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\title{Math 129: Algebraic Number Theory\\
{\bf Lecture 6}}
\author{William Stein}
\date{Tuesday, February 24, 2004}
\begin{document}
\maketitle

First we will learn how, if $p\in \Z$ is a prime and $\O_K$ is
the ring of integers of a number field, to write $p\O_K$ as a product
of primes of $\O_K$.  Then I will sketch the main results and
definitions that we will study in detail during the next 2 or 3
lectures.  We will cover discriminants and norms of ideals, define the
class group of $\O_K$ and prove that it is finite and computable, and
define the group of units of $\O_K$, determine its structure,
and prove that it is also computable.

\section{Factoring Primes}
\begin{center}
\shadowbox{
\includegraphics[width=20em]{spec.eps}}\\
A diagram from \cite{lenstras:nfs}.
\end{center}

\begin{tabular}{lr}
{\begin{minipage}{4in}
``The obvious mathematical breakthrough would be development of an easy
way to factor large prime numbers.''
\mbox{ }\hfill --{Bill Gates, {\em The Road Ahead}, pg. 265}
\end{minipage}}
&
\begin{minipage}{2in}
\mbox{}\vspace{3ex}
\shadowbox{\includegraphics[width=1in]{gates.eps}}
\end{minipage}
\\
\end{tabular}

Let $K=\Q(\alpha)$ be a number field, and let $\O_K$ be the ring of
integers of $K$.  To employ our geometric intuition, as the Lenstras
did on the cover of \cite{lenstras:nfs}, it is helpful to
view $\O_K$ as a one-dimensional scheme
$$
  X = \Spec(\O_K) = \{\text{ all prime ideals of $\O_K$ }\}
$$
over
$$
  Y=\Spec(\Z) = \{ (0) \} \union \{ p\Z : p \in\Z \text{ is prime }\}.
$$ 
There is a natural map $\pi :X\ra Y$ that sends a prime ideal $\p\in X$ to
$\p\intersect \Z\in Y$.   For much more on this point of view,
see \cite[Ch.~2]{eisenbud_harris:geometry}.

Ideals were originally introduced by Kummer because, as we proved last
Tuesday, in rings of integers of number fields ideals factor uniquely
as products of primes ideals, which is something that is not true for
general algebraic integers.  (The failure of unique factorization for
algebraic integers was used by Liouville to destroy Lam\'{e}'s
purported 1847 ``proof'' of Fermat's Last Theorem.)

If $p\in\Z$ is a prime number, then the ideal $p\O_K$ of $\O_K$
factors uniquely as a product $\prod \p_i^{e_i}$, where the $\p_i$ are
maximal ideals of $\O_K$.  We may imagine the
decomposition of $p\O_K$ into prime ideals geometrically as 
the fiber $\pi^{-1}(p\Z)$ (with multiplicities).

How can we compute $\pi^{-1}(p\Z)$ in practice?

\begin{example}
The following \magma{} session shows the commands needed to compute 
the factorization of $p\O_K$ in \magma{} for~$K$ the number field
defined by a root of $x^5+7x^4+3x^2-x+1$.
\begin{verbatim}
   > R<x> := PolynomialRing(RationalField());
   > K<a> := NumberField(x^5 + 7*x^4 + 3*x^2 - x + 1);
   > OK := MaximalOrder(K);
   > I := 2*OK;
   > Factorization(I);
   [
   <Principal Prime Ideal of OK
   Generator:
   [2, 0, 0, 0, 0], 1>
   ]
   > J := 5*OK;
   > Factorization(J);
   [
   <Prime Ideal of OK
   Two element generators:
   [5, 0, 0, 0, 0]
   [2, 1, 0, 0, 0], 1>,
   <Prime Ideal of OK
   Two element generators:
   [5, 0, 0, 0, 0]
   [3, 1, 0, 0, 0], 2>,
   <Prime Ideal of OK
   Two element generators:
   [5, 0, 0, 0, 0]
   [2, 4, 1, 0, 0], 1>
   ]
   > [K!OK.i : i in [1..5]];
   [ 1, a, a^2, a^3, a^4 ]
\end{verbatim}
Thus $2\O_K$ is already a prime ideal, and 
$$ 
  5\O_K = (5,2+a)\cdot(5,3+a)^2\cdot(5,2+4a+a^2).
$$ 
Notice that in this example $\O_K=\Z[a]$.  (Warning: There are
examples of $\O_K$ such that $\O_K\neq \Z[a]$ for any $a\in\O_K$, as
Example~\ref{ex:dedekind} below illustrates.)  When $\O_K=\Z[a]$ it is
very easy to factor $p\O_K$, as we will see below.  The following
factorization gives a hint as to why:
$$
  x^5+7x^4+3x^2-x+1 \con (x+2)\cdot (x+3)^2 \cdot (x^2+4x+2)\pmod{5}.
$$

