Equivalence of Definitions of P-adic Norms

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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:

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

$\blacksquare$