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

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

Corollary 3

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

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

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


Necessary Condition

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

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

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


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 $n \in \N$:

\(\displaystyle \norm {x + 1_R}^n\) \(=\) \(\displaystyle \norm {\sum_{i \mathop = 0}^n \binom n i \cdot x^i}\) 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 $i$, $0 \le i \le n$:

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



\(\displaystyle \norm {x + 1_R}^n\) \(\le\) \(\displaystyle \sum_{i \mathop = 0}^n \norm x^i\) 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 $1$.


By the Multiple Rule for Real Sequences:

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

By Inequality Rule for Real Sequences:

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

The result follows.