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\title{Math 129: Algebraic Number Theory\\
{\bf Lecture 16: Examples of Valuations, Topology}}
\author{William Stein}
\date{Tuesday, April 6, 2004}
\begin{document}
\maketitle
HOMEWORK REMINDER: Because of the senior thesis crunch, several people
received extensions on homework.  This
homework is all due {\em this Thursday}.  I will not accept any of the
homework due Thursday later than Thursday. 

Projects: Discuss now.  The list was as follows, but other
projects are possible.

\begin{enumerate}
\itemsep -.03in
\item How to compute class groups of number fields.
\item How to compute the unit group of a number field (we
didn't even prove the unit group is computable in class).
\item How to solve the norm equation $\Norm_{K/\Q}(x)=d$.
\item Explore relations between quadratic reciprocity and class
field theory for $\Q$.
\item The Chebotarev Density Theorem: read about it and explain
what the point is, and something about why it is true (e.g., for 
quadratic fields).
\item Give a proof of Dirichlet's theorem on primes in an arithmetic 
progression (connected to the Chebotarev project above).
\item Connection between ideal class groups of quadratic
imaginary fields and classes of positive definite binary
quadratic forms.  Gauss's class number problem.
\item The conjecture that there are infinitely many number fields
of class number 1.  What is known?  What do the Cohen-Lenstra
heuristics predict?  Why is this problem so hard?
\item Quadratic imaginary fields and complex multiplication
elliptic curves.
\item Elements of Shafarevich-Tate groups: give complete examples with
proofs of equations like $3x^3+4y^3+5z^3=0$ that have a solution (with
not all $x,y,z$ zero) over every $p$-adic field $\Q_p$ and over $\R$,
but not over $\Q$.
\end{enumerate}

Now the lecture.  First, we finish section 2.
\setcounter{section}{1}
\section{Types of Valuations (continued)}
Let $K$ be a field with a valuation $\abs{\mbox{ }}$.
Suppose the valuation is discrete.  Then
$$
\O = \{a \in K : |a| \leq 1\}
$$
is a subring of $K$ called the ring of integers with
respect to $\abs{\mbox{ }}$.

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 the discrete set~$S$ is 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}\label{ex:padic_valuation}
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}\label{lem:nonarch}
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}

\setcounter{section}{2}
\section{Examples of Valuations}
The archetypal example of an archimedean valuation is the absolute
value on the complex numbers.  It is essentially the only one:
\begin{theorem}[Gelfand-Tornheim] Any field~$K$ with
an archimedean valuation is isomorphic to a subfield of~$\C$,
the valuation being equivalent to that induced by the usual
absolute value on~$\C$.
\end{theorem}
We do not prove this here as we do not need it.  For a proof,
see [E. Artin, {\em Theory of Algebraic Numbers}, pages 45 and 67].

There are many non-archimedean valuations.  On the rationals $\Q$
there is one for every prime $p>0$, the $p$-adic valuation, as
in Example~\ref{ex:padic_valuation}.

\begin{theorem}[Ostrowski] The nontrivial valuations
on $\Q$ are those equivalent to $\absspc_p$, for some
prime $p$, and the usual absolute value $\absspc_\infty$.
\end{theorem}
\begin{remark}
Before giving the proof, we pause with a brief remark about
Ostrowski.  According to
{\small
\begin{verbatim}
   http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Ostrowski.html
\end{verbatim}}
\noindent{}Ostrowski was a Ukrainian mathematician who lived
1893--1986.  Gautschi writes about Ostrowski as follows: ``... you are
able, on the one hand, to emphasise the abstract and axiomatic side of
mathematics, as for example in your theory of general norms, or, on
the other hand, to concentrate on the concrete and constructive
aspects of mathematics, as in your study of numerical methods, and to
do both with equal ease. {\em You delight in finding short and
succinct proofs, of which you have given many examples} ...'' [italics mine]
\end{remark}
We will now give an example of one of these short and succinct proofs.
\begin{proof}                                     
Suppose $\absspc$ is a nontrivial valuation on $\Q$.
\par{\em Nonarchimedean case:} 
Suppose $\abs{c}\leq 1$ for all $c\in\Z$, so by
Lemma~\ref{lem:nonarch}, $\absspc$ is nonarchimedean.  
Since $\absspc$ is nontrivial, the set 
$$
  \p=\{a\in\Z : \abs{a}<1\}
$$
is nonzero.  Also $\p$ is an ideal and if $\abs{ab}<1$,
then $\abs{a}\abs{b}=\abs{ab}<1$, so $\abs{a}<1$ or $\abs{b}<1$,
so $\p$ is a prime ideal of~$\Z$.  Thus $\p=p\Z$, for some prime
number~$p$.  Since every element of $\Z$ has valuation at most
$1$, if  $u\in\Z$ with $\gcd(u,p)=1$, then $u\not\in\p$,
so $\abs{u}=1$.  Let $\alpha=\log_{\abs{p}}\frac{1}{p}$, so 
$\abs{p}^\alpha = \frac{1}{p}$.    Then for any $r$ and any $u\in\Z$
with $\gcd(u,p)=1$, we have
$$
\abs{up^r}^{\alpha} = \abs{u}^{\alpha}\abs{p}^{\alpha r}
   = \abs{p}^{\alpha r} = p^{-r} = \abs{up^r}_p.
$$
Thus $\absspc^{\alpha} =  \absspc_p$ on $\Z$, hence on $\Q$
by multiplicativity, so $\absspc$ is equivalent to $\absspc_p$,
as claimed.

