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\begin{document}

\chapter*{September 24:}

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

\bigskip

\begin{itemize}
\item
Primes
\item 
Pell's equation
\item 
Embeddings
\item
Primitive Element Theorem
\item
Traces and Norms
\end{itemize}

\section{Primes}

Recall, that to find primes in $\Z[i]$ it suffices to
factor (rational) prime numbers~$p\in\Z$.

Indeed, if $P$ is a prime in $\Z[i]$ then $N(P)$ is an integer.
Since $P$ is irreducible it must divide one of the primes that
divides~$N(P)$.

If the ring of integers~$\Z_F$ is a PID the same story holds.
If~$\Z_F$ isn't a PID, then in turns out that the same story
holds if we consider \emph{ideals}~$P$: Every ideal
factors uniquely into a produce of prime ideals, and these
ideals can be determined by factoring the ideals $(p) := p\Z_F$
generated by rational primes~$p$.

\section{Pell's equation}

The existence of nontrivial solutions to Pell's equation $x^2-dy^2= \pm 1$ 
can be proved in several ways.  One is to explicitly 
give an algorithm using
continued fractions.  Another is to apply a special case of a result
that will be proved later in the course for arbitrary number fields.
Another is to use an argument about approximation of algebraic
numbers by rational numbers; specially one shows that
for all irrational~$\alpha$ and all positive integers~$n$ there
are $y,z$ such that
\begin{displaymath}
\left| \alpha - \frac{y}{z} \right| < \frac{1}{zn}.
\end{displaymath}
This can be used to show that there are infinitely many approximations 
$$|\alpha - (x/y)| < \frac{1}{yz},$$ which in turn implies that there exist $x$, $y$ such
that $|x^2 - d y^2| < 2 \sqrt{d}$, which finally implies that there
exist $x$, $y$ such that $|x^2 - d y^2| = 1$.

It's not clear which of these arguments Neukirch had in mind in the exercise
earlyl on in his text, or whether there is a different and better argument.

\section{Embeddings}

Our investigations into galois theory culminated in being able to
count embeddings of separable field extensions into splitting fields.
This did \emph{not} require any of the results of galois theory that
we didn't prove (e.g., the fundamental theorem of galois theory or
its immediate precursors, such as the independence of characters).  

We'll need the following result in several contexts.

\begin{theorem} (Embedding Theorem) Let $E$ be a separable
extension of degree~$n$ of a field~$F$.  Let $K$ be an extension
of~$E$ that is the splitting
field over~$F$ of a separable polynomial (with coefficients in~$F$).
Then there are $n$ embeddings of $E$ into~$K$, i.e.,
$$\left | \Hom_F(E,K) \right | = n.$$
\end{theorem}

\begin{remark}
The theorem was proved by generalizing to a ``relative'' version
that involved extending a given automorphism of ground fields,
and then proving the generalized theorem with an easy proof by
induction.  
\end{remark}

\begin{remark}
On several occasions we will number the embeddings
$\sigma_1, \cdots , \sigma_n$; it is customary to let $\sigma_1$ be
the identity map.
\end{remark}

We will also be interested in the set
$\mathrm{Hom}_{\Q}(F,\C)$ of field embeddings
of~$F$ into the complex numbers.  We can, without loss of
generality, assume that $F$ is contained in the complex numbers,
and that a splitting field~$K$ that contains it is also contained
in the complex numbers.  Then all of the $n = [F:\Q]$ embeddings given in
the above theorem take their image in~$K$.

An embedding of a number field~$F$ into~$\C$ is said to be 
\defn{real} (resp. \defn{complex}) if its
image is contained (resp., not contained) in the real numbers~$\R\subset\C$.
Complex embeddings come in pairs (if $\sigma$ is an embedding so is
its complex conjugate $x \mapsto \overline{\sigma(x)})$.
So if $r_1$ is the number of real embeddings and $r_2$ is the number of
complex conjugate pairs of embeddings then
$$r_1 + 2 r_2 = n.$$

