Exercise 1.1.8

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}}

Use Cardan's formulas to solve $y^3+3\omega y+1 = 0$.

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 equation $y^3+3\omega y +1 =0$,  with the substitution $$t = \omega y, y =\omega^2 t,$$ becomes

$$t^3+3t+1 = 0.$$

With $p/3 = 1, q/2 = 1/2$, the Cardan's formulas give
\begin{align*}
z_1 &= \sqrt[3]{-\frac{q}{2}+\sqrt{\left(\frac{q}{2}ight)^2 + \left(\frac{p}{3}ight)^3}}\
z_1 &=\sqrt[3]{\frac{-1+\sqrt{5}}{2}}\
z_2 &=\sqrt[3]{\frac{-1-\sqrt{5}}{2}}\
\end{align*}

The roots of $t^3+3t+1$ are  $z_1+z_2,\omega z_1+\omega^2 z_2,\omega^2 z_1 + \omega z_2$.

Thus the roots of  $y^3+3\omega y + 1$ are $\omega^2 z_1 + \omega^2 z_2,z_1 + \omega z_2, \omega z_1 + z_2$.

The solutions of  $y^3+3\omega y +1 =0$ are
\begin{align*}
&y_1 = \omega^2  \sqrt[3]{\frac{-1+\sqrt{5}}{2}} + \omega ^2 \sqrt[3]{\frac{-1-\sqrt{5}}{2}},\
&y_2 =  \sqrt[3]{\frac{-1+\sqrt{5}}{2}} + \omega \sqrt[3]{\frac{-1-\sqrt{5}}{2}},\
&y_3 = \omega \sqrt[3]{\frac{-1+\sqrt{5}}{2}} + \sqrt[3]{\frac{-1-\sqrt{5}}{2}}.\
\end{align*}
\end{proof}

Exercise 1.1.8

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Exercise 1.1.8

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