\documentclass{article}
\include{macros}
\usepackage{amssymb}

%%%%%%%%%% Start TeXmacs macros
\newtheorem{lemma}{Lemma}
\newenvironment{proof}{
  \noindent\textbf{Proof}\ }{\hspace*{\fill}
  \begin{math}\Box\end{math}\medskip}
\newenvironment{proof*}[1]{
  \noindent\textbf{#1\ }}{\hspace*{\fill}
  \begin{math}\Box\end{math}\medskip}
\newtheorem{definition}{Definition}
\newtheorem{proposition}{Proposition}
\newtheorem{varexample}{Example}
\newenvironment{example}{\begin{varexample}\em}{\em\end{varexample}}
\newtheorem{theorem}{Theorem}
\newcommand{\tmdummy}{$\mbox{}$}
\newtheorem{corollary}{Corollary}
%%%%%%%%%% End TeXmacs macros

\begin{document}

\title{Math 129: Algebraic Number Theory\\
\textbf{Lecture 8}} \author{William Stein} \maketitle \section{The
Discriminant} \label{sec:disc}Suppose $w_1, \ldots, w_n$ are a basis for a
number field $K$, which we view as a $\Q$-vector space. Let $\sigma : K \hra
\C^n$ be the embedding $\sigma ( a ) = ( \sigma_1 ( a ), \ldots, \sigma_n ( a
) )$, where $\sigma_1, \ldots, \sigma_n$ are the distinct embeddings of $K$
into $\C$. Let $A$ be the matrix whose rows are $\sigma ( w_1 ), \ldots,
\sigma ( w_n )$. The quantity $\det ( A )$ depends on the ordering of the
$w_i$, and need not be an integer.

If we consider $\det ( A )^2$ instead, we obtain a number that is a
well-defined integer which can be either positive or negative. Note that
\[ \det ( A )^2 = \det ( AA ) = \det ( AA^t ) = \det \left( \sum_{k = 1,
   \ldots, n} \sigma_k ( w_i ) \sigma_k ( w_j ) \right) = \det ( \Tr ( w_i w_j
   )_{1 \leq i, j \leq n} ), \]
so $\det ( A )^2$ can be defined purely in terms of the trace without
mentioning the embeddings $\sigma_i$. Also, changing the basis for $\O_K$ is
the same as left multiplying $A$ by an integer matrix $U$ of determinant $\pm
1$, which does not change the squared determinant, since $\det ( UA )^2 = \det
( U )^2 \det ( A )^2 = \det ( A )^2$. Thus $\det ( A )^2$ is well defined, and
does not depend on the choice of basis.

If we view $K$ as a $\Q$-vector space, then $( x, y ) \mapsto \Tr ( xy )$
defines a bilinear pairing $K \times K \to \Q$ on $K$, which we call the
\textit{trace pairing}. The following lemma asserts that this pairing is
nondegenerate, so $\det ( \Tr ( w_i w_j ) ) \neq 0$ hence $\det ( A ) \neq 0$.

\begin{lemma}
  \label{lem:tracenondegen}The trace pairing is nondegenerate.
\end{lemma}

\begin{proof}
  If the trace pairing is degenerate, then there exists $a \in K$ such that
  for every $b \in K$ we have $\Tr ( ab ) = 0$. In particularly, taking $b =
  a^{- 1}$ we see that $0 = \Tr ( aa^{- 1} ) = \Tr ( 1 ) = [ K : \Q ] > 0$,
  which is absurd.
\end{proof}

\begin{definition}
  [Discriminant] Suppose $a_1, \ldots, a_n$ is any $\Q$-basis of $K$. The
  \textit{discriminant} of $a_1, \ldots, a_n$ is
  \[ \Disc ( a_1, \ldots, a_n ) = \det ( \Tr ( a_i a_j )_{1 \leq i, j \leq n}
     ) \in \Q . \]
  The \textit{discriminant} $\Disc ( \O )$ of an order $\O$ in $\O_K$ is the
  discriminant of any basis for $\O$. The \textit{discriminant} $d_K = \Disc (
  K )$ of the number field $K$ is the discrimimant of $\O_K$.
\end{definition}

Note that the discriminants defined above are all nonzero by Lemma
\ref{lem:tracenondegen}.