{\em Archimedean case:} By replacing $\absspc$ by a power of
$\absspc$, we may assume without loss that $\absspc$ satisfies the
triangle inequality.  We first make some general remarks about any
valuation that satisfies the triangle inequality.  
Suppose $a\in\Z$ is greater than $1$.  Consider, for any $b\in\Z$
the base-$a$ expansion of $b$:
$$
  b = b_m a^m + b_{m-1} a^{m-1} + \cdots + b_0,
$$
where 
$$
  0 \leq b_j < a \qquad (0\leq j \leq m),
$$  
and $b_m\neq 0$.
Since $a^m\leq b$, taking logs we see that
$m\log(a)\leq \log(b)$, so 
$$m \leq \frac{\log(b)}{\log(a)}.$$
Let $\ds M=\max_{1\leq d<a}\abs{d}$.  Then by the triangle
inequality for $\absspc$, we have
\begin{align*}
\abs{b}&\leq \abs{b_m}{a}^m + \cdots + \abs{b_1}\abs{a} + \abs{b_0}\\
    & \leq M\cdot (\abs{a}^m + \cdots + \abs{a} + 1)\\
   & \leq M\cdot (m+1)\cdot \max(1,\abs{a}^m)\\
   & \leq M\cdot\left(\frac{\log(b)}{\log(a)} + 1\right)
        \cdot \max\left(1,\abs{a}^{\log(b)/\log(a)}\right),
\end{align*}
where in the last step we use that $m\leq \frac{\log(b)}{\log(a)}$.
Setting $b=c^n$, for $c\in\Z$, in the above inequality and
taking $n$th roots, we have 
\begin{align*}
\abs{c} &\leq \left( M\cdot 
\left(\frac{\log(c^n)}{\log(a)}+1\right)\cdot
  \max(1,\abs{a}^{log(c^n)/\log(a)})\right)^{1/n}\\
  & = M^{1/n}\cdot\left(
       \frac{\log(c^n)}{\log(a)}+1\right)^{1/n}\cdot
      \max\left(1,\abs{a}^{\log(c^n)/\log(a)}\right)^{1/n}.
\end{align*}
The first factor $M^{1/n}$ converges to~$1$ as $n\to\infty$,
since $M\geq 1$ (because $\abs{1}=1$).  The second factor
is
$$
  \left(\frac{\log(c^n)}{\log(a)}+1\right)^{1/n}
   = 
\left(n \cdot \frac{\log(c)}{\log(a)}+1\right)^{1/n}
$$
which also converges to $1$, for the
same reason that $n^{1/n}\to 1$ 
(because $\log(n^{1/n})=\frac{1}{n}\log(n)\to 0$ as
$n\to\infty$).
The third factor is
$$
\max\left(1,\abs{a}^{\log(c^n)/\log(a)}\right)^{1/n}
  = \begin{cases}
   1 & \text{if }\abs{a}<1,\\
\abs{a}^{\log(c)/\log(a)} & \text{if }\abs{a}\geq 1.
\end{cases}
$$
Putting this all together, we see that
$$
\abs{c} \leq \max\left(1,\abs{a}^{\frac{\log(c)}{\log(a)}}\right).
$$

