Understanding the correspondence between $\mathbb{Q}_p^\times$ and $ Gal(\mathbb{Q}_p^{ab}/\mathbb{Q}_p)$ I started studying local class field theory by reading the book "Number theory 2" by Kato, Kurokawa and Saito. I want to understand some examples for the correspondence between the  multiplicative group of $K$ and the galoisgroup of its abelian closure to get the main idea before reading the proof. $\mathbb{C}$ and $\mathbb{R}$ are pretty easy so I tried the p-adic numbers $\mathbb{Q}_p$. Here I have problems with finding examples for the correspondence. For example what subgroup of $\mathbb{Q}_p^\times$ would correspont to $\mathbb{Q}_p(\zeta_{n})$, where $n$ is a $p$-th power root of unity. Or to some $n$ that is not divisible by $p$.
I have also problems with calculating  $ N_{\mathbb{Q}_p(\zeta_{n})/\mathbb{Q}}\mathbb{Q}_p(\zeta_{n})^\times $.  While it is very easy to calculate $N_{\mathbb{C}/\mathbb{R}}\mathbb{C^\times}$ I do not really know how to do this here.  
I would not only appreciate answers but also hints where to find explicit calculations and examples. 
 A: Here are my 'hints'--it's most of the solution. I would suggest reading and trying to preempt the next line if you want to solve it on your own. $\newcommand \Q{\mathbb{Q}}$ $\newcommand \Art{\mathrm{Art}}$ $\newcommand \ab{\mathrm{ab}}$ $\newcommand \Gal{\mathrm{Gal}}$ $\newcommand \Z{\mathbb{Z}}$
1) Let's break the calculation of $N(\Q_p(\zeta_n)^\times)\subseteq\Q_p^\times$ into steps. Namely, let's factor $n=p^k r$ with $(r,p)=1$. Figure out why it suffices to compute $N(\Q_p(\zeta_r)^\times)$ and $N(\Q_p(\zeta_{p^k})^\times)$.
2) Let's handle the case of $\Q_p(\zeta_r)$. Now, since this extension is unramified (why?) we can uniquely write it as $\Q_p(\zeta_{p^m-1})$ for some $m$ (namely $m=[\Q_p(\zeta_r):\Q_p]$). Consider now the Artin homomorphism composed with the quotient map to $\Gal(\Q_p(\zeta_{p^m-1})/\Q_p)$:
$$\varphi:\Q_p^\times\xrightarrow{\Art}\Gal(\Q_p^\ab/\Q_p)\twoheadrightarrow\Gal(\Q_p(\zeta_{p^m-1})/\Q_p)\cong \Z/m\Z$$
Note that since $\Art$ induces an isomorphism $\Z_p^\times\xrightarrow{\approx}I_{\Q_p}$ and $\Q_p(\zeta_{p^m-1})/\Q_p$ is unramified we must have that $\Z_p^\times\subseteq\ker\varphi$. But, note that $\Q_p^\times/\Z_p^\times\cong\Z$ (with a splitting given by choosing any uniformizer in $\Z_p$) and thus our map $\varphi$ must factor as a map $\Z\to\Z/m\Z$ and thus its kernel is precisely $m\Z$. Thus, given the splitting corresponding to the uniformizer $p$ we get that $\ker\varphi = p^{m\Z}\times\Z_p^\times$. Since $\ker\varphi=N(\Q_p(\zeta_{p^m-1})^\times)$ by CFT this allows us to conclude that $N(\Q_p(\zeta_{p^m-1})^\times)=p^{m\Z}\times \Z_p^\times\subseteq\Q_p^\times$.
3) Let's now consider the totally ramified case $\Q_p(\zeta_{p^k})/\Q_p$. Again, let's consider the composition
$$\Q_p^\times\xrightarrow{\Art}\Gal(\Q_p^\ab/\Q_p)\to\Gal(\Q_p(\zeta_{p^k})/\Q_p)$$
Now, let's choose the uniformizer $p$ in $\Z_p$ to give us the decomposition $\Q_p^\times=p^{\Z}\times\Z_p^\times$. Now, note that since $\Q_p(\zeta_{p^k})/\Q_p$ is totally ramified that the entire $p^\Z\subseteq\ker\varphi$. Moreover, note that $\Z_p^\times$ decomposes as $\zeta_{p-1}\times U^1$ where $U^1$ is the $1$-units. Finally, note that $U^1\cong\Z_p$ as a topological group. Thus, we have figured out that we have a continuous surjection 
$$\varphi:p^{\Z}\times\mu_{p^-1}\times \Z_p\twoheadrightarrow \Gal(\Q_p(\zeta_{p^k})/\Q_p)\cong \Z/p^{k-1}(p-1)\Z$$
and, as already noted, $p^{\Z}\subseteq\ker\varphi$. Since $\Z_p$ only has continuous finite quotients $\Z/p^\ell\Z$ for all $\ell$ it's easy to deduce that the $\mu_{p-1}$ surjects (and thus injects) onto the $\Z/(p-1)\Z$ part of $\Z/p^k(p-1)\Z$ leaving $\Z_p$ to surject onto the $\Z/p^k\Z$ part and thus having kernel $p^k\Z_p$. But, we want to write this canonically so note that under the canonical '$\log$' isomorphism $U^1\cong\Z_p$ one has that $p^k\Z$ corresponds to the '$k$-units' $1+p^k\Z_p$. Thus, all-in-all we see that $\ker\varphi =p^\Z\times (1+p^k\Z_p)$. So, again, using CFT we conclude that 
$$N(\Q_p(\zeta_{p^k})^\times)=p^\Z(1+p^k\Z_p)\subseteq\Q_p^\times$$

So, what other sort of examples are you after? I can likely provide them if you can be more specific about what you're looking for.

PS, note that we've incidentally proved local Kronecker-Weber by the above calculation. Namely, suppose that $K/\Q_p$ is any abelian extension. Then, $N(K^\times)\subseteq\Q_p^\times$ is finite index and thus contains $p^{n\Z}(1+p^m\Z_p)$ for some $n,m\geqslant 0$ (this is evident from the decomposition $\Q_p^\times=p^\Z \mu_{p-1}(1+\Z_p)$) and thus if $N(K^\times)\supseteq p^{n\Z}(1+p^m\Z_p)=N(\Q_{\zeta_{(p^n-1)p^m)}}^\times)$ and thus, from CFT, we know that $K\subseteq\Q_p(\zeta_{(p^n-1)p^m)})$.  
