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\author{Jason Wojciechowski}
\title{Notebook problem about an abelian Galois group}

\newtheorem*{thm}{Theorem}

\begin{document}
\maketitle

\begin{thm}
  The Galois group of the splitting field of $x^n-a$ over the
  splitting field of $x^n-1$ is abelian.
\end{thm} 

\begin{proof}
  Let $K$ be the splitting field of $x^n-1$ and let $L$ be the
  splitting field of $x^n-a$. Let $\omega \in K$ be a zero of $x^n-1$
  and let $\alpha \in L$ be a zero of $x^n-a$.  Note that for any
  $\omega$, $\omega^n=1$, so that $(\omega \alpha)^n = \omega^n
  \alpha^n = 1 \alpha^n$.  Thus every zero $\alpha$ of $x^n-a$ can
  also be expressed as $\omega \alpha$ for some $\omega$, so $L$ can
  be expressed as $K(\alpha)$.  Since $L$ is a simple extension of
  $K$, the members of the Galois group of $L:K$, the automorphisms
  that fix $K$, are fully determined by their action on $\alpha$.
  
  Consider two automorphisms, $\psi _1$ and $\psi _2$ such that $\psi
  _1 (\alpha) = \omega_1 \alpha$ and $\psi _2 (\alpha) = \omega_2
  \alpha$, where the $\omega_i$'s are zeros of $x^n-1$. Let us examine
  $\psi_1 \circ \psi_2 (\alpha)$.  $$\psi_1 \circ \psi_2
  (\alpha) = \psi_1 (\omega_2 \alpha) = \omega_2 \omega_1 \alpha =
  \omega_1 \omega_2 \alpha = \psi_2 (\omega_1 \alpha) = \psi_2
  \circ \psi_1 (\alpha).$$
  
  $\omega_1$ commutes with $\omega_2$ because $e^{\frac{a\cdot2\pi i
      \theta}{n}} e^{\frac{b\cdot 2 \pi i \theta}{n}} =
  e^{\frac{(a+b)2\pi i \theta}{n}} = e^{\frac{b\cdot2\pi i \theta}{n}}
  e^{\frac{a\cdot 2 \pi i \theta}{n}}$. Thus we have that the Galois
  group is abelian.
\end{proof}

\end{document}