\section{Algebraic Number Fields}
We can look at the factorisation problem as the study of the group of
non-zero rationals; writing every element in terms of the generators
(the prime numbers and $-1$) and taking into account the relation
$(-1)^2=1$. The study of the unit group in $\bbZ/N\bbZ$ can be
identified with the study of a suitable quotient of a suitable
subgroup (elements prime to $N$) of this group. We now ask how
this group can be generalised. One natural idea is to use algebraic
number fields. An algebraic number is an ``object'' (we will be more
specific later) that satisfies a polynomial equation with rational
(equivalently integer) coefficients (we should actually insist on
irreducibility of the equation). We can represent such objects as we
will see below.  However, it turns out that studying groups of
algebraic numbers is not quite the same as studying the generalised
factorisation problem; that involves the study of {\em divisors} or
{\em ideals} and their groups.

\subsection{Algebraic Numbers}
How concisely can we specify an algebraic number? Since every equation
in one variable with complex coefficients can be solved completely
with complex numbers as solutions (Gauss's Fundamental theorem of
algebra), one way to specify an algebraic number is to specify it as a
complex number. However, a real (or complex) number is (in general)
only specified by its {\em entire} decimal expansion which cannot be
stored in a finite space. On the other hand it is enough to specify an
algorithm that produces, on sufficient iteration, an arbitrarily close
approximation to the complex number that represents it.

Thus, one way to specify an algebraic number $\alpha$ is as follows.
First we give $P(t)$ which is the non-zero polynomial (with rational
or integer coefficients) of least degree such that $P(\alpha)=0$ (by
Euclidean division applied to polynomials it follows that $P$ divides
any other $Q$ for which $Q(\alpha)=0$).  Further, we need to specify a
number $x_0$ of the form $r+s\cdot\sqrt{-1}$ with $r$ and $s$ rational
so that successive iterations of Newton's method
\[  x_{k+1} = x_k - \frac{P(x_k)}{P'(x_k)} \]
(where $P'(T)$ denotes the (entirely formal) derivative of $P(T)$ with
respect to $T$) converge to the complex number representing $\alpha$.
There is some (minor) ambiguity in this due the to the ``choice'' of
$\sqrt{-1}$ (which we cannot ``specify'' by this method). To quote
Abhyankar ``which is $i$ and which $-i$, perhaps only a physicist can
tell!''

Another way is to make use of Hensel's lemma. We will define below the
discriminant $D_P$ for a polynomial $P$. For now it suffices that if a
prime $p$ does not divide $D_P$ then for any $n$ so that $p$ divides
$P(n)$, we have that $p$ does not divide $P'(n)$. In other words $D_P$
is the least common multiple of $\gcd(P(n),P'(n))$ as $n$ varies over
all integers. Now, for $n$ sufficiently large it is clear that there
is such a prime $p$ (i.~e.\ not dividing $D_P$) so that $p$ divides
$P(n)$. We can now specify $\alpha$ by saying that is should be {\em
  congruent to} $n$ modulo $p$. Because of Hensel's lemma (which is
Newton's iteration done modulo powers of $p$!) we can then produce
$n_k$ so that $\alpha-n_k$ is divisible by $p^k$ for every $k$. In
modern language, we are replacing the approximation by complex numbers
given above by a $p$-adic approximation. We can actually, find a
suitable $p$ so that this can be done for {\em all} roots of the
polynomial $P$. (This is a particular case of Chebychev's density
theorem).

An entirely less obvious problem is how we can perform common
arithmetic operations on algebraic numbers when they are represented
in this fashion. For that reason, and for the reason mentioned at the
beginning of this section we now turn to the matrix representation of
algebraic numbers.

\subsection{Algebraic Number Fields as Matrix Algebras}
Let $n$ be any positive integer and consider a sub-algebra $K$ of the
algebra of $n\times n$ matrices with rational entries; by this we mean
that $K$ contains scalar multiples of the identity matrix and is
closed under matrix addition, subtraction and multiplication. To
handle division we also insist that non-zero matrices in $K$ are
invertible (this actually implies that the inverses are also in $K$
but it is not entirely trivial to prove this). Finally, one knows that
the algebra of matrices is not commutative for $n\geq 2$. So we put in
an additional hypothesis that matrix multiplication between elements
of $K$ is commutative.

