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\title{MATH0037 Galois Theory 2002, Sheet 9}
\author{Geoff Smith, http://www.bath.ac.uk/$\sim$masgcs}
\date{}
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\maketitle
\begin{enumerate}
\item Consider the field $M = \mathbb Q (\omega, \alpha)$ discussed
in question 1 of sheet 8.
\begin{enumerate}
\item[(a)] {\em List all 6 subgroups of $G = \mbox{Gal}_{\mathbb Q}M$.}
\newline \noindent
{\bf Solution} The group $G$ is a copy of $S_3$. In subgroups of a group
of order 6 must have size 1,2,3 or 6 by Lagrange's theorem. 
In standard notation the subgroups are 
\[ \{id\}, \{id, (1,2)\}, \{id, (1,3)\}, \{id, (2,3)\}, \{ id, (1,2,3), (1,3,2)\}
\mbox{ and } G.\]
To realise this group, 1 denotes $\alpha$, 2 denotes $\alpha \omega$
and $3$ denotes $\alpha \omega^2$. 

\item[(b)] For each subgroup $H$ in the list you found in part (a),
describe the field $H^\dagger = \{ m \mid m \in M, (m)h = m \forall h \in H\}$
by exhibiting an element $\alpha_H \in H^\dagger$ such that 
$H^\dagger = \mathbb Q(\alpha_H)$. {\em Perhaps I should be writing
star instead of dagger. I have forgotten which notation was used in
lectures.} \newline \noindent {\bf Solution}\ 
Taking the six subgroups in sequence,
the fixed fields are as follows:
\begin{enumerate}
\item[(i)] $M$.
\item[(ii)] $\mathbb Q(\alpha\omega^2)$ is fixed elementwise by this group,
and is an extension of the rationals of degree 3. The Galois correspondence
tells us that the fixed field should be an extension of $\mathbb Q$ of degree
3, so the solution is $\mathbb Q(\alpha \omega^2)$. Similar arguments
deal with the remaining cases.
\item[(iii)] $\mathbb Q(\alpha \omega)$.
\item[(iv)] $\mathbb Q(\alpha)$.
\item[(v)] $\mathbb Q(\omega)$.
\item[(vi)] $\mathbb Q$.
\end{enumerate}
\end{enumerate}
\item {\em Suppose that $K \leq L$ is a field extension
and that $L$ is a splitting field for
some polynomial $f \in K[X]$. Prove that there are only finitely
many fields $T$ such that $K \leq T \leq L$. {\rm You may assume the
truth of the fundamental theorem of Galois theory}.} \newline
\noindent {\bf Solution}\ Let $\Omega$ denote the set of roots of 
$f$, and the the Galois group embeds and a subgroup of the symmetric
groupon $\Omega$. A finite group has only finitely many subgroups,
so we are done. 
\item {\em Suppose that $K \leq L$ is a field extension of finite degree.
Prove that there are only finitely
many fields $T$ such that $K \leq T \leq L$.} {\rm You may use the result
stated in the previous question}\newline
\noindent {\bf Solution}\ As shown in lectures, we simply build a field
$\Sigma$ with $L \leq \Sigma$ and $\Sigma$ is a splitting field of a polynomial
with coefficients in $K$.
\item {\em Let $f = X^n -1 \in \mathbb Q[X]$. Let $\omega = e^{2\pi i/n} \in
\mathbb C$. Show that $Q(\omega)$ is a splitting field for $f$.
By considering what happens to $\omega$ under the action of elements
of $G = \mbox{Gal}_{\mathbb Q} Q(\omega)$, show that $G$ is
an abelian group.}\newline
\noindent {\bf Solution}\ The roots of $f$ in $\mathbb C$ are 
$1, \omega, \omega^2, \ldots, \omega^{n-1}.$ All these roots are elements
of $\mathbb Q(\omega)$ so $Q(\omega)$ is a splitting field for $f$.
The elements of the Galois group are determiney their action on $\omega$,
and of course $\omega$ must be sent to a power of itself by each element of 
the Galois group. Consider the Galois group elements $\theta$, $\psi$ which
send $\omega$ to $\omega^i$ and $\omega^j$ respectively. Now
$(\omega^i)^j = \omega^{ij} = (\omega^j)^i$ so $\theta$ and $\psi$ commute.
It follows that the Galois group is abelian.
\end{enumerate}


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