## Absolute values on Number Fields

One fundamental construction in modern algebraic number theory is that of the completion of a number field with respect to an absolute value, the p-adic numbers being a particular case. I will introduce the idea behind this construction and explore some of its consequences.

1. Introduction

Let ${\mathop{\mathbb P}}$ denote the set of prime numbers. For now lets fix ${p\in \mathop{\mathbb P}}$. If ${x\equiv y \mod p^2}$, then clearly ${x\equiv y \mod p}$, so the first relation is stronger than the second and ${x}$ and ${y}$ will be more similar or closer to each other if the first relation holds. This is the basic idea to define the metric ${|.|_p}$ on ${{\mathbb Q}}$:

Definition 1 For each prime ${p}$ define the function ${|.|_p:{\mathbb Q}\rightarrow {\mathbb R}}$ the following way. For ${x\in {\mathbb Q}}$, write it as ${\displaystyle x=\prod_{q \in \mathop{\mathbb P}} q^{\alpha_q}}$ where each ${\alpha_q}$ is a (possibly negative) integer and zero for all but finitely many ${q}$. Then ${|x|_p=p^{-\alpha_p}}$. This induces a metric on ${{\mathbb Q}}$ by ${d_p(x,y)=|x-y|_p}$.

While so far it seems intuitive, lets think about the special case ${x=0}$. Then we are saying that ${p^2}$ is closer to ${0}$ than ${p}$, and the sequence ${p^n}$ is converging to ${0}$ as ${n}$ goes to infinity.

More generally we can define in an arbitrary number field (which is just a name for finite field extensions of ${{\mathbb Q}}$) the notion of absolute value:

Definition 2 Let ${K}$ be a field and ${|.|:K\rightarrow {\mathbb R}}$ a function which satisfies the following properties for all ${x,y\in K}$:

• 1 ${|x|\geq 0}$ and ${|x|=0\iff x=0}$
• 2 ${|xy|=|x|.|y|}$
• 3 ${|a+b|\leq |a|+|b|}$

We call ${|.|}$ an absolute value. If instead of condition ${3}$ it satisfies

• 3′ ${|a+b|\leq \max(|a|,|b|)}$

the absolute value is called non-archimedean, otherwise it is called archimedean.

It is easy to prove that the usual absolute value in ${{\mathbb R}}$ and ${{\mathbb C}}$ satisfy this definition. Also when restricted to a subfield, like ${{\mathbb Q}}$ or a number field, the usual absolute value still are archimedean absolute values.

A more important example for us here is the above defined ${|.|_p}$ on ${{\mathbb Q}}$. It is called the p-adic absolute value and it is not hard to verify that it is indeed a non-archimedean absolute value. More generally, in any number field ${k}$, we can define a similar absolute value, but instead of using prime numbers, we must use prime ideals. Since the set of integral elements on ${k}$, call it ${\mathfrak{o}}$, is a Dedekind domain and ${k}$ is the field of fractions of ${\mathfrak{o}}$, each element defines a principal fractional ideal, which is then uniquely factorized as product of prime ideals, almost the same way rational numbers are factorized uniquely into prime numbers. We can then define an absolute value on ${k}$ for each prime ideal using the same idea used to define the ${\|.\|_p}$ absolute value on ${{\mathbb Q}}$.

It is easy to verify that, for each absolute value, the distance defined by ${d(x,y)=|x-y|}$ is indeed a metric. And since we have a metric, we can now complete our field with respect to that metric. In the case of the p-adic absolute value, the obtained complete field is denoted by ${{\mathbb Q}_p}$ and is called the p-adic field.

In the rest of the post I try to give good reasons why this construction is useful. First I will show how the well-known Chinese Remainder Theorem looks like in this setting. Second I will explain how the Newton’s method to approximate roots of regular functions can be applied to this metric.

