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\bec {\Large\bf Fields to Groups and Back Again} \ece
Let us fix some finite extension $F\subset K$ of subfields of $C$
and set $G$ to be the Galois Group $\Gamma(K:F)$. We will be
interested in intermediate fields $L$, that is, $F\subset L \subset K$,
and in subgroups $H$ of $G$.
We will describe first a mapping from
fields $L$ to groups $H$
\begin{definition}\label{stardef} Let $F\subset L \subset K$ be an
intermediate field. We define $L^*$, a subgroup of $G$, by
\beq \label{stareq}
L^* = \{ \sig\in G: \sig(\ah)=\ah \mbox{ for all } \ah\in L \}
\eeq
That is, $L^*$ is those automorphisms of $L$ which fix all elements
of $L$.
\end{definition}
Why is $L^*$ a subgroup of $G$? Well, suppose $\sig,\tau$ were two
automorphims of $K$ over $F$. Then, as we have discussed before,
so is $\sig\tau$. But further, if $\sig(\ah)=\ah$ and $\tau(\ah)=\ah$ for all
$\ah\in L$ then
\[ (\sig\tau)(\ah)= \sig(\tau(\ah)) = \sig(\ah) = \ah \]
for all $\ah\in L$ and so $\sig\tau\in L^*$. Similarly $\sig^{-1}\in L^*$.
Finally the identity $e\in L^*$ as $e$ fixes all elements.
We now will describe first a mapping from groups $H$ to fields $L$.
\begin{definition}\label{dagdef} Let $H \subset G$ be a
subroup of the Galois Group. We define $H^{\dag}$, an intermediate
field, by
\beq \label{dageq}
H^{\dag} = \{ \ah \in K: \sig(\ah)=\ah \mbox{ for all } \sig \in H \}
\eeq
That is, $H^{\dag}$ is those elements of $K$ which are fixed by all automorphisms
$\sig\in H$.
\end{definition}
Set $L=H^{\dag}$. Why is $L$ an intermediate field? First of all, as all
automorphisms $\sig\in \Gamma$ fix all elements $c\in F$, any element $c\in F$
will be fixed by all $\sig\in H$, so that $F\subset L$. Now suppose $\ah,\beta\in L$
and take any $\sig\in H$. As $\sig(\ah)=\ah$ and $\sig(\beta)=\beta$,
we must have $\sig(\ah+\beta)=\sig(\ah)+\sig(\beta)=\ah+\beta$. Thus $\ah+\beta\in L$
and similarly $\ah\beta, -\ah, \ah^{-1}\in L$.
{\tt An Extended Example:} Take ground field $F=Q$ and extension $K=Q(\sqrt{2},\sqrt{3})$.
The four elements of the Galois Group $G$
are $e,\sig_1,\sig_2,\sig_3$ where (as done earlier)
\[ e(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \]
\[ \sig_1(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6} \]
\[ \sig_2(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= a+b\sqrt{2}-c\sqrt{3}-d\sqrt{6} \]
\[ \sig_3(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6} \]
$G$ is the Vierergruppe. There are five subgroups (we count the trivial ones
here) of $G$:
\[ \{e\}, H_1=\{e,\sig_1\}, H_2=\{e,\sig_2\}, H_3=\{e,\sig_3\}, \mbox{ and } G
\mbox{ itself.} \]
\par Lets start in the ``middle" with $H_1=\{e,\sig_1\}$. What is $H_1^{\dag}$.
That is,
which $\ah=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ are fixed by all $e,\sig_1$.
Well, $e$ is the identity so it fixes everything so we can ignore it. If we think
of $\sig_1(\ah)=\ah$ as an (easy!) equation it is true precisely when $b=d=0$.
So $\ah\in H_1^{\dag}$ if and only if we can write $\ah=a+c\sqrt{3}$. That is,
$H_1^{\dag}= Q(\sqrt{3})$.
Similarly, for $\ah\in H_2^{\dag}$ the necessary and
sufficient condition is that $c=d=0$ so $\ah=a+b\sqrt{2}$ and $H_2^{\dag}=Q(\sqrt{2})$.
