Exercise 14.2.12

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}}

Let $f$ be an irreducible imprimitive polynomial of degree 6,8 or 9 over a field $F$ of characteristic 0. Prove that $f$ is solvable by radicals over $F$.

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}
Write $L$ the splitting field of $f$. Since the characteristic of $F$ is 0, the irreducible polynomial $f$ is separable, so $f$ is separable, irreducible and imprimitive.  By Corollary 14.2.10, $G = \Gal(L/F)$ is isomorphic to a subgroup of $S_k \wr S_l$, where $n = kl$ is a nontrivial factorization. The only nontrivial factorizations of 6,8 or 9 are
$$6 = 2 	imes 3 = 3 	imes 2, \qquad 8 = 2 	imes 4 = 4 	imes 2, \qquad 9 = 3 	imes 3.$$
Thus $G$ is isomorphic to a subgroup of the list
$$S_2 \wr S_3, S_3\wr S_2,S_4 \wr S_2,S_2 \wr S_4, S_3\wr S_3,$$
whose cardinalities are
\begin{align*}
|S_2 \wr S_3| &= 2!3^2 = 2 	imes 3^2,\
 |S_3\wr S_2| &= 3!2^3 = 2^4 	imes 3,\
 |S_4 \wr S_2| &= 4!2^4 = 2^7 	imes 3,\
 |S_2 \wr S_4| &= 2!4^2 = 2^5,\
 |S_3\wr S_3| &=3!3^3 = 2 	imes 3^4. 
\end{align*}
So $S_k \wr S_l$ has only two prime factors 2 and 3. By Burnside's Theorem (Theorem 8.1.8), $S_k \wr S_l$ solvable for these values of $k,l$, thus the subgroup $G$ is solvable. Since the characteristic of $F$ is 0, this proves that $f$ is solvable by radicals over $f$.
\end{proof}

Exercise 14.2.12

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Exercise 14.2.12

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