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\title{Math 129: Algebraic Number Theory\\
{\bf Lecture 8}}
\author{William Stein}
\date{Tuesday, March 2, 2004}
\begin{document}
\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
\[
\det(A)^2 = \det(AA) = \det(A A^t) 
 = \det\left(\sum_{k=1,\ldots,n} \sigma_k(w_i)\sigma_k(w_j)\right)
 = \det(\Tr(w_i w_j)_{1\leq i,j\leq n}),
\] 
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 {\em 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 $0=\Tr(a a^{-1})=\Tr(1)=[K:\Q]>0$, which is
absurd.
\end{proof}

\begin{definition}[Discriminant]
Suppose $a_1,\ldots, a_n$ is any $\Q$-basis of $K$.  The {\em 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 {\em discriminant} $\Disc(\O)$ of an order $\O$ in $\O_K$ is
the discriminant of any basis for~$\O$.
The {\em 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}
%A quadratic field $K$ is determined  determines that field,
%because $K=\Q(\sqrt{\Disc(K)})$.


\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 {\em
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 {\em 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 
\[
  \Norm(a I) = |\Norm_{K/\Q}(a)| \Norm(I).
\]
\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 {\em
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
$$
  0 \to \O_K^* \to K^* \to \Div(K) \to C_K \to 0.
$$

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 {\em convex}
if whenever $x,y\in S$ then the line connecting $x$ and $y$
lies entirely in $S$.  We say that $S$ is {\em symmetric about
the origin} if whenever $x\in S$ then $-x\in S$ also. 
If $L$ is a lattice in $V$, then the {\em 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 $2S$ 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) = \big(&\sigma_1(x), \sigma_2(x),\ldots, \sigma_r(x),\\
     &\quad \Re(\sigma_{r+1}(x)), \ldots, \Re(\sigma_{r+s}(x)), 
      \Im(\sigma_{r+1}(x)), \ldots, \Im(\sigma_{r+s}(x))\big),
\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 {\em not} exactly the same as the one
at the beginning of Section~\ref{sec:disc}.

\begin{lemma}\label{lem:volok}
\[
  \Vol(V/\sigma(\O_K)) = 2^{-s} \sqrt{|d_K|}.
\]
\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 {\em 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 $-2i$, thus changing the determinant by $(-2i)^s$.
\item Add each columns with entries $\Re(\sigma_{r+j}(w_i))$
to the the column with entries $-2i\Im(\sigma_{r+j}(w_i))$.
\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
\[
  \Vol(V/L) = |\Det(A)| = |(-2i)^{-s}\cdot \Det(B)| = 2^{-s}\sqrt{|d_K|}.
\]
\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(aI) &= |\Norm_{K/\Q}(a)|\cdot \Norm(I)\\
    &\leq \lambda^n M\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 [1..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;
  1, 2, 3, 7, 11, 19, 43, 67, 163
\end{verbatim}
It is a theorem that the above list of $9$ fields is the complete list
with class number $1$.  More generally, 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}
