Exercise 12.1.2

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}}

Work out the details of Example 12.1.2.

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 $F = \Q(\omega)$, $z_1 = \frac{1}{3}(x_1+\omega^2 x_2+\omega x_3) \in K=\Q(\omega)(x_1,x_2,x_3)$,  and $	heta(z) \in \Q(\omega)[z]$ be the resolvent polynomial of $z_1$.
The orbit of $z_1$ under the action of $S_n$ is composed of 
\begin{align*}
z_1& =\frac{1}{3}(x_1+\omega^2 x_2+\omega x_3),\
  (2,3)\cdot z_1 &= \frac{1}{3}(x_1+\omega^2 x_3+\omega x_2) = \frac{1}{3}(x_1+\omega x_2+\omega^2 x_3)=z_2\
  (1,3) \cdot z_1 &=\frac{1}{3}(x_3+\omega^2 x_2+\omega x_1) =  \frac{1}{3}(\omega x_1+\omega^2 x_2+ x_3) = \omega z_2\
  (1,2) \cdot z_1 &=\frac{1}{3}(x_2+\omega^2 x_1+\omega x_3) =  \frac{1}{3}(\omega^2 x_1+ x_2+ \omega x_3) = \omega^2 z_2\
 (1,2,3)\cdot z_1 &= \frac{1}{3}(x_2+\omega^2 x_3+\omega x_1) =  \frac{1}{3}(\omega x_1+ x_2+\omega^2 x_3) = \omega z_1\
 (1,3,2)\cdot z_1 &= \frac{1}{3}(x_3+\omega^2 x_1+\omega x_2) =  \frac{1}{3}(\omega^2 x_1+ \omega x_2+ x_3) = \omega^2 z_1.\
\end{align*}
So the orbit of $z_1$ is
$${\cal O}_{z_1} = \{z_1,z_2,\omega z_1,\omega z_2,\omega^2 z_1,\omega^2 z_2\},$$
and these six elements are distinct in $F(x_1,x_2,x_3)$.

Moreover,
\begin{align*}
	heta(z) &= (z-z_1)(z-z_2)(z-\omega z_1)(z-\omega z_2) (z- \omega^2 z_1)(z-\omega^2 z_2)\
&=(z^3-z_1^3)(z^3 - z_2^3)\
&=z^6 - (z_1^3+z_2^3)z^3 +(z_1z_2)^3
\end{align*}
and 
\begin{align*}
z_1z_2 &= \frac{1}{9}(x_1+\omega^2 x_2+\omega x_3)(x_1+\omega x_2+\omega^2 x_3)\
&= \frac{1}{9}(x_1^2+x_2^2+x_3^2-x_1x_2-x_2x_3-x_1x_3)\
&= \frac{1}{9}[(x_1+x_2+x_3)^2 - 3(x_1x_2+x_2x_3+x_1x_3)]\
&=\frac{1}{9}(\sigma_1^2 - 3 \sigma_2),
\end{align*}
so 
$$z_1^3z_2^3 = \frac{1}{3^6}(\sigma_1^2-3\sigma_1)^3 = -\frac{1}{27}\left(-\frac{\sigma_1^2}{3} + \sigma_2 ight)^3 = -\frac{p^3}{27},	ext{ where } p = -\frac{\sigma_1^2}{3} + \sigma_2.$$

\begin{align*}
z_1^3 + z_2^3 &= \frac{1}{27}\left[2 (x_{1}^{3} +  x_{2}^{3} +  x_{3}^{3}) - 3 (x_{1}^{2} x_{2} + x_{1} x_{2}^{2} +x_{1}^{2} x_{3} + x_{2}^{2} x_{3} + x_{1}x_{3}^{2} +x_{2} x_{3}^{2}) + 12 x_{1} x_{2} x_{3}ight]
\end{align*}
\begin{align*}
s &= x_{1}^{2} x_{2} + x_{1} x_{2}^{2} +x_{1}^{2} x_{3} + x_{2}^{2} x_{3} + x_{1}x_{3}^{2} +x_{2} x_{3}^{2}\
&= (x_1x_2+x_2x_3+x_1x_3)(x_1+x_2+x_3)-3x_1x_2x_3\
&= \sigma_2 \sigma_1 - 3 \sigma_3
\end{align*}

\begin{align*}
x_1^3+x_2^3+x_3^3 &= (x_1^2+x_2^2+x_3)^2(x_1+x_2+x_3) - (x_{1}^{2} x_{2} + x_{1} x_{2}^{2} +x_{1}^{2} x_{3} + x_{2}^{2} x_{3} + x_{1}x_{3}^{2} +x_{2} x_{3}^{2})\
&=(\sigma_1^2-2\sigma_2)\sigma_1 - (\sigma_2 \sigma_1 - 3 \sigma_3)\
&=\sigma_1^3 -3\sigma_1\sigma_2+3\sigma_3.
\end{align*}
Thus
\begin{align*}
z_1^3 + z_2^3 &= \frac{1}{27}\left[2(\sigma_1^3 -3\sigma_1\sigma_2+3\sigma_3) -3(\sigma_1 \sigma_2 - 3 \sigma_3) + 12 \sigma_3ight]\
&= \frac{1}{27} (2 \sigma_1^3-9\sigma_1 \sigma_2 + 27 \sigma_3)\
&=  \frac{2\sigma_1^3}{27} - \frac{\sigma_1\sigma_2}{3} + \sigma_3
\end{align*}
Finally,
$$	heta(z) = z^6 +q z^3 -\frac{p^3}{27},$$
where
$$p = -\frac{\sigma_1^2}{3} + \sigma_2,\quad q = -\frac{2\sigma_1^3}{27} + \frac{\sigma_1\sigma_2}{3}- \sigma_3.$$
\end{proof}

Exercise 12.1.2

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Exercise 12.1.2

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