Exercise 2.4.7

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}}

Explain how the third property of (2.31) was used (implicitly) in (2.28) in the proof of Proposition 2.4.1.

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}
Knowing that $	au \cdot \sqrt{\Delta} = -\sqrt{\Delta}$ for a transposition $	au\in S_n$, we show by induction on  $l$ that
$$(	au_l  \cdots 	au_1) \cdot \sqrt{\Delta} = (-1)^l\sqrt{\Delta}.$$
By the induction hypothesis $(	au_l  \cdots 	au_1). \sqrt{\Delta}= (-1)^l\sqrt{\Delta}$, we deduce, using 2.31
\begin{align*}
(	au _{l+1} 	au_l  \cdots 	au_1) \cdot \sqrt{\Delta}&= 	au _{l+1} \cdot [( 	au_l  \cdots 	au_1) \cdot \sqrt{\Delta}]\
&= 	au _{l+1}\cdot ((-1)^l \sqrt{\Delta})\
&=(-1)^l	au_{l+1} \cdot \sqrt{\Delta}\
&= (-1)^{l+1}\sqrt{\Delta}.
\end{align*}
\end{proof}

Exercise 2.4.7

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Exercise 2.4.7

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