# Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition

## Theorem

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

Then:

$\forall n \in \N_{>0}: \norm {n \cdot 1_R} \le 1 \implies \norm {\,\cdot\,}$ is Definition:Non-Archimedean Division Ring Norm

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

## Proof

Let:

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

Let $x, y \in R$.

Let $y = 0_R$ where $0_R$ is the zero of $R$.

Then $\norm {x + y} = \norm x = \max \set {\norm x, 0} = \max \set {\norm x, \norm y}$

#### Lemma 1

Let $y \neq 0_R$ where $0_R$ is the zero of $R$.

Then:

$\norm{x + y} \le \max \set{ \norm{x}, \norm{y} } \iff \norm{x y^{-1} + 1_R} \le \max \set{ \norm{x y^{-1}}, 1 }$

$\Box$

Hence to complete the proof it is sufficient to prove:

$\forall x \in R: \norm {x + 1_R} \le \max \set {\norm x, 1}$

For $n \in \N$:

 $\displaystyle \norm {x + 1_R}^n$ $=$ $\displaystyle \norm {\sum_{i \mathop = 0}^n \binom n i \cdot x^i}$ Binomial Theorem $\displaystyle$ $\le$ $\displaystyle \sum_{i \mathop = 0}^n \norm {\binom n i \cdot x^i}$ Norm axiom (N3) (Triangle Inequality) $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop = 0}^n \norm {\binom n i \cdot 1_R} \norm x^i$ Norm axiom (N2) (Multiplicativity) $\displaystyle$ $\le$ $\displaystyle \sum_{i \mathop = 0}^n \norm x^i$ $\forall n \in \N_{>0}: \norm {n \cdot 1_R} \le 1$

#### Lemma 2

Then for all $i$, $0 \le i \le n$:

$\norm {x}^i \le \max \set {\norm {x}^n , 1}$

$\Box$

Hence

 $\displaystyle \norm {x + 1_R}^n$ $\le$ $\displaystyle \sum_{i \mathop = 0}^n \norm x^i$ continuing from above $\displaystyle$ $\le$ $\displaystyle \sum_{i \mathop = 0}^n \max \set {\norm x^n , 1}$ Lemma 2 $\displaystyle$ $=$ $\displaystyle \paren {n + 1} \max \set {\norm x^n , 1}$

Taking $n$th roots yields:

$\norm {x + 1_R} \le \paren {n + 1}^{1/n} \max \set {\norm x, 1}$

#### Lemma 3

Let $\sequence {x_n}$ be the real sequence defined as $x_n = \paren {n + 1}^{1/n}$, using exponentiation.

Then $\sequence {x_n}$ converges with a limit of $1$.

$\Box$

$\displaystyle \lim_{n \mathop \to \infty} \paren {n + 1}^{1/n} \max \set {\norm x, 1} = \max \set {\norm x, 1}$
$\norm {x + 1_R} \le \max \set {\norm x, 1}$

The result follows.

$\blacksquare$