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. [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{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. [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.