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\begin{document}

\chapter*{November 2:}

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

\bigskip

\begin{itemize}
\item
Remarks on class group computations
\item
Pythagoras, via elementary means
\item
Pythagoras, via Gaussian integers
\item
Pythagoras, via algebraic geometry
\item
Splitting of primes
\end{itemize}

\section{Remarks on class group computations}

The last phase of computing a class group involves verifying that various
ideals are nonprincipal.  In the case of imaginary quadratic fields this
is usually quite easy, since one has to show that a positive definite
quadratic form doesn't have a given value; usually the value is so small
that the verification is trivial and in any case it is a finite problem.

Things can be a little harder for real quadratic fields, since the corresponding
norm forms are indefinite.  For instance, in the case of the field
$\Q(\sqrt{70})$ we had to show that the equation
$$x^2-70y^2 = 2$$
was impossible.   Fortunately, this particular case could 
be handled by congruence considerations
(e.g., looking at the equation modulo~$5$).  Sometimes more is needed,
and we will return to this after we understand the structure of units
a little better.

In the case of quadratic fields, it is sometimes handy to know the following
lower bound on the power of 2 that divides the class number.  Suppose that
$s$ distinct primes divide the discriminant of a quadratic field~$F$.  Then
the class number is divisible at least by $2^{r-1}$ if $F$ is imaginary
and $2^{r-2}$ if $F$ is real.

\section{Pythagoras, via elementary means}

We now begin a discussion of Diophantine equations, with the primary 
goal being
to see how algebraic number theory can be of use.

Let's start with the famous Pythagorean equation $x^2+y^2 = z^2$.  Here are
three possible approaches.

First, one can find all solutions by elementary number theory.  We can 
divide out by a common factor if necessary to assume that $x$, $y$, and~$z$
are pairwise relatively prime.  It is impossible for all three to be odd,
and impossible for $x$ and~$y$ to be odd (the square of an odd number
$(2k+1)^2$ is congruent to 1 modulo~4, and the perfect square $z^2$
can not be congruent to 2 modulo~4).  Thus one of $x$ or $y$ is even,
and we can assume, to fix the ideas, that $y$ is even.  Then the equation
can be written
$$y^2 = z^2-x^2 = (z-x)(z+x).$$
Dividing by 4 gives
$$\left( \frac{y}{2} \right) = \left(\frac{z-x}{2}\right)
\left(\frac{z+x}{2}\right).$$
We claim that the factors on the RHS are relatively prime.  Indeed, if $p$
divides both of them then it divides their sum, $z$, and their difference, $x$,
which is impossible.

By unique factorization (in the rational integers!) if the product of two
relatively prime integers is a perfect square then each factor is a
perfect square.  If $(z+x)/2 = m^2$ and $(z-x)/2 = n^2$ then we find that
$$z = m^2 + n^2, \quad x = m^2-n^2, \quad y = 2mn.$$
In order to meet our congruence conditions, $m$ and $n$ must have opposite
parity.

The conclusion is that if $x^2 + y^2 = z^2$ where $x$, $y$, and~$z$ are pairwise
relatively prime integers, and $y$ is even, then there are relatively prime
integers $m$ and $n$, of opposite parity, such that $x$, $y$, and~$z$ are
described as above.

\section{Pythagoras, via Gaussian integers}

Perhaps a more natural, but more advanced, approach to $x^2+y^2=z^2$ is
to observe that the LHS factors in the number ring $\Z[i]$.
Assuming that $x$, $y$, and $z$ are relatively prime, with $y$ even, as
above, we write
$$x^2+y^2 = (x+iy)(x-iy) = z^2.$$
Any divisor of the two factors on the LHS divides their sum, $2x$, and
their difference $2y$, and hence divides~2.  Since $x$ and $y$ have
opposite parity, neither factor is divisible by the prime $1+i$ that
divides~2, and the two factors are relatively prime.  It follows
from unique factorization of ideals that $(x+iy)$ is a perfect
square, i.e., that
$$x+iy = u(m+ni)^2$$
where $u$ is a unit.  It follows by taking the conjugate that
$$x-iy = \overline{u}(m-ni)^2.$$
Since the square of $m+ni$ has even imaginary part it follows that
$u = \pm 1$ (rather than $u = \pm i$) and by switching $m$ and $n$
if necessary we can take $u = 1$.  Thus
$$x = m^2-n^2, \quad y = 2mn$$
as above.

