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\begin{document}

\chapter*{September 14:}

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

\bigskip

\begin{itemize}
\item
Galois theory
\item
Examples
\end{itemize}


\section{Galois theory}

We know that the following are equivalent:
\begin{itemize}
\item $E$ is a galois extension of~$F$.
\item $E$ is the splitting field over~$F$ of a separable polynomial.
\item $\#\Aut_F(E) = [E:F].$
\end{itemize}
       
If $E/F$ is a galois extension then the group
$G(E/F) := \Aut_F(E)$ is called the \defn{galois} group of
the extension, and the basic idea is that the knowledge
of this group conveys a lot of information about the fields
involved.

One of the most basic way in which this is manifested is in
the (famous) correspondence between subfields and subgroups,
sometimes called the fundamental galois correspondence.

Let $E/F$ be a (finite) galois extension, and let
$G := G(E/F) = \Aut_F(E)$ be its galois groups.  If $H$
is a subgroup of~$G$ then let
$$E^H := 
\left\{x \in E \mid h(x) = x, \quad \forall \; h \in H \right\} .$$
One checks in a very straightforward fashion that the fixed
fields $E^H$ of~$H$ is in fact a subfield, and that the
map from subgroups to subfields is inclusion reversing:
$$H \subset H' \Longrightarrow E^{H} \supset E^{H'}.$$

Now suppose that $K$ is an intermediate field, i.e., that
$F \subset K \subset E$.  Define 
$$\stab(K) := \{ g \in G : g(x) \in K \mbox{ for all }
x \in K \}.$$
One checks in a very straightforward fashion that the stabilizer
of~$K$ is a subgroup of~$G$, and that the
map from subfields to subgroups is inclusion reversing:
$$K \subset K' \Longrightarrow \stab(K) \supset \stab(K').$$

One also checks in a straightforward fashion that
\begin{eqnarray*}
K & \subset & E^{\stab(K)} \\
H & \subset & \stab(E^H).
\end{eqnarray*}
The Fundamental Theorem of Galois Theorem says
that these inclusions are equalities, and that the fixed field
and stabilizer operations are mutually inverse bijections.

\begin{theorem}
There is a one-to-one correspondence between subgroups
of~$G$ and intermediate fields between $F$ and~$E$. 
Specifically,
\begin{eqnarray*}
K = E^{\stab(K)} \\
H = \stab(E^H)
\end{eqnarray*}
so that ``taking the fixed field'' and ``taking the stabilizer''
are inverse bijections.
Normal subgroups of $G$ correspond to
subfields of~$E$ that are themselves galois extensions of~$F$.
\end{theorem}

\section{Examples}

\begin{example}
Let $F$ be a field of characteristic not equal to~2, and
let $E/F$ be an extension of degree~2, i.e., a quadratic
extension.  Then $E = F(\alpha)$ where the minimal polynomial
of~$\alpha$ has degree~2.  This polynomial splits in~$E$, and has
distinct roots, so $E$ is a galois extension.
\end{example}

In fact, it is easy to see that any quadratic extension in
characteristic not equal to~2 has the form $F(\sqrt{d})$.

\begin{example}
Some cubic extensions are galois, and others aren't.
In a later homework exercise we'll learn how to find
equations satisfied by $\cos(2\pi/n)$, and in particular will
find that
$$(x- \cos(2\pi/7)) (x- \cos(4\pi/7)) (x- \cos(6\pi/7)) = x^3+x^2-2x-1.$$
By simple trigonometric identities, the second two roots are
polynomials in the first one, so that $\Q(2\pi/7)$ is actually a
splitting field for the cubic polynomial on the right.  Thus,
there is a cubic galois extension.
\Eex

\begin{example}
Let $f(x) = x^3-2$, and let $E$ be the splitting field of~$f$
over~$\Q$.  Since two of the roots are complex, the
root field $\Q(\sqrt[3]{2})$ isn't a splitting field.
The polynomial $f$ factors as a linear times a quadratic over
this field, and consequently $E$ has degree~6 over~$\Q$.
The galois group $G = G(E/\Q)$ is of order~6, and it permutes
the three roots of~$f$, so in fact $G$ is isomorphic to
the symmetric group of degree~3, $S_3$.  
The isomorphism is obvious: a permutation
of three things corresponds to the automorphism of~$E$ that
effects that permutation on the roots.
\Eex

According to the Fundamental Theorem, the subgroups of~$G$
correspond to the intermediate fields.  The trivial subgroup
corresponds to~$E$, and the entire group corresponds to~$\Q$.

There are three (conjugate) subgroups of order 2, so there
are three cubic fields inside~$E$, namely each of the fields
generated by a root of~$f$; since these fields are all root
fields they are isomorphic, but they are distinct subfields.

There is a unique subgroup of order~3, namely the alternating
group $A_3$ of order~3.  Its subfields turns out to be
$\Q(\sqrt{-3})$.


\end{document}