Now, consider the map $\alpha\mapsto\alpha\cdot v$ where $v$ is any
(fixed) non-zero column vector such as the transpose of
$(1,0\dots,0)$. When $\alpha$ and $\beta$ are an elements of $K$ with
$\alpha\cdot v=\beta\cdot v$, we obtain $(\alpha-\beta)\cdot v=0$. But
we have assumed that every non-zero element of $K$ is invertible so we
must have $\alpha-\beta=0$. In other words this map is {\em
  one-to-one} on $K$. Thus $K$ is actually isomorphic to a vector
space of rank at most $n$ over the rationals. By a suitable change of
basis (and restricting to a submatrix) we may as well assume that the
space $K\cdot v$ contains {\em all} column vectors or equivalently
that $K$ has rank $n$. Then $K\cdot w$ is the space of all column
vectors for {\em any} non-zero vector $w$.  We will henceforth make
this additional assumption as well.

For any $n\times n$ matrix $\alpha$ we have (the Cayley-Hamilton
theorem) that {\em characteristic} polynomial $\ch_{\alpha}(T)$ of
degree $n$ and $\ch_{\alpha}(\alpha)=0$. (In the words of one
mathematician {\em khudh kaa nahi satisfy karega to kiska satisfy
  karega?}\/(Hindi); if it doesn't satisfy its' own then whose will it
satisfy?). On the other hand, we have the minimal
polynomial $\min_{\alpha}(T)$, which is the polynomial of least degree
with rational coefficients that is satisfied by $\alpha$. If
$\min_{\alpha}(T)=P(T)Q(T)$, then $P(\alpha)Q(\alpha)=0$. Since,
$P(\alpha)$ and $Q(\alpha)$ are in $K$ at least one of them must be
zero thus one of them must be a constant; in other words the minimal
polynomial is {\em irreducible}. It also follows as before that it
divides the characteristic polynomial. One can show that, under the
hypothesis of the previous paragraph (and the fact the we are working
over rationals; a {\em perfect} field), there is an element $\alpha$
in $K$ whose characteristic polynomial is {\em irreducible}, i.~e.\ 
its characteristic polynomial equals its minimal polynomial. In
particular, the field $K$ has a basis over the field $\bbQ$ of
rationals of the form 1, $\alpha$, \dots, $\alpha^{n-1}$.
\begin{proof}[(Sketch of Proof)] Let $\alpha_1$, $\alpha_2$, \dots,
  $\alpha_n$ be a basis of $K$ over the field $\bbQ$. Consider the
  characteristic polynomial of $T_1\alpha_1 + \dots + T_n\alpha_n$ as
  a function of the {\em variables} $T_1$, \dots, $T_n$. The condition
  that this is reducible will impose certain non-trivial polynomial
  relations between the $T_k$'s. Thus all we need to do is to find
  rational numbers $r_k$ that do not satisfy these relations. Then the
  characteristic polynomial of $\alpha=r_1\alpha_1 + \dots +
  r_n\alpha_n$ will be irreducible (and of degree $n$. It follows
  that, the elements 1, $\alpha$, \dots ,$\alpha^{n-1}$ will be
  independent over $\bbQ$.
\end{proof}
To summarise, we will henceforth think of an algebraic number field as
a sub-algebra of the ring of $n\times n$ matrices which is commutative,
with all non-zero elements being invertible. Moreover, this algebra
contains an element $\alpha$ whose characteristic polynomial $P_(T)$
is equal to its minimal polynomial. An further extension of the above
argument then shows that {\em any} invertible matrix $g$ that commutes
with every element of $K$ is contained in $K$; we will use this in
later sections.

As an example, let us consider the ``construction'' of the field
associated with an irreducible polynomial
$P(T)=T^n+a_1T^{n-1}+\dots+a_n$. We consider the matrix
\[
\alpha_P =   \begin{pmatrix}
     0 & 1 &  \cdots & 0 \\
     0 & 0 &  \cdots & 0 \\
     \vdots & \vdots & \ddots & \vdots \\
     -a_n & -a_{n-1} & \cdots & -a_1
   \end{pmatrix}
\]
This has minimal polynomial and characteristic polynomial equal to
$P(T)$. The sub-algebra of matrices generated by $\alpha_P$ is the
required field $\bbQ(\alpha_p)$, sometimes also denoted by
$\bbQ[T]/(P(T))$ (one uses Euclid's algorithm for polynomials to show
that every non-zero element of this is invertible). The above
discussion says that any field under consideration is {\em isomorphic}
to a field of this form for some irreducible polynomial $P(T)$.

