# 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 $$\begin{eqnarray*}\mathrm{ord}_p\left(\frac{a}{b}-\frac{c}{d}\right) &=& \mathrm{ord}_p\left(\frac{ad-bc}{bd}\right) \\ &=& \mathrm{ord}_p(ad-bc)-\mathrm{ord}_p(bd) \\ &\geq& \min(\mathrm{ord}_p(ad), \mathrm{ord}_p(bc)) – \mathrm{ord}_p(bd) \\ &=& \min(\mathrm{ord}_p(ad/bd), \mathrm{ord}_p(bc/bd)).\end{eqnarray*}$$ Therefore, we obtain $$|a/b-c/d|_p \leq \max(|a/b|_p, |c/d|_p) \leq |a/b|_p+|c/d|_p$$. QED

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 $$\begin{eqnarray*}|n| & \leq & |a_0| + |a_1||n_0| + \ldots + |a_s||n_0|^s \\ & \leq & 1 + |n_0| + \ldots |n_0|^s \\ & \leq & n_0^{s\alpha}\left(1+n_0^{-\alpha}+n_0^{-2\alpha}+\ldots\right) \\ & \leq & Cn^\alpha,\end{eqnarray*}$$ 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 $$\begin{eqnarray*}|n| & \geq & |n_0^{s+1}| – |n_0^{s+1}-n| \\ & \geq & n_0^{(s+1)\alpha} – (n_0^{s+1}-n_0^s)^\alpha\\ & = & n_0^{(s+1)\alpha}\left[1-(1-1/n_0)^\alpha\right] \\ & > & C’n^\alpha.\end{eqnarray*}$$ 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 $$|\cdot|_p$$. QED

#### 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 $$|y| \leq |x\pm y|$$. QED.

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 $$D(c, r^-) = \{x : |x-c|<r\}$$ 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 $$|a-b|_p \leq p^{-i}$$. qed

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 $$i<j$$, 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 $$a_i – a_j$$.

QED

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.

3 (this post is made with love)