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.


(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 $G_{m,n}\subseteq G$ ?

Thanks in advance !

  • $\begingroup$ What do you mean by ${O_K}^n$ ? $\endgroup$ – nguyen quang do Jul 13 at 17:30
  • $\begingroup$ 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.? $\endgroup$ – Torsten Schoeneberg Jul 13 at 17:31
  • $\begingroup$ Sorry, I have modified my questions just now. $\endgroup$ – Sssss Jul 14 at 17:13

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.

  • $\begingroup$ 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 $O_K^*$. $\endgroup$ – Sssss Jul 13 at 6:11
  • 1
    $\begingroup$ I'm saying that no basic neighbourhood of $1$ is contained in $O_K^*$. $\endgroup$ – Lord Shark the Unknown Jul 13 at 7:17
  • $\begingroup$ Why not look at the groups $N_{L/K}(O_L^*)$ instead of $N_{L/K}(L^*)$ ? $\endgroup$ – reuns Jul 14 at 19:00

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