## Background and Motivation

Local Class Field Theory says that abelian extensions of a finite extension $K/\mathbb{Q}_p$ are parametrized by the open subgroups of finite index in $K^\times$. The correspondence takes an abelian extension $L/K$ and sends it to $N_{L/K}(L^\times)$, and this correspondence is bijective.

If one starts instead with a galois extension $L/K$ that isn't abelian, one can then ask "What abelian extension does $N_{L/K}(L^\times)$ correspond to?" The answer is the maximal abelian extension of $K$ contained in $L$.

The hypothesis of being galois isn't necessary in the statement of the non-abelian theorem: both the question and the answer still make sense. I am thus asking

Assume that $L/K$ as above. Is the abelain extension of $K$ corresponding to $N_{L/K}(L^\times)$ the maximal abelian extension of $K$ contained in $L$?

A couple examples to illustrate this problem (including the example that I was told would sink this):

If $p > 2$, then consider $L = \mathbb{Q}_p(\sqrt[p]{p})$. The norm subgroup that I am anticipating is all of $\mathbb{Q}_p^\times$. Moving into the galois closure and using the theorem, one gets that $N_{L/\mathbb{Q}_p}(L^\times)$ contains $p^\mathbb{Z} \times (1 + p\mathbb{Z}_p)$. Moreover, the norm of a number $a \in \mathbb{Z}_p$ will just be $a^p$, which is congruent to $a$ mod $p$, so the norm will also hit something congruent to any given root of unity in $\mathbb{Q}_p$, and that was all that I was missing from the earlier note.

One can also, using the same idea (moving into the galois closure and getting a lot of information from that, and then using the explicit structure of the field that one started with) show that this works for $K(\sqrt[n]p)$ with $(n, p^{f(K/\mathbb{Q}_p)}) = (n,p) = 1$ as well.