\subsection{Orders and Maximal orders}
We now look a the subring $R$ of $K$ consisting of matrices with {\em
  integer} entries; $R$ is called an {\em order} in $K$. Now, for any
invertible $n\times n$ matrix $g$ it is clear that $gKg^{-1}$ is
isomorphic to $K$; {\em but} $g$ or $g^{-1}$ may have entries with
denominators. So the ring $R_g$ consisting of matrices in $gKg^{-1}$
with integer entries need not be that same as $R$. Thus, one can look
for a {\em maximal} order. We will see below that one such exists and
is unique. It is usually called the ring of integers in $K$ and is
denoted by $\Oh_K$.

There is a natural symmetric pairing on $n\times n$ matrices given by
\[     \langle A,B\rangle = \Trace(A\cdot B) \]
We study the restriction of this to $K$. This pairing is {\em
  non-degenerate}; i.~e.\ for any non-zero $A$ there is a $B$ so that
$<A,B>\neq 0$. For example, if $\alpha$ in $K$, then
$\langle\alpha,\alpha^{-1}\rangle=\Trace(1)=n$ which is non-zero!
(Clearly, a different argument is required when the base field is not
$\bbQ$ but a finite field). From the non-degeneracy it also follows
that for {\em any} additive map from $K$ to the rationals $\bbQ$ there
is an $\alpha$ in $K$ so that the map is precisely
$\beta\mapsto\langle\alpha,\beta\rangle$.

Now, $R$ is a subgroup of the finitely-generated free abelian group of
$n\times n$ matrices with integer coefficients; thus $R$ is a
finitely-generated free abelian group as well. If $\alpha$ is any
element of $K$ we can clear denominators to find an integer $d$ so
that $d\alpha$ is a matrix with integer entries. It follows that $R$
contains a basis of $K$ as a vector space over $\bbQ$. Thus $R$ is of
the form $\bbZ\cdot w_1 + \dots + \bbZ\cdot w_n$; moreover,
$K=\bbQ\cdot w_1+\dots+\bbQ\cdot w_n$. Let $\check{R}$ denote the
collection of all elements $\alpha$ in $K$ so that $<\alpha,\beta>$ is
an {\em integer} for all $\beta$ in $R$. Finding such an $\alpha$ is
clearly equivalent to solving the system of equations
\[
 \begin{array}{ccccccc}
  r_1\cdot\langle w_1,w_1\rangle& + &\cdots &+ &
             r_n\cdot\langle w_n,w_1\rangle &=& p_1\\
  \vdots &&&&\vdots&&\vdots  \\
  r_1\cdot\langle w_1,w_n\rangle& +&\cdots&+ &
             r_n\cdot\langle w_n,w_n\rangle &=& p_n
 \end{array}
\]
By Cramer's rule, this requires the inversion of the matrix $(\langle
w_i,w_j\rangle)_{i=1,j=1}^{n,n}$. The determinant of this matrix is
called the {\em discriminant} of the order $R$ and is denoted by
$D_R$. Note that if $v_1$, \dots, $v_n$ is another basis for $R$ and
$A$ is the matrix that gives the ``change of co-ordinates'', then the
determinant of $(\langle v_i,v_j\rangle)_{i=1,j=1}^{n,n}$ differs from
the earlier determinant by $\det(A)^2$. Since $A$ and $A^{-1}$ have
integer entries, $\det(A)=\pm 1$. Hence $D_R$ is independent of the
choice of basis. Clearly, $R\subset \check{R}$ and $\check{R}/R$ is a
finite group with $|D_R|$ elements.

