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\begin{document}

\chapter*{October 29:}

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

\begin{itemize}
\item
Finiteness of the class group
\item
Computing a class group
\item
The Minkowski bound
\end{itemize}

\section{Finiteness of the class group}

The Minkowski bound says that every ideal $I$ in a number ring $\Z_F$
contains an element $\alpha$ such that
$$|N(\alpha)| \le C_F N(I)$$
where $C_F$ depends only on the signature $(r_1,r_2)$, degree $n$, and
discriminant $D_F$ of~$F$:
$$C_F = \frac{n!}{n^n} \left( \frac{4}{\pi}\right)^{r_2} \sqrt{|D_F|}.$$

It is an interesting exercise to show that
$$ N(I) = \gcd( \{ | N(\alpha)|  \ : \alpha\in I \} ). $$
Thus the term $C_F \sqrt{|D_F|}$ measures how far off we can be
in comparing $N(I)$ with the smallest possible $N(\alpha)$, for
$\alpha \in I$.  If the class group is trivial then of course
one can find $\alpha$ with $|N(\alpha)| = N(I)$.  Thus it is plausible
that this result says something about the class group.

\begin{theorem}
The class group $\Cl(F)$ of a number field is finite.
\end{theorem}

\begin{proof}
We show that any
class contains an ideal of norm at most $C_F .$ 
To see this, let $I$
be an ideal in $\Z_F$, and denote
its class by $[I]$.  Then $I$ is invertible so there is an ideal
$I'$ such that $II' = (\beta)$ for some~$\beta$. Now apply
the Minkowski bound to $I'$ to obtain an element $\alpha$ such that
$$ |N(\alpha)|\leq C_F \cdot N(I). $$
Since $(\alpha)\subset I'$, $(\alpha) = I'J$ for some $J$.
This implies that $I(\alpha) = II'J = (\beta)J $. Thus
$I$ and $J$ are  in the same ideal class.
However
$$ N(J) = {N((\alpha))\over N(I')} = {|N(\alpha)|\over N(I')} 
\leq C_F .$$
There are finitely many rational primes less than $N(J)$, and hence
finitely many prime ideals $P$ in $\Z_F$ of norm less than a given
bound, and hence finitely many ideals of norm less than $C_F \sqrt{|D|}$.
Since every ideal class has a representative in a finite set, the ideal
class group is finite.
\end{proof}

\section{Computing a class group}

Let $F = \Q(\sqrt{-15})$.  Then we know that $\Z_F = \Z[\delta]$ where
$\delta = (1+\sqrt{-15})/2$, $D_F = -15$, and $N(x+y\delta) = x^2+xy+4y^2$.
From the proof of the finiteness of the class number using the Minkowski
bound we know that every ideal class contains an ideal of norm at most
$$\frac{2!}{2^2} \left( \frac{4}{\pi}\right) \sqrt{15} < 4.$$
It turns out that class groups can be computed by computing sufficiently
many norms, and then drawing suitable inferences!

So we start by computing $N(x+\delta) = x^2+x+4$ for $0 \le x < 12$.
(Since $N(-x+\delta) = N(x-1-\delta)$  it is unnecessary to compute this
function for negative~$x$.)