The exponent~$2$ of $(5,3+a)^2$ in the factorization of $5\O_K$ above
suggests ``ramification'',
in the sense that the cover $X\ra Y$ has less points (counting their ``size'', i.e.,
their residue class degree) in its fiber over~$5$ than
it has generically.  Here's a suggestive picture:
\begin{center}
\psfrag{p1}{$(5,2+4a+a^2)$}
\psfrag{p2}{$(5,3+a)^2$}
\psfrag{p3}{$(5,2+a)$}
\psfrag{p}{$5\Z$}
\psfrag{q1}{$2\O_K$}
\psfrag{q}{$2\Z$}
\psfrag{zero}{$(0)$}
\psfrag{r}{$3\Z$}
\psfrag{s}{$7\Z$}
\psfrag{t}{$11\Z$}
\shadowbox{\includegraphics[width=21em]{cover1.eps}}

Diagram of $\Spec(\O_K) \ra \Spec(\Z)$
\end{center}

\end{example}

\subsection{A Method for Factoring that Often Works}
Suppose $a\in\O_K$ is such that $K=\Q(a)$, and let
$g(x)$ be the minimal polynomial of~$a$.  Then
$\Z[a]\subset \O_K$, and we have a diagram of schemes
$$\xymatrix{
  {(??)\, }\ar@{^(->}[r]\ar[d]               &{\Spec(\O_K)}\ar[d]  \\
{\bigcup\Spec(\F_p[x]/(\overline{g}_i^{e_i}))\,}\ar@{^(->}[r] \ar[d]
         &{\Spec(\Z[a])}\ar[d] \\%\ar@{=}[r] & \Spec(\Z[x]/g(x))\\
{\Spec(\F_p)\,}\ar@{^(->}[r]&{\Spec(\Z)}
}$$
where $\overline{g} = \prod_i \overline{g}_i^{e_i}$
is the factorization of the image of $g$ in $\F_p[x]$.

The cover $\pi:\Spec(\Z[a])\ra \Spec(\Z)$ is easy to understand
because it is defined by the single equation $g(x)$.  To give a
maximal ideal $\p$ of $\Z[a]$ such that $\pi(\p) = p\Z$ is the
same as giving a homomorphism $\vphi:\Z[x]/(g) \ra \Fbar_p$ (up to
automorphisms of the image), which is in turn the same as giving a
root of~$g$ in $\Fbar_p$ (up to automorphism), which is the same
as giving an irreducible factor of the reduction of~$g$ modulo~$p$.

