P-adic Norm forms Non-Archimedean Valued Field/Rational Numbers

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Theorem

The $p$-adic norm $\norm{\,\cdot\,}_p$ forms a non-Archimedean norm on the rational numbers $\Q$.


The rational numbers $\struct{\Q, \norm{\,\cdot\,}_p}$ with the $p$-adic norm is a valued field with a non-Archimedean norm.


Proof

First we note that the $p$-adic norm is a norm.

Let $\nu_p$ denote the $p$-adic valuation on the rational numbers.


Recall the definition of the $p$-adic norm:

$\forall q \in \Q: \norm q_p := \begin{cases} 0 & : q = 0 \\ p^{-\map {\nu_p} q} & : q \ne 0 \end{cases}$


We must show the following holds for all $x, y \in \Q$:

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


If $x = 0$ or $y = 0$, or $x + y = 0$, the result is trivial, as follows:

Let $x = 0$.

Then:

\(\ds x\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \norm x_p\) \(=\) \(\ds 0\) Definition of $p$-adic Norm
\(\ds \leadsto \ \ \) \(\ds \max \set {\norm x_p, \norm y_p}\) \(=\) \(\ds \norm y_p\) as $\norm y_p \ge 0 = \norm x_p$ from Non-Archimedean Norm Axiom $\text N 1$: Positive Definiteness
\(\ds \) \(=\) \(\ds \norm {x + y}_p\)

and so $\norm {x + y}_p \le \max \left( \norm x_p, \norm y_p \right)$

The same argument holds for $y = 0$.


Let $x + y = 0$.

\(\ds x + y\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \norm {x + y}_p\) \(=\) \(\ds 0\) Definition of $p$-adic Norm
\(\ds \) \(\le\) \(\ds \max \set {\norm x_p, \norm y_p}\) as $\norm x_p \ge 0$ and $\norm y_p \ge 0$ from Non-Archimedean Norm Axiom $\text N 1$: Positive Definiteness


Let $x, y, x + y \in \Q_{\ne 0}$.

From $p$-adic Valuation is Valuation:

$\map {\nu_p} {x + y} \ge \min \set {\map {\nu_p} x, \map {\nu_p} y}$

Then:

\(\ds \norm {x + y}_p\) \(=\) \(\ds p^{-\map {\nu_p} {x + y} }\) Definition of $p$-adic Norm
\(\ds \) \(\le\) \(\ds \max \set {p^{- \map {\nu_p} x}, p^{-\map {\nu_p} y} }\)
\(\ds \) \(=\) \(\ds \max \set {\norm x_p, \norm y_p}\) Definition of $p$-adic Norm

$\blacksquare$


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