Warning: In MAGMA $\Disc ( K )$ is defined to be the discriminant of the
polynomial you happened to use to define $K$, which is (in my opinion) a poor
choice and goes against most of the literature.

The following proposition asserts that the discriminant of an order $\O$ in
$\O_K$ is bigger than $\disc ( \O_K )$ by a factor of the square of the index.

\begin{proposition}
  Suppose $\O$ is an order in $\O_K$. Then
  \[ \Disc ( \O ) = \Disc ( \O_K ) \cdot [ \O_K : \O ]^2 . \]
  
\end{proposition}

\begin{proof}
  Let $A$ be a matrix whose rows are the images via $\sigma$ of a basis for
  $\O_K$, and let $B$ be a matrix whose rows are the images via $\sigma$ of a
  basis for $\O$. Since $\O \subset \O_K$ has finite index, there is an
  integer matrix $C$ such that $CA = B$, and $| \det ( C ) | = [ \O_K : \O ]$.
  Then
  \[ \Disc ( \O ) = \det ( B )^2 = \det ( CA )^2 = \det ( C )^2 \det ( A )^2 =
     [ \O_K : \O ]^2 \cdot \Disc ( \O_K ) . \]
  
\end{proof}

This result is enough to give an algorithm for computing $\O_K$, albeit a
potentially slow one. Given $K$, find some order $\O \subset K$, and compute
$d = \Disc ( \O )$. Factor $d$, and use the factorization to write $d = s
\cdot f^2$, where $f^2$ is the largest square that divides $d$. Then the index
of $\O$ in $\O_K$ is a divisor of $f$, and we (tediously) can enumerate all
rings $R$ with $\O \subset R \subset K$ and $[ R : \O ] \mid f$, until we find
the largest one all of whose elements are integral.

\begin{example}
  Consider the ring $\O_K = \Z*{( 1 + \sqrt{5} ) / 2}$ of integers of $K = \Q
  ( \sqrt{5} )$. The discriminant of the basis $1, a = ( 1 + \sqrt{5} ) / 2$
  is
  \[ \Disc ( \O_K ) = \left| \mtwo{2}{1}{1}{3} \right| = 5. \]
  Let $\O = \Z*{\sqrt{5}}$ be the order generated by $\sqrt{5}$. Then $\O$ has
  basis $1, \sqrt{5}$, so
  \[ \Disc ( \O ) = \left| \mtwo{2}{0}{0}{10} \right| = 20 = [ \O_K : \O ]^2
     \cdot 5. \]
  
\end{example}

\section{Norms of Ideals} In this section we extend the notion of norm to
ideals. This will be helpful in proving of class groups in the next section.
For example, we will prove that the group of fractional ideals modulo
principal fractional ideals of a number field is finite by showing that every
ideal is equivalent to an ideal with norm at most some a priori bound.

\begin{definition}
  [Lattice Index] If $L$ and $M$ are two lattices in vector space $V$, then
  the \textit{lattice index} $[ L : M ]$ is by definition the absolute value
  of the determinant of any linear automorphism $A$ of $V$ such that $A ( L )
  = M$.
\end{definition}

The lattice index has the following properties:
\begin{itemize}
  \item If $M \subset L$, then $[ L : M ] = \# ( L / M )$.
  
