# 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:

$\norm {a^{p^{n + 1} } - a^{p^n} }_p = \norm {a^{p^n} (a^{p^n \left({p - 1}\right)} - 1) }_p$

From the corollary to Euler's Theorem:

$a^{p^n \left({p - 1}\right)} - 1 \equiv 0 \pmod {p^n}$

so:

$\norm {a^{p^n} \paren {a^{p^n \paren {p - 1} } - 1} }_p \le p^{-n} \xrightarrow {n \to \infty} 0$

That is:

$\displaystyle \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n}_p = 0$
$\sequence {x_n }$ 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:

$x = \displaystyle \lim_{n \mathop \to \infty} x_n$
$\displaystyle \lim_{n \mathop \to \infty} \norm {x_n }_p = \norm {x }_p$

Since $\forall n, p \nmid a^{p^n} = x_n$, then:

$\norm {x_n }_p = 1$

So:

$\norm x_p = \displaystyle \lim_{n \mathop \to \infty} \norm {x_n}_p = 1$

By Axiom (N1) of a norm on a division ring then:

$x \ne 0$.

Since:

 $\displaystyle x$ $=$ $\displaystyle \lim_{n \mathop \to \infty} x_n$ Definition of $x$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} x_{n + 1}$ Limit of Subsequence equals Limit of Sequence $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \paren {x_n}^p$ Definition of $x_n$ $\displaystyle$ $=$ $\displaystyle \paren {\lim_{n \mathop \to \infty} x_n}^p$ Product rule for Normed Division Rings $\displaystyle$ $=$ $\displaystyle x^p$ 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 $n > N$:

 $\displaystyle \norm {x - a}_p$ $=$ $\displaystyle \norm {x - a^{p^n} + a^{p^n} - a}_p$ $\displaystyle$ $\le$ $\displaystyle \max \set {\norm {x - a^{p^n} }_p, \norm {a^{p^n} - a}_p}$ P-adic Norm is Non-Archimedean Norm

As $\norm {x - a^{p^n} }_p < \norm {x - a}_p$:

 $\displaystyle \norm {x - a}_p$ $=$ $\displaystyle \norm {a^{p^n} - a}_p$ Three Points in Ultrametric Space have Two Equal Distances $\displaystyle$ $=$ $\displaystyle \norm a_p \norm {a^{p^n - 1} - 1}_p$ Axiom (N2) of a norm on a division ring $\displaystyle$ $=$ $\displaystyle \norm {a^{p^n - 1} - 1}_p$ as $\norm a_p = 1$ $\displaystyle$ $<$ $\displaystyle 1$ corollary 4 to Fermat's Little Theorem

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 $3$

$\blacksquare$

## Proof 2

Hensel's Lemma is used to prove the existence of a Cauchy sequence that does not converge.

#### Lemma 1

$\exists x \in \Z_{>0}: p \nmid x, x \ge \dfrac {p + 1} 2$

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

$a \in \Z_{> 0}: \nexists \,c \in \Z : c^k = a$

Let $\map f X \in \Z \sqbrk X$ be the polynomial:

$X^k - a$

#### Lemma 3

$\map f {x_1} \equiv 0 \pmod p$

Let $\map {f'} X \in \Z \sqbrk X$ be the formal derivative of $\map f X$.

#### Lemma 4

$\map {f'} {x_1} \not \equiv 0 \pmod p$

By 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

$\displaystyle \lim_{n \mathop \to \infty} x_n^k = a$ in $\struct {\Q, \norm{\,\cdot\,}_p}$
$\sequence {x_n}$ 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 $c \in \Q$:

$\displaystyle \lim_{n \mathop \to \infty} x_n = c$

in $\struct {\Q, \norm {\,\cdot\,}_p}$

$\displaystyle \lim_{n \mathop \to \infty} x_n^k = c^k$

Hence:

$c^k = a \in \Z$.
$c \in \Z$

So the sequence $\sequence {x_n}$ does not converge in $\struct {\Q, \norm{\,\cdot\,}_p}$.

The result follows.

$\blacksquare$

## Proof 3

Let $\mathcal C$ be the commutative ring of Cauchy sequences over $\struct {\Q, \norm {\,\cdot\,}_p}$.

Let $\mathcal N$ be the set of null sequences in $\struct {\Q, \norm {\,\cdot\,}_p}$.

The $p$-adic numbers $\Q_p$ is the quotient ring $\mathcal C \, \big / \mathcal N$ by definition.

By Embedding Division Ring into Quotient Ring of Cauchy Sequences, the mapping $\phi: \Q \to \Q_p$ defined by:

$\map \phi r = \sequence {r, r, r, \dotsc} + \mathcal N$

where $\sequence {r, r, r, \dotsc} + \mathcal N$ is the left coset in $\mathcal C \, \big / \mathcal N$ 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:

$\exists \sequence {x_n} \in \mathcal C: \sequence {x_n} + \mathcal N \not \in \map \phi \Q$

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.

$\blacksquare$