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 definitions of the concept of Non-Archimedean Division Ring Norm are equivalent:

Definition 1

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

\((\text N 4)\)   $:$   Ultrametric Inequality:      \(\ds \forall x, y \in R:\)    \(\ds \norm {x + y} \)   \(\ds \le \)   \(\ds \max \set {\norm x, \norm y} \)      

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:

\((\text N 1)\)   $:$   Positive Definiteness:      \(\ds \forall x \in R:\)    \(\ds \norm x = 0 \)   \(\ds \iff \)   \(\ds x = 0_R \)      
\((\text N 2)\)   $:$   Multiplicativity:      \(\ds \forall x, y \in R:\)    \(\ds \norm {x \circ y} \)   \(\ds = \)   \(\ds \norm x \times \norm y \)      
\((\text N 4)\)   $:$   Ultrametric Inequality:      \(\ds \forall x, y \in R:\)    \(\ds \norm {x + y} \)   \(\ds \le \)   \(\ds \max \set {\norm x, \norm y} \)      


Proof

Definition 1 implies Definition 2

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

\((\text N 4)\)   $:$   Ultrametric Inequality:      \(\ds \forall x, y \in R:\)    \(\ds \norm {x + y} \)   \(\ds \le \)   \(\ds \max \set {\norm x, \norm y} \)      

It remains only to show that $\norm {\,\cdot\,}$ satisfies $(\text N 1)$ and $(\text N 2)$.

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: $(\text N 1)$, $(\text N 2)$ and $(\text N 4)$.

To show that $\norm{\,\cdot\,}$ is a norm on a division ring satisfying $(\text N 4)$, it remains to show that $\norm{\,\cdot\,}$ satisfies:

\((\text N 3)\)   $:$   Triangle Inequality:      \(\ds \forall x, y \in R:\)    \(\ds \norm {x + y} \)   \(\ds \le \)   \(\ds \norm x + \norm y \)      


Let $x, y \in R$.

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

From Non-Archimedean Norm Axiom $\text N 1$: Positive Definiteness:

$0 \le \norm x$


Then:

\(\ds \norm{x + y}\) \(\le\) \(\ds \max \set {\norm x, \norm y}\) Non-Archimedean Norm Axiom $\text N 4$: Ultrametric Inequality
\(\ds \) \(=\) \(\ds \norm y\) as $\norm x \le \norm y$ by assumption
\(\ds \) \(\le\) \(\ds \norm x + \norm y\) as $0 \le \norm x$

The result follows.

$\blacksquare$


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