# Equivalence of Definitions of P-adic Norms

## Theorem

Let $p \in \N$ be a prime.

Let $\Q$ denote the rational numbers.

The following definitions of the concept of $p$-adic norm on $\Q$ are equivalent:

### Definition 1

Let $\nu_p: \Q \to \Z \cup \set {+\infty}$ be the $p$-adic valuation on $\Q$.

The $p$-adic norm on $\Q$ is the mapping $\norm {\,\cdot\,}_p: \Q \to \R_{\ge 0}$ defined as:

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

### Definition 2

The $p$-adic norm on $\Q$ is the mapping $\norm {\,\cdot\,}_p: \Q \to \R_{\ge 0}$ defined as:

$\forall r \in \Q: \norm r_p = \begin {cases} 0 & : r = 0 \\ \dfrac 1 {p^k} & : r = p^k \dfrac m n: k, m, n \in \Z, p \nmid m, n \end {cases}$

## Proof

From Negative Powers of Group Elements, Definition 2 can be rewritten as:

$\forall r \in \Q: \norm r_p = \begin {cases} 0 & : r = 0 \\ p^{-k} & : r = p^k \dfrac m n: k, m, n \in \Z, p \nmid m, n \end {cases}$

Hence if follows that Definition 1 and Definition 2 are equivalent if it is shown:

$\forall r \in \Q_{\ne 0}: \map {\nu_p} r = k : r = p^k \dfrac m n: k, m, n \in \Z, p \nmid m, n$

Let $r \in \Q_{\ne 0}$.

Let $r = \dfrac a b : a, b \in \Z_{\ne 0}$

We have:

 $\ds \map {\nu_p} r$ $=$ $\ds \map {\nu_p} {\dfrac a b}$ $\ds$ $=$ $\ds \map {\nu_p} a - \map {\nu_p} b$ Definition of P-adic Valuation

Let:

$k_a := \map {\nu_p} a$
$k_b := \map {\nu_p} b$

### Lemma 1

$\forall x \in Z_{\ne 0}: \map {\nu_p} x = k : x = p^k y : p \nmid y$

From Lemma 1:

$a = p^{k_a} m$
$b = p^{k_b} n$
$p \nmid m, n$

Hence:

 $\ds r$ $=$ $\ds \dfrac a b$ $\ds$ $=$ $\ds \dfrac {p^{k_a} m} {p^{k_b} n}$ From lemma 1 $\ds$ $=$ $\ds p^{k_a - k_b} \dfrac m n$ Sum of Powers of Group Elements

Let:

$k = k_a - k_b$.

It follows that:

$\map {\nu_p} r = k : r = p^k \dfrac m n: k, m, n \in \Z, p \nmid m, n$

Since $r$ was arbitrary:

$\forall r \in \Q_{\ne 0}: \map {\nu_p} r = k : r = p^k \dfrac m n: k, m, n \in \Z, p \nmid m, n$

$\blacksquare$