Our assumption that $\absspc$ is nonarchimedean implies
that there is $c\in\Z$ with $c>1$ and $\abs{c}>1$.
Then for all $a\in\Z$ with $a>1$ we have
\begin{equation}\label{eqn:pow}
  1 < \abs{c} \leq
  \max\left(1,\abs{a}^{\frac{\log(c)}{\log(a)}}\right),
\end{equation}
so $1<\abs{a}^{\log(c)/\log(a)}$, so 
$1<\abs{a}$ as well (i.e., any $a\in\Z$ with
$a>1$ automatically satisfies $\abs{a}>1$).  Also, taking the
$1/\log(c)$ power on both sides of (\ref{eqn:pow})
we see that 
\begin{equation}\label{eqn:ineqac}
  \abs{c}^{\frac{1}{\log(c)}}
    \leq   \abs{a}^{\frac{1}{\log(a)}}.
\end{equation}
Because, as mentioned above, $\abs{a}>1$, we can interchange the roll
of $a$ and $c$ to obtain the reverse inequality of (\ref{eqn:ineqac}).
We thus have 
$$
  \abs{c}
    =   \abs{a}^{\frac{\log(c)}{\log(a)}}.
$$ Letting $\alpha=\log(2)\cdot \log_{\abs{2}}(e)$ and setting $a=2$,
we have
$$
  \abs{c}^{\alpha} = \abs{2}^{\frac{\alpha}{\log(2)}\cdot \log(c)}
      = \left(\abs{2}^{\log_{\abs{2}}(e)}\right)^{\log(c)} =
   e^{\log(c)} = c = \abs{c}_\infty.
$$
Thus for all integers $c\in\Z$ with $c>1$ we have
$\abs{c}^{\alpha} = \abs{c}_{\infty}$, which implies
that $\absspc$ is equivalent to $\absspc_\infty$.
\end{proof}

Let $k$ be any field and let $K=k(t)$, where $t$
is transcendental.  Fix a real number $c>1$.
If $p=p(t)$ is an irreducible
polynomial in the ring $k[t]$, we define a valuation
by 
\begin{equation}\label{eqn:ffabsp}
  \abs{p^a \cdot \frac{u}{v}}_p = c^{-\deg(p)\cdot a},
\end{equation}
where $a\in\Z$ and $u,v\in k[t]$ with
$p\nmid u$ and $p\nmid v$.
\begin{remark}
This definition differs from the one page 46 of [Cassels-Frohlich,
Ch. 2] in two ways.   First, we assume that $c>1$ instead
of $c<1$, since otherwise $\absspc_p$ does not satisfy
Axiom 3 of a valuation.  Also, we write $c^{-\deg(p)\cdot a}$
instead of $c^{-a}$, so that the product formula will
hold. 
\end{remark}
In addition there is a a non-archimedean valuation
$\absspc_\infty$ defined by 
\begin{equation}\label{eqn:ffabsoo}
  \abs{\frac{u}{v}}_\infty = c^{\deg(u)-\deg(v)}.
\end{equation}
\begin{remark}
In [Cassels-Frohlich, Ch. 2] page 46 Cassels writes
$c^{\deg(v)-\deg(u)}$, which is not correct, because
the product formula would not hold.
\end{remark}
Note the (albeit imperfect) analogy between $K=k(t)$ and $\Q$.
If $s=t^{-1}$, so $k(t)=k(s)$, the valuation $\absspc_{\infty}$
is of the type (\ref{eqn:ffabsp}) belonging to the irreducible
polynomial $p(s)=s$.

The reader is urged to prove the following lemma as a homework
problem.
\begin{lemma}
The only nontrivial valuations on $k(t)$ which are trivial
on $k$ are equivalent to the valuation (\ref{eqn:ffabsp})
or (\ref{eqn:ffabsoo}).
\end{lemma}
For example, if $k$ is a finite field, there are no
nontrivial valuations on $k$, so the only
nontrivial valuations on $k(t)$ are equivalent to 
(\ref{eqn:ffabsp}) or (\ref{eqn:ffabsoo}).

\setcounter{section}{3}
\section{Topology}\label{sec:topology}
A valuation $\absspc$ on a field $K$ induces a topology in which a
basis for the neighborhoods of $\alpha$ are the {\em open balls}
$$
  B(\alpha,d) = \{x\in K : \abs{x-\alpha} < d\}
$$
for $d>0$.
\begin{lemma}
Equivalent valuations induce the same topology.
\end{lemma}
\begin{proof}
If $\absspc_1=\absspc_2^r$, then 
$\abs{x-\alpha}_1 < d$ if and only if
$\abs{x-\alpha}_2^r<d$ if and only if
$\abs{x-\alpha}_2<d^{1/r}$
so $B_1(\alpha,d) = B_2(\alpha,d^{1/r})$.
Thus the basis of open neighborhoods of $\alpha$
for $\absspc_1$ and $\absspc_2$ are identical.
\end{proof}

