Exercise 9.6

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}}

Prove that
\begin{enumerate}
\item[(a)] $\chi_\gamma(\lambda) = 1$ for $\gamma \equiv 8, 8 + 3\omega, 8 + 6 \omega \ [9]$.
\item[(b)] $\chi_\gamma(\lambda) = \omega$ for $\gamma \equiv 5, 5 + 3\omega, 5 + 6 \omega \ [9]$.
\item[(c)]$\chi_\gamma(\lambda) = \omega^2$ for $\gamma \equiv 2, 2 + 3\omega, 2 + 6 \omega \ [9]$.
\end{enumerate}

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}
Here $\gamma = -1 + 3(m+n\omega)$ is a primary prime, and $\chi_\gamma(\lambda) = \omega^{2m}$.
\begin{align*}
\chi_\gamma(\lambda)=1 &\iff m\equiv0\ [3] \Rightarrow \gamma \equiv 8+3n\omega\ [9]\Rightarrow \gamma\equiv8,8+3\omega,8+6\omega\ [9]\
\chi_\gamma(\lambda)=\omega&\iff m\equiv2\ [3] \Rightarrow \gamma \equiv5+3n\omega\ [9]\Rightarrow \gamma\equiv5,5+3\omega,5+6\omega\ [9]\
\chi_\gamma(\lambda)=\omega^2& \iff m\equiv1\ [3] \Rightarrow \gamma \equiv2+3n\omega\ [9]\Rightarrow \gamma\equiv2,2+3\omega,2+6\omega\ [9]\
\end{align*}

As $\chi_\gamma(\lambda) \in \{1,\omega,\omega^2\}$, these 9 cases are the only possibilities. Moreover these 9 cases are mutually exclusive, since $9$ doesn't divide  any difference. Thus the reciprocals are true.
\begin{align*}
\chi_\gamma(\lambda)=1 &\iff \gamma\equiv8,8+3\omega,8+6\omega\ [9]\
\chi_\gamma(\lambda)=\omega&\iff \gamma\equiv5,5+3\omega,5+6\omega\ [9]\
\chi_\gamma(\lambda)=\omega^2&\iff \gamma\equiv2,2+3\omega,2+6\omega\ [9]\
\end{align*}
\end{proof}

Exercise 9.6

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Exercise 9.6

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