Trivial Norm on Division Ring is Non-Archimedean

Theorem

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

Then the trivial norm $\norm {\, \cdot \,}: R \to \R_{\ge 0}$, which is given by:

$\norm x = \begin{cases}

 0 & : x = 0_R \\
 1 & : \text{ otherwise}

\end{cases}$

is non-archimedean:

$\norm {x + y} \le \max \set {\norm x, \norm y}$

Proof

Let $x, y = 0_R$.

Then:

$\norm x, \norm y = 0$

Therefore:

$\max \set {\norm x, \norm y} = 0$.

Hence:

\(\ds \norm {x + y}\)

\(=\)

\(\ds \norm {0_R + 0_R}\)

\(\ds \)

\(=\)

\(\ds \norm {0_R}\)

\(\ds \)

\(=\)

\(\ds 0\)

\(\ds \)

\(=\)

\(\ds \max \set {\norm x, \norm y}\)

Let $x \ne 0_R$ or $y \ne 0_R$.

Then:

$\norm x = 1$ or $\norm y = 1$

Therefore:

$\max \set {\norm x, \norm y} = 1$.

Hence:

\(\ds \norm {x + y}\)

\(\le\)

\(\ds 1\)

Definition of Trivial Norm on Division Ring

\(\ds \)

\(=\)

\(\ds \max \set {\norm x, \norm y}\)

$\blacksquare$