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\begin{document}

\chapter*{November 12:}

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

\begin{itemize}
\item
\section{The class number of $\Q(\sqrt[3]{6})$}
\item
Finding the ring of integers
\item
Finding the class number of a cubic field
\end{itemize}


\section{The class number of $\Q(\sqrt[3]{6})$}

In discussing the infamous group quiz, it was determined that
there was a typo in the second problem, and the the problem
on the class number of $\Q(\sqrt[3]{6})$ should perhaps
have included a hint about the ring of integers.
(Later, it was also revealed that the peaceful drunken rooks
puzzle actually was well-posed, and that the puzzle is available
commercially.)

Let $F=\Q(\sqrt[3]{6})$.  Then, for $\alpha = \sqrt[3]{6}$, $m_\alpha
= f$, and $f = x^3 - 6$,  we know that $\disc(f) = \disc(1, \alpha,
\alpha^2) = [\Z_F : \Z[\alpha]]^2\cdot D_F$.  From the formula for the
discriminant of a cubic we also know $\disc(1,
\alpha, \alpha^2) = -27\cdot 6^2 = -3^5\cdot 2^2$. 

Therefore $\Z[\alpha]$ is of index $1, 2, 3, 6, 9, 18$ in $\Z_F$
(all of the integers whose square divides the discriminant).  In fact,
$\Z_F = \Z[\alpha]$ but to prove this gracefully it is easiest to use
results on algebraic integers.

\section{Finding the ring of integers}

Let $R$ be a ring such that
$$ \Z[\alpha] \subset R \subset \Z_F.$$
The basic idea is to start with $R = \Z[\alpha]$ and gradually enlarge $R$ until
we (provably) have reached $\Z_F$.  The tools needed to do this are
loosely summarized as follows.

\begin{theorem}
\begin{itemize}
\item[a)] the primes of $\Z[\alpha]$ are of the form $P = (p, g(\alpha))$, 
where $g \in \F_p[x]$ is irreducible and $g \mid f \bmod p)$.
\item[b)] If $P$ is a prime ideal in~$R$ then there is an $x \in F - R$ such that
$xP \subset R$.
\item[c)] The question of whether or not $xP \subset P$ is true is independent of
which $x$ is chosen in part b).
\item[d)] If $xP \subset P$ then $x$ is an algebraic integer, and $R$ can be
enlarged, and the prime ideals can be recomputed.
\item[e)] If $xP$ is not a subset of~$P$, then we say that $x$ is a ``certificate
of invertibility'' for~$P$.  If all prime ideals in~$R$ have certificates of
invertibility then $R = \Z_F$.
\end{itemize}
\end{theorem}

The only prime ideals that are not invertible are those lying over primes~$p$
such that $p^2$ divides the discriminant of $\alpha$.  Thus there are finitely
many prime ideals in play at any given time, and since the discriminant of~$R$
goes down every time it is enlarged, the algorithm implicit in the above results
clearly terminates.

In the example of current we start with let $R = \Z[\alpha]$, 
where $\alpha = \sqrt[3]{6}$.
Then we only have to worry about $p = 2$ and $p =3$ and the only prime ideals
above those primes are
$P_2 = (2, \alpha)$ and $P_3 = (3, \alpha)$.  
We want to find certificates of invertibility for those ideals; if we fail then
we can enlarge~$R$.

Consider $P_2$.  We need $2x$ and $\alpha x$ to be in $R$.  The former implies
that $x = y/2$, where $y = A + B\alpha+C\alpha^2$.  From 
$$y\alpha = 6C + A \alpha + B \alpha^2$$
we see that $x = \alpha^2/2$ works just fine.  Then $x\alpha = 3 \notin P_2$
so $x$ is a certificate of invertibility of~$P_2$.

Similar arguments show that for $P_3$ we can also use $x = \alpha^2/3$,
and that $xP_3$ is not contained in $P_3$.  Thus $P_3$ is invertible as well,
and we conclude that $\Z[\alpha] = \Z[\sqrt[3]{6}$ is the full ring of
integers~$\Z_F$. 

\section{Finding the class number of a cubic field}

Now we can return to our regularly scheduled program of computing the
class group.  We know that 
$$N(\alpha-x) = -f(x)$$
for integers~$x$, and this enables us to compute a table of norms.

\begin{center}

\begin{tabular}{|r|r|r|}
\hline
$x$ & $x^2 - 6$ & prime ideals \\ \hline
0 & -6 = $-2\cdot 3$ & $P_2 P_3$ \\
1 & -5 = $-5$ & $P_5$ \\
2 & 2 = $2$ & $P_2$ \\
3 & 21 = $3\cdot 7$ & $P_3 P_7$ \\
-1 & -7 = $-7$ & $P'_7$ \\
-2 & -14 = $-2\cdot 7$ & $P_2 P_7''$ \\
\end{tabular}
\end{center}

We can calculate the Minkowski bound to be 
$$B=\frac{n!}{n^n}\left(\frac{4}{\pi}\right)^{r_2}
\sqrt{|D_F|}=\frac{6}{3^3}\left(\frac{4}{\pi}\right)
\sqrt{3^5 \cdot 2^2}<9$$
so we know that every element of the class group contains an ideal of norm less
than~9, 
and, in particular, that the class group is generated by prime ideals of norm less than~9.

By the above table, 
$$(7) = P_7P'_7P"_7, \quad (5) = P_5Q_5, \quad (3) = P_3^3, \quad
 (2) = P_2^3.$$  
By the $x = -1$ table entry, we know that one of
the $P_7$'s is principal. Similarly, we know that $P_2$ and $P_5$ are
principal (by the $x = 2$ and $x = 1$ entries).  So, $P_3$ is principal
(by $x = 0$), and consequently, both of the other $P_7$'s are principal.
Therefore, $h = 1$ and we have a PID.

Note that in fact, we did not have to use the result from the previous
section about the ring of integers.  Namely, the norm calculations are
valid no matter what the ring of integers actually is; e.g.,
if an element has prime
norm then it generates a prime ideal of degree 1 in $\Z_F$; finally, since
we found that the class number was~1, we didn't have to prove that any ideals
were non-principal and didn't need to fully calculate the ``norm form.''

We also remark that
the norm table could be used to find a nontrivial unit, by finding
a principal ideal that was trivial, as a principal ideal.  A famous theorem
due to Dirichlet that we will cover next week says that the units of a
number field~$F$ consist of roots of unity, together with $s$ independent
elements, where $s = r_1+r_2-1$, i.e.,
$$\Z_F^* \simeq U \times \Z^s$$
where $U$ is a finite group.  For the cubic field above, $r_1+r_2-1 = 1$
and there is a ``fundamental unit'' of infinite order, much as in the
case of real quadratic fields.



\end{document}

