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 if and only if:

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

where:

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

Corollary 1

$\norm {\,\cdot\,}$ is non-Archimedean if and only if:

$\sup \set {\norm {n \cdot 1_R}: n \in \N_{> 0}} = 1$.

Corollary 2

Let $\sup \set {\norm {n \cdot 1_R}: n \in \N_{> 0} } = C < +\infty$.

Then $\norm {\,\cdot\,}$ is non-Archimedean and $C = 1$.

Corollary 3

$\norm {\,\cdot\,}$ is Archimedean if and only if:

$\sup \set {\norm {n \cdot 1_R}: n \in \N_{>0} } = +\infty$

Corollary 4

Let $R$ have characteristic $p > 0$.

Then $\norm {\,\cdot\,}$ is a non_Archimedean norm on $R$.

Corollary 5

If $\norm {\, \cdot \,}$ is non-Archimedean then:

$\sup \set {\norm {n \cdot 1_R}: n \in \Z} = 1$

where $n \cdot 1_R = \begin{cases} \underbrace {1_R + 1_R + \dots + 1_R}_{\text {$n$times} } & : n > 0 \\ 0 & : n = 0 \\ \\ -\underbrace {\paren {1_R + 1_R + \dots + 1_R} }_{\text {$-n$times} } & : n < 0 \\ \end{cases}$

Proof

Necessary Condition

Let $\norm {\,\cdot\,}$ be non-Archimedean.

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

 $\ds \forall n \in \N_{>0}: \,$ $\ds \norm {n \cdot 1_R}$ $=$ $\ds \norm {1_R + \dots + 1_R}$ ($n$ summands) $\ds$ $\le$ $\ds \max \set {\norm {1_R}, \ldots, \norm {1_R} }$ $\ds$ $=$ $\ds 1$ because $\norm {1_R} = 1$

$\Box$

Sufficient Condition

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 $y \ne 0_R$ 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 $n \in \N$:

 $\ds \norm {x + 1_R}^n$ $=$ $\ds \norm {\sum_{i \mathop = 0}^n \binom n i \cdot x^i}$ Binomial Theorem $\ds$ $\le$ $\ds \sum_{i \mathop = 0}^n \norm {\binom n i \cdot x^i}$ Norm Axiom $\text N 3$: Triangle Inequality $\ds$ $=$ $\ds \sum_{i \mathop = 0}^n \norm {\binom n i \cdot 1_R} \norm x^i$ Norm Axiom $\text N 2$: Multiplicativity $\ds$ $\le$ $\ds \sum_{i \mathop = 0}^n \norm x^i$ $\forall n \in \N_{>0}: \norm {n \cdot 1_R} \le 1$

Lemma 2

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

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

$\Box$

Hence

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

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 $1$.

$\Box$

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

The result follows.

$\blacksquare$