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\begin{document}

\chapter*{October 31:}

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

\begin{itemize}
\item
The class group of $\Q(\sqrt{-14})$
\item
The class group of $\Q(\sqrt{70})$
\end{itemize}


\section{The class group of $\Q(\sqrt{-14})$}

Let $F=\Q(\sqrt{-14})$.  Then $\Z_F=\Z[\sqrt{-14}]$ since $-14
\equiv 2 \bmod 4$.  This means that if $x+y\sqrt{-14} \in F$ then
$N(x+y\sqrt{-14})=x^2+14y^2$.  To find the prime ideals of degree one
(prime ideals such that $\Z_F/P \cong \Z/p\Z$) we will check the following
table where we have set $y=1$.

\begin{center}

\begin{tabular}{|r|r|r|}
$x$ & $x^2 + 14$ & prime ideals \\ \hline
0 & 14 = $2\cdot 7$ & $P_2 P_7$ \\
1 & 15 = $3\cdot 5$ & $P_3 P_5$ \\
2 & 18 = $2\cdot 3^2$ & $P_2 (P'_3)^2$ \\
3 & 23 = $23$ & $P_23$ \\
4 & 30 = $2\cdot 3\cdot 5$ & $P_2 P_3 P'_5$ \\
5 & 39 = $3\cdot 13$ & $P'_3 P_{13}$ \\
6 & 50 = $2\cdot  5^2$ & $P_2 (P_5)^2$ \\
7 & 63 = $3^2\cdot 7$ & $(P_3)^2 P_7$ \\
\end{tabular}
\end{center}

Since we can calculate the Minkowski bound to be 
$$B=\frac{2}{\pi} \sqrt{|D_F|}=\frac{2}{\pi}\sqrt{56}<6$$
we know that our class group
is generated by the ideals lying over primes below six.  
Thus the above table might be sufficient to determine the
class group.

How were the prime
ideal factorizations obtained?  Given our prime factorization of a given
ideal's norm, we have a factorization of that ideal into prime ideals
lying over those primes.  This leaves the question of when those primes
are the same and when they are conjugates.  For example, if we have a
factor of 7 in the norm of our ideal, should this be represented by $P_7$
or $P'_7$?  Let our ideal in this list be $I=(m+\delta)$ since we have
$y=1$.  We know that for any prime ideal, $P$, $\Z_F/P\cong \Z/p\Z$, so
in $P$ we can work mod $p$ where $p$ is the prime underlying our prime
ideal $P$.  Thus we know that $\delta\equiv m \bmod p$.  Thus for any
prime $p\in \Z$ if $p$ appears in the factorization of principal ideals
generated by two numbers which are not equivalent modulo p we would get
$1\in P$ which is clearly impossible.  Thus if $p$ appears only once
in the factorization of the norms of the principal ideals $(\delta)$,
$(1+\delta)$, ..., $(p-1+\delta)$ we know that $p$ splits in $\Z_F$ or
$(p)=P_p^2$.  If it appears twice, then we have $(p)=P_pP'_p$.  If $p$
fails to appear at all, we know that $p$ is inert in $\Z_F$.

From the above table, we now know that $(2)=P_2^2$, $(3)=P_3P'_3$, and
$(5)=P_5P'_5$.  Since principal ideals are trivial in the class group,
this table also gives us relations between these prime ideals, namely
$P_3P_5 \equiv P_2(P'_3)^2 \equiv P_2P_3P_5 \equiv P_2P_5^2 \equiv 1$.
We can derive from these relations that $P_3^2 \equiv P_2$, $P_3^4 \equiv
1$, $P_3^3 \equiv P_5$.  Therefore our conjecture is that the class
group is  generated by the class $[P_3]$, and that that element has
order 4; in terms of generators and relations used in finite group
theory one might write
$$\Cl(F)=<a: a^4=1>, \quad a = [P_3].$$  
This is true unless $P_2$ is trivial.
Suppose that $(a+b\delta)=P_2$.  Then $N(a+b\delta)=2=a^2+14b^2$ which
is clearly impossible in the integers.  Thus we have $Cl_F\cong \Z/4\Z$.

\section{The class group of $\Q(\sqrt{70})$}

Now let $F=\Q(\sqrt{70})$.  In this case, the Minkowski bound is
$B=\frac{2}{pi}\sqrt{280}<17$.  Thus we need only look at primes below 17.
Here, letting $y=1$ we get the following table of norms.  (Any norm which
contains a prime larger than 13 will be disregarded and not factored.)

\newpage

\begin{center}

\begin{tabular}{|r|r|r|}
$x$ & $x^2-70$ & Prime ideals \\ \hline
0 & -70 & $P_2P_5P_7$ \\
1 & -69 &  \\
2 & -66 & $P_2P_3P_11$ \\
3 & -61 &  \\
4 & -54 & $P_2(P'_3)^3$ \\
5 & -45 & $P_3^2P_5$ \\
6 & -34 &  \\
7 & -21 & $P'_3P_7$ \\
8 & -6 & $P_2P_3$ \\
9 & 11 & $P'_11$ \\
10 & 30 & $P_2P'_3P_5$ \\
11 & 51 &  \\
12 & 74 &  \\
13 & 99 & $(P'_3)^2P_11$ \\
\end{tabular}

\end{center}

Here we can observe that 2, 5, and 7 ramify, 3 and 11 split and nothing
is inert.  Thus we have $P_2^2 \equiv P_5^2 \equiv P_7^2 \equiv 1$.
However from the relations given by the table we discover that $P_2
\equiv P_3 \equiv P'_3 \equiv P_7$ and also that $P_5 \equiv 1$.
Thus our class group is 
$$Cl_F=<[P_2]:P_2^2=1>$$
unless $P_2$ is trivial.
Suppose that there is some $(a+b\delta)=P_2$.  Then $a^2-70b^2=2$, which
looked at mod 5 tells us that in $\Z/5\Z$ 5 is a square.  Since this is
not true, we have $Cl_F\cong \Z/2\Z$.  

\end{document}
