Exercise 8.5.4

Question

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\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}\,}

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

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

Let $f$ be the minimal polynomial over $\Q$ of $\sqrt[5]{\sqrt[3]{17} + \sqrt[4]{37}}$ over $\Q$, where all of the indicated radicals are real. Prove that $f$ is solvable by radicals over $\Q$.

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}
Let $f$ be the minimal polynomial over $\Q$ of 
$$\alpha = \sqrt[5]{\sqrt[3]{17} + \sqrt[4]{37}}.$$

$\alpha \in K = \Q( \sqrt[5]{\sqrt[3]{17} + \sqrt[4]{37}})$, and we obtain the inclusion chain
$$F_0 = \Q \subset F_1 = \Q(\sqrt[3]{17} ) \subset F_2 = \Q(\sqrt[3]{17}, \sqrt[4]{37}) \subset F_3 = \Q\left(\sqrt[3]{17}, \sqrt[4]{37}, \sqrt[5]{\sqrt[3]{17} + \sqrt[4]{37}}ight);$$
that is
$$F_0 \subset F_1=F_0(\gamma_1) \subset F_2 = F_1(\gamma_2) \subset F_3 = F_3(\gamma_3)=K,$$
where $\gamma_1 = \sqrt[3]{17} \in F_1,\gamma_2 = \sqrt[4]{37} \in F_2, \gamma_3 = \sqrt[5]{\sqrt[3]{17} + \sqrt[4]{37}}$ satisfy $$\gamma_1^3 = 17 \in \Q=F_0, \gamma_2^4 = 37 \in \Q \subset F_1, \gamma_3^5 = \gamma_1+\gamma_2 \in F_2.$$

This proves that $\Q \subset K$ is a radical extension, with $\alpha$ in $K$, so $\alpha$ is expressible by radicals over $\Q$ according to Definition 8.5.1. By Proposition 8.5.2, as the irreducible polynomial $f$ has a root expressible by radicals over $\Q$, $f$ is solvable by radicals. So all the roots of $f$ are expressible by radicals over $\Q$.
\end{proof}

Exercise 8.5.4

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Exercise 8.5.4

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