Equivalence of Definitions of Non-Archimedean Division Ring Norm

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Theorem

Let $\struct {R, +, \circ}$ be a division ring whose zero is denoted $0_R$.


The following are equivalent:

Definition 1

A norm $\norm {\, \cdot \,}$ on $R$ is non-Archimedean if and only if $\norm {\, \cdot \,}$ satisfies the axiom:

\((N4)\)   $:$   Ultrametric Inequality:      \(\displaystyle \forall x, y \in R:\)    \(\displaystyle \norm {x + y} \)   \(\displaystyle \le \)   \(\displaystyle \max \set {\norm x, \norm y} \)             


Definition 2

A non-Archimedean norm on $R$ is a mapping from $R$ to the non-negative reals:

$\norm {\,\cdot\,}: R \to \R_{\ge 0}$

satisfying the non-Archimedean norm axioms:

\((N1)\)   $:$   Positive Definiteness:      \(\displaystyle \forall x \in R:\)    \(\displaystyle \norm x = 0 \)   \(\displaystyle \iff \)   \(\displaystyle x = 0_R \)             
\((N2)\)   $:$   Multiplicativity:      \(\displaystyle \forall x, y \in R:\)    \(\displaystyle \norm {x \circ y} \)   \(\displaystyle = \)   \(\displaystyle \norm x \times \norm y \)             
\((N4)\)   $:$   Ultrametric Inequality:      \(\displaystyle \forall x, y \in R:\)    \(\displaystyle \norm {x + y} \)   \(\displaystyle \le \)   \(\displaystyle \max \set{\norm x, \norm y} \)             


Proof

Definition 1 implies Definition 2

Let $\norm{\,\cdot\,} : R \to \R_{\ge 0}$ be a norm on a division ring satisfying:

\((N4)\)   $:$   Ultrametric Inequality:      \(\displaystyle \forall x, y \in R:\)    \(\displaystyle \norm {x + y} \)   \(\displaystyle \le \)   \(\displaystyle \max \set {\norm x, \norm y} \)             

It remains only to show that $\norm{\,\cdot\,}$ satisfies $(N1)$ and $(N2)$.

This follows from the definition of a norm on a division ring.

$\Box$

Definition 2 implies Definition 1

Let $\norm{\,\cdot\,} : R \to \R_{\ge 0}$ satisfy the non-Archimedean norm axioms: $(N1)$, $(N2)$ and $(N4)$.

To show that $\norm{\,\cdot\,}$ is a norm on a division ring satisfying $(N4)$, it remains to show that $\norm{\,\cdot\,}$ satisfies:

\((N3)\)   $:$   Triangle Inequality:      \(\displaystyle \forall x, y \in R:\)    \(\displaystyle \norm {x + y} \)   \(\displaystyle \le \)   \(\displaystyle \norm x + \norm y \)             


Let $x, y \in R$.

Without loss of generality, suppose $\norm x \le \norm y$.

From non-Archimedean norm axiom $(N1)$ : Positive Definiteness:

$0 \le \norm x$


Then:

\(\displaystyle \norm{x + y}\) \(\le\) \(\displaystyle \max \set{ \norm x, \norm y}\) non-Archimedean norm axiom $(N4)$ : Ultrametric Inequality
\(\displaystyle \) \(=\) \(\displaystyle \norm y\) as $\norm x \le \norm y$ by assumption
\(\displaystyle \) \(\le\) \(\displaystyle \norm x + \norm y\) as $0 \le \norm x$

The result follows.

$\blacksquare$

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