\begin{lemma}
Suppose the index of $\Z[a]$ in $\O_K$ is coprime to~$p$.  Then
the primes~$\p_i$ in the factorization of $p\Z[a]$ do not
decompose further going from $\Z[a]$ to $\O_K$, so finding the
prime ideals of $\Z[a]$ that contain~$p$ yields the factorization
of $p\O_K$.
\end{lemma}
\begin{proof}
\comment{{\em Hi-brow argument:} By hypothesis we have an exact sequence
of abelian groups
$$ 0 \to \Z[a]\to \O_K \to H \to 0,$$
where $H$ is a finite abelian group of order coprime 
to~$p$.  Tensor product is right exact, and there is
an exact sequence
$$
  \Tor_1(H,\F_p) \to \Z[a]\tensor\F_p \to \O_K\tensor\F_p \to H\tensor\F_p \to 0,
$$
and $\Tor_1(H,\F_p)= H\tensor\F_p = 0$,
so $\Z[a]\tensor\F_p\isom \O_K\tensor\F_p$.\\
{\em Low-brow argument:}}
The inclusion map $\Z[a]\hra \O_K$
is defined by a matrix over $\Z$ that has determinant
$\pm [\O_K:\Z[a]]$, which is coprime to~$p$.  The reduction of
this matrix modulo~$p$ is invertible, so it defines an isomorphism
$\Z[a]\tensor\F_p \to \O_K\tensor\F_p$.
Any homomorphism $\O_K\to \Fpbar$ is the composition of a homomorphism
$\O_K \to \O_K\tensor\F_p$ with a homomorphism $\O_K\tensor\F_p \to
\Fpbar$.  Since $\O_K\tensor\F_p\isom \Z[a]\tensor\F_p$, the
homomorphisms $\O_K\to \Fpbar$ are in bijection with the homomorphisms
$\Z[a]\to \Fpbar$, which proves the lemma.
\end{proof}

As suggested in the proof of the lemma, we find all homomorphisms
$\O_K\to \Fpbar$ by finding all homomorphism $\Z[a]\to \Fpbar$.  In
terms of ideals, if $\p=(g(a),p)\Z[a]$ is a maximal ideal of $\Z[a]$,
then the ideal $\p'=(g(a),p)\O_K$ of $\O_K$ is also maximal, since
$$\O_K/\p'\isom (\O_K\tensor\F_p)/(g(\tilde{a}))
\isom (\Z[a]\tensor\F_p) / (g(\tilde{a})) \subset \Fpbar.$$

We formalize the above discussion in the following theorem:
\begin{theorem}\label{thm:fac1}
Let $f(x)$ denote the minimal polynomial of~$a$ over~$\Q$.
Suppose that~$p\nmid [\O_K:\Z[a]]$ is a prime.
Let
$$
 \overline{f} = \prod_{i=1}^t \overline{f}_i^{e_i} \in \F_p[x]
$$
where the $\overline{f}_i$ are distinct monic irreducible
polynomials.  
Let 
$
  \p_i = (p,f_i(a))
$
where $f_i\in\Z[x]$ is a lift of $\overline{f}_i$ in $\F_p[X]$.
Then
$$
  p\O_K = \prod_{i=1}^t \p_i^{e_i}.
$$
%Geometrically, the fiber of $\Spec(\O_K) \to p\Z$ 
%contains the points $\{\p_1,\p_2,\ldots, \p_t\}$
%with multiplicites $e_i$.
\end{theorem}

We return to the example from above, in which $K=\Q(a)$, where~$a$ is
a root of $x^5+7x^4+3x^2-x+1$.  According to MAGMA, the maximal
order $\O_K$ has discriminant $2945785$:
\begin{verbatim}
   > Discriminant(MaximalOrder(K));
   2945785
\end{verbatim}
The order $\Z[a]$ has the same discriminant as $\O_K$, so
$\Z[a]=\O_K$ and we can apply the above theorem.
\begin{verbatim}
   > Discriminant(x^5 + 7*x^4 + 3*x^2 - x + 1);
   2945785
\end{verbatim}
We have
$$
  x^5+7x^4+3x^2-x+1 \con (x+2)\cdot (x+3)^2 \cdot (x^2+4x+2)\pmod{5},
$$
which yields the factorization of $5\O_K$ given before the theorem.

If we replace $a$ by $b=7a$, then the index of $\Z[b]$
in $\O_K$ will be a power of $7$, which is coprime to $5$,
so the above method will still work.
\begin{verbatim}
   > f:=MinimalPolynomial(7*a);
   > f;
   x^5 + 49*x^4 + 1029*x^2 - 2401*x + 16807
   > Discriminant(f);
   235050861175510968365785
   > Discriminant(f)/Discriminant(MaximalOrder(K));
   79792266297612001    // coprime to 5
   > S<t> := PolynomialRing(GF(5));
   > Factorization(S!f);
   [
       <t + 1, 2>,
       <t + 4, 1>,
       <t^2 + 3*t + 3, 1>
   ]
\end{verbatim}
Thus $5$ factors in $\O_K$ 
as 
$$
  5\O_K = (5, 7a+1)^2 \cdot (5, 7a+4) \cdot (5, (7a)^2 + 3(7a) + 3).
$$
If we replace $a$ by $b=5a$ and try the above algorithm with $\Z[b]$,
then the method fails because the index of $\Z[b]$ in $\O_K$ is divisible
by~$5$.
\begin{verbatim}
   > f:=MinimalPolynomial(5*a);
   > f;
   x^5 + 35*x^4 + 375*x^2 - 625*x + 3125
   > Discriminant(f) / Discriminant(MaximalOrder(K));
   95367431640625    // divisible by 5
   > Factorization(S!f);
   [
       <t, 5>
   ]
\end{verbatim}