2. Chinese Remainder Theorem

From basic point set topology we know that from a metric we can define a natural topology. We say that two absolute values are equivalent if they give the same topology. It is not hard to see that if ${p,q}$ are primes (in ${{\mathbb Z}}$), then the absolute values ${|.|_p}$ and ${|.|_q}$ are non-equivalent.

Theorem 3 (Chinese Remainder Theorem for valued fields) Let ${|.|_1,...,|.|_n}$ be non-equivalent absolute values on a field ${K}$ and let ${\epsilon>0}$ and ${a_1,...,a_n}$ arbitrary elements in ${K}$. Then there is some ${x\in K}$ such that ${|x-a_i|_i<\epsilon}$ for all ${1\leq i\leq n}$.

Actually this is usually called the Weak Approximation Theorem and, as stated, is not quite equivalent to the Chinese Remainder theorem (well, logically all theorems are equivalent, but of course that’s not what I mean here). For instance, the classic Chinese Remainder Theorem in ${{\mathbb Z}}$ would provide an integer satisfying some integer congruences, while the Weak Approximation Theorem, with ${K={\mathbb Q}}$ would provide a rational number satisfying some set of rational congruences.

Another way I like to think about this theorem when applied to ${{\mathbb Q}}$ is as follows. We consider a finite set of primes, call it ${P}$, and define, for a function ${f:P\rightarrow {\mathbb Q}}$, ${\displaystyle |f|=\sup_{p\in P} |f(p)|_p}$. Then the constant functions are dense, for the induced metric, in the set of all functions ${f:P\rightarrow {\mathbb Q}}$. Of course instead of a set of primes we can consider, more generally, a set of non-equivalent absolute values in any field. Recently I learned that it is in this form that the weak approximation theorem has a stronger version, called (unsurprisingly) the strong approximation theorem. I plan to write about this in a future post.

3. Non-Archimedean Newton’s Method

Now we try to understand why we consider the completion of the fields under this metrics. The first example is, of course ${{\mathbb R}}$ as the completion of ${{\mathbb Q}}$ in the usual absolute value. By the fundamental theorem of the algebra, every polynomial with real coefficients can be split as product of linear or quadratic factors in ${{\mathbb R}[x]}$. This implies that “most” of the algebraic numbers are in ${{\mathbb R}}$. If we complete ${{\mathbb Q}}$ with respect to other, namely non-archimedean, absolute values we also expect to create roots for many polynomials. This can be achieved approximating such roots of polynomials by rational numbers (where the approximation is in the metric we are completing with respect to). Then the root will be in the completion.

A way to approximate roots of polynomials is to use Newton’s method. We now state a version of the Newton’s method that can be applied to the p-adic fields.

Theorem 4 (Hensel’s lemma) Let ${p}$ be a fixed prime. Let ${f(x)\in {\mathbb Z}[x]}$ and ${\alpha_o\in {\mathbb Z}}$ be such that ${|f(\alpha_0)|_p<| f'(\alpha_0)|_p^2}$. Then the sequence defined by

$\displaystyle \alpha_{n+1}=\alpha_n-\frac{f(\alpha_n)}{ f^\prime(\alpha_n)}$

converges quadratically to a root of ${f}$ (in the p-adic metric).

The Hensel Lemma can be generalized to any complete field with a non-archimedean metric. In that case, instead of ${{\mathbb Z}}$ we have the ring ${\mathfrak{o}:=\{x\in K: |x|\leq 1\}}$. It’s easy to see that ${K}$ is then the field of fractions of ${\mathfrak{o}}$ and that ${\mathfrak{o}}$ is integrally closed. Also to replace ${p{\mathbb Z}}$ we consider ${\mathfrak{p}:=\{x\in K: |x|<1\}}$. It is easy to see that ${\mathfrak{p}}$ is a prime ideal in the ring ${\mathfrak{o}}$.