  \item If $M, L, N$ are lattices then $[ L : N ] = [ L : M ] \cdot [ M : N
  ]$.
\end{itemize}
\begin{definition}
  [Norm of Fractional Ideal] Suppose $I$ is a fractional ideal of $\O_K$. The
  \textit{norm} of $I$ is the lattice index
  \[ \Norm ( I ) = [ \O_K : I ] \in \Q_{\geq 0}, \]
  or $0$ if $I = 0$.
\end{definition}

Note that if $I$ is an integral ideal, then $\Norm ( I ) = \# ( \O_K / I )$.

\begin{lemma}
  \label{lem:aI}Suppose $a \in K$ and $I$ is an integral ideal. Then
  \[ \Norm ( aI ) = | \Norm_{K / \Q} ( a ) | \Norm ( I ) . \]
  
\end{lemma}

\begin{proof}
  By properties of the lattice index mentioned above we have
  \[ [ \O_K : aI ] = [ \O_K : I ] \cdot [ I : aI ] = \Norm ( I ) \cdot |
     \Norm_{K / \Q} ( a ) |. \]
  Here we have used that $[ I : aI ] = | \Norm_{K / \Q} ( a ) |$, which is
  because left multiplication $\ell_a$ is an automorphism of $K$ that sends
  $I$ onto $aI$, so $[ I : aI ] = | \det ( \ell_a ) | = | \Norm_{K / \Q} ( a )
  |$.
\end{proof}

\begin{proposition}
  If $I$ and $J$ are fractional ideals, then
  \[ \Norm ( IJ ) = \Norm ( I ) \cdot \Norm ( J ) . \]
  
\end{proposition}

\begin{proof}
  By Lemma \ref{lem:aI}, it suffices to prove this when $I$ and $J$ are
  integral ideals. If $I$ and $J$ are coprime, then Theorem \ref{thm:chinese}
  (Chinese Remainder Theorem) implies that $\Norm ( IJ ) = \Norm ( I ) \cdot
  \Norm ( J )$. Thus we reduce to the case when $I = \p^m$ and $J = \p^k$ for
  some prime ideal $\p$ and integers $m, k$. By Proposition \ref{prop:quopow}
  (consequence of CRT that $\O_K / \p \isom \p^n / \p^{n + 1}$), the
  filtration of $\O_K / \p^n$ given by powers of $\p$ has successive quotients
  isomorphic to $\O_K / \p$, so we see that $\# ( \O_K / \p^n ) = \# ( \O_K /
  \p )^n$, which proves that $\Norm ( \p^n ) = \Norm ( \p )^n$.
\end{proof}

\begin{lemma}
  \label{lem:finitewithnorm}Fix a number field $K$. Let $B$ be a positive
  integer. There are only finitely many integral ideals $I$ of $\O_K$ with
  norm at most $B$.
\end{lemma}

\begin{proof}
  An integral ideal $I$ is a subgroup of $\O_K$ of index equal to the norm of
  $I$. If $G$ is any finitely generated abelian group, then there are only
  finitely many subgroups of $G$ of index at most $B$, since the subgroups of
  index dividing an integer $n$ are all subgroups of $G$ that contain $nG$,
  and the group $G / nG$ is finite. This proves the lemma.
\end{proof}

\section{Finiteness of the Class Group via Geometry of Numbers} We have seen
examples in which $\O_K$ is not a unique factorization domain. If $\O_K$ is a
principal ideal domain, then it is a unique factorization domain, so it is of
interest to understand how badly $\O_K$ fails to be a principal ideal domain.
The class group of $\O_K$ measures this failure. As one sees in a course on
Class Field Theory, the class group and its generalizations also yield deep
insight into the possible abelian Galois extensions of $K$.

\begin{definition}
  [Class Group] Let $\O_K$ be the ring of integers of a number field $K$. The
  \textit{class group} $C_K$ of $K$ is the group of nonzero fractional ideals
  modulo the sugroup of principal fractional ideals $( a )$, for $a \in K$.
\end{definition}

Note that if we let $\Div ( K )$ denote the group of nonzero fractional
ideals, then there is an exact sequence
\[ 0 \to \O_K^{\ast} \to K^{\ast} \to \Div ( K ) \to C_K \to 0. \]
A basic theorem in algebraic number theory is that the class group $C_K$ is
finite, which follows from the first part of the following theorem and the
fact that there are only finitely many ideals of norm less than a given
integer.

