# On Extension of Rationals

This note is based on my talk An Expedition to the World of $p$-adic Numbers at Carnegie Mellon University on January 15, 2014.

#### Construction from Cauchy Sequences

A standard approach to construct the real numbers from the rational numbers is a procedure called completion, which forces all Cauchy sequences in a metric space by adding new points to the metric space.

Definition. A norm, denoted by $|\cdot|$, on the field $F$ is a function from $F$ to the set of nonnegative numbers in an ordered field $R$ such that (1) $|x|=0$ if and only if $x=0$; (2) $|xy|=|x||y|$; (3) $|x+y|\leq|x|+|y|$.

Remark. The ordered field is usually chosen to be $\mathbb{R}$. However, to construct of $\mathbb{R}$, the ordered field is $\mathbb{Q}$.

A norm on the field $F$ naturally gives rise to a metric $d(x,y)=|x-y|$ on $F$. For example, the standard metric on the rationals is defined by the absolute value, namely, $d(x, y) = |x-y|$, where $x,y\in\mathbb{Q}$, and $|\cdot|$ is the standard norm, i.e., the absolute value, on $\mathbb{Q}$.

Given a metric $d$ on the field $F$, the completion procedure considers the set of all Cauchy sequences on $F$ and an equivalence relation
$$(a_n)\sim (b_n)\text{ if and only if }d(a_n, b_n)\to 0\text{ as }n\to\infty.$$

Definition. Two norms on $F$, $|\cdot|_1, |\cdot|_2$, are equivalent if and only if for every sequence $(a_n)$, it is a Cauchy sequence with respect to $d_1$ if and only if it is so with respect to $d_2$, where $d_1, d_2$ are the metrics determined by $|\cdot|_1, |\cdot|_2$ respectively.

Remark. It is reasonable to worry about the situation in which two norms $|\cdot|_1$ and $|\cdot|_2$ are equivalent, but they introduce different equivalent relationships on the set of Cauchy sequences. However, given two equivalent norms, $|a_n|_1$ converges to 0 if and only if it does so with respect to $d_2$. (Hint: prove by contradiction and consider the sequence $(1/a_n)$.)

Definition. The trivial norm on $F$ is a norm $|\cdot|$ such that $|0|=0$ and $|x|=1$ for $x\neq 0$.

Since we are interested in norms that generate different completions of the field, it would be great if we can classify all nontrivial norms modulo the norm equivalence.

#### Alternative Norms on Rationals

Definition. Let $p$ be any prime number. For any non-zero integer $a$, let $\mathrm{ord}_pa$ be the highest power of $p$ which divides $a$. For any rational $x=a/b$, define $\mathrm{ord}_px=\mathrm{ord}_pa-\mathrm{ord}_pb$. Further define a map $|\cdot|_p$ on $\mathbb{Q}$ as follows: $$|x|_p = \begin{cases}p^{-\mathrm{ord}_px} & \text{if }x\neq 0 \\ 0 & \text{if }x=0\end{cases}.$$

Proposition. $|\cdot|_p$ is a norm on $\mathbb{Q}$. We call it the $p$-adic norm on $\mathbb{Q}$.

Proof (sketch). We only check the triangle inequality. Notice that \begin{aligned}\mathrm{ord}_p\left(\frac{a}{b}-\frac{c}{d}\right) & = \mathrm{ord}_p\left(\frac{ad-bc}{bd}\right) \\ & = \mathrm{ord}_p(ad-bc)-\mathrm{ord}_p(bd) \\ & \geq \min(\mathrm{ord}_p(ad), \mathrm{ord}_p(bc)) - \mathrm{ord}_p(bd) \\ &= \min(\mathrm{ord}_p(ad/bd), \mathrm{ord}_p(bc/bd)).\end{aligned} Therefore, we obtain $|a/b-c/d|_p \leq \max(|a/b|_p, |c/d|_p) \leq |a/b|_p+|c/d|_p$.

Remark. Some counterintuitive fact about the $p$-adic norm are
The following theorem due to Ostrowski classifies all possible norms on the rationals up to norm equivalence. We denote the standard absolute value by $|\cdot|_\infty$.

Theorem (Ostrowski 1916). Every nontrivial norm $|\cdot|$ on $\mathbb{Q}$ is equivalent to $|\cdot|_p$ for some prime $p$ or for $p=\infty$.

Proof. We consider two cases (i) There is $n\in\{1,2,\ldots\}$ such that $|n|>1$; (ii) For all $n\in\{1,2,\ldots\}$, $|n|\leq 1$. As we shall see, in the 1st case, the norm is equivalent to $|\cdot|_\infty$, whereas, in the 2nd case, the norm is equivalent to $|\cdot|_p$ for some prime $p$.

