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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 $L$?

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    $\begingroup$ Ad 1: No; the extensions at the bottom might be unramified. Take $K = {\mathbb Q}(\sqrt{-6}\,)$ and ${\mathfrak p} = (2, \sqrt{-6})$. $\endgroup$ Dec 5, 2022 at 14:11
  • $\begingroup$ 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? $\endgroup$ Dec 5, 2022 at 23:11

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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.

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  • $\begingroup$ Thank you for your answer Professor. $\endgroup$ Dec 6, 2022 at 20:38
  • $\begingroup$ Franz Lemmermeyer: For question one, what if the class number of $K$ be one? $\endgroup$ Dec 9, 2022 at 12:26
  • $\begingroup$ 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. $\endgroup$ Dec 9, 2022 at 12:56
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    $\begingroup$ Yes, if $K$ has class number $1$ and $L/K$ is abelian of prime power conductor, then $p$ is totally ramified. $\endgroup$ Feb 24 at 20:09
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    $\begingroup$ 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. $\endgroup$ Feb 28 at 17:11

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