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\begin{document}
\chapter*{November 5:}

\begin{center}
{\Huge \bf Math 431 Class Lecture Notes}
\end{center}

\bigskip

\begin{itemize}
\item
Factoring primes
\item
Diophantine equations, continued
\item
Integer points on an elliptic curve
\end{itemize}


\section{Factoring primes}

Suppose $F$ is a number field and let $\alpha \in F$. Let $p$ be a
rational prime not dividing $[\Z_F : \Z[\alpha]]$. Then a very
useful theorem, stated last time, asserts that the principal
ideal generated by $p$ factors as 
%
\[
(p) := p \Z_F = \prod P_i^{e_i},
\]
%
where $P_i = (p, f_i(\alpha))$ and $m_{\alpha} \equiv \prod f_i^{e_i}
\bmod{p}$.  In addition, the residue class degree of $P_i$ is the degree of
$f_i$, and in fact, $\Z_F/P_i = \F_p[x]/(f_i(x))$.

We outline a proof of this result, mostly by reducing it to a sequence of
``elementary'' algebra isomorphisms; the reader will have to draw on their
abstract algebra brain cells to verify the details.

The behavior of $p$ i $\Z_F$ is determined by the structure of the quotient
$\Z_F/(p)$ where $(p)$ denotes $p\Z_F$.  We start by showing that our hypothesis
on the index of $\Z[\alpha]$ in $\Z_F$ implies that this quotient
is isomorphic to $\Z[\alpha]/p\Z[\alpha]$.

The kernel of the map from $\Z[\alpha]/p\Z[\alpha]$ to 
$\Z_F/p\Z_F$ is $(p\Z_F \cap \Z[\alpha])/p\Z[\alpha]$, and the
cokernel (the quotient of the domain by the image) is 
$\Z_F/(p\Z_F+\Z[\alpha])$.  In the language of abstract algebra,
we have an exact sequence
$$0 \to \frac{p\Z_F \cap \Z[\alpha]}{p\Z[\alpha]} \to
\frac{\Z[\alpha]}{p\Z[\alpha]} \to
\frac{\Z_F}{p\Z_F} \to \frac{\Z_F}{p\Z_F+\Z[\alpha]} \to 0.$$
Consider the following diagram of inclusions of abelian groups (in
which all upward arrows are inclusions).
\begin{center}
\begin{tabular}{ccccc}
&&$ \Z_F$ &&\\
&&$\uparrow$&&\\
&&$p\Z_F+\Z[\alpha]$&&\\
&$\nearrow$&&$\nwarrow$&\\
$p \Z_F $ &&& $\Z[\alpha]$\\
&$\nwarrow$&&$\nearrow$&\\
&&$p\Z_F \cap \Z[\alpha]$&&\\
&&$\uparrow$&&\\
&&$p\Z[\alpha]$&&\\
\end{tabular}
\end{center}
The index $a := [\Z_F:p\Z_F+\Z[\alpha]]$ is a divisor
both of $[\Z_F:p\Z_F]$, which has order a power of~$p$, and of
$[Z_F:\Z[\alpha]]$, which is prime to~$p$ by assumption.  Therefore
$a = 1$.  Similarly the index $b := [p\Z_F \cap \Z[\alpha]$ is
a divisor of $[p\Z_F:p\Z[\alpha]] = [\Z_F:\Z[\alpha]$ and of
$[\Z[\alpha]:p\Z[\alpha]]$ and we conclude that $b = 1$.
Thus the kernel and cokernel of the map $\phi$ are trivial, and
the map from $\Z[\alpha]/p\Z[\alpha]$ to $\Z_F/p\Z_F$
is an isomorphism, as claimed.

With this in hand, we study
the ring $\Z[\alpha]/p\Z[\alpha]$.  There are
natural isomorphisms
$$\frac{\Z[\alpha]}{p\Z[\alpha]} \simeq
\frac{\Z[x]}{(p,f(x))} \simeq
\frac{\F_p[x]}{\overline f(x)} \simeq
\bigoplus \frac{\F_p[x]}{\overline f_i(x)^{e_i}}.$$
The quotient $\F_p[x]/\overline f_i^{e_i}$ is a ring with a unique
prime ideal, name the principal ideal $P'_i := (f_i)$.  By chasing 
back through the isomorphisms, one finds that this corresponds to
the prime ideal $P_i = (p,f_i(\alpha))$ in $\Z[\alpha]$.  Since
the prime ideals in a direct sum of rings correspond to prime ideals
in each summand, we conclude that the $P_i$ are the unique prime ideals
in $\Z[\alpha]$ that contain~$p$.  Moreover, the quotient rings
$\Z[\alpha]/P_i$ are isomorphic to $\F_p[x]/\overline f_i(x)$ so
that the residue class degree of $P_i$ is the degree of~$f_i$.  

