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\bec {\Large\bf Fields to Groups and Back Again II} \ece
Let us fix some finite extension $F\subset K$ of subfields of $C$
and set $G$ to be the Galois Group $\Gamma(K:F)$. However, we
now assume $K$ is a Normal Extension of $F$. Recall that we
have already defined the map $*$ from intermediate fields to
subgroups and the map $\dag$ from subgroups to intermediate
fields.
\begin{theorem} Let $F\subset K$ be subfields of $C$ with $K$ a
Normal extension of $F$ and set $G$ to be the Galois Group $\Gamma(K:F)$.
Then for any intermediate field $L$
\beq\label{e1} (L^*)^{\dag} = L \eeq
\end{theorem}
{\tt Proof:} We already know $L\subset (L^*)^{\dag}$. Now suppose $\beta\in K$
and $\beta\not\in L$. Our goal is to show $\beta\not\in (L^*)^{\dag}$. Recall
that as $K$ is a normal extension of $F$, $K$ is a normal extension of $L$.
Let $p(x)$ be the minimal polynomial for $\beta\in L[x]$ and let $\beta_1$ be
another root of $p(x)$. As $K$ is a normal extension of $L$, $\beta_1\in K$.
Thus there is an isomorphism $\sig: L(\beta)\ra L(\beta_1)$ which fixed $L$
and has $\sig(\beta)=\beta_1$. Applying the Full Isomorphism Extension Theorem
we extend $\sig$ to an isomorphism $\sig^{++}$ with domain $K$. But as
$\sig^{++}$ fixes $L$ and $K$ is normal over $L$, the range of $\sig^{++}$ must
be $K$. That is, $\sig^{++}$ is an automorphism of $K$ which fixes all $\ah\in L$
but does not fix $\beta$. So $\beta\not\in (L^*)^{\dag}$. End of Proof.
This has a perhaps surprising followup.
\begin{theorem}\label{e2} Let $F\subset K$ be subfields of $C$ with $K$ a
normal extension of $F$. Then there are only {\em finitely many}
intermediate fields $L$.
\end{theorem}
{\tt Proof:} From Theorem \ref{e2}, $L$ is determined by $L^*$ but as
$G=\Gamma(K:F)$ is finite there can be only finitely many subgroups
$H$, only finitely many possible $L^*$.
\begin{theorem}\label{e3}
Let $K$ be a finite extension of $F$, both
subfields of $C$.
Then there are only {\em finitely many}
intermediate fields $L$.
\end{theorem}
{\tt Proof:} Extend $K$ to $K^+$ so that $K^+$ is a normal extension of
$F$. From Theorem \ref{e2} there are only finitely many intermediate
fields between $F$ and $K^+$ and thus only finitely many intermediate
fields between $F$ and the smaller $K$.
\begin{theorem}\label{e4} Let $F$ be a subfield of $C$ and $\ah,\beta\in C$,
both algebraic over $F$. Then there exists $\gam\in C$ with
\beq\label{twotoone} F(\gam)=F(\ah,\beta) \eeq
\end{theorem}
{\tt Proof:} As $\ah,\beta$ are algebraic over $F$, $F(\ah,\beta)$ is a
finite extension of $F$. Now for each integer $i$ set $F_i=F(\ah+i\beta)$.
Each of these are subfields of $F(\ah,\beta)$ but by Theorem \ref{e3}
there are only finitely many such subfields so there must be $i\neq j$
with $F_i=F_j$. Thus $F_i$ contains $\ah+i\beta$ {\em and} $\ah+j\beta$.
But then it contains $\ah = \frac{1}{j-i}j(\ah+i\beta)-i(\ah+j\beta)$ and
$\beta = \frac{1}{j-i}((\ah+j\beta)-(\ah+i\beta))$. Thus $F_i$ must be
all of $F(\ah,\beta)$ and so we can take $\gam=\ah+i\beta$.
\begin{theorem}\label{e5} {\tt Single Generator Theorem.}
Let $K$ be a finite extension of $F$, both subfields of $C$.
Then there is an element $\gam\in K$ such that $K=F(\gam)$.
\end{theorem}
{\tt Proof:} We claim that for any $\ah_1,\ldots,\ah_r\in C$, all
algebraic over $F$, there exists a $\gam\in C$ with
$F(\gam)=F(\ah_1,\ldots,\ah_r)$. This comes from repeatedly
applying Theorem \ref{e4} to replace two of the generators by one.
(Formally we apply induction on $r$.) Now as $K$ is a finite extension
of $F$ we can write $K=F(\ah_1,\ldots,\ah_r)$ for some finite set of
$\ah$'s and then replace them by a single $\gam$.
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