Exercise 15.5.2

Question

\def\imp{\Rightarrow}
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\def\dcro{\mbox{]\hspace{-.15em}]}}

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ewcommand{\ssi} {si et seulement si }

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As in the proof of Theorem 15.5.1, let $u_0 = \varphi \left(\frac{\varpi}{n} ight)$, and assume that $\sigma \in \Gal\left(L/\Q[i]ight)$ satisfies $\sigma(u_0) = \varphi\left(\alpha 
\frac{\varpi}{n} ight)$, where $\alpha \in \Z[i]$ is odd. Use the multiplication formula for $\beta \in \Z[i]$ odd to prove (15.65):
$$\sigma \left(\varphi\left(\beta \frac{\varpi}{n} ight) ight) = \varphi\left(\alpha \beta \frac{\varpi}{n} ight).$$

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} $\sigma$ fixes the elements of $\Q[i]$, thus $\sigma(i^\varepsilon) = i^\varepsilon$, and since $P_\beta(u), Q_\beta(u) \in \Z[i][u]$, for all $\zeta  \in L$,
$$\sigma \left(\frac{P_\beta(\zeta)}{Q_\beta(\zeta) } ight)= \frac{P_\beta(\sigma(\zeta))}{Q_\beta(\sigma(\zeta))}.$$
Starting from the multiplication formula for $\beta \in \Z[i]$ odd, we obtain
$$\varphi\left(\beta \frac{\varpi}{n}ight) = i^\varepsilon \varphi\left(\frac{\varpi}{n}ight) \frac{P_\beta\left(\varphi^4(\frac{\varpi}{n})ight))}{Q_\beta\left(\varphi^4(\frac{\varpi}{n})ight))}.$$
Since $\sigma \left(\varphi \left(\frac{\varpi}{n} ight)ight) = \varphi\left(\alpha \frac{\varpi}{n} ight)$, using the multiplication formula for $\beta$ anew,
\begin{align*}
\sigma \left(\varphi\left(\beta \frac{\varpi}{n} ight) ight)&= \sigma(i^\varepsilon)\sigma\left( \varphi\left(\frac{\varpi}{n}ight)ight) \sigma \left( \frac{P_\beta\left(\varphi^4(\frac{\varpi}{n})ight)) }{Q_\beta\left(\varphi^4(\frac{\varpi}{n})ight))} ight)\
&= i^\varepsilon \varphi\left(\alpha \frac{\varpi}{n}ight) \frac{P_\beta\left(\varphi^4(\alpha \frac{\varpi}{n})ight))}{Q_\beta\left(\varphi^4( \alpha \frac{\varpi}{n})ight))}\
&= \varphi\left( \beta\left ( \alpha  \frac{\varpi}{n} ight)ight) 
\end{align*}
We have proved
$$\sigma \left(\varphi\left(\beta \frac{\varpi}{n} ight) ight) = \varphi\left(\alpha \beta \frac{\varpi}{n} ight).$$
\end{proof}

Exercise 15.5.2

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Exercise 15.5.2

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