Let $K$ be a number field, $\mathfrak{p}$ be a prime of it, and $L=K(\mathfrak{p}^n)$ be the ray class field of $K$ with finite conductor $\mathfrak{p}^n$ (we do not care about the infinite part of the conductor).


1. Is it true that $L/K$ is totally ramified at $\mathfrak{p}$?
2. (Here we may assume $L/K$ is cyclic.) Is it true that the order of the zeroth Tate cohomology group $\smash{\hat{H}}^0(\operatorname{Gal}(L/K), U\_L)$ is equal to $2^t$, where $U\_L$ is the group of units of the ring of integers of $L$ and $t$ is the number of infinite places of $K$ which ramified in $L$?



Ramification of primes and order of $\smash{\hat{H}}^0$ in ray class fields with one finite prime divisor
The answer to Question 1 is negative. Take $K = {\mathbb Q}(\sqrt{-6})$
and let ${\mathfrak p} = (2,\sqrt{-6})$ denote the prime ideal above $2$.
The class number of $K$ is $2$, its maximal unramified (abelian) extension
is $L = {\mathbb Q}(\sqrt{-3},\sqrt{2})$. The ray class number formula shows that $h\{\mathfrak m\} = h \cdot \Phi(\mathfrak m)/(E:E^{(1)})$; here $\Phi$ is Euler's Phi function in $K$. In the present case, $E$ is generated by $-1$, hence the index in the denominator is $2$ for all ideals with norm $> 4$, and we simply have $h\{\mathfrak m\} = \Phi(\mathfrak m)$. We now compute the ray class numbers $h\{\mathfrak p^m\}$:
$$ \begin{array}{c|c}
 m & h\{\mathfrak p^m\} \\ \hline 
 1 & 2 \\
 2 & 4 \\
 3 & 4 \\
 4 & 8 
\end{array} $$
Observe that $\Phi(\mathfrak p^m) = \Phi(\mathfrak p) \cdot N(\mathfrak p)^{m-1} = 2^{m-1}$ in our case.


This shows that the ray class field defined modulo $\mathfrak p^2 = (2)$ has conductor $(2)$ (because the ray class field defined modulo $\mathfrak p$ is strictly smaller) and that it contains the Hilbert class field of $K$ with conductor $(1)$.


In Question 2 there are conditions missing. The order $2^t$ is something I would expect for quadratic extensions.


