Exercise 9.2.14

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Consider the quotation from Galois given at the end of the Historical Notes.
\be
\item[(a)] Show that the permutations obtained by mapping the first line in the displayed table to the other lines give a cyclic group of order $n-1$. Also explain how these permutations relate to the Galois group.
\item[(b)] Explain what Galois is saying in the last sentence of the quotation.
\ee

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
This group of permutations is generated by the cycle $$(a,b,c,\cdots,k) = (r,r^g,r^{g^2}, \cdots,r^{g^{n-2}}).$$ It is a cyclic subgroup of order $n-1$ in the group of permutation of the $n-1$ roots of $\Phi_n(x)$. The Galois group of $\Phi_n(x)$, as a permutation group of the roots, is indeed a cyclic group of order $n-1$, if $n$ is prime: 
$$\Gal_{\Q}(\Phi_n) = \Gal(\Q(\zeta_n)/\Q) \simeq (\Z/n\Z)^* \simeq C_{n-1}.$$
For such a Galois extension,
$$\vert \Gal(\Q(\zeta_n)/\Q) \vert = [\Q(\zeta_n):\Q] = n-1 = \deg(\Phi_n(x)).$$

\item[(b)] If all the roots are rational function of one fixed root $\alpha$ of $f$, then the extension $\Q \subset \Q(\alpha)$ is Galois, so  the equality $\vert \Gal(\Q(\alpha)/\Q) \vert = [\Q(\alpha) : \Q]  = \deg(f)$ is true for the minimal polynomial $f$ of $\alpha$ over $\Q$.
\end{proof}

Exercise 9.2.14

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Exercise 9.2.14

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