Exercise 2.4.9

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}}

For $n=4$, the variables $x_1,x_2,x_3,x_4$ have discriminant
$$\Delta = (x_1-x_2)^2(x_1-x_3)^2(x_1-x_4)^2(x_2-x_3)^2(x_2-x_4)^2(x_3-x_4)^2.$$
Let $y_1 = x_1x_2+x_3x_4,y_2 = x_1x_3+x_2x_4,y_3=x_1x_4+x_2x_3$, and consider
$$	heta(y) = (y-y_1)(y-y_2)(y-y_3).$$
This is a cubic polynomial in $y$. As in the text, the discriminant of $	heta$ will be denoted $\Delta(	heta)$. Show that $\Delta(	heta) = \Delta$.

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}
\begin{align*}
y_1-y_2 &= x_1x_2+x_3x_4-x_1x_3-x_2x_4=x_1(x_2-x_3)-x_4(x_2-x_3) =(x_1-x_4)(x_2-x_3)\
y_1-y_3&= x_1x_2+x_3x_4-x_1x_4-x_2x_3=x_1(x_2-x_4)-x_3(x_2-x_4) =(x_1-x_3)(x_2-x_4)\
y_2-y_3&= x_1x_3+x_2x_4-x_1x_4-x_2x_3=x_1(x_3-x_4)-x_2(x_3-x_4) =(x_1-x_2)(x_3-x_4),\
\end{align*}
Therefore
\begin{align*}
\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.
\end{align*}
\end{proof}

Exercise 2.4.9

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Exercise 2.4.9

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