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\title{Math 129: Algebraic Number Theory\\
{\bf Lecture 15: Valuations}}
\author{William Stein}
\date{Thursday, March 25, 2004}
\begin{document}
\maketitle I don't know about you, but Swinnerton-Dyer's book is
getting on my nerves (I will avoid more passionate words), so we're
switching to the venerable and famous book by Cassels-Frohlich.  In
particular, we're going to systematically go through the article {\em
Global Fields} by Cassels, which is chapter 2 of the book.  The topics
are similar to the ones in chapter 2 of Swinnerton-Dyer, but Cassels's
article is amazingly well written.  Also, you are well prepared to
read and appreciate it given what you've learned so far in this
course.

A scan of the article is available on the web page for the course,
and you can get a photocopy from me.

The notes for the rest of the course will be a rewrite of {\em Global
Fields} meant to make it more accessible.  I will copy Cassels's
article closely, except I will fix any typos found, reword things in a
way consistent with the rest of these notes, and add exercises and
comments you might have.  I will also add the details of the implicit
exercises and remarks that are left to the reader.

\setcounter{section}{0}
\section{Valuations}

\begin{definition}[Valuation]
A {\em valuation} $\abs{\mbox{ }}$ on a field $K$ is a function
defined on $K$ with values in $\R_{\geq 0}$ satisfying
the following axioms:
\begin{enumerate}
\item[(1)] $\abs{a} = 0$ if and only if $a = 0$,
\item[(2)] $\abs{ab}=\abs{a}\abs{b}$, and
\item[(3)] there is a constant $C\geq 1$ such that 
$\abs{1+a}\leq C$ whenever $\abs{a}\leq 1$.
\end{enumerate}
\end{definition}

The {\em trivial valuation} is the valuation for which
$\abs{a}=1$ for all $a\neq 0$.  We will often tacitly
exclude the trivial valuation from consideration.

From (2) we have 
$$
  \abs{1} = \abs{1}\cdot \abs{1},
$$
so $\abs{1} = 1$ by (1).
If $w\in K$ and $w^n=1$, then $|w|=1$ by (2).
In particular, the only valuation of a finite field
is the trivial one.    The same argument shows that $|-1|=|1|$,
so 
$$
  |-a| = |a|\qquad \text{all }a \in K.
$$

\begin{definition}[Equivalent]
Two valuations $\abs{\mbox{ }}_1$ and $\abs{\mbox{ }}_2$ on the
same field are equivalent if there exists $c>0$ such
that $$\abs{a}_2 = \abs{a}_1^c$$
for all $a\in K$.
\end{definition}
Note that if $\abs{\mbox{ }}_1$ is a valuation, then 
$\abs{\mbox{ }}_2=\abs{\mbox{ }}_1^c$ is also a valuation.
Also, equivalence of valuations is an equivalence relation.

