Is finite index subgroup of unit group of local field is always open ?
Open subgroup and finite index subgroup of topological group says this does not hold in general topological group, but what about local field case ?
In general case, unit group of profinite completion of $ \Bbb{Z}$ fives counter example, but I couldn't find counter example in local field case. Thank you in advance.
For finite extensions of $\Bbb{Q}_p$ yes because $1+\pi_K O_K$ is a $\Bbb{Z}_p$-module so finite index subgroups have index $p^r$ and $(1+\pi_K O_K)^{p^r}$ is open.
For finite extensions of $\Bbb{F}_p((t))$ I'd say no, but I don't have a proof.
$1+\pi_K O_K$ stays a $\Bbb{Z}_p$-module but $(1+\pi_K O_K)^{p^r}$ doesn't have finite index anymore.
$H=(1+t\Bbb{F}_p[[t]])/(1+t^p\Bbb{F}_p[[t^p]])$ is an infinite-dimensional $\Bbb{F}_p$-vector space, I think there is some $f\in H$ which is not in the span of $1+p\Bbb{F}_p[t]$ (not sure how to construct it) which would imply that (under the axiom of choice) the kernel of a linear form $H\to \Bbb{F}_p$ sending $f$ to $1$ and $1+p\Bbb{F}_p[t]$ to $0$ would have index $p$ and not be open.