\begin{theorem}
  [Finiteness of the Class Group]\label{thm:finiteclassgrp}Let $K$ be a number
  field. There is a constant $C_{r, s}$ that depends only on the number $r$,
  $s$ of real and pairs of complex conjugate embeddings of $K$ such that every
  ideal class of $\O_K$ contains an integral ideal of norm at most $C_{r, s}
  \sqrt{|d_K |}$, where $d_K = \Disc ( \O_K )$. Thus by Lemma
  \ref{lem:finitewithnorm} the class group $C_K$ of $K$ is finite. One can
  choose $C_{r, s}$ such that every ideal class in $C_K$ contains an integral
  ideal of norm at most
  \[ \sqrt{|d_K |} \cdot \left( \frac{4}{\pi} \right)^s \frac{n!}{n^n} . \]
  
\end{theorem}

The explicit bound in the theorem is called the Minkowski bound, and I think
it is the best known unconditional general bound (though there are better
bounds in certain special cases).

Before proving Theorem \ref{thm:finiteclassgrp}, we prove a few lemmas. The
strategy of the proof will be to start with any nonzero ideal $I$, and prove
that there is some nonzero $a \in K$, with very small norm, such that $aI$ is
an integral ideal. Then $\Norm ( aI ) = \Norm_{K / \Q} ( a ) \Norm ( I )$ will
be small, since $\Norm_{K / \Q} ( a )$ is small. The trick is to determine
precisely how small an $a$ we can choose subject to the condition that $aI$ be
an integral ideal, i.e., that $a \in I^{- 1}$.

Let $S$ be a subset of $V = \R^n$. Then $S$ is \textit{convex} if whenever $x,
y \in S$ then the line connecting $x$ and $y$ lies entirely in $S$. We say
that $S$ is \textit{symmetric about the origin} if whenever $x \in S$ then $-
x \in S$ also. If $L$ is a lattice in $V$, then the \textit{volume} of $V / L$
is the volume of the compact real manifold $V / L$, which is the same thing as
the absolute value of the determinant of any matrix whose rows form a basis
for $L$.

\begin{lemma}
  \label{lem:gettwo}Let $L$ be a lattice in $V = \R^n$, and let $S$ be a
  bounded closed convex subset of $V$ that is symmetric about the origin.
  Assume that $\Vol ( S ) \geq 2^n \Vol ( V / L )$. Then $S$ contains a
  nonzero element of $L$.
\end{lemma}

\begin{proof}
  First assume that $\Vol ( S ) > 2^n \cdot \Vol ( V / L )$. If the map $\pi :
  \frac{1}{2} S \to V / L$ is injective, then
  \[ \frac{1}{2^n} \Vol ( S ) = \Vol \left( \frac{1}{2} S \right) \leq \Vol (
     V / L ), \]
  a contradiction. Thus $\pi$ is not injective, so there exist $P_1 \neq P_2
  \in \frac{1}{2} S$ such that $P_1 - P_2 \in L$. By symmetry $- P_2 \in
  \frac{1}{2} S$. By convexity, the average $\frac{1}{2} ( P_1 - P_2 )$ of
  $P_1$ and $- P_2$ is also in $\frac{1}{2} S$. Thus $0 \neq P_1 - P_2 \in S
  \meet L$, as claimed.
  
  Next assume that $\Vol ( S ) = 2^n \cdot \Vol ( V / L )$. Then for all $\eps
  > 0$ there is $0 \neq Q_{\eps} \in L \meet ( 1 + \eps ) S$, since $\Vol ( (
  1 + \eps ) S ) > \Vol ( S ) = 2^n \cdot \Vol ( V / L )$. If $\eps < 1$ then
  the $Q_{\eps}$ are all in $L \meet{} 2 S$, which is finite since $2 S$ is
  bounded and $L$ is discrete. Hence there exists $Q = Q_{\eps} \in L \meet{}
  ( 1 + \eps ) S$ for arbitrarily small $\eps$. Since $S$ is closed, $Q \in L
  \cap S$.
\end{proof}

\begin{lemma}
  \label{lem:latticevolchange}If $L_1$ and $L_2$ are lattices in $V$, then
  \[ \Vol ( V / L_2 ) = \Vol ( V / L_1 ) \cdot [ L_1 : L_2 ] . \]
  
