Exercise 8.5.5

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 $F$ have characteristic 0, and assume that we have fields $F \subset K \subset L$. Also suppose that $\alpha \in L$ is expressible by radicals over $K$ and that the extension $F \subset K$ is a solvable extension. Prove carefully that the minimal polynomial of $\alpha$ over $F$ 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}
$F$ has characteristic 0, and $F \subset K \subset L$.

Since $\alpha \in L$ is expressible by radicals over $K$,  there exists by definition a radical extension $K \subset N$ such that $\alpha\in N$. 
 
By hypothesis $F\subset K$ is solvable, so there exists a radical extension $F\subset M$ such that $K \subset M$.

As $K \subset N$ is radical (and $K \subset M$), then  $M \subset MN$ is also radical by Lemma 8.2.7(b).

So $F \subset M$ and $M \subset MN$ are radical extensions, so $F \subset MN$ is a radical extension (Lemma 8.2.7(a)).

As $\alpha \in N \subset MN$, with $F \subset MN$ radical, by definition $\alpha$ is expressible par radicals over $F$ and by Proposition 8.5.2, its minimal polynomial $f$ over $F$ is solvable by radicals over $F$.
\end{proof}

Exercise 8.5.5

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Exercise 8.5.5

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