Now suppose $R\subset S$, where $S$ is another order (i.~e.\ an $R_g$
for some $g$). We clearly have the sequence of inclusions $R\subset
S\subset \check{S}\subset\check{R}$. It follows that $D_S$ {\em
  divides} $D_R$; by decreasing induction we see that there is a
maximal order. We also note that by duality, $S/R$ and
$\check{R}/\check{S}$ have the same order, so that $D_R$ is the
multiple of $D_S$ by the {\em square} of an integer. Let $\Oh_K$ be
the collection of all elements of $K$ whose characteristic polynomials
have integer coefficients; one can show that this is closed under
addition and multiplication. It is clear that $\Oh_K$ contains $R$
since very matrix with integer entries has a characteristic polynomial
with integer coefficients. By the above, we see that $\Oh_K$ is
contained in $\check{R}$, hence it is finitely generated; let
$\Oh_k=\bbZ\cdot u_1+\dots+\bbZ\cdot u_n$.  Let $v$ be any non-zero
column vector and consider the basis $u_i\cdot v$ of the space of
column vectors. With this change of basis, each each element of
$\Oh_K$ is represented by a matrix with integer entries. Thus $\Oh_K$
is an order and the {\em unique} maximal order.

An extension of the example we looked at for fields is to associate an
order with an irreducible polynomial $P(T)=T^n+a_1T^{n-1}+\dots+a_n$
where the $a_i$ are all integers. We continue the notation of the
previous subsection. It follows that $\alpha_P$ is a matrix with
integer coefficients; with a little effort one can also show that the
natural order $R_P$ in $\bbQ(\alpha_P)$ is precisely the collection of
all integer linear combinations of the powers 1, $\alpha_P$, \dots,
$\alpha_P^{n-1}$.  The discriminant of this order is also the
discriminant of the polynomial $P(T)$ and is denoted as $D_P$. Unlike
the case of fields, however, it is {\em not} true that every order has
the form $R_P$ for some polynomial $P(T)$.

\subsection{Lattices and ideals}
To generalise one step further, we can consider any finitely-generated
subgroup $M$ of $K$ which contains a basis of $K$ over $\bbQ$; such an
$M$ is called a {\em lattice}. Standard arguments then show that $M$
is of the form $\bbZ\cdot m_1+\dots+\bbZ\cdot m_n$ for some basis
$m_i$ of $K$. For any fixed column vector $v$, let $g$ be the
invertible matrix that makes $m_i\cdot v$ the standard basis of the
space of column vectors.  Then after applying $g$, we see that $M\cdot
v$ becomes the standard lattice of column vectors with integer
entries.  The collection $R(M)$ of all matrices in $K$ that take $M$
to itself, is thus identified with the ring which we denoted as $R_g$
above. In the following paragraphs we assume that we have made this
change of co-ordinates (\ie that $g$ is the identity matrix). In that
case $R=R(M)$ is precisely the order consisting of integer matrices.
Moreover, there is a non-zero vector $v$ so that $M$ is precisely the
collection of all $\alpha$ so that $\alpha\cdot v$ is a vector with
integer entries.

By collecting the denominators of the generators of $M$ we can find a
non-zero integer $d$ so that $d\cdot M$ is contained in $R$. Since
this is a subgroup of $R$ that is closed under multiplication by $R$,
it is an {\em ideal} $I$ in $R$. Thus $M=d^{-1}I$ is a {\em fractional
  ideal} for $R$. It is clear that $R(d\cdot M)=R(M)=R$.  More
generally, for any non-zero $\alpha$ in $K$, we have $R(\alpha\cdot
M)=R$. Moreover, $\alpha\cdot M$ is obtained by replacing the $v$ in
the previous paragraph by $\alpha^{-1}v$, which is just another
non-zero vector.

Conversely, let $I$ be a non-zero ideal in the ring $R$. Let $\alpha$
be a non-zero element of $I$. Then $\alpha^{-1}$ is in $K$ and by
collecting the denominators we find a non-zero integer $d$ so that
$d\cdot\alpha^{-1}$ has integer coefficients so is in $R$. But then
$d=d\alpha^{-1}\cdot\alpha$ is in $I$; thus $I$ contains $d\cdot R$.
In particular, $I$ contains a basis of $K$ and is a free group of rank
$n$; in other words $I$ is a lattice. Clearly $R$ is contained in
$R(I)$ but in general the latter could be bigger.

