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\title{Handout -Version 2}
\author{GCS 2000}
\date{}
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Thanks to Elizabeth Crewdson for pointing out that Version1
was broken. Please discard it, and replace it by Version 2.
If you find anything wrong with this document, please let me know
immediately and a replacement will go up at the web site (which you
should monitor during January).

Recall that at this stage we know that an extension
$K \leq L$ of finite degree of subfields of $\mathbb C$
has the property that if $G = \mbox{Gal}_K L$, then
$|G| \leq |L:K|$. We need to improve this result
to replace the inequality by an equality.
At this stage we did not yet have the theorem that
(in this context) finite normal extensions 
are the same things as extensions which are splitting fields.

\vspace{0.3cm}

\noindent{\bf Proposition}\ Let $k$ be a subfield of $\mathbb C$. Suppose that $\Sigma$
is a splitting field for $f \in k[X]$ (so we may assume that
$\Sigma \leq \mathbb C$). Let $n = |\Sigma : k|$ and
$G = \mbox{Gal}_k\Sigma$.
It follows that $|G| = n$.\newline

\noindent{\bf Proof}\ 
Since we are working with subfields of $\mathbb C$ and
$|\Sigma : k| < \infty,$ we may assume that $\Sigma = k(\theta)$.
Now $\theta$ is algebraic over $k$ with minimum polynomial
$h \in k[X]$.

Let $\Lambda \leq \mathbb C$ be the splitting field of $hf \in k[X]$.
Let $G_\Lambda = \mbox{Gal}_k\Lambda$ and suppose that
$g_\Lambda \in G_\Lambda.$ Now $g_\Lambda$ moves the roots
of $f$ among themselves and so maps $\Sigma$ to $\Sigma$.
In particular, $(\theta)\sigma_\Lambda \in \Sigma$ for all
$\sigma_\Lambda \in G_\Lambda$.

Suppose that $\theta'$ is any root of $h$ in $\Lambda$.
Since $h \in k[X]$ is irreducible, the fields $k(\theta)$, 
$k(\theta')$ are isomorphic via a map fixing the
elements of $k$ and sending $\theta$ to $\theta'$. 
As we showed in a previous
proposition, there is an element of $G_\Lambda$ which 
induces the given isomorphism $k(\theta) \rightarrow k(\theta')$ 
which sends $\theta$ to $\theta'$. The previous paragraph now yields that
$\theta' \in \Sigma$. Thus $h$ splits into linear factors
in $\Sigma[X]$ and so $\Sigma = k(\theta)$ contains a splitting 
field for $h$. However, $\theta$ must be contained in 
the splitting field and so $\Sigma$ is the splitting field of $h \in k[X]$
and so $\Lambda = \Sigma$ and $G_\Lambda = G$.

Now if $\theta'$ is any root of $h$ in $\Sigma$,
then $k(\theta) \simeq k(\theta')$ via an isomorphism fixing
the elements of $k$ and sending $\theta$ to $\theta'$.
Moreover we have $|k(\theta') : k| = \deg h = n$ and so $k(\theta') = \Sigma$
for each $\theta'$. Now $h \in k[X]$ is irreducible of degree $n$
and so has exactly $n$ distinct complex roots.
We conclude that $G$ has exactly $n$ elements.
The proof is complete.\newline

\newpage

Inspired by this, we make a definition:
\vspace{0.3cm}

\noindent{\bf Definition}\ 
Suppose that $K \leq L \leq \mathbb C$. We say that the 
extension of fields $K \leq L$ is {\em normal} if whenever
$f \in K[X]$ is irreducible and $f$ has at least one
root in $L$, then $f$ splits into linear factors in $L[X]$
(so all the complex roots of $f$ are in $L$).\newline

Note that Ian Stewart calls this the ``Trade Union Definition'' because
you can remember it by the phrase ``one out, all out!''.

\noindent{\bf Proposition}\ Suppose that $K \leq L \leq \mathbb C$ and
$|L:K| < \infty$. The following are equivalent:
\begin{enumerate}
\item[(i)] The extension $K \leq L$ is a normal extension.
\item[(ii)] $L$ is the splitting field of some $h \in K[X]$.
\end{enumerate}
\vspace{0.5cm}
\noindent{\bf Proof} First we show that (i) implies (ii).
Choose $\theta \in \mathbb C$ such that $L = K(\theta).$
Let $h \in K[X]$ be the minimum polynomial of $\theta,$ so
$\mbox{deg }h = n = |L:K|.$ Now the extension $K \leq L$
is a normal extension so $h$ splits into linear factors
in $L[X]$. However, the splitting field of $h$ must contain $\theta$
and so $L$ is the splitting field of $h$.

Next we address (ii) implies (i). This proof is very similar
to that of the previous proposition. Choose any irreducible $f \in K[X]$
which has a root in $L$, and assume that $\Sigma$ is the splitting field
of $h \in K[X]$. Let $\Lambda \leq \mathbb C$ be the splitting field
of $fh$. Now if $\theta'$ is any complex root of $f$, the fields
$K(\theta)$ and $K(\theta')$ are isomorphic via a map
fixing the elements of $K$ and sending $\theta$ to $\theta'$.
This isomorphism can be induced by an element of $G_\Lambda
= \mbox{Gal}_K\Lambda$. However, elements of $\mbox{Gal}_K\Lambda$ 
move the roots of $h$ among themselves and hence map
$L$ into itself. Thus each $\theta' \in L$ so $f$ splits in $L[X].$
Thus the extension $K \leq L$ is normal. We are done.
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