Characterisation of Non-Archimedean Division Ring Norms

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

Corollary 1

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

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

Corollary 2

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

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

Corollary 3

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

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

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 $n \cdot 1_R = \begin{cases} \underbrace {1_R + 1_R + \dots + 1_R}_{n \, times} &\mbox{if } n \gt 0 \\ 0 &\mbox{if } n = 0 \\ \\ -\underbrace {\paren {1_R + 1_R + \dots + 1_R}}_{-n \, times} &\mbox{if } n \lt 0 \\ \end{cases}$


Necessary Condition

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

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

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


Sufficient Condition


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

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 $R$.


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


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 $n \in \N$:

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

Lemma 2

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

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



\(\displaystyle \norm{x + 1_R}^n\) \(\le\) \(\displaystyle \sum_{i = 0}^n \norm{x}^i\) $\quad$ Continuing from above $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \sum_{i = 0}^n \max \set {\norm {x}^n , 1}\) $\quad$ Lemma 2 $\quad$
\(\displaystyle \) \(=\) \(\displaystyle (n+1) \max \set {\norm {x}^n , 1}\) $\quad$ $\quad$

Taking $n$th roots yields:

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

Lemma 3

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

Then $\left \langle {x_n} \right \rangle$ converges with a limit of $1$.


By multiple rule for real sequences then:

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

By Inequality Rule for Real Sequences then:

$\norm{x + 1_R} \le \max \set {\norm {x}, 1}$

The result follows.