Exercise 12.1.6

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}}

Explain why the polynomial $	heta(t)$ (12.13) has coefficients in $K = F(\sigma_1,\sigma_2,\sigma_3,\sigma_4)$.

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}
$$	heta(t) = (t^2 - 4y_1-\sigma_1^2+4\sigma_2)(t^2 - 4y_2-\sigma_1^2+4\sigma_2)(t^2 - 4y_3-\sigma_1^2+4\sigma_2).$$
Recall that
\begin{align*}
y_1 &= x_1x_2+x_3x_4\
y_2 &= x_1x_3+x_2x_4\
y_3 &= x_1x_4 + x_2x_3
\end{align*}
Let $	au = (1\, 2), \sigma = (1\,2\,3\,4)$. Then
$$	au \cdot y_1 = x_2 x_1 + x_3x_4 =  y_1,\quad 	au \cdot y_2 = x_2x_3 + x_1x_4= y_3,\quad 	au \cdot y_3 = x_2x_4+x_1x_3= y_2,$$
and of course $	au \cdot \sigma_1 = \sigma_1,	au \cdot \sigma_2 = \sigma_2$.

Therefore $	au\cdot 	heta(t) = 	heta(t)$.

Similarly,
$$\sigma\cdot y_1 = x_2x_3+x_4x_1 = y_3,\quad \sigma\cdot y_2=x_2x_4+x_3x_1 = y_2,\quad \sigma \cdot y_3 = x_2x_1+x_3x_4 = y_1.$$
Therefore $\sigma \cdot 	heta(t) = 	heta(t)$.

Since $S_4 = \langle \sigma, 	au angle$, every permutation in $S_4$ lets the coefficients of $	heta(t)$ unchanged, therefore $	heta(t)$ has coefficients in $K = F(\sigma_1,\sigma_2,\sigma_3,\sigma_4)$ and $	heta(t) \in K[t]$.
\end{proof}

Exercise 12.1.6

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Exercise 12.1.6

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