P-adic Norm of p-adic Number is Power of p
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 1
From Rational Numbers are Dense Subfield of P-adic Numbers $\Q$ is dense in $\Q_p$.
By the definition of a dense subset then $\map \cl \Q = \Q_p$.
By Closure of Subset of Metric Space by Convergent Sequence then:
That is:
- $\ds \lim_{n \mathop \to \infty} x_n = x$
From Modulus of Limit:
- $\ds \lim_{n \mathop \to \infty} \norm{x_n}_p = \norm x_p$
By Convergent Sequence in Normed Division Ring is Cauchy Sequence, $\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q_p, \norm {\,\cdot\,}_p}$.
From Sequence is Cauchy in P-adic Norm iff Cauchy in P-adic Numbers, $\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$.
From Lemma:
- $\exists v \in \Z: \norm x_p = \lim_{n \mathop \to \infty} \norm{x_n}_p = p^{-v}$
$\blacksquare$
Proof 2
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:
- $\norm x_p = \ds \lim_{n \mathop \to \infty} \norm{x_n}_p = p^{-v}$
$\blacksquare$