A valuation satisfying the triangle inequality gives a metric
for the topology on defining the distance from $\alpha$
to $\beta$ to be $\abs{\alpha-\beta}$.
\begin{lemma}
A field with the topology induced by a valuation is
a topological field, i.e., the operations sum, product, 
and reciprocal are continuous.
\end{lemma}
\begin{proof}
For example (product) the triangle inequality implies that
$$
  \abs{(\alpha+\eps)(\beta+\delta) - \alpha\beta}
   \leq \abs{\eps}\abs{\delta} + \abs{\alpha}\abs{\delta}
     + \abs{\beta}\abs{\eps}
$$
     is small when $\abs{\eps}$ and $\abs{\delta}$ are
small (for fixed $\alpha, \beta$).
\end{proof}

\begin{lemma}\label{lem:absvalconv}
Suppose two valuations $\absspc_1$ and $\absspc_2$ on the same
field $K$ induce the same topology. Then
for any sequence $\{x_n\}$ in~$K$ we
have 
$$
  \abs{x_n}_1 \to 0 \iff \abs{x_n}_2 \to 0.
$$
\end{lemma}
\begin{proof}
It suffices to prove that if $\abs{x_n}_1\to 0$
then $\abs{x_n}_2\to 0$, since the proof of the
other implication is the same. 
Let $\eps>0$.  The topologies induced by the two absolute
values are the same, so $B_2(0,\eps)$ can be covered by
open balls $B_1(a_i,r_i)$.  One of these open balls
$B_1(a,r)$ contains~$0$, and %using the triangle inequality
we see that there is $\eps'>0$ such that
$$
  B_1(0,\eps') \subset B_1(a,r)\subset B_2(0,\eps).
$$
Since $\abs{x_n}_1\to 0$, there exists $N$ such
that for $n\geq N$ we have $\abs{x_n}_1 <\eps'$.
For such~$n$, we have $x_n\in B_1(0,\eps')$, so $x_n\in B_2(0,\eps)$,
so $\abs{x_n}_2<\eps$.  Thus $\abs{x_n}_2\to 0$.
\end{proof}

\begin{proposition}
If two valuations $\absspc_1$ and $\absspc_2$ on the same
field induce the same topology, then they are equivalent in
the sense that there is a positive real $\alpha$ such that 
$\absspc_1 = \absspc_2^{\alpha}$.
\end{proposition}
\begin{proof}
If $x\in K$ and $i=1,2$, then $\abs{x^n}_i \to 0$
if and only if $\abs{x}_i^n\to 0$, which is the
case if and only if $\abs{x}_i<1$.   Thus 
Lemma~\ref{lem:absvalconv} implies that 
$\abs{x}_1<1$ if and only if $\abs{x}_2<1$.
On taking reciprocals we see that $\abs{x}_1>1$
if and only if $\abs{x}_2>1$, so finally
$\abs{x}_1 = 1$ if and only if $\abs{x}_2=1$.

Let now $w,z\in K$ both nonzero and with $\abs{w}_i, 
\abs{z}_i\neq 1$.  On
applying the foregoing to
$$
  x = w^m z^n \qquad (m,n\in\Z)
$$
we see that
$$
  m\log\abs{w}_1 + n\log\abs{z}_1 \geq 0
$$
according as
$$
  m\log\abs{w}_2 + n\log\abs{z}_2 \geq 0.
$$
Dividing through by $\log\abs{z}_i$, and rearranging,
we see that for every rational number $\alpha=-n/m$, 
$$
  \frac{\log\abs{w}_1}{\log \abs{z}_1} \geq \alpha
\iff
  \frac{\log\abs{w}_2}{\log \abs{z}_2} \geq \alpha.
$$
Thus 
$$
  \frac{\log\abs{w}_1}{\log \abs{z}_1} =  
                \frac{\log\abs{w}_2}{\log \abs{z}_2},
$$
so 
$$
  \frac{\log\abs{w}_1}{\log \abs{w}_2} =  
                \frac{\log\abs{z}_1}{\log \abs{z}_2}.
$$
Since this equality does not depend on the choice of~$z$,
we see that there is a constant $c$ ($=\log\abs{z}_1/\log \abs{z}_2$)
such that $\log\abs{w}_1/\log \abs{w}_2 = c$ for all $w$.
Thus $\log\abs{w}_1 = c\log\abs{w}_2$, so 
$\abs{w}_1 = \abs{w}_2^c$, which implies that $\absspc_1$
is equivalent to $\absspc_2$.
\end{proof}

%\bibliography{biblio}
\end{document}
