Exercise 9.26

Question

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

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{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Combine the previous two exercises to conclude that $\chi_\pi(1-\omega) = \omega^{2m}$.

Richard Ganaye · Accepted Answer

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

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{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Since $\pi$ and $a+b$ are primary elements of $D$, by Exercise 9.20,
$$\chi_\pi(a+b) = \chi_{a+b}(\pi).$$
By Exercises 9.24 and 9.25,
\begin{align*}
\chi_\pi(a+b) &= \omega^{2n} \chi_\pi(1-\omega)\
\chi_{a+b}(\pi) &= \omega^{2(m+n)}
\end{align*}
Thus $\omega^{2n} \chi_\pi(1-\omega) = \omega^{2(m+n)}$.
Consequently
$$\chi_\pi(1-\omega) = \omega^{2m}.$$
\end{proof}

Exercise 9.26

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Exercise 9.26

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