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\title{MATH0037 Galois Theory 2000}
\author{Fundamental Theorem of Galois Theory}
\date{Version 1}
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\maketitle

We work with subfields of $\mathbb C$ throughout, so 
field extensions $K \leq L$ of finite degree (i.e.
$| L:K| = dim_K L < \infty$) must be simple (i.e.
there is $\theta \in L$ such that $L = K(\theta)$.
Also recall that an {\em algebraic number field} $M$ is 
a subfield if $\mathbb C$ such that $|M:\mathbb Q| < \infty$.

At this stage we assume that we have proved that
if $G$ is the Galois group of finite normal field
extension inside $\mathbb C$, then $|G|$ is the degree
of the extension.

Let $\Sigma$ be an algebraic number field which
is a splitting field of some $f \in \mathbb Q[X]$
(so the extension $\mathbb Q \leq \Sigma$ is normal).
Let $G = \mbox{Gal}_{\mathbb Q} \Sigma$.
To each subfield $L$ of $\Sigma$ we associate the 
subgroup $L^* = \mbox{Gal}_L\Sigma$ of $G$. To each
subgroup $H$ of $G$ we associate the subfield $H^\dagger$
of $\Sigma$ consisting of the elements of $\Sigma$ which
are fixed by each element of $H$. We have already proved
that $|G| = |\Sigma : \mathbb Q|$.


\noindent{\bf Fundamental Theorem}
The maps $^*$ and $^\dagger$ are mutually inverse bijections
between the collection of all fields intermediate between 
$\mathbb Q$ and $\Sigma$ and the set of all subgroups of $G$.
An intermediate field $M$ is a normal extension of $\mathbb Q$
if and only if $M^*$ is a normal subgroup of $G$. In this
event $\mbox{Gal}_{\mathbb Q} M \simeq G/M^*$.

Furthermore the correspondence is such that if $M$ is
an intermediate field then $|\Sigma:M| = |M^*|$ and 
$|M:\mathbb Q| = |G:M^*|.$

\noindent{\bf Proof}\ 
Suppose that $L$ is an intermediate field. We prove that
that ${L^*}^\dagger = L$. For formal reasons $L \leq {L^*}^\dagger 
\leq \Sigma$. Now the field extension $L \leq \Sigma$
is a normal extension (since $\Sigma$ can be obtained by adjoining
the roots of $f$ to $L$). If $L = \Sigma$, then ${L^*}^\dagger =L$
so we are done. Thus we may assume that $L$ is a proper subfield
of $\Sigma$. Choose $\theta \in \Sigma - L$. Let $\theta$
have minimum polynomial $h \in L[X]$ with deg $h \geq 2$. Now
Let $\theta'$ be a different root of $h$ in $\Sigma$. The fields
$L(\theta)$ and $L(\theta')$ are isomorphic via an isomorphism
fixing $L$ and sending $\theta$ to
$\theta'$. This isomorphism then lifts to an element $g \in
\mbox{Gal}_L\Sigma = L^*$ sending $\theta$ to
$\theta'$ (as was discussed earlier in the course).
Thus the fixed field of $L^*$ is $L$ and so
$L = {L^*}^\dagger$ as required.

Now we begin again, and start with $H$ a subgroup of $G$.
The fixed field of $H$ is $H^\dagger$. Select $\psi$ so
that $\Sigma = H^\dagger(\psi)$. Choose $id = g_1, \ldots, g_m$
to be a sequence of elements of $H$ of maximal length so that
the quantities $\psi g_i$ are distinct. Now if $g \in H$
this must be the same list as $\psi g_ig$ as $i$ ranges
from 1 to $m$ else we violate the maximality of $m$.
Now consider $p = \prod_i [X - \psi g_i]$. This polynomial
has coefficients which are unchanged by the application of
any element of $H$, so are in the fixed field
of $H$ which is $H^\dagger$. Thus the minimum polynomial
of $\psi$ in $H^\dagger[X]$ must divide $p$ and so
$|\Sigma : H^\dagger| \leq m \leq |H|$.
However, $|\Sigma : H^\dagger[X]| = |{H^\dagger}^*|$
since the extension $H^\dagger \leq \Sigma$ is finite and normal.
Thus $|{H^\dagger}^*| \leq |H|$. However, $H \leq {H^\dagger}^*$
for formal reasons so $H = {H^\dagger}^*$.

We now move on to questions of normality. Suppose that
$L$ is an intermediate field and that $\alpha \in G$.
It is a formality to check that the fixed field of $L\alpha$
is $\alpha^{-1}L\alpha$. If the extension $\mathbb Q \leq L$
is normal then $L$ is a splitting field of some rational
polynomial, the roots of which must be permuted by any $\alpha$
in $G$, so each $L\alpha = L$ and so each $\alpha^{-1}L\alpha = L$.
On the other hand, if $L$ is not a normal extension, then
there exists an irreducible rational polynomial $q$ with
one root $\zeta \in L$ and another root $\zeta' \in \Sigma - L$. The 
isomorphism $\mathbb Q(\zeta) \rightarrow \mathbb Q(\zeta')$
which sends $\zeta \mapsto \zeta'$ extends to an element $\beta
\in G$ such that $L\beta \not = L$, and so
$\beta^{-1}L^*\beta \not = L$. Thus $L$ is a normal
extension of $\mathbb Q$ if and only if $L$ is a normal
subgroup of $G$.

Now suppose that $L$ is a normal subgroup of $G$, so
$\mathbb Q \leq L$ is a normal field extension. If $g \in G$
then $Lg = L$ and we teh action of $g$ on $L$ is 
an element of $\mbox{Gal}_{\mathbb Q}L.$ It is easy
to verify that $g \mapsto g \mid_L$ is a group homomorphism
from $G$ to $\mbox{Gal}_{\mathbb Q}L.$ The kernel
is $L^*$ and the image is isomorphic to $G/L^*$ by
the first isomorphism theorem for groups. However,
 $|\mbox{Gal}_{\mathbb Q}L| = |L: \mathbb Q| = |G : L^*|$
by the Galois correspondence. Thus $\mbox{Gal}_{\mathbb Q}L$
and $G/L^*$ are isomorphic groups.

The final part of the theorem about degrees of field extensions,
orders of subgroups and indices of subgroups is straightforward.
we have proved the first assertion already, since $M^*$ is the 
Galois group of a splitting field extension. The rest follows
from the fact that $\mathbb Q^* = G$ and $\Sigma^* = 1$, and the
fact that \[|G| = |\Sigma : M | \cdot |M : \mathbb Q| = |G:M^*| \cdot
|M*|.\]

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