If $\abs{\mbox{ }}$ is a valuation and $C$ is the constant from Axiom
(3), then there is a $c>0$ such that $C^c=2$ (i.e.,
$c=\log(C)/\log(2)$).  Then we can take $2$ as constant for the
equivalent valuation $\abs{\mbox{ }}^c$.  Thus every valuation is
equivalent to a valuation with $C=2$. Note that if $C=1$, e.g.,
if $\abs{\mbox{ }}$ is the trivial valuation, then we could
simply take $C=2$ in Axiom (3).
\begin{proposition}
Suppose $\abs{\mbox{ }}$ is a valuation with $C=2$.
Then for all $a, b\in K$ we have 
\begin{equation}\label{val3p}
  |a + b| \leq |a| + |b|\qquad \text{(triangle inequality)}.
\end{equation}
\end{proposition}
\begin{proof}
Suppose $a_1, a_2\in K$ with $|a_1|\geq|a_2|$.  Then $a=a_2/a_1$
satisfies $|a|\leq 1$.  By Axiom (3) we have $|1+a|\leq 2$, so
multiplying by $a_1$ we see that
$$|a_1+ a_2|\leq 2|a_1| = 2\cdot\max\{|a_1|,|a_2|\}.$$
Also we have 
$$|a_1+ a_2 + a_3 + a_4|\leq 2\cdot\max\{|a_1+a_2|,|a_3+a_4|\}
   \leq 4\cdot \max\{|a_1|,|a_2|,|a_3|,|a_4|\},
$$
and inductively we have for any $r>0$ that
$$|a_1 + a_2 + \cdots  + a_{2^r}| \leq 2^r\cdot\max{|a_j|}.$$
If $n$ is any positive integer, let $r$ be such
that $2^{r-1}\leq n\leq 2^r$. Thenn
$$|a_1 + a_2 + \cdots + a_{n}| \leq 2^r\cdot \max\{|a_j|\} 
 \leq 2n\cdot \max\{|a_j|\},$$
since $2^r\leq 2n$.  In particular,
\begin{equation}\label{eqn:absn}
  |n| \leq 2n\cdot |1| = 2n \qquad\text{(for $n>0$)}.
\end{equation}
Applying (\ref{eqn:absn}) to $\ds\abs{\binom{n}{j}}$ and using
the binomial expansion, we have for any $a,b\in K$ that
\begin{align*}
|a+b|^n &= \abs{\sum_{j=0}^n \binom{n}{j} a^j b^{n-j}}\\
   &\leq  2(n+1)\max_j\left\{ \abs{\binom{n}{j}} \abs{a}^j\abs{b}^{n-j}\right\}\\
   &\leq  2(n+1)\max_j\left\{ 2 \binom{n}{j} \abs{a}^j\abs{b}^{n-j}\right\}\\
   &\leq  4(n+1)\max_j\left\{ \binom{n}{j} \abs{a}^j\abs{b}^{n-j}\right\}\\
   &\leq  4(n+1)(\abs{a}+\abs{b})^n.
\end{align*}
Now take $n$th roots of both sides to obtain
$$
|a+b| \leq \sqrt[n]{4(n+1)}\cdot (|a| + |b|).
$$
We have by elementary calculus that
$$
  \lim_{n\to \infty} \sqrt[n]{4(n+1)} = 1,
$$ so $|a+b| \leq |a|+|b|$.  
(The ``elementary calculus'': We instead prove that $\sqrt[n]{n}\to 1$, since
the argument is the same and the notation is simpler.  First, for any
$n\geq 1$ we have $\sqrt[n]{n}\geq 1$, since upon taking $n$th powers
this is equivalent to $n\geq 1^n$, which is true by hypothesis.
Second, suppose there is an $\eps>0$ such that $\sqrt[n]{n}\geq
1+\eps$ for all $n\geq 1$.  Then taking logs of boths sides we see
that $\frac{1}{n}\log(n)\geq \log(1+\eps) > 0$.  But 
 $\log(n)/n\to 0$, so there is no such $\eps$.  Thus
$\sqrt[n]{n}\to 1$ as $n\to \infty$.)
\end{proof}
Note that Axioms (1), (2) and Equation (\ref{val3p}) imply Axiom (3)
with $C=2$.  We take Axiom (3) instead of Equation (\ref{val3p}) for
the technical reason that we will want to call the square of the
absolute value of the complex numbers a valuation.

\begin{lemma}
Suppose $a, b \in K$, and $\abs{\mbox{ }}$ is a valuation on $K$
with $C\leq 2$.
Then 
$$
  \Bigl||a| - |b|\Bigr| \leq \abs{a-b}.
$$
(Here the big absolute value on the outside of the left-hand
side of the inequality is the usual absolute value on 
real numbers, but the other absolute values are a valuation
on an arbitrary field~$K$.)
\end{lemma}
\begin{proof}
We have 
$$|a| = |b + (a-b)| \leq |b| + |a-b|,$$
so $|a|-|b|\leq \abs{a-b}$.  The same argument
with $a$ and $b$ swapped implies that
$|b|-|a|\leq \abs{a-b}$, which proves the lemma.
\end{proof}

\setcounter{section}{1}
\section{Types of Valuations}
We define two important properties of valuations, both of which
apply to  equivalence classes of valuations (i.e., the property
holds for $\abs{\mbox{ }}$ if and only if it holds for a valuation
equivalent to $\abs{\mbox{ }}$).
\begin{definition}[Discrete]
A valuation $\abs{\mbox{ }}$ is {\em discrete} if there is a $\delta>0$
such that for any $a\in K$ 
$$
   1-\delta < \abs{a} < 1+\delta \implies |a|=1.
$$
Thus the absolute values are bounded away from $1$.
%then $|a|=1$.
\end{definition}
To say that $\abs{\mbox{ }}$ is discrete is the same as saying
that the set
$$G=\bigl\{
  \log\abs{a} : a \in K, a\neq 0
\bigr\} \subset \R
$$
forms a discrete subgroup of the reals under addition (because
the elements of the group $G$ are bounded away from $0$).
\begin{proposition}\label{prop:discrete}
A nonzero discrete subgroup $G$ of $\R$ is free on one generator.
\end{proposition}
\begin{proof}
Since $G$ is discrete there is a positive  $m\in G$
such that for any positive $x\in G$ we have $m\leq x$.
Suppose $x\in G$ is an arbitrary positive element.
By subtracting off integer multiples of~$m$, we
find that there is a unique $n$ such that
$$
   0\leq x-nm <m.
$$
Since $x-nm\in G$ and $0<x-nm<m$, it follows
that $x-nm=0$, so $x$ is a multiple of $m$.
\end{proof}
By Proposition~\ref{prop:discrete}, the set 
of $\log\abs{a}$ for nonzero $a\in K$
is free on one generator, so there
is a $c<1$ such that $\abs{a}$, for $a\neq 0$,
runs precisely through the set $$c^\Z = \{c^m : m\in \Z\}$$
(Note: we can replace $c$ by $c^{-1}$ to see that we
can assume that $c<1$).

