Inverse image of norm map on principal units for an unramified extension

Question

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!

Answer 1

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.