Similarly, for $\ah\in H_3^{\dag}$ the necessary and
sufficient condition is that $b=c=0$ so $\ah=a+d\sqrt{d}$ and $H_3^{\dag}=Q(\sqrt{6})$.
\par The case $\{e\}^{\dag}$ is always the same, since the identity preserves everything
$\{e\}^{\dag}=K$. Finally, what about $G^{\dag}$. That is,
which $\ah=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ are fixed by all $e,\sig_1,\sig_2,\sig_3$.
To be fixed by $\sig_1$ forces $b=d=0$, to be fixed by $\sig_2$ forces $c=d=0$, to be
fixed by $\sig_3$ is now redundant as it forces $b=c=0$. So all of $b,c,d$ must be
zero but $a$ can be an arbitrary rational and so $G^{\dag}=Q$.
\par Now set
\[ L_1=Q(\sqrt{2}), L_2=Q(\sqrt{3}), L_3=Q(\sqrt{6}) \]
and consider
the groups associated (via $*$) with the fields $Q,L_1,L_2,L_3,K$. The easiest
is $Q^*=G$, which is to say that all $\sig\in G$ fix all $\ah\in Q$ which is
true as $G$ was {\em defined} as all automorphisms $\sig$ of $K$ which fix
all $\ah\in Q$.
\par How about $L_1^*$? Clearly $e\in L_1^*$ as $e$ fixes everything.
Also $\sig_2\in L_1^*$ as $\sig_2(a+b\sqrt{2})=a+b\sqrt{2}$. But
$\sig_1,\sig_3\not\in L_1^*$ as they send $\sqrt{2}$ to $-\sqrt{2}$.
So $L_1^*=\{e,\sig_2\}$
\par How about $L_2^*$? Clearly $e\in L_2^*$ as $e$ fixes everything.
Also $\sig_1\in L_2^*$ as $\sig_1(a+c\sqrt{3})=a+c\sqrt{3}$. But
$\sig_2,\sig_3\not\in L_2^*$ as they send $\sqrt{3}$ to $-\sqrt{3}$.
So $L_2^*= \{e,\sig_1\}$
\par How about $L_3^*$?
Clearly $e\in L_2^*$ as $e$ fixes everything.
Also $\sig_3\in L_3^*$ as $\sig_3(a+d\sqrt{6})=a+d\sqrt{6}$. But
$\sig_1,\sig_2\not\in L_3^*$ as they send $\sqrt{6}$ to $-\sqrt{6}$.
So $L_3^*= \{e,\sig_3\}$
\par Finally, how about $K^*$.
Clearly $e\in K^*$ as $e$ fixes everything. But the other $\sig_1,\sig_2,\sig_3$
do not fix everything and so are not in $K^*$. Thus $K^*=\{e\}$.
\par We can put this all in tabular form.
\bec
\begin{tabular}{rr}
Field & Group \\ \hline
$K=Q(\sqrt{2},\sqrt{3})$ & $\{e\}$ \\
$L_1=Q(\sqrt{2})$ & $H_2=\{ e,\sig_2\}$ \\
$L_2=Q(\sqrt{3})$ & $H_1=\{ e,\sig_1\}$ \\
$L_3=Q(\sqrt{6})$ & $H_3=\{ e,\sig_3\}$ \\
$Q$ & $G=\{e,\sig_1,\sig_2,\sig_3\}$
\end{tabular}
\ece
We see we have a one-to-one correspondence. We can go from fields to groups by applying
$*$. And we can go from groups to fields by applying $\dag$. And $\dag$ and $*$ are
inverses as maps, if we apply one and then the other we get back where we started.
Does this always work? No. But it works {\em in important cases} and that will be
the substance of the major theorem of Galois Theory. Indeed, not to keep you in
suspence, here is that theorem. However, we haven't yet defined what a normal
extension is! The normal extensions are precisely those extensions for which
the correspondence works and they will be defined in the next section.
\begin{theorem}\label{bigkahuna}{\tt The Galois Correspondence Theorem}
Let $F\subset K$ be subfields of $C$ with $K:F$ a {\em normal} extension.