\end{lemma}

\begin{proof}
  Let $A$ be an automorphism of $V$ such that $A ( L_1 ) = L_2$. Then $A$
  defines an isomorphism of real manifolds $V / L_1 \to V / L_2$ that changes
  volume by a factor of $| \det ( A ) | = [ L_1 : L_2 ]$. The claimed formula
  then follows.
\end{proof}

Fix a number field $K$ with ring of integers $\O_K$. Let $\sigma : K \to V =
\R^n$ be the embedding

\begin{align*}
  {\sigma}(x)= \left( \right. &
  {\sigma}_1(x),{\sigma}_2(x),{\ldots},{\sigma}_r(x),\\
  & \left. \quad
  {\Re}({\sigma}_{r+1}(x)),{\ldots},{\Re}({\sigma}_{r+s}(x)),{\Im}({\sigma}_{r+1}(x)),{\ldots},{\Im}({\sigma}_{r+s}(x))
  \right),
\end{align*}

where $\sigma_1, \ldots, \sigma_r$ are the real embeddings of $K$ and
$\sigma_{r + 1}, \ldots, \sigma_{r + s}$ are half the complex embeddings of
$K$, with one representative of each pair of complex conjugate embeddings.
Note that this $\sigma$ is \textit{not} exactly the same as the one at the
beginning of Section \ref{sec:disc}.

\begin{lemma}
  \label{lem:volok} {\tmdummy}
  
  \[ \Vol ( V / \sigma ( \O_K ) ) = 2^{- n} \sqrt{|d_K |} . \]
  
\end{lemma}

\begin{proof}
  Let $L = \sigma ( \O_K )$. From a basis $w_1, \ldots, w_n$ for $\O_K$ we
  obtain a matrix $A$ whose $i$th row is
  \[ ( \sigma_1 ( w_i ), \cdots, \sigma_r ( w_i ), \Re ( \sigma_{r + 1} ( w_i
     ) ), \ldots, \Re ( \sigma_{r + s} ( w_1 ) ), \Im ( \sigma_{r + 1} ( w_i )
     ), \ldots, \Im ( \sigma_{r + s} ( w_1 ) ) ) \]
  and whose determinant has absolute value equal to the volume of $V / L$. By
  doing the following three column operations, we obtain a matrix whose rows
  are exactly the images of the $w_i$ under \textit{all} embeddings of $K$
  into $\C$, which is the matrix that came up when we defined $d_K$.
  \begin{enumerate}
    \item Add $i = \sqrt{- 1}$ times each column with entries $\Im ( \sigma_{r
    + j} ( w_i ) )$ to the column with entries $\Re ( \sigma_{r + j} ( w_i )
    )$.
    
    \item Multiply all columns $\Im ( \sigma_{r + j} ( w_i ) )$ by $- 2 i$,
    thus changing the determinant by $( - 2 i )^s$.
    
    \item Add each columns with entries $\Re ( \sigma_{r + j} ( w_i ) )$ to
    the the column with entries $- 2 i \Im ( \sigma_{r + j} ( w_i ) )$.
  \end{enumerate}
  Recalling the definition of discriminant, we see that if $B$ is the matrix
  constructed by the above three operations, then $\Det ( B )^2 = d_K$. Thus
  \[ \Vol ( V / L ) = | \Det ( A ) | = | ( - 2 i )^{- s} \cdot \Det ( B ) | =
     2^{- s} \sqrt{|d_K |} . \]
  
\end{proof}

\begin{lemma}
  \label{lem:volfracideal}If $I$ is a nonzero fractional ideal for $\O_K$,
  then $\sigma ( I )$ is a lattice in $V$, and
  \[ \Vol ( V / \sigma ( I ) ) = 2^{- s} \sqrt{|d_K |} \cdot \Norm ( I ) . \]
  
