Is $K^\times/ F^\times$ compact for local fields? 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?
 A: $O_K^\times$ 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 $$K^\times/F^\times \cong \pi_K^{\Bbb{Z/eZ}} \times (O_K^\times)_{tors}/(O_F^\times)_{tors} \times H\times O_K/O_F$$
A: Let $K\subset L$ be a finite extension of normed non-discrete locally compact fields. Let $r>1$ be the norm of some element of $K$. Then every nonzero element of $L$ can be written as $r^nw$ with $\|w\|\in [1,r]$ and $n\in\mathbf{Z}$. Since $\{w\in L:\|w\|\in [1,r]\}$ is compact, it immediately follows that $K^*/L^*$ is compact.
(Actually, this shows that for every locally compact normed field $L$ and $w\in L$ with $\|w\|>1$, $L^*/\langle w\rangle$ is compact. In particular, in the above setting with $w\in K$, the group $L^*/K^*$ is a quotient of $L^*/\langle w\rangle$ and inherits compactness.)
A: N.B. The result is true without the assumption that the characteristic of $F$ is $0$.
You moreover have to assume that $F$ is locally compact.
The standard proof is reuns's one. If you already know that the projective space ${\mathbb P}(F^n )$ is compact for any $n\geqslant 0$, you can observe that
$K^\times /F^\times$ identifies to ${\mathbb P}(F^n )$, where $n=[K:F]$, as a topological space.