Now, for any non-zero ideal in $R$ we have the {\em restriction}
$I\cap\bbZ=a\bbZ$. By the above discussion this is a non-zero ideal in
$\bbZ$. We also see that $R/I$ is a quotient of the finite group
$R/aR$; the latter group has order $a^n$.  The order of $R/I$ is
called the {\em norm} of the ideal and denoted as $\Nm(I)$. The norm
of an element $\alpha$ is $\det(\alpha)$; these two definitions are
related since $\Nm(\alpha\cdot R)=|\det(\alpha)|$ (Exercise).

Now, we noted above that $\Nm(d\cdot R)=d^n$ for any positive integer
$d$ so we can extend the above definition by defining for $M=d^{-1}I$,
$\Nm(M)=d^{-n}\Nm(I)$. Similarly, the restriction of $d\cdot R$ is
clearly $d$, so we define the restriction of $M$ to be
$d^{-1}(I\cap\bbZ)$. When $M$ is contained in (\ie $M$ is an ideal)
$R$, the two definitions are consistent.

\subsection{Groups of invertible fractional ideals}
The above definitions depended on a choice of ring $R\subset R(M)$,
but the following definition does not. As before, let $\check{M}$
denote the collection of all $\alpha$ in $K$ for which
$\langle\alpha,\beta\rangle$ is an integer for every $\beta$ in $M$.
Now, the non-degeneracy of the pairing $\langle,\rangle$ means that
for every additive map from $K$ to $\bbQ$ there is an $\alpha$ in $K$
so that the additive map is given by
$\beta\mapsto\langle\alpha,\beta\rangle$. It follows that $\check{M}$
can also be identified with the collection of {\em all} additive maps
from $M$ to $\bbZ$. By the usual double-duality result it follows that
$M=(\check{M}\check{)}$. In particular, we see that $\check{M}$ is
also a lattice and $R(\check{M})=R(M)$.

%It follows from the above definitions that $\check{M}$ is equivalently
%defined as the collection of all $\alpha$ so that $\alpha\cdot M$ is
%contained in $\check{R}$.

%Now let $R=R(M)$. We now relate $\check{M}$ to a certain object
%associated with $M$ and $R$.

Let $[M:R]$ denote the collection of all $\alpha$ in $K$ so that
$\alpha\cdot M$ is contained in $R$.  Clearly, $d\cdot R$ is contained
in $[M:R]$. On the other hand $\check{M}=[M:\check{R}]$ was shown
above to contain all $\alpha$ that send $M$ into $\check{R}$. The
latter contains $R$ so we see that $[M:R]$ is contained in
$\check{M}$. Thus $[M:R]$ is also a lattice.  Specifically, we define
$C_R$ as $[\check{R}:R]$.
\begin{defn}
  Let $M$ be a lattice in $K$ and $R\subset R(M)$. Then we say that
  $M$ is {\rm projective} over $R$ if $M\cdot[M:R]=R$. When $R=R(M)$
  and we have $[M:R]=C_R\cdot\check{M}$ then we say that $M$ is a {\em
    Gorenstein} $R$ module. Here the product of lattices $L_1\cdot
  L_2$ is the collection of all linear combinations of products
  $\alpha\beta$ with $\alpha$ in $L_1$ and $\beta$ in $L_2$.
\end{defn}

Armed with this result, we now consider the collection of all lattices
$M$ with the property that $R(M)=R$ for a fixed {\em Gorenstein} order
$R$. This collection of lattices includes $R$, $\check{R}$ and $C_R$.
For any such $M$, the above lemma says that $M\cdot [M:R]=R$. If we
define the product of $M$ and $N$ as $M\cdot N$, then this shows that
we have a group with $R$ playing the role of identity. It is further
clear that $M$ and $\alpha\cdot M$ are naturally isomorphic for any
non-zero $\alpha$ in $K$. We may further consider lattices modulo such
isomorphisms. This gives us the {\em class group} of invertible
fractional ideals modulo isomorphism which is denoted by $\Cl(R)$. We
noted above that there could be ideals (and fractional ideals) $M$ for
$R$ such that $R$ is a proper subring of $R(M)$. In this case we do
not necessarily have $M[M:R]=R$; we do not include such $M$ in the
class group. However, since $R(M)$ is an order as well, this situation
cannot arise if $R$ is the maximal order $\Oh_K$. The corresponding
class group is sometimes loosely referred to as the class group of $K$
and denoted $\Cl(K)$.

