Characterisation of Non-Archimedean Division Ring Norms

Theorem
Let $\struct{R, \norm{\,\cdot\,}}$ be a normed division ring with unity $1_R$.

Then $\norm{\,\cdot\,}$ is non-archimedean :


 * $\forall n \in \N_{>0}: \norm{n \cdot 1_R} \le 1$.

where $n \cdot 1_R = \underbrace {1_R + 1_R + \dots + 1_R}_{n \, times}$

Necessary Condition
Let $\norm{\,\cdot\,}$ is non-archimedean.

Then by the definition of a non-archimedean norm, for $n \in \N$,

Sufficient Condition
Let:
 * $\forall n \in \N_{>0}: \norm{n \cdot 1_R} \le 1$

First note that for all $x, y \in R$:

Hence it is sufficient to prove:
 * $\forall x \in R: \norm{ x + 1_R} \le \max \set{\norm{x }, 1 }$

For $n \in \N$:

If $\norm{x} \gt 1$ then for all $i$, $0 \le i \le n$:
 * $\norm {x}^i \le \norm {x}^n \le \max \set {\norm {x}^n, 1}$

If $\norm{x} \le 1$ then for all $i$, $0 \le i \le n$:
 * $\norm {x}^i \le 1 \le \max \set {\norm {x}^n, 1}$

In either case for all $i$, $0 \le i \le n$, then:
 * $\norm {x}^i \le \max \set {\norm {x}^n, 1}$

Hence

Taking $n$th roots yields:
 * $\norm{x + 1_R} \le \paren {\sqrt [n] {n+1} } \max \set {\norm {x}, 1}$