Exercise 1.2.5

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}}

Since $\Delta = (x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2$, we can define the square root of $\Delta$ to be $\sqrt{\Delta} = (x_1-x_2)(x_1-x_3)(x_2-x_3)$. Prove that an even permutation of the roots takes $\sqrt{\Delta}$ to $\sqrt{\Delta}$ while an odd permutation takes $\sqrt{\Delta}$ to $-\sqrt{\Delta}$. In section 2.4 we will see that this generalizes nicely to the case of degree $n$.

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}
By definition,
$$\sqrt{\Delta} = (x_1-x_2)(x_1-x_3)(x_2-x_3).$$

If $\sigma \in S_3$, we define $$\sigma \cdot \sqrt{\Delta}  = (x_{\sigma(1)}-x_{\sigma(2)})(x_{\sigma(1)}-x_{\sigma(3)})(x_{\sigma(2)}-x_{\sigma(3)}).$$

If $	au = (1\ 2)$, 
 $$	au \cdot \sqrt{\Delta}  = (x_2-x_1)(x_2-x_3)(x_1-x_3) = -\sqrt{\Delta}.$$
 
Same result if $	au = (1\ 3)$, or $	au = (2\ 3)$.

If $\sigma = (1\  2\  3)$,
$$\sigma \cdot \sqrt{\Delta}  = (x_2-x_3)(x_2-x_1)(x_3-x_1) = \sqrt{\Delta},$$
with the  same result if $\sigma = (1 3 2)$ (and also if $\sigma$ is identity).

In conclusion,

$\sigma \cdot \sqrt{\Delta}  = \sqrt{\Delta} $ if $\sigma$ is even,

$\sigma \cdot \sqrt{\Delta}  =- \sqrt{\Delta} $ if $\sigma$ is odd.
\end{proof}

Exercise 1.2.5

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Exercise 1.2.5

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