\begin{center}

\begin{tabular}{|r|r|r|}
\hline\hline
$x$ & $x^2+x+4$ & $(x+\delta)$\\ \hline\hline
0 & $4 = 2^2$ & $P_2^2$\\ \hline
1 & $6 = 2 \cdot 3$ & $P_2' P_3$\\ \hline
2 & $10 = 2\cdot 5$ & $P_2 P_5$\\ \hline
3 & $16 = 2^4$ & $P_2'^4 $\\ \hline
4 & $24 = 2^3 \cdot 3$ & $P_2^3 P_3$\\ \hline
5 & $34 = 2\cdot 17$ & $P_2' P_{17}$\\ \hline
6 & $46 = 2\cdot 23$ & $P_2 P_{23}$\\ \hline
7 & $60 = 2^2 \cdot 3 \cdot 5$ & $P_2'^2 P_3 P_5$\\ \hline
8 & $74 = 2 \cdot 37$ & $P_2 P_{37}$\\ \hline
9 & $94 \cdot 47$ & $P_2' P_{47}$\\ \hline
10 & $114 = 2\cdot 3 \cdot 19$ & $P_2 P_3 P_{19}$\\ \hline
11 & $119 = 2^3 \cdot 17$ & $P_2'^3 P_{17}'$\\ \hline
12 & $160= 2^5 \cdot 5$ & $P_2^5 P_5$\\ \hline
\end{tabular}
\end{center}

After computing the norm we factor the resulting integer; 
the idea is that this gives
us lots of information about the factorization of the principal ideal
$(x+\delta)$ into prime ideals.

For instance, the factorization $N(2+\delta) = 6 = 2 \cdot 3$ tells us that
the factorization of the principal ideal $(2+\delta)$ must involve a prime
ideal of norm~2 and a prime ideal of norm~3. 

If $P$ is a prime ideal of degree 1 lying over a rational prime~$p$, 
then the residue field $\Z_F/P$ is equal
to $\Z/p\Z$, and $\delta$ is congruent to an integer modulo~$P$.  If
$\delta \equiv a \bmod P$ then $N(a-\delta)$ is divisible by~$p$.
Thus any prime of degree 1 ``occurs'' sooner or later in the factorization
table above.

On the other hand, $\delta \bmod P$ generates $\Z_F/P$ as an extension of
$\Z/p\Z$, so if $\delta \equiv a \bmod P$ then this extension is of
degree~1.  Thus a rational prime occurs in a factorization of $(x+\delta)$
if and only if the corresponding prime ideal is of degree~1.  

(Warning:
this relies on the fact that the ring of integers has a power basis, i.e.,
that $\Z_F = \Z[\delta]$, so in fields of higher degree this sort of
argument might not hold for primes dividing the index of $\Z[\alpha]$ in
$\Z_F$.)

Next, we note that a prime ideal $P$ occurs in the factorization of
$(x+\delta)$ and $(x'+\delta)$ if and only if $x \equiv x' \bmod p$.
Indeed, if $x+\delta \in P$ and $x'+\delta \in P$ then
$$x+\delta-x'-\delta = x-x' \in P \cap \Z = (p).$$

These remarks enables us to make many inferences from the table of norms.
First, $2$ splits, $3$ and $5$ are ramified, and $7$ is inert.
Next, we can factor each principal ideal into prime ideals as indicated.

What do the factorizations tell us about the class group?  From the equations
$$(2) = P_2P_2', \; (3) = P_3^2 , \; (5) = P_5^2$$
we see that the primes $P_2$ and $P_2'$ are inverses in the class group, and
that the classes of $P_3$ and $P_5$ have order~2.  From the factorization
$(1+\delta) = P_2' P_3$, $(2+\delta) = P_2 P_5$ we see that $P_2$ and $P_2'$
have order~2 (so that $[P_2] = [P_2']$ and that $[P_3] = [P_2] = [P_5]$.
In fact, the only class groups that are consistent with all of the above
information are: the trivial group (every ideal is principal), or the
group of order~2, in which, for instance, $[P_2]$ is the nontrivial element.

Morally speaking, we are sure that the latter is probably true.  To verify
this, it suffices to show that $P_2$ is nontrivial.  If $P_2$ were principal,
say $P_2 = (x+y\delta)$, then there would be an element of order~2:
$$N(x+y\delta) = x^2+xy+4y^2 = N(P_2) = 2.$$
This equation is unsolvable in integers $x,y$ (e.g., implies 
$(2x+y)^2 + 15y^2 = 8)$.  So we conclude that $Cl(F)$ is of order~2.


\end{document}
