# Norm topology and usual topology in a p-adic field

I define a p-adic field is a field $$K$$ which is a finite extension of $$Q_p$$ and $$\pi$$ is a uniformizer of $$K$$. And we define the norm topology on $$K$$ is given that the norm groups form a fundamental system of neighborhoods of 1.

Questions:

(1) Why $$O_K^*$$ is not open for the norm toplogy ?

(2) Let $$m,n$$ be two arbitary positive integers, Why $$G_{m,n}:=(1+\pi^nO_K)\times \pi^m$$ is a finite index open subgroup of $$K^*$$ for the usual metric topology on $$K^*$$ ? And for any finite index open subgroup $$G$$ in $$K^*$$ for the metric topology, does there exist a $$G_{m,n}$$ such that $$G_{m,n}\subseteq G$$ ?

• What do you mean by ${O_K}^n$ ? – nguyen quang do Jul 13 at 17:30
• Do you mean $(1+\pi^n O_K)$ where you write $(1+O^n_K)$? If so, are you aware that these so-called "higher principal units" give a filtration of $O^*_K$, that each of them is of finite index in $O^*_K$, etc.? – Torsten Schoeneberg Jul 13 at 17:31
Each norm group, that is the nonzero norms from some finite extension $$L/K$$, will contain elements of nonzero valuation, so will not be contained in $$O_K^*$$, the set of elements of zero valuation.
• Thanks for your answer! But I want to know what can be deduced from your answer. I think I don't need that the norm groups are contained in $O_K^*$. – Sssss Jul 13 at 6:11
• I'm saying that no basic neighbourhood of $1$ is contained in $O_K^*$. – Lord Shark the Unknown Jul 13 at 7:17
• Why not look at the groups $N_{L/K}(O_L^*)$ instead of $N_{L/K}(L^*)$ ? – reuns Jul 14 at 19:00