\end{lemma}

\begin{proof}
  We know that $[ \O_K : I ] = \Norm ( I )$ is a nonzero rational number.
  Lemma \ref{lem:volok} implies that $\sigma ( \O_K )$ is a lattice in $V$,
  since $\sigma ( \O_K )$ has rank $n$ as abelian group and spans $V$, so
  $\sigma ( I )$ is also a lattice in $V$. For the volume formula, combine
  Lemmas \ref{lem:latticevolchange}--\ref{lem:volok} to get
  \[ \Vol ( V / \sigma ( I ) ) = \Vol ( V / \sigma ( \O_K ) ) \cdot [ \O_K : I
     ] = 2^{- s} \sqrt{|d_K |} \Norm ( I ) . \]
  
\end{proof}

\begin{proof}
  [Proof of Theorem \ref{thm:finiteclassgrp}] Let $K$ be a number field with
  ring of integers $\O_K$, let $\sigma : K \hra V \isom \R^n$ be as above, and
  let $f : V \to \R$ be the function defined by
  \[ f ( x_1, \ldots, x_n ) = |x_1 \cdots x_r \cdot ( x_{r + 1}^2 + x_{( r + 1
     ) + s}^2 ) \cdots ( x_{r + s}^2 + x_n^2 ) . \]
  Notice that if $x \in K$ then $f ( \sigma ( x ) ) = | \Norm_{K / \Q} ( x )
  |$.
  
  Let $S \subset V$ be any closed, bounded, convex, subset that is symmetric
  with respect to the origin and has positive volume. Since $S$ is closed and
  bounded,
  \[ M = \max \{ f ( x ) : x \in S \} \]
  exists.
  
  Suppose $I$ is any nonzero fractional ideal of $\O_K$. Our goal is to prove
  there is an integral ideal $aI$ with small norm. We will do this by finding
  an appropriate $a \in I^{- 1}$. By Lemma \ref{lem:volfracideal},
  \[ c = \Vol ( V / I^{- 1} ) = \frac{2^{- s} \sqrt{|d_K |}}{\Norm ( I )} . \]
  Let $\lambda = 2 \cdot \left( \frac{c}{v} \right)^{1 / n}$, where $v = \Vol
  ( S )$. Then
  \[ \Vol ( \lambda S ) = \lambda^n \Vol ( S ) = 2^n \frac{c}{v} \cdot v = 2^n
     \cdot c = 2^n \Vol ( V / I^{- 1} ), \]
  so by Lemma \ref{lem:gettwo} there exists $0 \neq a \in I^{- 1} \meet
  \lambda S$. Since $M$ is the largest norm of an element of $S$, the largest
  norm of an element of $I^{- 1} \cap \lambda S$ is at most $\lambda^n M$, so
  \[ | \Norm_{K / \Q} ( a ) | \leq \lambda^n M. \]
  Since $a \in I^{- 1}$, we have $aI \subset \O_K$, so $aI$ is an integral
  ideal of $\O_K$ that is equivalent to $I$, and
  
  \begin{align*}
    {\Norm}(a*I) & =|{\Norm}_{K/{\Q}}(a)|{\cdot} {\Norm}(I)\\
    & {\leq} {\lambda}^nM{\cdot} {\Norm}(I)\\
    & {\leq}2^n {\frac{c}{v}}M{\cdot} {\Norm}(I)\\
    & {\leq}2^n {\cdot}2^{-s} {\sqrt{|d_K|}} {\cdot}M{\cdot}v^{-1}\\
    & =2^{r+s} {\sqrt{|d_K|}} {\cdot}M{\cdot}v^{-1}.
  \end{align*}
  
