For a local field $E$, denote by $U(E)$ the units of the corresponding valuation ring $\mathcal{O}\_E$, and denote by $U\_n(E)$ the prinicipal $n$-units, i.e. $U\_n(E)=1+M\_E^n$ where $M\_E$ is the maximal ideal of $\mathcal{O}\_E$.


Now let $L/K$ be a finite unramified extension of local fields, of characteristic $0$, say. Then we have the norm map $N\_{L/K}:U(L)\rightarrow U(K)$. 


**The Question:** is there a simple description of $N\_{L/K}^{-1}(U\_n(K))$?


I know it has to contain $U\_n(L)$, and by Local Class Field Theory, I know that $[U(L):N\_{L/K}^{-1}(U\_n(K))]=(q-1)q^{n-1}$, where $q$ is the size of the residue field of $K$. It is also not hard to find which roots of unity of $L$ are contained in $N\_{L/K}^{-1}(U\_n(K))$. Apart from that, I haven't been able to say much. I thought that maybe there is some simple description of this group, given that this is the unramified case, but couldn't find a reference.


Thank you!



Inverse image of norm map on principal units for an unramified extension
Let $\pi$ be a uniformizer for $K$ and $L$.


For sufficiently large $m$, the $\log$ and $\exp$ maps give isomorphism between $U\_m(L)$ and $\pi^m \mathcal{O}\_L$. Same for $K$. This $m$ should be thought as a fixed "small" integer.


These maps commute with Galois action, i.e. for any $\sigma \in Gal(\overline{K}/K)$, viewed as a map $\sigma: L \rightarrow L^\sigma$, we have: $\log(x^\sigma) = (\log(x))^\sigma$.


Consequently, for the norm map $N=N\_{L/K}$, we have: $\log(N(x)) = Tr(\log(x))$, where $Tr=Tr\_{L/K}$ is the trace map.


So the question is essentially: what is the inverse image of $\pi^n \mathcal{O}\_K$ in $\pi^m \mathcal{O}\_L$ under the trace map?




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Example: $L = K[\sqrt{d}]$ is the unramified quadratic extension. Then the thing we are looking for is the set $\{a + b\sqrt{d}: a \in \pi^n\mathcal{O}\_K, b \in \pi^m\mathcal{O}\_K\}$. It is an $\mathcal{O}\_K$ submodule, but not an $\mathcal{O}\_L$ submodule.




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I'm not sure if this answers your question, since "a simple description" is hard to define. It then all depends on what kind of particular question you have in mind.


