Exercise 1.1.3

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}}

Show that Cardan's formulas give the roots of $y^3+py+q$ when $p=0$.

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}
We choose a root of $q^2$ by writing $\sqrt{q^2} = q$ (the other choice $-q$ only exchanges the two terms of the sum giving $y_1$, and exchanges  $y_2,y_3$).

If $p=0$, 
\begin{align*}
\sqrt[3]{-\frac{q}{2}+\sqrt{\left(\frac{q}{2}ight)^2 + \left(\frac{p}{3}ight)^3}}=\sqrt[3]{-\frac{q}{2}+\frac{q}{2}} &= 0,\
\sqrt[3]{-\frac{q}{2}-\sqrt{\left(\frac{q}{2}ight)^2 + \left(\frac{p}{3}ight)^3}}=\sqrt[3]{-\frac{q}{2}-\frac{q}{2}} &= \sqrt[3]{-q}.\
\end{align*}

Thus $y_1 = \sqrt[3]{-q}, y_2 = \omega^2 \sqrt[3]{-q}, y_3 = \omega \sqrt[3]{-q}$.

These are the roots of $y^3 + q=0$ : the Cardan's formulas, established for $p
eq 0$, remain true if $p=0$.
\end{proof}

Exercise 1.1.3

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Exercise 1.1.3

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