  Notice that the right hand side is independent of $I$. It depends only on
  $r$, $s$, $|d_K |$, and our choice of $S$. This completes the proof of the
  theorem, except for the assertion that $S$ can be chosen to give the claim
  at the end of the theorem, which we leave as an exercise.
\end{proof}

\begin{corollary}
  Suppose that $K \neq \Q$ is a number field. Then $|d_K | > 1$.
\end{corollary}

\begin{proof}
  Applying Theorem \ref{thm:finiteclassgrp} to the unit ideal, we get the
  bound
  \[ 1 \leq \sqrt{|d_K |} \cdot \left( \frac{4}{\pi} \right)^s \frac{n!}{n^n}
     . \]
  Thus
  \[ \sqrt{|d_K |} \geq \left( \frac{\pi}{4} \right)^s \frac{n^n}{n!}, \]
  and the right hand quantity is strictly bigger than $1$ for any $s \leq n /
  2$ and any $n > 1$ (exercise).
\end{proof}

\subsection{An Open Problem} \begin{conjecture}
  There are infinitely many number fields $K$ such that the class group of
  $K$ has order $1$. 
\end{conjecture} For example, if we consider real quadratic fields $K = \Q (
\sqrt{d} )$, with $d$ positive and square free, many class numbers are
probably $1$, as suggested by the MAGMA output below. It looks like 1's will
keep appearing infinitely often, and indeed Cohen and Lenstra conjecture that
they do. Nobody has found a way to prove this yet.
\begin{verbatim}
   > for d in [2..1000] do 
        if d eq SquareFree(d) then 
           h := ClassNumber(NumberField(x^2-d));
           if h eq 1 then
              printf "%o, ", d;
           end if;
        end if;
      end for;
   
   2, 3, 5, 6, 7, 11, 13, 14, 17, 19, 21, 22, 23, 29, 31, 33, 37, 
   38, 41, 43, 46, 47, 53, 57, 59, 61, 62, 67, 69, 71, 73, 77, 83, 
   86, 89, 93, 94, 97, 101, 103, 107, 109, 113, 118, 127, 129, 131, 
   133, 134, 137, 139, 141, 149, 151, 157, 158, 161, 163, 166, 167, 
   173, 177, 179, 181, 191, 193, 197, 199, 201, 206, 209, 211, 213, 
   214, 217, 227, 233, 237, 239, 241, 249, 251, 253, 262, 263, 269, 
   271, 277, 278, 281, 283, 293, 301, 302, 307, 309, 311, 313, 317, 
   329, 331, 334, 337, 341, 347, 349, 353, 358, 367, 373, 379, 381, 
   382, 383, 389, 393, 397, 398, 409, 413, 417, 419, 421, 422, 431, 
   433, 437, 446, 449, 453, 454, 457, 461, 463, 467, 478, 479, 487, 
   489, 491, 497, 501, 502, 503, 509, 517, 521, 523, 526, 537, 541, 
   542, 547, 553, 557, 563, 566, 569, 571, 573, 581, 587, 589, 593, 
   597, 599, 601, 607, 613, 614, 617, 619, 622, 631, 633, 641, 643, 
   647, 649, 653, 661, 662, 669, 673, 677, 681, 683, 691, 694, 701, 
   709, 713, 717, 718, 719, 721, 734, 737, 739, 743, 749, 751, 753, 
   757, 758, 766, 769, 773, 781, 787, 789, 797, 809, 811, 813, 821, 
   823, 827, 829, 838, 849, 853, 857, 859, 862, 863, 869, 877, 878, 
   881, 883, 886, 887, 889, 893, 907, 911, 913, 917, 919, 921, 926, 
   929, 933, 937, 941, 947, 953, 958, 967, 971, 973, 974, 977, 983, 
   989, 991, 997, 998, 
\end{verbatim}


In contrast, if we look at class numbers of quadratic imaginary fields, only a
few at the beginning have class number $1$.
\begin{verbatim}
   > for d in [2..1000] do 
        if d eq SquareFree(d) then 
           h := ClassNumber(NumberField(x^2+d));
           if h eq 1 then
              printf "%o, ", d;
           end if;
        end if;
     end for;
   2, 3, 7, 11, 19, 43, 67, 163
\end{verbatim}
It is possible (in theory), using deep work of Gross, Zagier, and Goldfeld
involving zeta functions and elliptic curves, to enumerate all quadratic
number fields with a given class number.



\end{document}
