbb{Q}$ associated to the modulus $p^r$ is $L=‎\mathbb{Q}(\zeta_{p^{r}})$. Then, by a generalization of classical principal ideal theorem for Hilbert class field (see for example this page of MO), any unramified prime of $\mathbb{Q}$ (all primes other than $p$) becomes principal in $L=‎\mathbb{Q}(\zeta_{p^{r}})$. But for $p$, which is ramified, we have $p\mathcal{O}_L = (1-\zeta_{p^{r}})^{[L:‎\mathbb{Q}]}$, i.e. $p\mathcal{O}_L$ is principal and totally ramified.

   I think in any number field $K$ instead of $\mathbb{Q}$, for any prime power modulus $\mathfrak{p}^r$, the ray class field modulo $\mathfrak{p}^r$, $L=K(\mathfrak{p}^r)$, has this property: the unique ramified prime of $K$ becomes principal and totally ramified in $L$, i.e. $\mathfrak{p}\mathcal{O}_L=(a)^{[L:K]}$ for some $a \in L$.
   But I can't prove this, or find a theorem that say this.

2. When we work over $‎\mathbb{Q}$, the ray class field for any modulus $m=p_1^{r_1}p_2^{r_2}...p_n^{r_n}$ is $L=\mathbb{Q}(\zeta_m)$. In this case ramified primes $p_i$ maybe don't become principal in $L$, but the product of all prime ideals of $L$ above $p_i$ is principal. In fact this product is equal to $(1-\zeta_{p_i^{r_i}})\mathcal{O}_L$. Remember that we saw unramified primes become principal in $L$.

   I think in any number field $K$ instead of $\mathbb{Q}$, for any modulus $\mathfrak{m}$, the ray class field modulo $\mathfrak{m}$, $L=K(\mathfrak{m})$, has this property: for any (finite) $\mathfrak{p}|\mathfrak{m}$, the product of all prime ideals of $L$ above $\mathfrak{p}$ is principal.
   But I can't prove this, or find a theorem that say this.

Answer 1

It is not true that the prime ${\mathfrak p}$ is totally ramified in the ray class field of a number field defined modulo ${\mathfrak p}$. For example, the ray class field modulo $3$ over the rationals is trivial, and the ray class field of $K = {\mathbb Q}(\sqrt{-6})$ modulo the prime ideal $(2, \sqrt{-6})$ above $2$ is equal to the Hilbert class field $K^1 = {\mathbb Q}(\sqrt{2}, \sqrt{-3})$ of $K$: it has conductor $1$, and $(2, \sqrt{-6})$ is not ramified.

For the second question, choose a number field $K$ and a prime ideal ${\mathfrak p}$ such that the prime ideal ${\mathfrak P}$ above ${\mathfrak p}$ in the Hilbert class field of $K$ is not principal. You are claiming that ${\mathfrak P}$ becomes principal in the ray class field modulo ${\mathfrak P}$ of $K$; to me it seems unlikely to be true for every choice of $K$ and ${\mathfrak P}$.