Case (i). Let $n_0$ be the least such $n\in\{1,2,\ldots\}$ such that $|n|>1$. Let $\alpha > 0$ be such that $|n_0|=n_0^\alpha$.

For every positive integer $n$, if $n_0^s \leq n < n_0^{s+1}$, then we can write it in $n_0$-base: $$n = a_0 + a_1n_0 + \ldots + a_sn_0^s.$$

By the choice of $n_0$, we know that $|a_i|\leq 1$ for all $i$. Therefore, we obtain \begin{aligned}|n| & \leq |a_0| + |a_1||n_0| + \ldots + |a_s||n_0|^s \\ & \leq 1 + |n_0| + \ldots |n_0|^s \\ & \leq n_0^{s\alpha}\left(1+n_0^{-\alpha}+n_0^{-2\alpha}+\ldots\right) \\ & \leq Cn^\alpha,\end{aligned} where $C$ does not depend on $n$. Replace $n$ by $n^N$ and get $|n|^N = |n^N| \leq Cn^{N\alpha}$, and so $|n|\leq \sqrt[N]{C}n^\alpha$. As we can choose $N$ to be arbitrarily large, we obtain $|n| \leq n^\alpha$.

On the other hand, we have \begin{aligned}|n| & \geq |n_0^{s+1}| - |n_0^{s+1}-n| \\ & \geq n_0^{(s+1)\alpha} - (n_0^{s+1}-n_0^s)^\alpha\\ & = n_0^{(s+1)\alpha}\left[1-(1-1/n_0)^\alpha\right] \\ & > C'n^\alpha.\end{aligned} Using the same trick, we can actually take $C'=1$.

Therefore $|n| = n^\alpha$. It is easy to see it is equivalent to $|\cdot|_\infty$.

Case (ii). Since the norm is nontrivial, let $n_0$ be the least $n$ such that $|n|<1$.

Claim 1. $n_0=p$ is a prime.

Claim 2. $|q|=1$ if $q$ is a prime other than $p$.

Suppose $|q| < 1$. Find $M$ large enough so that both $|p^M|$ and $|q^M|$ are less than $1/2$. By Bézout’s lemma, there exists $a,b\in\mathbb{Z}$ such that $ap^M + bq^M = 1$. However, $$1 = |1| \leq |a||p^M| + |b||q^M| < 1/2 + 1/2 = 1,$$ a contradiction.

Therefore, we know $|n|=|p|^{ord_p n}$. It is easy to see it is equivalent to $|\cdot|_p$.

#### Non-Archimedean Norm

As one might have noticed, the $p$-adic norm satisfies an inequality stronger than the triangle inequality, namely $|a\pm b|_p\leq \max(|x|_p, |y|_p)$.

Definition. A norm is non-Archimedean provided $|x\pm y|\leq \max(|x|, |y|)$.

The world of non-Archimedean norm is good and weird. Here are two testimonies.

Proposition (no more scalene triangles). If $|x|\neq |y|$, then $|x\pm y| = \max(|x|, |y|)$.

Proof. Suppose $|x| < |y|$. On one hand, we have $|x\pm y| \leq |y|$. On the other hand, $|y| \leq \max(|x\pm y|, |x|)$. Since $|x| < |y|$, we must have $|y| \leq |x\pm y|$.

Proposition (all points are centers). $D(a, r^-) = D(b, r^-)$ for all $b\in D(a, r^-)$ and $D(a, r) = D(b,r)$ for all $b\in D(a,r)$, where $D(c, r^-) = \{x : |x-c| and $D(c,r)=\{x:|x-c|\leq r\}$.

#### Construction of $p$-adic Numbers

The $p$-adic numbers are the completion of $\mathbb{Q}$ via the $p$-adic norm.

Definition. The set of $p$-adic numbers is defined as $$\mathbb{Q}_p = \{\text{Cauchy sequences with respect to }|\cdot|_p\} / \sim_p,$$ where $(a_n)\sim_p(b_n)$ iff $|a_n-b_n|_p\to 0$ as $n\to\infty$.

We would like to extend $|\cdot|_p$ from $\mathbb{Q}$ to $\mathbb{Q}_p$. When extending $|\cdot|_\infty$ from $\mathbb{Q}$ to $\mathbb{R}$, we set $|[(a_n)]|_\infty$ to be $[(|a_n|)]$, an element in $\mathbb{R}$. However, the values that $|\cdot|_p$ can take, after the extension, are still in $\mathbb{Q}$.

