Exercise 13.1.3

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 carefully why (13.6) follows from Exercise 9 of section 2.4.

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}
By definition, 
$$y_1 = x_1x_2+x_3x_4, \qquad y_2 = x_1x_3+x_2x_4,\qquad y_3 = x_1x_4+x_2x_3.$$
By Exercise 2.4.9, we know that
$$\Delta(	heta) = (y_1-y_2)^2(y_1-y_3)^2(y_2-y_3)^2
=[(x_1-x_4)(x_2-x_3)(x_1-x_3)(x_2-x_4)(x_1-x_2)(x_3-x_4)]^2
=\Delta
$$
As the evaluation is a ring homomorphism, if we applied the evaluation defined by $x_1 \mapsto \alpha_1,\ldots,x_4\mapsto \alpha_4$ to this equality in $F[x_1,x_2,x_3,x_4]$, we obtain
that the roots
$$\beta_1 = \alpha_1\alpha_2+\alpha_3\alpha_4, \qquad \beta_2 = \alpha_1\alpha_3+\alpha_2\alpha_4,\qquad \beta_3 = \alpha_1\alpha_4+\alpha_2\alpha_3,$$
are the images of $y_1,y_2,y_3$ and satisfy
\begin{align*}
\Delta(	heta_f) &= (\beta_1-\beta_2)^2(\beta_1-\beta_3)^2(\beta_2-\beta_3)^2\
&=[(\alpha_1-\alpha_4)(\alpha_2-\alpha_3)(\alpha_1-\alpha_3)(\alpha_2-\alpha_4)(\alpha_1-\alpha_2)(\alpha_3-\alpha_4)]^2\
&=\Delta(f)
\end{align*}
\end{proof}

Exercise 13.1.3

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Exercise 13.1.3

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