Let $K/F$ be a finite extension of local fields (of characteristic 0). Is it true that the quotient group $K^\times/ F^\times$ is always compact?


I understand that if the extension is cyclic, it is compact by Hilbert 90. But does it hold in general?



Is $K^\times/ F^\times$ compact for local fields?
$O\_K^\times$ is compact thus so is $$K^\times/ \pi\_F^\Bbb{Z}=\pi\_K^{\Bbb{Z/eZ}} \times O\_K^\times, \qquad e=\frac{v(\pi\_F)}{v(\pi\_K)}$$ Being a quotient of a compact group by a closed subgroup $$K^\times/F^\times=(K^\times/ \pi\_F^\Bbb{Z})/(F^\times/ \pi\_F^\Bbb{Z})$$ is compact.


Otherwise you can use the isomorphism $\log : 1+ p^2 O\_K\to p^2 O\_K$ to make it more concrete, as $(1+p^2 O\_K)/(1+p^2O\_F)$ is a finite index subgroup of $O\_K^\times /O\_F^\times$ and $K^\times/F^\times$.


The structure theorem is that $O\_K^\times / (O\_K^\times)\_{tors}$ is a torsion free $\Bbb{Z}\_p$ module with $1+p^2O\_K$ as a finite index subgroup, thus isomorphic to $O\_K$, so that $K^\times \cong \pi\_K^\Bbb{Z} \times (O\_K^\times)\_{tors}\times O\_K$


The torsion of $H=(O\_K^\times / (O\_K^\times)\_{tors})/(O\_F^\times / (O\_F^\times)\_{tors})$ is not obvious, as it is non-trivial for $\Bbb{Q}\_3((1+3)^{1/3})/\Bbb{Q}\_3$, and $$K^\times/F^\times \cong \pi\_K^{\Bbb{Z/eZ}} \times (O\_K^\times)\_{tors}/(O\_F^\times)\_{tors} \times H\times O\_K/O\_F$$