Moreover, in the quotient $\F_p[x]/\overline f$ the product
$\prod (P'_i)^{e_i} $ vanishes.  Chasing back through the isomorphisms, we
find that this means that in $\Z[\alpha]$ the product
$$\prod P_i^{e_i}$$
contains the ideal $p\Z[\alpha]$.  If $p\Z[\alpha] = \prod P_i^{E_i}$
then by the ``$\sum e_i f_i$'' theorem, we have
$$n = \sum e_i \deg(f_i) \ge \sum E_i \deg(f_i) = n$$
so equality holds throughout.

Putting all of this together gives us the desired result!

\section{Diophantine equations, continued}

Many applications of algebraic number theory to specific Diophantine
equations start by factoring a homogeneous polynomial into two
variables into a product of linear forms, with coefficients in an
algebraic number field.  Typically, the other side of the
equation is a perfect power, or a constant times a perfect power.

For instance:
\begin{itemize}
\item
The parametrization of all Pythagorean triples over $\Z$,
can be done by carefully considering the factorization
$$(x + iy)(x - iy) = z^2,$$
in $\Z[i]$. 
\item
The integral points of the elliptic
curve $y^2 = x^3 - 355$ can be found by factoring $y^2+355$ and setting
it equal to a perfect cube.
\item All solutions to $x^2 + y^2 = z^3$ 
can be found by factoring the LHS and working in $\Z[i]$.
\item As we will see shortly, the nonexistence of nontrivial solutions
to $x^3+y^3 = z^3$ can be proved by writing this as
$$(x+y)(x+\omega y)(x+\overline\omega  y) = z^3,$$
where $\omega = \exp^{2\pi i/3}$, and then working in $\Z[\alpha]$.
\end{itemize}

The problem of finding which numbers are values of a given homogeneous
polynomial sometimes succumbs to the same techniques.  For instance,
solutions to the equation $x^2 + y^2 = n$ boil down to the question of which
integers are norms of elements of $\Z[i]$; the equation has solutions
if and only if all primes $p \mid n$, $p \equiv 3 \bmod{4}$ have even exponents in
the prime factorization of $n$.

\section{Integer points on an elliptic curve}

One of the famous open conjectures in number theory asserts that the
rank of an elliptic curve (the number of independent points of
infinite order) is equal to the order of vanishing at $s = 1$ of the L-series
of the elliptic curve.  This conjecture is due to Bryan Birch and
Peter Swinnerton-Dyer.  Their conjecture has led to an immense amount
of research.  It is somehow comforting that this conjecture was
originally motivated by calculations on many specific curves.

One of the most famous curves that Birch and Swinnerton-Dyer used was the
curve
$$y^2 + m = x^3.$$ 
As above, this leads naturally to consideration of the 
field $\Q(\sqrt{-m}) =: F$.  To treat a simple general case, let's suppose
that $-m \equiv 2$ or $3 \bmod{4}$, that $m$ is positive and squarefree, 
and that $3 \nmid h_F$.

The equation factors as $(y + \sqrt{-m})(y - \sqrt{-m}) =
x^3$.  Considering this modulo~$8$ we see that $x$ must be odd; also,
$(y,m) = 1$, since if $p \mid y$, $p \mid m$, then $p \mid x$; but
then $m$ is not squarefree. 

The ideals $y \pm \sqrt{-m}$ are relatively prime. If $IJIK = x^3$
then $y \pm \sqrt{-m} \in I \Rightarrow 2 \sqrt{-m}, 2y \in
I$. $(y,m)=1 \Rightarrow (x,m) = 1$. So we have $(y + \sqrt{-m}) = I^3
= (a + b \sqrt{-m})^3$; moving to elements, $y + \sqrt{-m} = (a + b
\sqrt{-m})^3$, since we can absorb the unit into $a$ and $b$. Thus,
%
\begin{align}
y + \sqrt{-m} &= \pm (a+b\sqrt{-m})^3 \\
	      &= (a^3 - 3 a b^2 m) + (3 a^2 b - b^3 m)\sqrt{-m}
\end{align}
%
Now, $(3a^2 - b^2 m)b = 1 \Rightarrow b = \pm 1$. So we have 
$3 a^2 - m = \pm 1$ which implies $m = 3a^2 \pm 1$.  The conclusion is 
as follows.

\begin{theorem}
With the above assumptions on~$m$, the equation
$$y^2 + m = x^3$$
has a solution if and only if $m$ is of the form $m = 3a^2 \pm 1$,
in which case we must have $x = (a^2+m)^3$, and
$y = a^3-3am$.
\end{theorem}

\end{document}