\begin{definition}[Order]
If $\abs{a} = c^m$, we call $m=\ord(a)$ the {\em order} 
of $a$. 
\end{definition}
Axiom (2) of valuations
translates into
$$
  \ord(ab) = \ord(a) + \ord(b).
$$

\begin{definition}[Non-archimedean]
A valuation $\abs{\mbox{ }}$ is {\em non-archimedean} 
if we can take $C=1$ in Axiom (3), i.e., if
\begin{equation}\label{eqn:na}
   |a + b| \leq \max\bigl\{|a|,|b|\bigr\}.
\end{equation}
If $\abs{\mbox{ }}$ is not non-archimedean then
it is {\em archimedean}.
\end{definition}
Note that if we can take $C=1$ for $\abs{\mbox{ }}$
then we can take $C=1$ for any valuation equivalent to
$\abs{\mbox{ }}$.
To see that (\ref{eqn:na}) is equivalent to Axiom (3) with
$C=1$, suppose $|b|\leq |a|$.  Then $|b/a|\leq 1$, so 
Axiom (3) asserts that $|1+b/a|\leq 1$, which implies
that $|a+b| \leq |a| = \max\{|a|,|b|\}$, and conversely.

We note at once the following consequence:
\begin{lemma}
Suppose $\abs{\mbox{ }}$ is a non-archimedean valuation.
If $a,b\in K$ with $|b|<|a|$, then
$
 |a+b|=|a|.
$
\end{lemma}
\begin{proof}
Note that $|a+b|\leq \max\{|a|,|b|\} = |a|$, which
is true even if $|b|=|a|$.  Also,
$$
  |a| = |(a+b) - b| \leq \max\{|a+b|, |b|\} = |a+b|,
$$
where for the last equality we have used that $|b|<|a|$
(if $\max\{|a+b|,|b|\} = |b|$, then $|a|\leq |b|$,
a contradiction).

\end{proof}

\begin{definition}[Ring of Integers]
Suppose $\abs{\mbox{ }}$ is a non-archimedean absolute
value on a  field $K$.  Then
$$
  \O = \{a\in K : |a|\leq 1\}
$$ is a ring called the {\em ring of integers} of $K$
with respect to $\abs{\mbox{ }}$.
\end{definition}

\begin{lemma}
Two non-archimedean valuations $\abs{\mbox{ }}_1$ and 
$\abs{\mbox{ }}_2$ are equivalent if and only if they
give the same $\O$.
\end{lemma}
We will prove this modulo the claim (to 
be proved next time\edit{in Section~\ref{sec:topology}}) that 
valuations are equivalent if (and only if) they induce the 
same topology.
\begin{proof}
Suppose suppose $\abs{\mbox{ }}_1$ is equivalent to
$\abs{\mbox{ }}_2$, so $\abs{\mbox{ }}_1 = \abs{\mbox{ }}_2^c$,
for some $c>0$.  Then $\abs{c}_1 \leq 1$ if and only if
$\abs{c}_2^c \leq 1$, i.e., if $\abs{c}_2 \leq 1^{1/c}=1$.
Thus $\O_1 = \O_2$.

Conversely, suppose $\O_1 = \O_2$. 
Then $|a|_1<|b|_1$ if and only if $a/b\in \O_1$
and $b/a\not\in \O_1$, so 
\begin{equation}\label{eqn:ineqiff}
  |a|_1<|b|_1 \iff |a|_2 < |b|_2.
\end{equation}
The topology induced by $|\mbox{ }|_1$ has as basis
of open neighborhoods the set of open balls
$$
 B_1(z,r) = \{x \in K : |x-z|_1<r \},
$$
for $r>0$, and likewise for $|\mbox{ }|_2$.  Since 
the absolute values $|b|_1$ get arbitrarily close
to $0$, the set $\mathcal{U}$ of open balls $B_1(z,|b|_1)$ also 
forms a  basis of the topology induced
by $|\mbox{ }|_1$ (and similarly for $|\mbox{ }|_2$).
By (\ref{eqn:ineqiff}) we have
$$
 B_1(z,|b|_1) = B_2(z,|b|_2),
$$
so the two topologies both have $\mathcal{U}$ as
a basis, hence are equal.  That equal topologies
implies equivalence of the corresponding valuations
will be proved later\edit{in Section~\ref{sec:topology}}.
\end{proof}

