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
- $\forall x \in Z_{\ne 0}: \map {\nu_p} x = k : x = p^k y : p \nmid y$
$\Box$
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:
- $\forall r \in \Q_{\ne 0}: \map {\nu_p} r = k : r = p^k \dfrac m n: k, m, n \in \Z, p \nmid m, n$
$\blacksquare$