P-adic Norm not Complete on Rational Numbers
Theorem
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
Then:
- the valued field $\struct {\Q, \norm {\,\cdot\,}_p}$ is not complete.
That is, there exists a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not converge to a limit in $\Q$.
Proof 1
Case: $p \gt 3$
Let $p > 3$.
Then there exists $a \in \Z: 1 < a < p-1$.
Consider the sequence $\sequence {x_n} \subseteq \Q$ where $x_n = a^{p^n}$ for some $a \in \Z: 1 < a < p-1$.
Let $n \in \N$.
Then:
- $\norm {a^{p^{n + 1} } - a^{p^n} }_p = \norm {a^{p^n} (a^{p^n \paren {p - 1} } - 1) }_p$
From Euler's Theorem (Number Theory): Corollary $1$:
- $a^{p^n \paren {p - 1} } - 1 \equiv 0 \pmod {p^n}$
so:
- $\norm {a^{p^n} \paren {a^{p^n \paren {p - 1} } - 1} }_p \le p^{-n} \xrightarrow {n \to \infty} 0$
That is:
- $\ds \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n}_p = 0$
By Characterisation of Cauchy Sequence in Non-Archimedean Norm
- $\sequence {x_n }$ is a cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p }$.
Aiming for a contradiction, suppose $\sequence {x_n}$ converges to some $x \in \Q$.
That is:
- $x = \ds \lim_{n \mathop \to \infty} x_n$
By Modulus of Limit on a Normed Division Ring:
- $\ds \lim_{n \mathop \to \infty} \norm {x_n }_p = \norm {x }_p$
Since $\forall n, p \nmid a^{p^n} = x_n$, then:
- $ \norm {x_n }_p = 1$
So:
- $\norm x_p = \ds \lim_{n \mathop \to \infty} \norm {x_n}_p = 1$
By Axiom (N1) of a norm on a division ring then:
- $x \ne 0$.
Since:
| \(\ds x\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} x_n\) | Definition of $x$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} x_{n + 1}\) | Limit of Subsequence equals Limit of Sequence | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {x_n}^p\) | Definition of $x_n$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\lim_{n \mathop \to \infty} x_n}^p\) | Product Rule for Sequences in Normed Division Ring | |||||||||||
| \(\ds \) | \(=\) | \(\ds x^p\) | Definition of $x$ |
and $x \ne 0$ then:
- $x^{p-1} = 1$
So:
- $x = 1$ or $x = -1$
and so $a-x$ is an integer:
- $0 < a - x < p$
It follows that:
- $p \nmid \paren {a - x}$
and so:
- $\norm {x - a}_p = 1$
Since $x_n \to x$ as $n \to \infty$ then:
- $\exists N: \forall n > N: \norm {x_n - x}_p < \norm {x - a}_p$
That is:
- $\exists N: \forall n > N: \norm {a^{p^n} - x}_p < \norm {x - a}_p$
Let $n > N$:
| \(\ds \norm {x - a}_p\) | \(=\) | \(\ds \norm {x - a^{p^n} + a^{p^n} - a}_p\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \max \set {\norm {x - a^{p^n} }_p, \norm {a^{p^n} - a}_p}\) | P-adic Norm on Rational Numbers is Non-Archimedean Norm |
As $\norm {x - a^{p^n} }_p < \norm {x - a}_p$:
| \(\ds \norm {x - a}_p\) | \(=\) | \(\ds \norm {a^{p^n} - a}_p\) | Three Points in Ultrametric Space have Two Equal Distances | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm a_p \norm {a^{p^n - 1} - 1}_p\) | Norm Axiom $\text N 2$: Multiplicativity | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {a^{p^n - 1} - 1}_p\) | as $\norm a_p = 1$ | |||||||||||
| \(\ds \) | \(<\) | \(\ds 1\) | Fermat's Little Theorem: Corollary $4$ |
This contradicts the earlier assertion that $\norm {x - a}_p = 1$.