The set of $a\in \O$ with $|a|<1$ forms an ideal $\p$ in $\O$.  The
ideal $\p$ is maximal, since if $a\in\O$ and $a\not\in\p$ then
$|a|=1$, so $|1/a| = 1/|a| = 1$, hence $1/a\in \O$, so $a$ is a unit.

\begin{lemma}
A non-archimedean valuation $\abs{\mbox{ }}$ is
discrete if and only if $\p$ is a principal ideal.
\end{lemma}
\begin{proof}
First suppose that $\abs{\mbox{ }}$ is discrete.
Choose $\pi \in \p$ with $|\pi|$ maximal, which
we can do since 
$$
  S=\{\log|a| : a \in \p\} \subset (-\infty,1],
$$
so~$S$ is discrete and bounded above.
Suppose $a\in \p$.   Then 
$$
  \abs{\frac{a}{\pi}} = \frac{\abs{a}}{\abs{\pi }} \leq 1,
$$
so $a/\pi\in \O$.
Thus $$a = \pi \cdot \frac{a}{\pi} \in \pi \O.$$

Conversely, suppose $\p=(\pi)$ is principal.  For any $a\in \p$
we have $a=\pi b$ with $b\in\O$.  Thus 
$$
  |a| = |\pi|\cdot |b| \leq |\pi| < 1.
$$
Thus $\{|a| : |a|<1\}$ is bounded away from $1$,
which is exactly the definition of discrete.
\end{proof}

\begin{example}
For any prime $p$, define the $p$-adic valuation
$\abs{\mbox{ }}_p:\Q\to\R$ as follows.  Write a nonzero $\alpha\in K$
as $p^n\cdot \frac{a}{b}$, where $\gcd(a,p)=\gcd(b,p)=1$.  Then
$$\abs{p^n\cdot \frac{a}{b}}_p := p^{-n} = \left(\frac{1}{p}\right)^{n}.$$
This valuation is both discrete and non-archimedean.
The ring $\O$ is the local ring
$$
  \Z_{(p)} = \left\{\frac{a}{b}\in\Q : p\nmid b\right\},
$$
which has maximal ideal generated by $p$.  Note that
$\ord(p^n\cdot \frac{a}{b}) = p^n.$
\end{example}

We will need the following lemma later.\edit{where?}
\begin{lemma}
A valuation $\abs{\mbox{ }}$ is non-archimedean if and only if 
$|n|\leq 1$ for all $n$ in the ring generated by $1$ in $K$.
\end{lemma}
Note that we cannot identify the ring generated by $1$ with~$\Z$ 
in general, because~$K$ might have characteristic $p>0$.
\begin{proof}
If $\abs{\mbox{ }}$ is non-archimedean, then $|1|\leq 1$,
so by Axiom (3) with $a=1$, we have  $|1+1|\leq 1$.  By
induction it follows that $|n|\leq 1$.

Conversely, suppose $|n|\leq 1$ for all integer multiples~$n$ of~$1$.
This condition is also true if we replace $\abs{\mbox{ }}$ by
any equivalent valuation, so replace $\abs{\mbox{ }}$ by
one with $C\leq 2$, so that the triangle inequality holds.
Suppose $a\in K$ with $|a|\leq 1$.  Then 
by the triangle inequality,
\begin{align*}
  \abs{1+a}^n &= \abs{(1+a)^n} \\
     \leq& \sum_{j=0}^n \abs{\binom{n}{j}} \abs{a}\\
     \leq& 1 + 1 + \cdots + 1 = n.
\end{align*}
Now take $n$th roots of both sides to get
$$\abs{1+a} \leq \sqrt[n]{n},$$
and take the limit as $n\to \infty$ to see
that $\abs{1+a} \leq 1$.  This proves that one
can take $C=1$ in Axiom (3), hence that $\abs{\mbox{ }}$
is non-archimedean.
\end{proof}
\comment{
\setcounter{section}{2}
\section{Examples of Valuations}

\setcounter{section}{3}
\section{Topology}\label{sec:topology}
}

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\end{document}
