Trivial Norm on Division Ring is Norm

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Theorem

Let $\struct {R, +, \circ}$ be a division ring, and denote its ring zero by $0_R$.


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

$\norm x = \begin{cases}
 0 & \text { if } x = 0_R\\
 1 & \text { otherwise}

\end{cases}$

defines a norm on $R$.


Proof

Proving each of the norm axioms one by one:


Proving Norm Axiom $\text N 1$: Positive Definiteness: $\forall x \in R: \norm x = 0 \iff x = 0_R$

This follows directly from the definition of the trivial norm.

$\Box$


Proving Norm Axiom $\text N 2$: Multiplicativity: $\forall x, y \in R: \norm {x \circ y} = \norm x \times \norm y$

If $x = 0_R$:

\(\ds \norm {x \circ y}\) \(=\) \(\ds \norm {0_R \circ y}\) by substitution
\(\ds \) \(=\) \(\ds \norm {0_R}\) Ring Product with Zero
\(\ds \) \(=\) \(\ds 0\) Definition of Trivial Norm on Division Ring
\(\ds \) \(=\) \(\ds 0 \times \norm y\)
\(\ds \) \(=\) \(\ds \norm x \times \norm y\) since $x = 0_R$

The reasoning is similar if $y = 0_R$.


If $x,y \ne 0_R$, then $x \circ y \ne 0_R$ by alternative definition $(3)$ of division rings. We get:

\(\ds \norm {x \circ y}\) \(=\) \(\ds 1\) since $x \circ y \ne 0_R$
\(\ds \) \(=\) \(\ds 1 \times 1\)
\(\ds \) \(=\) \(\ds \norm x \times \norm y\) since $x, y \ne 0_R$

$\Box$


Proving Norm Axiom $\text N 3$: Triangle Inequality: $\forall x, y \in R: \norm {x + y} \le \norm x + \norm y$

If $x = 0_R$:

\(\ds \norm {x + y}\) \(=\) \(\ds \norm {0_R + y}\)
\(\ds \) \(=\) \(\ds \norm y\)
\(\ds \) \(=\) \(\ds 0 + \norm y\)
\(\ds \) \(=\) \(\ds \norm x + \norm y\) since $x = 0_R$

The reasoning is similar if $y = 0_R$.


If $x, y \ne 0_R$:

\(\ds \norm {x + y}\) \(\le\) \(\ds 1\)
\(\ds \) \(<\) \(\ds 1 + 1\)
\(\ds \) \(=\) \(\ds \norm x + \norm y\) since $x, y \ne 0_R$

$\blacksquare$

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