The idea of this proof can be used to solve the equation in $\Z[i]$.

\section{Pythagoras, via algebraic geometry}

It is interesting to thing of the solutions as an algebraic curve,
a simple geometric idea shows how to find all such solutions.

Go to Ben's talk to find out about this!

\section{Splitting of primes}

The factorization of most (all but finitely many) rational primes in $\Z_F$
can be determined by factoring the minimal polynomial of a generator modulo
the corresponding primes.

\begin{theorem}
Let $F = \Q(\alpha)$, where $\alpha$ is an algebraic integer with minimal
polynomial~$f(x)$, and let $p$ be a rational prime.  If
$$f(x) \equiv \prod_i f_i(x)^{e_i} 
\bmod p$$
where the $f_i$ are distinct irreducible polynomials mod~$p$, then
$$(p) := p\Z_F  = \prod_i P_i^{e_i}$$
where the $P_i$ are distinct prime ideals defined by
$$P_i = (p,f_i(\alpha)).$$
\end{theorem}

\begin{example}
If $f(x) = x^3+x+1$ then
$$f(x) \equiv (x-14)^2 (x-3) \bmod 31$$
so $(31) = P^2 Q$ in $\Z_F$, where $P = (31,\alpha-14)$, and
$Q = (31,\alpha-3)$.
\end{example}
\begin{example}
If $f(x) = x^3-x^2-2x-8$ then we have seen then there are three distinct
homomorphisms from $\Z_F$ to $\F_2$.  Thus there are three prime ideals
of degree one over~$2$  (the kernels of those three maps) and $2$ splits
completely $(2) = PP'P''$ in the field.  This means that there is no
$\alpha$ such that $\Z_F = \Z[\alpha]$!  I.e., no power bases exists
for this field.  The reason is simple: if such an $\alpha$ did exist
then its minimal polynomial would have to factor modulo~2 into the
product of three distinct linear polynomials modulo~2; the only irreducible
linear polynomials modulo~2 are $x$ and $x+1$.
\end{example}

The following theorem provides useful information on how primes split.
A prime $p$ is said to \defn{ramify} in $\Z_F$ if and only  if its
factorization into primes in $\Z_F$ involves repeated factors, i.e., if
and only if some $e_i$ is bigger than~1.

\begin{theorem}
A prime $p$ ramifies if and only if it divides the discriminant $D_F$.
\end{theorem}

The fact that ramify primes divide the discriminant isn't too hard, but
the reverse implication (that a prime dividing the discriminant does
in fact ramify) is harder.

\end{document}

Thm: p ramifies (some ei > 1) iff p|DF.
Prime ideals containing p
Prime ideals of ZF/pZF
Prime ideals of Z[?]/pZ[?]
Need ?: Z[?]/pZ[?] ? ZF/pZF is an isomorphism.
Looking at the cokernel of this map, we have a surjective map ZF/pZF ? ZF/(pZF + Z[?]).  The order of this map is a power of p.  But we also have a surjective map ZF/Z[?] ? ZF/(pZF + Z[?]) with order coprime to p.  Thus the cokernel is trivial, and ? is surjective.
Now say g(?) = p?
Then g(?)/p is in ZF.  Since p is comprime to the index [ZF: Z[?]], all coefficients are multiples of p.  Therefore ? is injective, so ? is an isomorphism.
Z[?]/pZ[?] = Z/pZ [?] = Fp[?] = Fp[x]/[f] = Fp[x]/?[f]ei = 
Pi = (p, fi(?))
?Piei is contained in pZF
PZF|?Piei
?eifi = n
All these imply that the intersection of the ideals generated by the fieis is 0, so we can legitimately apply the Chinese Remainder Theorem.
Now we trace back the prime ideals through the isomorphisms to get the theorem.