In conclusion:
- $\sequence {x_n}$ is a Cauchy sequence that does not converge in $\struct {\Q, \norm {\,\cdot\,}_p }$.
$\Box$
Case: $p = 2$ or $3$
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$\blacksquare$
Proof 2
Hensel's Lemma is used to prove the existence of a Cauchy sequence that does not converge.
Lemma 1
- $\exists x \in \Z_{>0}: p \nmid x, x \ge \dfrac {p + 1} 2$
$\Box$
Let $x_1 \in \Z_{>0}: p \nmid x_1, x_1 \ge \dfrac {p + 1} 2$
Let $k$ be a positive integer such that $k \ge 2, p \nmid k$.
Let $a = x_1^k + p$.
Lemma 2
- $a \in \Z_{> 0}: \nexists \,c \in \Z : c^k = a$
$\Box$
Let $\map f X \in \Z \sqbrk X$ be the polynomial:
- $X^k - a$
Lemma 3
- $\map f {x_1} \equiv 0 \pmod p$
$\Box$
Let $\map {f'} X \in \Z \sqbrk X$ be the formal derivative of $\map f X$.
Lemma 4
- $\map {f'} {x_1} \not \equiv 0 \pmod p$
$\Box$
From Hensel's Lemma there exists a sequence of integers $\sequence {x_n}$ such that:
- $(1) \quad \forall n: \map f {x_n} \equiv 0 \pmod {p^n}$
- $(2) \quad \forall n: x_{n + 1} \equiv x_n \pmod {p^n}$
Lemma 5
- $\ds \lim_{n \mathop \to \infty} {x_n}^k = a$ in $\struct {\Q, \norm {\,\cdot\,}_p}$
$\Box$
From Sequence of Consecutive Integers Modulo Power of p is Cauchy in P-adic Norm then:
- $\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$
Aiming for a contradiction, suppose $\sequence {x_n}$ is a sequence such that for some $c \in \Q$:
- $\ds \lim_{n \mathop \to \infty} x_n = c$
in $\struct {\Q, \norm {\,\cdot\,}_p}$
From Product Rule for Sequences in Normed Division Ring then:
- $\ds \lim_{n \mathop \to \infty} x_n^k = c^k$
Hence:
- $c^k = a \in \Z$.
From Nth Root of Integer is Integer or Irrational then:
- $c \in \Z$
This contradicts Lemma 2.
So the sequence $\sequence {x_n}$ does not converge in $\struct {\Q, \norm{\,\cdot\,}_p}$.
The result follows.
$\blacksquare$
Proof 3
By Rational Numbers are Countably Infinite, the set of rational numbers is countably infinite.
By P-adic Numbers are Uncountable, the set of $p$-adic numbers $\Q_p$ is uncountably infinite.
Let $\CC$ be the commutative ring of Cauchy sequences over $\struct {\Q, \norm {\,\cdot\,}_p}$.
Let $\NN$ be the set of null sequences in $\struct {\Q, \norm {\,\cdot\,}_p}$.
The $p$-adic numbers $\Q_p$ is the quotient ring $\CC \, \big / \NN$ by definition.
By Embedding Division Ring into Quotient Ring of Cauchy Sequences, the mapping $\phi: \Q \to \Q_p$ defined by:
- $\map \phi r = \sequence {r, r, r, \dotsc} + \NN$
where $\sequence {r, r, r, \dotsc} + \NN$ is the left coset in $\CC \, \big / \NN$ that contains the constant sequence $\sequence {r, r, r, \dotsc}$, is a distance-preserving monomorphism.
By Corollary to Surjection from Natural Numbers iff Countable then $\phi$ is not a surjection.
Hence:
- $\exists \sequence {x_n} \in \CC: \sequence {x_n} + \NN \not \in \map \phi \Q$
By Cauchy Sequence Converges Iff Equivalent to Constant Sequence then $\sequence {x_n}$ is not convergent in $\struct {\Q, \norm {\,\cdot\,}_p}$.
The result follows.
$\blacksquare$