Deduce unramifiedness from surjection on units

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Suppose $F$ is a finite extension of some $\mathbb Q_p$, and $L$ is a finite tamely ramified extension of $F$. I need to prove the following assertion:

If $Nm(\mathcal O_L^{\times})=\mathcal O_F^{\times}$, then $L/F$ is unramified.

I can see that in this case $Nm(L^{\times})$ is a subgroup of finite index in $F^{\times}$. But I have no idea how to get the assertion... Could you offer some help?

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I think that’s false. Take $L=F(\mu_e,\varpi^{1/e})$ where $\varpi$ is a uniformizer of $F$, $e \geq 1$ is an integer coprime to $q(q-1)$, where $q$ is the cardinality of the residue field of $F$.

It’s easy to see by Hensel that $O_F^{\times e}=O_F^{\times}$. But, if $K=F(\mu_e)$, then $N_{L/K}O_L^{\times} \supset O_K^{\times e}$, thus $N_{L/K}O_L^{\times} \supset N_{K/F}O_K^{\times e} =(N_{K/F}O_K^{\times})^e=O_F^{\times e}=O_F^{\times}$.


The result is true if you assume in fact that $L/F$ is Galois, totally ramified and tamely ramified (so that, in fact, $L=F$).

Indeed, let $k$ be the residue field and $q$ its cardinality, $e$ the ramification degree of $L/F$.

Because the norm is surjective, and the norm is the $e$-th power on the residue field, we see that the $e$-th power is a surjective function $k \rightarrow k$. Thus, by Hensel, $O_L^{\times e}=O_L^{\times}$.

It follows that there is a uniformizer $\varpi_L$ of $L$ such that $\varpi_L^e=\varpi_F$ is a uniformizer of $F$. Thus, for every $\sigma \in Gal(L/F)$, $\sigma(\varpi_L)/\varpi_L$ is a $e$-th root of unity in $L$, so by the above is one, so that $\varpi_L \in L^{Gal(L/F)}=F$ hence $e=1$ and $L=F$.


I always forget the name of this fact, but the idea is that if $L/K$ is a finite extension of $p$-adic fields, by class field theory $K^{\times}/N_{L/K}L^{\times}$ is an abelian invariant of the extension $L/K$, that is, its cardinality is the degree of the largest abelian extension of $K$ contained in $L$. The part of this quotient that comes from the valuation comes from the residual degree – so that the cardinality of $O_K^{\times}/N_{L/K}O_L^{\times}$ comes from the ramification degree of the largest abelian extension of $K$ contained in $L$.