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 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.
$\blacksquare$
Sources
- 1997: Fernando Q. Gouvea: p-adic Numbers: An Introduction ... (previous) ... (next): $\S 2.1$: Absolute Values on a Field: Definition $2.1.1 \, (iv)$
- 2007: Svetlana Katok: p-adic Analysis Compared with Real ... (previous) ... (next): $\S 1.2$ Normed fields: Remark $1.13$