Exercise 6.5.5

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 $f$ the minimal polynomial of $\sqrt{2+\sqrt{2}}$ over $\Q$. Show that $f=0$ is an Abelian equation.

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}

By Exercises 5.1.6 and 6.3.4,
\begin{align*}
f&= x^4-4x^2+2\
&= \left(x-\sqrt{2+\sqrt{2}}ight)\left(x+\sqrt{2+\sqrt{2}}ight)\left(x-\sqrt{2-\sqrt{2}}ight)\left(x+\sqrt{2-\sqrt{2}}ight)\
&=(x-\alpha)(x+\alpha)(x-\beta)(x+\beta)\
\end{align*}
where $\beta= \frac{1}{2}\left(\alpha- \frac{2}{\alpha^3}ight)  = \alpha^3-3\alpha$.

The 4 roots of $f$ are of the form $\alpha = 	heta_1(\alpha), -\alpha = 	heta_2(\alpha), \beta=	heta_3(\alpha), -\beta = 	heta_4(\alpha)$, where
$$	heta_1(x) = x, 	heta_2(x) = -x, 	heta_3(x) = x^3-3x,	heta_4(x) = -x^3+3x.$$

As $	heta_1 = x, 	heta_2 = -x, 	heta_4 = -	heta_3$ and $	heta_3,	heta_4$ are odd functions, as in Exercise 2,

$	heta_1(	heta_i(\alpha)) = 	heta_i(\alpha) = 	heta_i(	heta_1(\alpha)),\ i=2,3,4$.

$	heta_2(	heta_i(\alpha)) = - 	heta_i(\alpha) = 	heta_i(-\alpha) = 	heta_i(	heta_2(\alpha)),\ i=3,4$.

$	heta_3(	heta_4(\alpha)) = 	heta_3(-	heta_3(\alpha)) = -	heta_3^2(\alpha) = -	heta_4^2(\alpha) = 	heta_4(-	heta_4(\alpha)) = 	heta_4(	heta_3(\alpha))$.

Thus $	heta_i(	heta_j(\alpha)) = 	heta_j(	heta_i(\alpha))$, for $1\leq i <j\leq 4$, thus also for $1\leq i,j\leq 4$.

$	heta_i(	heta_j(\alpha)) = 	heta_j(	heta_i(\alpha)),\ 1\leq i,j\leq 4$. 
 \begin{center}
  $x^4-4x^2+2=0$ is an Abelian equation.
  \end{center}
\end{proof}

Exercise 6.5.5

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Exercise 6.5.5

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