\subsection{A Method for Factoring that Always Works}
There are numbers fields $K$ such that $\O_K$ is not of the form
$\Z[a]$ for any $a\in K$.  Even worse, Dedekind found a
field~$K$ such that $2\mid [\O_K : \Z[a]]$ for {\em all}
$a\in \O_K$, so there is no choice of $a$ such that
Theorem~\ref{thm:fac1} can be used to factor~$2$ for $K$ (see
Example~\ref{ex:dedekind} below).

Most algebraic number theory books do not describe an algorithm for
decomposing primes in the general case.  Fortunately, Cohen's book
\cite[\S6.2]{cohen:course_ant}) describes how to solve the general
problem.  The solutions are somewhat surprising, since the algorithms
are much more sophisticated than the one suggested by
Theorem~\ref{thm:fac1}.  However, these complicated algorithms all run
very quickly in practice, even without assuming the maximal order is
already known.

For simplicity we consider the following slightly easier problem whose
solution contains the key ideas:  {\em Let $\O$ be any order in $\O_K$
and let~$p$ be a prime of $\Z$.  Find the prime ideals of $\O$ that
contain~$p$.}

To go from this special case to the general case, given a prime~$p$
that we wish to factor in $\O_K$, we find a $p$-maximal order $\O$,
i.e., an order~$\O$ such that $[\O_K:\O]$ is coprime to~$p$.  A
$p$-maximal order can be found very quickly in practice using the
``round 2'' or ``round 4'' algorithms.  (Remark: Later we will see
that to compute $\O_K$, we take the sum of $p$-maximal orders, one for
every~$p$ such that $p^2$ divides $\Disc(\O_K)$.  The time-consuming
part of this computation of $\O_K$ is finding the primes~$p$ such that
$p^2\mid \Disc(\O_K)$, not finding the $p$-maximal orders.  Thus a
fast algorithm for factoring integers would not only break
many cryptosystems, but would massively speed up computation of the ring of
integers of a number field.)


\begin{algorithm}
Suppose $\O$ is an order in the ring $\O_K$ of integers of a number
field~$K$.  For any prime $p\in\Z$, the following (sketch of an)
algorithm computes the set of maximal ideals of~$\O$ that contain~$p$.

