Exercise 1.3.6

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}}

Prove (1.24) : $z_1z_2 = -\frac{p}{3} \Rightarrow z_2 = \overline{z_1}$.

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}
In the context of paragraph 1.3.A, 
$$z_1 = \sqrt[3]{\frac{1}{2}\left(-q+i\sqrt{\frac{\Delta}{27}} ight)}, z_2 = \sqrt[3]{\frac{1}{2}\left(-q-i\sqrt{\frac{\Delta}{27}} ight)}, \qquad p,q \in \R,\  \Delta>0.$$

We know $z_1z_2 = -\frac{p}{3}$.

Then $z_1^3 = \overline{z_2}^3$, so $z_1 = \omega^k \, \overline{z_2},\ k=0,1,2$. Therefore

$$-\frac{p}{3} = z_1 z_2 = \omega^k \, \overline{z_2} z_2 =\omega^k \vert z_2 \vert ^2.$$

As $z_2 
eq 0$, $\omega^k = -\frac{p}{3} \frac{1}{\vert z_2 \vert ^2} \in \mathbb{R}$, which implies $k=0$, therefore $z_2 = \overline{z_1}$.
\end{proof}

Exercise 1.3.6

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Exercise 1.3.6

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