Hensel lemma can be applied not only to find roots, but also to factor polynomials, or, more precisely, to lift factorizations from quotients. Given a complete field ${K}$ with an absolute value, we can define the ring of integers ${\mathfrak{o}}$ as above and the prime ${\mathfrak{p}}$. Let ${\mathbb{F}_\mathfrak{p}=\mathfrak{o}/\mathfrak{p}}$. ${\mathbb{F}_\mathfrak{p}}$ is a field (because this prime ideal is maximal). Also, every polynomial ${f\in K[x]}$ can be “projected” into ${\overline f\in\mathbb{F}_\mathfrak{p}[x]}$ in a natural way.

Theorem 5 (Full Hensel’s lemma)Let ${f\in \mathfrak{o}[x]}$ be monic. If the projection ${\overline f\in\mathbb{F}_\mathfrak{p}[x]}$  factors as ${\overline f=\overline g\overline h}$ where ${\overline g}$ and ${\overline h}$ are coprimes, then there are lifts ${g,h}$ of ${\overline g, \overline h}$ in ${\mathfrak{o}[x]}$ such that ${f=gh}$ in ${\mathfrak{o}[x]}$.

Since ${\mathbb{F}_\mathfrak{p}}$ is a field it is much easier to study polynomials there and thanks to the Hensel lemma we can now lift factorizations to the original ring.

4. Krasner’s Lemma

Yet another nice feature of complete fields is the Krasner’s Lemma. It essentially asserts that given a natural metric in the space of polynomials, the roots as a function of the polynomials is a continuous function. As a consequence we can construct a field containing a given field ${K}$ which is both algebraic closed and complete with respect to a metric.

To state it we will first agree on some standard notation: Given a field ${K}$ with an absolute value we denote by ${\hat K}$ its (metric) completion and by ${\overline K}$ its algebraic closure.

Theorem 6 (Krasner’s Lemma)Let ${K}$ be a complete field with respect to a non-archimedean absolute value. Let ${\alpha, \beta\in \overline K}$ be such that ${\alpha}$ is separable over ${K(\beta)}$. Suppose that for each embedding ${\sigma:K(\alpha)\rightarrow \overline K}$ over ${K}$ (this simply means that ${\sigma(x)=x}$ for all ${x\in K}$) if ${\sigma\neq id}$ then ${|\alpha-\sigma(\alpha)|>|\alpha-\beta|}$.

Then ${K(\alpha)\subset K(\beta)}$ (or equivalently ${\alpha\in K(\beta)}$).

It is a well known result that in a finite dimensional vector space over a complete field ${K}$ (for instance ${K={\mathbb R}}$), all metrics are equivalent (for topological proposes), so for any ${x\in \overline K}$, the absolute value in ${K}$ extends essentially uniquely to the algebraic extension ${K(x)}$ and in particular ${|x|}$ is well defined. Therefore we can extend the metric on a complete field to its algebraic closure in a essentially unique way. It makes sense then to consider ${\widehat{\overline K}}$. This is, by construction, a complete field, but is it algebraically closed? And is ${\overline{\widehat K}}$ complete with respect to the metric under consideration? The answer to the second question is, unfortunately, no. The answer to the first is also no in general, but yes if ${K}$ is complete. In general we have the result:

Corollary 7 Let ${K}$ be a field with a non-archimedean absolute value. Then the field ${\displaystyle \widehat{\overline{\widehat K}}}$ is both complete and algebraically closed.

PhD Student at OSU in Mathematics. I'm portuguese.
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### 2 Responses to Absolute values on Number Fields

1. João Nuno says:

Very nice how it all feels kind of natural but also such strange phenomena arise (I’m thinking of the CRT…natural but weird, in a way).

There’s a tricky detail with the naming there…a non-Archimedean absolute value is also Archimedean? Or Archimedean ones have to fail property 3′ ?

• Joel Moreira says:

I hope it’s clear that a non-Archimedean absolute value is not Archimedean.
The naming is related to the Archimedean property: For any real $x$ there is a natural $n$ with $|n|>|x|$.
An equivalent form would be that for any $x$ in the field there is a natural $n$ such that $|nx|=|x+x+...+x|>1$, and now we can see that this contradicts the property 3′.