Background and Motivation
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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 $L$.




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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.



The norm of a non-Galois extension of local fields
Yup, it is. This is Theorem III.3.5 (norm limitation theorem) of Milne's Class field theory notes (available [here](http://jmilne.org/math/CourseNotes/CFT.pdf)). The global analogue is Theorem VIII.4.8.


