# Equivalence of Definitions of Non-Archimedean Division Ring Norm

## 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 $\norm x \le \norm y$.

$0 \le \norm x$

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.

$\blacksquare$