P-adic Norm is Norm/Proof 1

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Theorem

The $p$-adic norm forms a norm on the rational numbers $\Q$.


Proof

Let $v_p$ be the $p$-adic valuation on the rational numbers.

Recall that the $p$-adic norm is defined as:

$\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 hold for all $x$, $y \in \Q$:

\((\text N 1)\)   $:$     \(\ds \forall x \in \Q:\)    \(\ds \norm x_p = 0 \)   \(\ds \iff \)   \(\ds x = 0 \)      
\((\text N 2)\)   $:$     \(\ds \forall x, y \in \Q:\)    \(\ds \norm {x y}_p \)   \(\ds = \)   \(\ds \norm x_p \times \norm y_p \)      
\((\text N 3)\)   $:$     \(\ds \forall x, y \in \Q:\)    \(\ds \norm {x + y}_p \)   \(\ds \le \)   \(\ds \norm x_p + \norm y_p \)      


Norm Axiom $\text N 1$: Positive Definiteness

By Power of Positive Real Number is Positive:

$\ds \forall s \in \R: \frac 1 {p^s} > 0$

By definition of the $p$-adic norm it follows that:

$\forall x \in \Q: \norm x_p = 0 \iff x = 0$

Thus the $p$-adic norm fulfils Norm Axiom $\text N 1$: Positive Definiteness.

$\Box$


Norm Axiom $\text N 2$: Positive Homogeneity

Let $x = 0$ or $y = 0$.

Then $\norm x_p = 0$ or $\norm y_p = 0$ from Norm Axiom $\text N 1$: Positive Definiteness, and:

\(\ds x y\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \norm {x y}_p\) \(=\) \(\ds 0\) Norm Axiom $\text N 1$: Positive Definiteness
\(\ds \) \(=\) \(\ds \norm x_p \times \norm y_p\)


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

Then:

\(\ds \norm {x y}_p\) \(=\) \(\ds \frac 1 {p^{\map {\nu_p} {x y} } }\) Definition of $p$-adic Norm
\(\ds \) \(=\) \(\ds \frac 1 {p^{\map {\nu_p} x + \map {\nu_p} y} }\) $p$-adic Valuation is Valuation
\(\ds \) \(=\) \(\ds \frac 1 {p^{\map {\nu_p} x} p^{\map {\nu_p} y} }\) Product of Powers
\(\ds \) \(=\) \(\ds \norm x_p \times \norm y_p\) Definition of $p$-adic Norm

Thus the $p$-adic norm fulfils Norm Axiom $\text N 2$: Positive Homogeneity.

$\Box$


Norm Axiom $\text N 3$: Triangle Inequality

Let $x = 0$ or $y = 0$, or $x + y = 0$, the result is trivial.

Let $x = 0$.

Then:

\(\ds x\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \norm x_p\) \(=\) \(\ds 0\) Definition of $p$-adic Norm
\(\ds \leadsto \ \ \) \(\ds \norm x_p + \norm y_p\) \(=\) \(\ds \norm y_p\)
\(\ds \) \(=\) \(\ds \norm {x + y}_p\)

and so $\norm {x + y}_p \le \norm x_p + \norm y_p$

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
\(\ds \) \(\le\) \(\ds \norm x_p + \norm y_p\) as $\norm x_p \ge 0$ and $\norm y_p \ge 0$ from 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
\(\ds \) \(\le\) \(\ds \norm x_p + \norm y_p\)

Thus the $p$-adic norm fulfils Norm Axiom $\text N 3$: Triangle Inequality.

$\Box$


All norm axioms are seen to be satisfied.

Hence the result.

$\blacksquare$


Sources

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