# Ramification of primes and order of $\smash{\hat{H}}^0$ in ray class fields with one finite prime divisor

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 $$L$$?

• Ad 1: No; the extensions at the bottom might be unramified. Take $K = {\mathbb Q}(\sqrt{-6}\,)$ and ${\mathfrak p} = (2, \sqrt{-6})$. Dec 5, 2022 at 14:11
• Franz Lemmermeyer: Is it possible to explain more about your answer Professor? I didn't get the point I think. And also, what is $K(\mathfrak{p})$ here? Dec 5, 2022 at 23:11

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.
• Franz Lemmermeyer: For question one, what if the class number of $K$ be one? Dec 9, 2022 at 12:26
• Then every abelian extension of $K$ is ramified; if the conductor is a power of a prime ideal, only this ideal can (and thus must) ramify. Dec 9, 2022 at 12:56
• Yes, if $K$ has class number $1$ and $L/K$ is abelian of prime power conductor, then $p$ is totally ramified. Feb 24 at 20:09
• Because an abelian extension is the compositum of cyclic extensions. At the bottom you have cyclic extensions of prime power conductor that must ramify somewhere since $K$ has class number $1$. But then the prime ramifies all the way up. Feb 28 at 17:11