\begin{example}
Suppose $F = \Q(\alpha)$. (We will soon see that any number field is
of this form.)  Any embedding takes
$\alpha$ to a root of $\m_{\alpha,\Q}$.  
Indeed, we apply $\sigma$ to
$$0 = \m_{\alpha,\Q}(\alpha) = \alpha^n + a_1 \alpha^{n-1} + \cdots, 
\qquad a_i \in \Q$$
we see that $\sigma(\alpha)$ is a root of the same polynomial.
So
$r_1$ is the number of real  roots of $\m_\alpha$ and
$r_2$ is the number of pairs of complex conjugate roots.
Therefore
\begin{itemize}
\item The field $F = \Q(\sqrt[3]{2})$ has $r_1 = 1$, $r_2 = 1$
since $x^3-2 = 0$ has one real root and a pair of complex conjugate
roots.
\item
The polynomial
$$(x-\cos 2\pi/7)(x - \cos 4 \pi/7 )(x - \cos 6 \pi/7) =
x^3+\frac{1}{2} x^2 - \frac{1}{2} x - \frac{1}{8}$$
has three real roots, so its root field has $r_1 = 3, r_2 = 0$.
\end{itemize}
\end{example}

\section{Primitive Element Theorem} 

In several circumstances it can be useful to know that in any
number field $F$ there is an element $\alpha$ that generates the
field, i.e., $F = \Q(\alpha)$.  This turns out to be any easy
application of the emedding theorem.

\begin{theorem} 
\textbf{Primitive Element Theorem.}  If $E$ is a finite separable
extension of a field~$F$ then there is an element $\alpha$ of~$E$
such that $E = F(\alpha)$.
\end{theorem}


\begin{proof}
If $E$ is finite, then it suffices to take $\alpha$ to be a
generator of the cyclic group~$E^*$.  So from now on we can
assume that $E$ is infinite.

By induction on the number of generators,
it suffices to find a single generator for a field
$E = F(\alpha,\beta)$.  Let $n := [E:F]$, and let
$\sigma_1, \cdots , \sigma_n$ be the embeddings of $E$ into a
normal extension $K$ of~$F$ that contains~$E$.
For $\lambda$ in~$F$, let
$$g(x) := \prod_{i<j} \sigma_i(\alpha) + x
\sigma_i(\beta) - \sigma_j(\alpha) - x \sigma_j(\beta).$$

Note that all factors are nonzero, since the embeddings are distinct,
and are determined by what they do to $\alpha$ and~$\beta$.
In addition, the coefficients are all in~$F$, since
they are symmetric functions of the conjugates of $\alpha$ and
$\beta$.

Thus the polynomial~$g$ is a nonzero polynomial with coefficients in~$F$, and 
there is a $\lambda \in F$ such that $g(\lambda) \ne 0$.
(Over a finite field it is in fact possible for a nonzero polynomial to 
have all of its values
equal to~0.)

Let $\gamma = \alpha+\lambda \beta$.  Then each $\sigma_i(\gamma)$ is
a root of~$g$, and they are all distinct, so the degree of $F(\gamma)$
over~$F$ is at least~$n$.  Since $F(\gamma)$ is contained in the
field $F(\alpha,\beta)$ of degree~$n$, it follows that the degree of~$\gamma$
is exactly~$n$, and
$$F(\alpha,\beta) = F(\gamma)$$
which finishes the proof. \end{proof}

\section{Traces and Norms}

If $E/F$ is a finite extension of fields, then for every element $x \in E$
we get a ``multiplication by~$x$'' mapping from $E$ to $E$ that is an
endomorphism of an $F$-vector space of dimension $n := [E:F]$.

We define the norm and trace of the element~$x$, from $E$ down to~$F$,
to be the determinant and trace of this endomorphism; let $mul_x$ denote
the multiplication by $x$ map, so that
$$N_{E/F}(x) := \det(mul_x), \qquad
\trr_{E/F}(x) := \trr(mul_x).$$
We will run into the trace and norm again and again, and will
need the following basic results (to be proved next time).

\begin{theorem} 
Let $x$ and $y$ range over elements of~$F$.  Then:
\begin{itemize}
\item $N_{E/F}(x) \in F$, \quad $\trr_{E/F}(x) \in F$
\item $N_{E/F}(xy) = N_{E/F}(x) N_{E/F}(y)$,  \quad
$\trr_{E/F}(x+y) = \trr_{E/F}(x) \trr_{E/F}(y)$.
\item 
$$N_{E/F}(x) = \prod_\sigma \sigma(x), \qquad
\trr_{E/F}(x) = \sum_\sigma \sigma(x)$$
where the sum and product range over all embeddings of $E$ into
a splitting field $K$ of $E$ over~$F$.
\end{itemize}
\end{theorem}
\end{document}