\subsection{Minkowski's Geometry of Numbers}
In order to decide whether or not a lattice $M$ is of the form
$\alpha\cdot R$ (and hence trivial in the class group) we need to find
elements $\alpha$ in $M$ whose norm is as close as possible to that of
$M$. This is achieved in the following section.

We now want to give a ``measure'' associated with an order $R$. The
space of $n\times n$ matrices with rational entries is naturally
contained in the space of $n\times n$ matrices with real entries. Thus
we can consider the ring $\bbR\cdot K$ of real linear combinations of
elements of $K$. This is an $n$-dimensional vector space over $\bbR$.
Thus, for any lattice $M$, the space $T_M=\bbR\cdot K/M$ is an
$n$-dimensional torus. Taking some translation invariant measure on
$\bbR\cdot K$ gives us a notion of volume for the tori $T_M$ with the
property that $\vol(T_M)=\vol(T_R)\Nm(M)$. Now, if $A$ is any (compact
measurable) subset of $\bbR\cdot K$ with the property that
$\vol(A)>\vol(T_M)$ then the map $A\to T_M$ cannot be one-to-one (with
a little thought it is clear that this is actually also true if
$\vol(A)\geq\vol(T_M)$). The difference between two points with the
same inverse image will give a non-zero element of $M$.

Now, one natural way to identify $\bbR\cdot K$ with $\bbR^n$ (and thus
put a measure on it) is to use ``simultaneous diagonalisation''. As
seen above $K$ is generated by a single $n\times n$ matrix $\alpha$
whose characteristic polynomial $P(T)$ is irreducible over rationals.
This means that this has distinct roots and so over real numbers can
be brought into a ``diagonal'' form as below by a suitable change of
co-ordinates. 
\[
     \begin{pmatrix}
       \alpha^{(1)} & 0 & \dots & 0 & 0 & 0 & \dots &0 & 0\\
       0 & \alpha^{(2)} & \dots & 0 & 0 & 0 & \dots &0 & 0\\
       \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & & \vdots & \vdots\\
       0 & 0 & \dots & \alpha^{(r_1)} & 0 & 0 & \dots &0 & 0\\
       0 & 0 & \dots & 0 &
           \hphantom{-}\Re\tilde{\alpha}^{(1)} & \Im\tilde{\alpha}^{(1)} & \dots &0 & 0\\
       0 & 0 & \dots & 0 &
           -\Im\tilde{\alpha}^{(1)} & \Re\tilde{\alpha}^{(1)} & \dots &0 & 0\\
       \vdots & \vdots & & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
       0 & 0 & \dots & 0 &
           0 & 0 & \dots & \hphantom{-}\Re\tilde{\alpha}^{(r_2)} & \Im\tilde{\alpha}^{(r_2)}\\
       0 & 0 & \dots & 0 &
           0 & 0 & \dots & -\Im\tilde{\alpha}^{(r_2)} & \Re\tilde{\alpha}^{(r_2)}
     \end{pmatrix}
\]  
Here $\alpha^{(i)}$ denote the real roots of $P(T)$, while
$(\tilde{\alpha}^{(j)},\overline{\tilde{\alpha}^{(j)}})$ are the pairs
of conjugate complex roots of $P(T)$. Now, every element of $K$ is a
linear combination of powers of $\alpha$ so that it too is brought
into the above form by the {\em same} change of co-ordinates. For
simplicity of notation we write the matrix associated with an element
$\beta$ of $K$ as
\(
   [\beta^{(1)},\dots,\beta^{(r_1)},
   \tilde{\beta}^{(1)},\dots,\tilde{\beta}^{(r_2)}]
\).
More generally, for any element $x$ in $\bbR\cdot K$ we have a
representation
$[x^{(1)},\dots,x^{(r_1)},\tilde{x}^{(1)},\dots,\tilde{x}^{(r_2)}]$.
This representation gives us an identification of $\bbR\cdot K$ with
$\bbR^n$.  If $R=\bbZ\cdot w_1+\dots+\bbZ w_n$, then the volume of
$T_R$, with respect to this identification is the determinant of the
$n\times n$ matrix $\Omega$ given by
 \[ \Omega = \left(
   w_i^{(1)},\dots,w_i^{(r_1)},\Re\tilde{w_i}^{(1)},\Im\tilde{w_i}^{(1)},
             \dots,\Re\tilde{w_i}^{(r_2)},\Im\tilde{w_i}^{(r_2)}
   \right)_{i=1}^n
 \]
Let the matrix $\tilde{\Omega}$ (complex entries) be given by
 \[ \tilde{\Omega} = \left(
   w_i^{(1)},\dots,w_i^{(r_1)},\tilde{w}_i^{(1)},\overline{\tilde{w}_i^{(1)}},
             \dots,\tilde{w}_i^{(r_2)},\overline{\tilde{w}_i^{(r_2)}}
   \right)_{i=1}^n
 \]
Standard rules for column operations on determinants show that the
determinant of $\tilde{\Omega}$ is $2^{r_2}$ times the determinant of
$\Omega$. On the other hand the $(i,j)$-th entry of the matrix
$\tilde{\Omega}\cdot\tilde{\Omega}^t$ is
\[
   \sum_{p=1}^{r_1} w_i^{(p)}w_j^{(p)} + 
      \sum_{q=1}^{r_2} \tilde{w}_i^{(q)}\tilde{w}_j^{(q)} +
      \sum_{q=1}^{r_2}
      \overline{\tilde{w}_i^{(q)}}\,\overline{\tilde{w}_j^{(q)}}
\]
which we immediately recognise as $\Trace(w_i\cdot w_j)$ when it
is expressed in the form given above. Combining these observations
we obtain the identity $\vol(T_R)=(1/2^{r_2})\sqrt{|D_R|}$.

