Exercise 12.1.7

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}}

Show that (12.15) implies the equations for $x_1,x_2,x_3,x_4$ given in the text.

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}
We know that
\begin{align*}
\sigma_1 &= x_1+x_2+x_3+x_4,\
t_1 &= x_1+x_2-x_3-x_4,\
t_2 &= x_1-x_2+x_3-x_4,\
t_3 &= x_1-x_2-x_3+x_4.
\end{align*}
The sum of these equations gives
$$\sigma_1 + t_1 + t_2 + t_3 = 4x_1,$$
so
$$x_1=\frac{1}{4} \left(\sigma_1 + t_1 + t_2 + t_3 ight).$$
We can compute similarly $\sigma_1+t_1-t_2-t_3$, ...

More conceptually, let $\sigma = (1\,2)(3\,4)$. Then 
$$\sigma \cdot x_1 = x_2,\quad  \sigma \cdot t_1 = t_1,\quad  \sigma \cdot t_2 = -t_2,\quad \sigma \cdot t_3 = -t_3.$$
Therefore 
$$x_2 = \frac{1}{4} \left(\sigma_1 + t_1 - t_2 - t_3 ight).$$
Similarly, if $	au = (1\,3)(2\,4)$,
$$\sigma \cdot x_1 = x_3,\quad  	au \cdot t_1 = -t_1,\quad  	au \cdot t_2 = t_2,\quad 	au \cdot t_3 = -t_3.$$
Therefore 
$$x_3 = \frac{1}{4} \left(\sigma_1 - t_1 + t_2 - t_3 ight).$$
Finally, if $\zeta = (1\,4)(2\,3)$,
$$\zeta \cdot x_1 = x_4,\quad  \zeta \cdot t_1 = -t_1,\quad  \zeta \cdot t_2 = -t_2,\quad \zeta \cdot t_3 = t_3.$$
Therefore 
$$x_4 = \frac{1}{4} \left(\sigma_1 - t_1 - t_2 + t_3 ight).$$
To conclude
\begin{align*}
x_1 &=\frac{1}{4} \left(\sigma_1 + t_1 + t_2 + t_3 ight),\
x_2 &= \frac{1}{4} \left(\sigma_1 + t_1 - t_2 - t_3 ight),\
x_3 &= \frac{1}{4} \left(\sigma_1 - t_1 + t_2 - t_3 ight),\
x_4 &= \frac{1}{4} \left(\sigma_1 - t_1 - t_2 + t_3 ight).
\end{align*}
\end{proof}

Exercise 12.1.7

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Exercise 12.1.7

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