Set $G=\Gamma(K:F)$.
Then there is a bijection between the intermediate fields
$L$, meaning that $F\subset L\subset K$ and the subgroups
$H$ of $G$. (We include $L=F$, $L=K$ as intermediate fields
and we include $\{e\}$ and $G$ itself as subgroups.) The
bijection is given by $*$ and $\dag$ as previously defined.
That is, $H=L*$ if and only if $L=H^{\dag}$. Thus
\beq\label{d10}
(L^*)^{\dag} = L \mbox{ and } (H^{\dag})^*=H
\eeq
Furthermore, the
correspondence {\em reverses} containment, making $L$ bigger
makes $H=L^*$ smaller and making $H$ bigger makes $L=H^{\dag}$
smaller.
The field $F$ is associated with all of $G$ while the field
$K$ is assoicated with $\{e\}$. Setting $n=[K:F]$ we have
$n=|G|$.
Further the sizes are connected, when $H=L^*$
\beq \label{size1}
[K:L] = |H|
\eeq
or, equivalently,
\beq \label{size2}
[L:F] = |G/H|
\eeq
\end{theorem}
{\tt A Powerful Application:} In the listing for $K=Q(\sqrt{2},\sqrt{3})$ above as $G$
has only four elements it doesn't take too much work (try it!) to show that
$\{e\},H_1,H_2,H_3,G$ are the {\em only}
subgroups. It is not at all clear that we have listed {\em all} of the subfields
of $K$. How do we know there isn't some other weird intermediate field between $Q$
and $K$? After all, these are infinite sets so we can't try everything. On the
other side, however, we are dealing with finite groups $G=\Gamma(K:F)$. These
can only have a finite number of subgroups $H$ and by some (perhaps laborious)
effort we can list them all. The Galois Correspondence Theorem then allows
us to give a {\em complete} list of the intermediate fields $L$.
{\tt A Cautionary Example:} Let the ground field $F=Q$ and the extension field
$K=Q(2^{1/3})$. Any automorphism $\sig:K\ra K$ must send $2^{1/3}$ to a root of
$x^3-2$ but $2^{1/3}$ is the {\em only} root of $x^3-2$ in $K$, as the other roots
are not real. Thus we must have $\sig(2^{1/3})=2^{1/3}$ and so $\sig$ must be the
identity. That is, $G=\Gamma(K:Q)=\{e\}$. As $[K:Q]=3$ there are no intermediate
fields except for $Q$ and $K$ themselves. So $Q^*=\{e\}$ and $K^*=\{e\}$. As
every element is fixed by $e$, $\{e\}^*=K$. So in this case we do {\em not} get
a bijection between subgroups of the Galois Group and intermediate fields.
For {\em any} extension $K:F$ the following ``easy" result is one part of
\ref{d10}, the main part of the Galois Correspondence Theorem, Theorem
\ref{bigkahuna}.
\begin{theorem}\label{d1} Let $F\subset K$ be subfields of $C$. Then for
any intermediate field $L$
\beq\label{d2} L\subset (L^*)^{\dag} \eeq
and for any subgroup $H$
\beq\label{d3} H\subset (H^{\dag})^* \eeq
\end{theorem}
{\tt Proof:}
$L^*$ is those automorphisms $\sig$ such that $\sig(\ah)=\ah$ for
all $\ah\in L$. That is, all $\sig\in L^*$ fix all $\ah\in L$. That is, all
$\ah\in L$ are fixed by all $\sig\in L^*$ and hence all $\ah\in L$ belong
to $(L^*)^{\dag}$.
Similarly, $H^{\dag}$ is those $\ah\in K$ such that $\sig(\ah)=\ah$ for all
$\sig\in H$. That is, all $\sig\in H$ fix all $\ah\in H^{\dag}$. That is,
all $\sig\in H$ are in $(H^{\dag})^*$.
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