Now consider the region $A$ consisting of all $x$ in $\bbR\cdot K$ so
that $|x^{(i)}|\leq a_i$ and $|\tilde{x}^{j}|\leq b_i$ for some
positive constants $a_i$ and $b_j$. We have 
\[
   \vol(A) = 2^{r_1}\pi^{r_2} \prod_{i=1}^{r_1} a_i
                \prod_{j=1}^{r_2} b_j^2
\]
Thus, in order to obtain a pair $(v_1,v_2)$ in $A$ so that $v=v_1-v_2$
is a non-zero element of $M$ we need the condition
\[ 
   2^{r_1}\pi^{r_2} \prod_{i=1}^{r_1} a_i \prod_{j=1}^{r_2} b_j^2
   = (1/2^{r_2})\sqrt{|D_R|}\Nm(M)
\]
Now the norm of the element $v$ is the product
\[ 
    \Nm(v) = \prod_{i=1}^{r_1} |v_1^{(i)}-v_2^{(i)}| \cdot
     \prod_{j=1}^{r_2} |\tilde{v}_1^{(j)}-\tilde{v}_2^{(j)}|^2
     \leq 2^{r_1} \prod_{i=1}^{r_1} a_i
                  \times 2^{2r_2}\prod_{j=1}^{r_2} b_j^2
\]
Hence, we have the following
\begin{lemma}
  For any ideal $I$ of an order $R$ in $K$ there is a non-zero element
  $v$ in $I$ so that
  \[ 
    \Nm(v) \leq 
        \left(\frac{2}{\pi}\right)^{r_2}\sqrt{|D_R|}\cdot|R/I|
  \]
  Here we have written $|R/I|$ instead of $\Nm(I)$ in order to make
  the dependence on $R$ clear.
\end{lemma}
\begin{proof}
  We just combine the inequality above with the condition that needs
  to be satisfied in order to obtain such a $v$.
\end{proof}
In particular, if $I$ is an invertible ideal, then $vR=I\cdot J$ where
$J=vR\cdot I^{-1}$ and $\Nm(J)\leq (2/\pi)^{r_2}\sqrt{|D_R|}$.  Now,
for any element of the class group $\Cl(R)$, let $I$ represent the
inverse of this class. The above argument produces a representative
$J$ of the class which has norm no more than
$(2/\pi)^{r_2}\sqrt{|D_R|}$. In particular, we have shown than the
class group is finite (an ideal $J$ of norm $n$ is a quotient of
cardinality $n$ the group $R/nR$; there are at most finitely many such
quotient groups).

While it is not too difficult to use this procedure to write all the
ideals $J$ satisfying the above condition, it is much harder to write
the ``multiplication table'' for the group $\Cl(R)$ on the basis of
what has gone so far. If $J_1$ and $J_2$ are two ideals as above and
the product no longer satisfies the above condition, then we need to
find the element $v$ in $(J_1\cdot J_2)$ that the lemma
guarantees. But the proof of the lemma gives us no way to find such
elements!