\hd{Sketch of algorithm.}
{\sf 
Let $K=\Q(a)$ be a number field given
by an algebraic integer~$a$ as a root of its
minimal monic polynomial~$f$ of degree~$n$.
We assume that an order $\O$ has been 
given by a basis $w_1,\ldots,w_n$
and that~$\O$ that contains $\Z[a]$.
%We do {\em not} assume that $p\nmid [\O:\Z[a]]$.
Each of the following steps can be carried out 
efficiently using little more
than linear algebra over $\F_p$.  
The details are in \cite[\S6.2.5]{cohen:course_ant}.
\begin{enumerate}
\item{} [Check if easy] If $p\nmid \disc(\Z[a]) / \disc(\O)$ (so 
$p\nmid [\O:\Z[a]]$), then by a 
slight modification of Theorem~\ref{thm:fac1}, we
easily factor~$p\O$.
\item{} [Compute radical] 
Let $I$ be the {\em radical} of $p\O$, which is the ideal of 
elements $x\in\O$ such that $x^m\in p\O$ 
for some positive integer~$m$.
Using linear algebra over the finite field 
$\F_p$, we can quickly compute a basis for $I/p\O$.
(We never compute $I\subset \O$.)
\item{} [Compute quotient by radical] 
Compute an $\F_p$ basis for
$$
  A = \O/I = (\O/p\O)/(I/p\O).
$$
The second equality comes from the fact that $p\O\subset I$,
which is clear by definition.  Note that $\O/p\O\isom \O\tensor\F_p$
is obtained by simply reducing the basis $w_1,\ldots, w_n$ modulo~$p$.
\item{} [Decompose quotient] The ring $A$ is a finite Artin ring with
no nilpotents, so it decomposes as a product 
$A \isom \prod \F_p[x]/g_i(x)$ of fields.  We can quickly find such
a decomposition explicitly, as described in 
\cite[\S6.2.5]{cohen:course_ant}.
\item{} [Compute the maximal ideals over $p$] Each maximal ideal $\p_i$ lying over~$p$ 
is the kernel of $\O \ra A \ra \F_p[x]/g_i(x)$.
\end{enumerate}
}
\end{algorithm}
The algorithm finds all primes of $\O$ that contain the radical $I$ of
$p\O$.  Every such prime clearly contains $p$, so to see that the
algorithm is correct, we must prove that the primes $\p$ of $\O$ that
contain~$p$ also contain~$I$.  If $\p$ is a prime of $\O$ that
contains~$p$, then $p\O \subset \p$.  If $x\in I$ then $x^m\in p\O$
for some $m$, so $x^m\in \p$ which implies that $x\in \p$ by primality
of $\p$.  Thus $\p$ contains $I$, as required.



\subsection{Essential Discriminant Divisors}
\begin{definition}
A prime $p$ is an {\em essential discriminant
divisor} if $p\mid [\O_K : \Z[a]]$ for 
{\em every} $a\in\O_K$.  
\end{definition}
Since $[\O_K : \Z[a]]$ is the absolute value of
$\Disc(f(x))/\Disc(\O_K)$, where $f(x)$ is the characteristic
polynomial of $f(x)$, an essential discriminant divisor divides the
discriminant of the characteristic polynomial of any element of
$\O_K$.

\begin{example}[Dedekind]\label{ex:dedekind}
Let $K=\Q(a)$ be the cubic field defined by a root $a$ of the polynomial
$f = x^3 + x^2 - 2x+8$.  We will use MAGMA, which implements the algorithm
described in the previous section, to show that~$2$ is an essential
discriminant divisor for~$K$.
\begin{verbatim}
   > K<a> := NumberField(x^3 + x^2 - 2*x + 8);
   > OK := MaximalOrder(K);
   > Factorization(2*OK);
   [
   <Prime Ideal of OK
   Basis:
   [2 0 0]
   [0 1 0]
   [0 0 1], 1>,
   <Prime Ideal of OK
   Basis:
   [1 0 1]
   [0 1 0]
   [0 0 2], 1>,
   <Prime Ideal of OK
   Basis:
   [1 0 1]
   [0 1 1]
   [0 0 2], 1>
   ]
\end{verbatim}
Thus $2\O_K=\p_1\p_2\p_3$, with the $\p_i$ distinct.
Moreover, one can check that
$\O_K/\p_i\isom \F_2$.  If $\O_K=\Z[a]$
for some $a\in\O_K$ with minimal polynomial~$g$, then 
$\overline{g}(x)\in\F_2[x]$ must be a product of three {\em distinct} 
linear factors, which is impossible.
\end{example}

\section{The Chinese Remainder Theorem}
Let $\O_K$ be the ring of integers of a number field.

\begin{theorem}[Chinese Remainder Theorem]
Suppose $I_1,\ldots, I_n$ are ideals of $\O_K$ such 
that $I_j+I_k=(1)$ for any $i\neq j$.  Then the natural
map $\O_K \to \prod \O_K/I_j$ induces an isomorphism
$$
\O_K/\left(\prod I_j\right) \to \prod \O_K/I_j.
$$
Thus if we choose $a_j \in I_j$ then there exists $a\in \O_K$
such that $a\con a_j\pmod{I_j}$ for $j=1,\ldots, n$.
\end{theorem}
%\begin{proof}\end{proof}

\begin{lemma}
Suppose $I, J$ are nonzero integral ideals in $\O_K$.  Then there
exists $a\in I$ such that $(a)/I$ is coprime to $J$.
\end{lemma}
%\begin{proof}\end{proof}