Definition. The extension of $|\cdot|_p$ on $\mathbb{Q}_p$ is defined by $|[(a_n)]|_p = \lim_{n\to\infty}|a_n|_p$.

Remark. Suppose $(a_n)\sim_p (a_n')$. Then $\lim_{n\to\infty}|a_n-a_n'|_p=0$, and so $\lim_{n\to\infty}|a_n|_p=\lim_{n\to\infty}|a_n'|_p$. Moreover, one can prove that $\lim_{n\to\infty}|a_n|_p$ always exists provided that $(a_n)$ is a Cauchy sequence. (Hint: Suppose $\lim_{n\to\infty}|a_n|_p > 0$. There exists $\epsilon > 0$ such that $|a_n|_p > \epsilon$ infinitely often. Choose $N$ enough so that $|a_m - a_n|_p < \epsilon$ for all $m,n>N$. Use ‘no more scalene triangles!’ property to deduce a contradiction.)

#### Representation of $p$-adic Numbers

Even though each real number is an equivalence class of Cauchy sequences, each equivalence class has a canonical representative. For instance, the canonical representative of $\pi$ is $(3, 3.1, 3.14, 3.141, 3.1415, \ldots)$. The analog for $\mathbb{Q}_p$ is the following.

Theorem. Every equivalence class $a$ in $\mathbb{Q}_p$ for which $|a|_p\leq 1$ has exactly one representative Cauchy sequence of the form $(a_i)$ for which (1) $0\leq a_i < p^i$ for $i=1,2,3,\ldots$; (2) $a_i = a_{i+1} (\pmod p^i)$ for $i=1,2,3,\ldots$.

Proof of uniqueness. Prove by definition chasing.

Proof of existence. We shall repeatedly apply the following lemma.

Lemma. For every $b\in\mathbb{Q}$ for which $|b|_p \leq 1$ and $i\in\mathbb{N}$, there exists $a\in\{0, \ldots, p^i-1\}$ such that $|a-b|_p \leq p^{-i}$.

Proof of Lemma. Suppose $b=m/n$ is in the lowest form. As $|b|_p\leq 1$, we know that $(n, p^i)=1$. By Bézout’s lemma, $an+a'p^i=m$ for some integers $a,a'$. We may assume $a\in\{0,\ldots,p^i-1\}$. Note that $a-b=a'p^i/n$, and so $|a-b|_p \leq p^{-i}$.

Suppose $(c_i)$ is a representative of $a$. As $(c_i)$ is a Cauchy sequence, we can extract a subsequence $(b_i)$ such that $|b_i - b_j| \leq p^{-i}$ for all $i < j$ which is still a representative of $a$. Using the lemma above, for each $b_i$, we can find $0 \leq a_i < q^i$ such that $|a_i-b_i|_p \leq q^{-i}$. Therefore $(a_i)$ is a representative of $a$ as well. For all $i, we have $|a_i-a_j|_p \leq \max(|a_i - b_i|_p, |b_i-b_j|_p, |b_j-a_j|_p) \leq q^{-i}$, Therefore $q^i$ divides $a_i - a_j$.

For $|a|_p\leq 1$, we write $a=b_0 + b_1p + b_2p^2 + \ldots$, where $(b_{n-1}b_{n-2}\ldots b_0)_p = a_n$.

What if $|a|_p > 1$? As $|ap^m|=|a|/p^m$, $|ap^m|\leq 1$ for some natural number $m$. By the representation theorem, we can write $ap^m = b_0 + b_1p + b_2p^2 + \ldots$, and $a = b_0p^{-m} + b_1p^{-m+1} + b_2p^{-m+2} + \ldots$.

Using the representation of $p$-adic numbers, one can perform arithmetic operations such as addition, subtraction, multiplication and division.

Like $\mathbb{R}$, $\mathbb{Q}_p$ is not algebraically closed. Though $\mathbb{C}$, the algebraic closure of $\mathbb{R}$, has degree $2$ over $\mathbb{R}$, and it is complete with respect to the absolute value, it is not so for $\overline{\mathbb{Q}_p}$, the algebraic closure of $\mathbb{Q}_p$. In fact, $\overline{\mathbb{Q}_p}$ has infinite degree over $\mathbb{Q}_p$ and is, unfortunately, not complete with respect to proper extension of $|\cdot|_p$. The good news is that the completion of $\overline{\mathbb{Q}_p}$, denoted by $\Omega$ is algebraically closed.