\subsection{Prime ideals}
Another way to write generators of groups of ideals is to use prime
ideals. An ideal $P$ of $R$ is prime if for every $a$ and $b$ in $R$
so that $ab$ lies in $P$ at least one of $a$ and $b$ lies in $P$; an
equivalent assertion is that $R/P$ is a domain---non-zero elements
give non-zero products. It is clear that the restriction $P\cap\bbZ$
satisfies the same conditions for integers $a$ and $b$. In other words
$P\cap\bbZ$ is generated by a prime number $p$. Thus $P$ determined by
the ideal $P/pR$ in $R/pR$. Now, if $R=\bbZ\cdot w_1+\dots+\bbZ\cdot
w_n$, then $R/pR$ is a vector space of rank $n$ over the finite field
$\bbZ/p\bbZ$. Instead of solving this specific problem, we can ask for
a structure theorem for commutative rings with identity with
underlying additive group a vector space of rank $n$ over
$\bbZ/p\bbZ$.
\begin{lemma}
  Any ring with underlying additive group a finite dimensional vector
  space over $\bbZ/p\bbZ$ is a direct sum of rings of the form
  $(\bbZ/p\bbZ)[T]/(P(T)^m)$, where $P(T)$ is an irreducible
  polynomial over the field $\bbZ/p\bbZ$.
\end{lemma}
This result follows from the Chinese Remainder theorem and Euclidean
division applied to the polynomial ring in one variable
$(\bbZ/p\bbZ)[T]$.  If we use the symbol $\bbF_q$ to denote the field
with $q$ elements (where $q$ is a prime power), then this result can
be refined further as follows
\begin{lemma}
  The ring $(\bbZ/p\bbZ)[T]/(P(T)^m)$, where $P(T)$ is an irreducible
  polynomial of degree $d$, is isomorphic to the ring
  $\bbF_{p^d}[h]/(h^n)$.
\end{lemma}
This result follows by constructing (using Newton's method of
successive approximations) a polynomial $\tilde{T}$ of the form
$T+P(T)Q_1(T)+\dots+P(T)^{n-1}Q_{n-1}(T)$ so that $P(\tilde{T})$ is
divisible by $P(T)^n$. Then $h=\tilde{T}-T$. Combining these results,
we see that $R/pR$ has the form
\[
  \frac{\bbF_{p^{f_1}}[h_1]}{(h_1^{e_1})} \oplus\cdots\oplus
  \frac{\bbF_{p^{f_g}}[h_g]}{(h_g^{e_g})}
\]
Corresponding to each factor we get a surjective ring homomorphism
$R/pR\to\bbF_{p^{f_i}}$. The kernel of this has the form $P_i/pR$ for
a prime ideal $P_i$ whose restriction is $p$. The number $f_i$ is
called the residual degree (or more simply the degree) of $P_i$ over
$p$, while the number $e_i$ is called the ramification degree (or order
of ramification) of $P_i$ over $p$.

Now, let $I$ be any ideal. Suppose first that the restriction $i$ of
$I$ is of the form $i_1i_2$, with $i_1$ and $i_2$ co-prime. We can
apply the Chinese Remainder theorem to write $R/I$ as a direct sum of
$R/I_1$ and $R/I_2$, where $I_1=I+i_1R$ and $I_2=I+i_2R$. Thus
$I=I_1\cap I_2$; if $I$ is invertible one easily shows that
$I=I_1\cdot I_2$. Thus we can reduce our study of groups of ideals to
the study of ideals $Q$ so that the restriction $q$ is the power of a
prime $p$. Now, $R/(Q+pR)$ is a quotient of the ring studied above and
is not zero. Thus there is a prime $P_i$ as above so that $Q$ is
contained in $P_i$. By successively removing such $P_i$ (if $Q$ is
invertible) we can write $Q$ as a product of various powers of the
$P_i$ considered above. Thus we have written any invertible ideal as a
product of (invertible) prime ideals. It is not difficult to show that
any such product expression is unique. Thus, we obtain the unique
factorisation of ideals in terms of prime ideals.

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