Norm elements in locally compact non-archimedean fields

Question

By a norm element, for example, I mean the elements $x \in \mathbb{R}$, such that there exists $y\in \mathbb{Q}_p$, with $\vert y \vert_{p}=x$. In brief, I am curious about the set $\{\vert y \vert_{\mathfrak{p}}: y \in L\}$, is it always discrete in $\mathbb{R}$? I think these questions would help me to understand their topology better.

I know that in $\mathbb{Q}_p$, the norm elements have the form $p^{\mathbb{Z}}$. I have three related questions.

1. What is the form of the norm elements in $\mathbb{C}_p$?

Let $K$ be a number field, and let $\mathfrak{p}$ be a prime ideal of the ring of integers $\mathcal{O}_K$.

2. What is the form of the norm elements in $\mathbb{K}_{\mathfrak{p}}$?
3. What is the form of the norm elements in $\bar{\mathbb{K}}_{\mathfrak{p}}$?

Edit: I received a satisfying answer for the first three questions, in the comments. Now I have this question:

4. What is the form of the norm elements in $K$, when $K$ is locally compact? I think it should have a form like $c^{\mathbb{Z}}$, but I can not prove or disprove it.

Answer 1

For $a\in \overline{\Bbb{Q}}_p$ the $p$-adic valuation extends to $\Bbb{Q}_p(a)$.

$\sum_{n=0}^d c_n a^n = 0$ (with $c_n\in \Bbb{Q}_p$) implies that $v(c_n a^n)=v(c_m a^m)$ for some $n\ne m, c_n\ne 0$, whence $v(a) \in \Bbb{Q}\cup \infty$.

As $v(p^{r/s}) = r/s$ we get that $v(\overline{\Bbb{Q}}_p) = \Bbb{Q}\cup \infty$.

$\Bbb{C}_p$ is the completion of $\overline{\Bbb{Q}}_p$ for the non-archimedian absolute value $|.|_v$ whence $v(\Bbb{C}_p)=v(\overline{\Bbb{Q}}_p)=\Bbb{Q}\cup \infty$.

Answer 2

This answer contains an argument why, as soon as the value group of a nonarchimedean valued field is not discrete, the field cannot be locally compact in the corresponding topology. It continues:

"For a field $K$ with non-archimedean valuation $\lvert \cdot\rvert$ to be locally compact, it is necessary that

• the residue field is finite, and
• the valuation is discrete (meaning, its value group $\lvert K^* \rvert$ is).

Both conditions are not satisfied e.g. for $\mathbb C_p$, and it's a good exercise to come up with many other fields which fail either, or both. -- An obvious third condition, which is met by $\mathbb C_p$, is that

• the field is complete.

Conversely, if a field with a non-archimedean valuation satisfies all three conditions above, it is locally compact. This is another good exercise; and a last good exercise is to show that such a field is necessarily a finite extension of some $\mathbb Q_p$ or $\mathbb F_p((T))$", as alluded to by reuns in the comments.