The next proposition asserts that every ideal of the ring of integers
of a number field can be generated, as an ideal, by two elements.
\begin{proposition}
Suppose $I$ is called an ideal in the ring $\O_K$ of integers of a number
field.  Then there exist $a,b\in\O_K$ such that $I=(a,b)$.
\end{proposition}
%\begin{proof}\end{proof}

\section{Discriminants}
Let~$K$ be a number field of degree $n$.  Then there 
are~$n$ embeddings 
$$\sigma_1,\ldots, \sigma_n:K\to \C.$$
Reorder these embeddings so the first $r$ have image $\R$
and the remaining $s=n-r$ have image $\C$.
Let $\sigma$ by the product map $a\mapsto 
(\sigma_1(a),\ldots,\sigma_n(a))$.  

\begin{definition}[Discriminant]
Suppose $a_1,\ldots, a_n$ is a $\Q$-basis of $K$.  The {\em discriminant}
of $a_1,\ldots, a_n$ is 
$$
  \Disc(a_1,\ldots,a_n) = \det(\Tr(a_i a_j)_{i,j=1,n}).
$$
\end{definition}


\section{Norms of Ideals}

\begin{definition}[Norm of Fractional Ideal]
Suppose $I$ is a fractional ideal of $\O_K$.  The {\em norm} of
$I$ is the lattice index
$$ 
  \Norm(I) = [\O_K : I] \in \Q_{\geq 0}.
$$
This lattice index is by definition the absolute value of the determinant
of any $\Q$-linear automorphism of~$K$ that sends $\O_K$ onto~$I$, or
$0$ if $I=(0)$. 
\end{definition}

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

\section{The Class Group}
\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}

If we let $\Div(K)$ denote the group of nonzero fractional
ideals, then there is an exact sequence
$$
  0 \to \O_K^* \to K^* \to \Div(K) \to C_K \to 0.
$$


\begin{theorem}[Finiteness of the Class Group]
Every ideal class in $C_K$ contains an integral
ideal of norm at most 
$$
  \sqrt{|\Disc(K)|}\cdot \left(\frac{4}{\pi}\right)^s\frac{n!}{n^n},
$$
where $s$ is the number of complex conjugate embeddings of~$K$.
Thus the class group $C_K$ of any number field $K$ is finite. 
\end{theorem}
The 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).
%\begin{proof}\end{proof}

\subsection{The Group of Units}
\begin{definition}[Unit Group]
The {\em group of units} $U_K$ associated to a number field~$K$ is the
group of elements of $\O_K$ that have an inverse in $\O_K$.
\end{definition}

\begin{proposition}
An element $a\in K$ is a unit if and only if $\Norm(a)=\pm 1$.
\end{proposition}
%\begin{proof}\end{proof}

\begin{theorem}\label{thm:units}
The group $U_K$ of units of $\O_K$ is the product
of a finite cyclic group of roots of unity with
a free abelian group of rank $r+s-1$, where~$r$
is the number of real embeddings of~$K$ and~$s$
is the number of complex conjugate pairs of embeddings.
\end{theorem}
%\begin{proof}\end{proof}

\begin{example}[Pell's Equation]
The classical Pell's equation is, given square-free $d>0$, to find all
positive integer solutions $(x,y)$ to the equation $x^2-dy^2 = 1$.
Note that if $x+y\sqrt{d}\in\Q(\sqrt{d})$, then
$$\Norm(x+y\sqrt{d}) = (x+y\sqrt{d})(x-y\sqrt{d}) = x^2 -dy^2.$$ The
solutions to Pell's equation thus form a finite-index subgroup of the
group of units in the ring of integers of $\Q(\sqrt{d})$.
Theorem~\ref{thm:units} implies that for any~$d$ the solutions to
Pell's equation form an infinite cyclic group, a fact that takes
substantial work to prove using only elementary number theory (for
example, using continued fractions).
\end{example}


\bibliography{biblio}
\end{document}
