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\title{Galois Theory Handout 2}
\author{Geoff Smith, http://www.bath.ac.uk/$\sim$masgcs}
\date{}
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\maketitle

\noindent {\bf Lemma} {\em Suppose that $K$ is a field of characteristic
$0$ (i.e. it contains a copy of $\mathbb Q$), and that $L$ is a spliting field
of some $f \in K[X]$. Suppose that $T_1, T_2$ are subfields of $L$ which both
contain $K$,
and that there is an isomorphism of fields $\varphi: T_1 \rightarrow T_2$ 
which fixes $K$ elementwise, then there is $g \in G = \mbox{Gal}_K L$
such that if $x \in T_1$, then $\varphi(x) = (x)g$.}

\noindent{\bf Proof} We induct on $|L:T_1|$. If this quantity is 1,
then $T_1 = L$. Now the isomorphism $\varphi$ ensures that 
$|T_1:K| = |T_2:K|$ so $T_2 = L$ and $\varphi \in \mbox{Gal}_K L$.

Suppose that $|L:T_1| > 1$. Choose $f_1$ a non-linear irreducible factor
of $f$ in $T_1[X]$. The field isomorphism $\varphi$ induces
a ring isomorphism $\widehat \varphi: T_1[X] \rightarrow 
T_2[X]$ which carries $f_1$ to $f_2$. Both $f_1$ and $f_2$ split in $L[X]$
since $f$ has a complete set of roots. Choose $a_1,a_2 \in L$ which are
roots of $f_1$, $f_2$ respectively. Now \[S_1 = T_1(a_1) \simeq
T_1[X]/(f_1) \simeq T_2[X]/(f_2) \simeq T_2(a_2) = S_2\]
and the composition $\theta$ of these maps agrees with $\varphi$ on $T_1$.
Now replace $T_1$ by $S_1$, $T_2$ by $S_2$ and $\varphi$ by $\theta$,
and observe that $|L:S_1| \cdot |S_1 : T_1| = |L:T_1|$ so
$|L:S_1| < |L:T_1|$. The result follows by induction.

\vskip 0.2cm

\noindent {\bf Corollary} {\em If $L$ is a splitting field of 
the irreducible $f \in K[X]$. The Galois group $G = \mbox{Gal}_K L$
acts transitively on the roots of $f$ (i.e. if $a,b$ are roots
of $f$ in $L$, then there is $g \in G$ such that $(a)g = b$).}

\vskip 0.2cm

\noindent {\bf Theorem} {\em Suppose that $K$ is a field of 
characteristic $0$ (i.e.
it contains a copy of $\mathbb Q$),and that $K\leq L$ is a field extension
with $|L:K| < \infty$. Then $L$ is a splitting field of some
polynomial $f \in K[X]$ if and only if the extension is normal.} 

\noindent {\bf Proof} Suppose that the extension is normal. Choose $c \in L$ such that
$L = K(c)$. Let $f$ be a monic irreducible element of $K[X]$ having
$c$ as a root. Now the normality of the extension forces all roots
of $f$ to be in $L$, so $L$ is the splitting field of $f$.

Conversely, suppose that $L$ is the splitting field of some
polynomial $h \in K[X]$ with $\Omega$ denoting the (finite) set of
roots of $h$ in $L$. Now suppose that $t \in K[X]$ is an irreducible polynomial
having a root $\alpha$ in $L$. Let $\Sigma$ be a splitting field for $f \in L[X]$,
so $L \leq \Sigma$. Suppose that $\alpha_1$ is a root of $f$ in $\Sigma$.
Since $f \in K[X]$ is irreducible, there is $g \in G = \mbox{Gal}_K \Sigma$
such that $(\alpha)g = \alpha_1$ by the lemma. Now $g$ must permute the roots 
of $h$ (which has coefficients in $K$) so must map $L$ into $L$.
It follows that $\alpha_1 \in L$ so the extension $K \leq L$ is normal.
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