Exercise 1.1.7

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}}

Cardan's formulas, as stated in the text, express the roots as sums of two cube roots. Each cube root has three values, so there are nine different possible values for the sum of the cube roots. Show hat these nine values are the roots of the equations $y^3+py+q=0, y^3+\omega py+q$, and $y^3+\omega^2py+q=0$,where as usual $\omega = \frac{1}{2}(-1 + i \sqrt{3})$.

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}
The nine possible sums of two cube roots are

\begin{align*}
&y_1 = z_1 + z_2, &y_4 = \omega z_1+\omega z_2 = \omega y_1,\hspace{1cm} &y_7 = \omega^2z_1+\omega^2 z_2 = \omega^2 y_1,\
&y_2=\omega z_1+\omega^2 z_2, &y_5 = \omega^2z_1+z_2=\omega y_2, \hspace{1cm}  &y_8 = z_1+\omega z_2=\omega^2 y_2,\
&y_3= \omega^2z_1+\omega z_2, &y_6 = z_1+\omega^2z_2 = \omega y_3 \hspace{1cm}  &y_9 = \omega z_1+z_2 = \omega^2 y_3.
\end{align*}

Let $t=\omega y$. Then
$y^3+py+q=0$ is equivalent to $\left ( \frac{t}{\omega} ight )^3+ p \left ( \frac{t}{\omega} ight )+q = 0$, so is equivalent to $t^3+\omega^2 p t +q=0$.

Consequently $y_4,y_5,y_6$ are the roots of the polynomial $y^3 + \omega^2 p y + q$.

For the same reason, $y_7,y_8,y_9$ are the roots of the polynomial $y^3 + \omega p y +q$.
\end{proof}

Exercise 1.1.7

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Exercise 1.1.7

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