P-adic Norm is Norm

Theorem

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

Proof 1

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.

$\blacksquare$

Proof 2

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

$\forall q \in \Q: \norm r_p := \begin {cases}

 0 & : r = 0 \\
 p^{- k} & : r \ne 0

\end {cases}$

where:

$r = p^k \dfrac m n$

and:

$k, n \in \Z, m \in \Z_{\ne 0} : p \nmid m, n$

where $\nmid$ stands for "does not divide".

We must show that the norm axioms for all $r_1$, $r_2 \in \Q$:

\((\text N 1)\)	$:$			\(\ds \forall r \in \Q:\)		\(\ds \norm r_p = 0 \)	\(\ds \iff \)	\(\ds x = 0 \)
\((\text N 2)\)	$:$			\(\ds \forall r_1, r_2 \in \Q:\)		\(\ds \norm {r_1 r_2} \)	\(\ds = \)	\(\ds \norm {r_1}_p \times \norm {r_2}_p \)
\((\text N 3)\)	$:$			\(\ds \forall r_1, r_2 \in \Q:\)		\(\ds \norm {r_1 + r_2}_p \)	\(\ds \le \)	\(\ds \norm {r_1}_p + \norm {r_2}_p \)

Norm Axiom $\text N 1$: Positive Definiteness

Let $r \in \Q : r \ne 0$.

Let $k, m\in \Z, n \in \Z_{\ne 0} : p \nmid m, n$.

Suppose $r = 0$.

By definition:

$\norm {r}_p = 0$

Suppose $r = p^k \dfrac m n \ne 0$

By definition:

$\norm r_p = \dfrac 1 {p^k} > 0$

Suppose $\norm r_p = 0$.

By definition:

$r = 0$

$\Box$

Norm Axiom $\text N 2$: Multiplicativity

Suppose $r_1 = 0$ or $r_2 = 0$.

From Norm Axiom $\text N 1$: Positive Definiteness, $\norm {r_1}_p = 0$ or $\norm {r_2}_p = 0$.

Suppose $r_1 \ne 0 \ne r_2$.

Let $k_1, k_2, m_1, m_2 \in \Z, n_1, n_2 \in \Z_{\ne 0} : p \nmid n_1, n_2, m_1, m_2$

Let $r_1 = p^{k_1} \dfrac {m_1} {n_1}, r_2 = p^{k_2} \frac {m_2} {n_2}$

Then:

$r_1 r_2 = p^{k_1 + k_2} \dfrac {m_1 m_2}{n_1 n_2}$

We have that $p \nmid m_1$, $p \nmid m_2$.

Since $p$ is prime:

$p \nmid m_1 m_2$.

Similarly:

$p \nmid n_1 n_2$.

Therefore:

\(\ds \norm {r_1 r_2}_p\)

\(=\)

\(\ds p^{-\paren {k_1 + k_2} }\)

\(\ds \)

\(=\)

\(\ds p^{-k_1} p^{-k_2}\)

\(\ds \)

\(=\)

\(\ds \norm {r_1}_p \norm {r_2}_p\)

$\Box$

Norm Axiom $\text N 3$: Triangle Inequality

Suppose one of the following is true:

$r_1 = 0$

$r_2 = 0$

$r_1 + r_2 = 0$

Then the result is straightforward.

Suppose $r_1 \ne 0$, $r_2 \ne 0$, $r_1 + r_2 \ne 0$.

Let $r_1 = p^{k_1} \dfrac {m_1} {n_1}, r_2 = p^{k_2} \dfrac {m_2} {n_2}$ where:

$k_1, k_2, m_1, m_2 \in \Z, n_1, n_2 \in \Z_{\ne 0} : p \nmid m_1, m_2, n_1, n_2$

Then:

\(\ds r_1 + r_2\)

\(=\)

\(\ds \frac {p^{k_1} m_1 n_2 + p^{k_2} m_2 n_1} {n_1 n_2}\)

\(\ds \)

\(=\)

\(\ds p^{\map \min {k_1, k_2} } \frac {p^{k_1 \mathop - \map \min {k_1, k_2} } m_1 n_2 + p^{k_2 \mathop - \map \min {k_1, k_2} } n_1 m_2} {n_1 n_2}\)

Definition of Min Operation

\(\ds \)

\(=\)

\(\ds p^{\map \min {k_1, k_2} } \frac {\tilde m} {n_1 n_2}\)

$\tilde m := p^{k_1 \mathop - \map \min {k_1, k_2} } m_1 n_2 + p^{k_2 \mathop - \map \min {k_1, k_2} } n_1 m_2$

By Fundamental Theorem of Arithmetic:

$\exists ! \tilde k \in \Z_{\ge 0} : \exists m \in \Z : p \nmid m : \tilde m = p^{\tilde k} m$

Obviously, $p \nmid n_1 n_2$

Hence:

\(\ds \norm {r_1 + r_2}_p\)

\(=\)

\(\ds \frac 1 {p^{\tilde k + \map \min {k_1, k_2} } }\)

\(\ds \)

\(\le\)

\(\ds \frac 1 {p^{\map \min {k_1, k_2} } }\)

\(\ds \)

\(=\)

\(\ds \map \max {p^{-k_1}, p^{-k_2} }\)

Definition of Max Operation

\(\ds \)

\(=\)

\(\ds \map \max {\norm {r_1}_p, \norm {r_1}_p}\)

\(\ds \)

\(\le\)

\(\ds \map \max {\norm {r_1}_p, \norm {r_2}_p} + \map \min {\norm {r_1}_p, \norm {r_2}_p}\)

\(\ds \)

\(=\)

\(\ds \norm {r_1}_p + \norm {r_2}_p\)

$\Box$

All norm axioms are seen to be satisfied.

Hence the result.

$\blacksquare$