P-adic Norm of p-adic Number is Power of p/Proof 2
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Theorem
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $x \in \Q_p: x \ne 0$.
Then:
- $\exists v \in \Z: \norm x_p = p^{-v}$
Lemma
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
Let $\sequence {x_n}$ be a Cauchy sequence in $\struct{\Q, \norm {\,\cdot\,}_p}$ such that $\sequence {x_n}$ does not converge to $0$.
Then:
- $\exists v \in \Z: \ds \lim_{n \mathop \to \infty} \norm{x_n}_p = p^{-v}$
$\Box$
Proof
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
That is, $\Q_p$ is the quotient ring $\CC \, \big / \NN$ where:
- $\CC$ denotes the commutative ring of Cauchy sequences over $\struct {\Q, \norm {\,\cdot\,}_p}$
- $\NN$ denotes the set of null sequences in $\struct {\Q, \norm {\,\cdot\,}_p}$.
Then $x$ is a left coset in $\CC \, \big / \NN$.
Let $\sequence{x_n}$ be any Cauchy sequence in $x$.
From Lemma:
- $\exists v \in \Z: \ds \lim_{n \mathop \to \infty} \norm{x_n}_p = p^{-v}$
By definition of the $p$-adic norm on the $p$-adic numbers:
- $\norm x_p = \ds \lim_{n \mathop \to \infty} \norm{x_n}_p = p^{-v}$
$\blacksquare$
Sources
- 2007: Svetlana Katok: p-adic Analysis Compared with Real ... (previous) ... (next): $\S 1.4$ The field of $p$-adic numbers $\Q_p$