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\title{MATH0037 Galois Theory 2000}
\author{GCS -- Sheet 7}
\date{}
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\maketitle
\begin{enumerate}
\item Suppose that $\alpha \in \mathbb C$ and that $\mathbb Q[\alpha]$
is a field. Prove that $\alpha$ is algebraic. \newline
\noindent{\bf Solution}\ If $\alpha = 0$ the result is
clear. If $\alpha \not = 0$ the $\alpha$ is invertible in the
field $Q[\alpha]$ so there is $f \in \mathbb Q[X]$ with $f \not = 0$ 
such that
$\alpha f(\alpha) = 1$. Now $\alpha$ is a root of
$X f -1$ a non-zero rational polynomial. Thus $\alpha$ is algebraic over
$\mathbb Q$. 
\item Suppose that $L$ is a field and that $\alpha: L \rightarrow L$
is an automorphism of $L$. Let 
\[ \hbox{Fix}(\alpha) = \{ \lambda \in L \mid (\lambda) \alpha = \lambda\}.\]
Show that $\hbox{Fix}(\alpha)$ is a subfield of $L$. \newline
\noindent{\bf Solution}\ This is entirely routine.
If $x, y \in \hbox{Fix}(\alpha)$, then $(x+y)\alpha
= (x)\alpha + (y)\alpha = x + y$. Similar remarks
apply to substraction, multiplication and division (where
defined). Note also that $0$ and $1$ must be fixed by
$\alpha$ (why?).
\item Suppose that $h \in \mathbb C[X]$. We define $h'$ to be the
polynomial which is the ``derivative of $h$ with respect to $X$''.
Now suppose that $K$ is a subfield of the complex numbers
and that $f \in K[X]$ is irreducible as an element of $K[X]$.
\begin{enumerate}
\item Prove that $(f) + (f') = K[X]$. \newline
\noindent{\bf Solution}\ Since $f$ is irreducible in 
$\mathbb Q[X]$ it follows that $(f)$ is a maximal ideal.
Now $f' \not = 0$ and so by degree considerations $f' \not \in (f).$
Thus the ideal $(f) + (f')$ is strictly larger than $(f)$ 
and so $(f) + (f') = \mathbb Q[X] \ni 1$.
\item Deduce that that there is no complex number $\alpha$
which is simultaneously a root of $f$ and of $f'$.\newline
\noindent{\bf Solution}\ Thus 
$1 = gf + hf'$ for some $g,h \in \mathbb Q[X]$.
Now if $\beta$ were a common root of $f$ and $f'$ in any extension
field of $\mathbb Q$ then $\beta$ would be a root of $1$
which is absurd.
\item Deduce that there is no complex number $\beta$ such that
$f = [X - \beta]^2 r$ for some $r \in K[X]$.\newline
\noindent{\bf Solution}\ Such a complex number $\beta$
would be a similultaneous root of $f$
and $f' = 2[X - \beta] r + [X-\beta]^2 r'$. 
\end{enumerate}
\item Show that for each odd natural
number $n$ there is a field extension
$\mathbb Q \leq K$ where 
$|K:\mathbb Q| = n$ and $\hbox{Gal}_{\mathbb Q}K = 1.$ \newline
\noindent{\bf Solution}\ Let $f = X^n - 2$. Now $f \in \mathbb Q[X]$
is irreducible by Eisenstein's criterion. Let $\alpha =
\sqrt[n] 2 \in \mathbb R$. Notice that $\alpha$ is the 
unique
real root of $f$. Let $K = \mathbb Q(\alpha) \leq \mathbb R$ so
any automorphism of $K$ must send $\alpha$ to a real $n$-th root
of $2$ and so must fix $\alpha$. Thus $G = \mbox{Gal}_{\mathbb Q} K
=1$.
\end{enumerate}
\end{document}