Exercise 9.30

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}}

Use the preceding two exercises to show $\chi_\pi(\overline{\pi}) = \chi_\pi(2) (-1)^{(a^2-1)/8}$.

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}
By Exercises 9.28, 9.29, and $\chi_\pi(-1) =(-1)^{(a-1)/2}$ (Prop. 9.8.3(d)),
\begin{align*}
\chi_{\pi}(\overline{\pi})&=  \chi_\pi(2) \chi_\pi(a)\
&=\chi_\pi(2) \chi_\pi(a(-1)^{(p-1)/4})(\chi_\pi(-1))^{(p-1)/4}\
&=\chi_\pi(2) (-1)^{(a^2-1)/8} ((-1)^{(a-1)/2})^{(p-1)/4}\
&=\chi_\pi (-2)(-1)^{(a^2-1)/8} ((-1)^{(a-1)/2})^{(p+3)/4}\
&=\chi_\pi(-2) (-1)^{(a^2-1)/8}  (-1)^{((a-1)/2)\,((p+3)/4)}.
\end{align*}

If $a \equiv 1 \pmod 4$, then $(-1)^{(a-1)/2}=1$.

If $a \equiv 3 \pmod 4$, then $b \equiv 2 \ [4]$ : $$a=4A+3,b=4B+2,p+3 = a^2+b^2+3 = (4A+3)^2+(4B+2)^2+3 \equiv 0 \ [8],$$ so $(p+3)/4 \equiv 0 \ [2].$

In both cases  $  (-1)^{((a-1)/2)\,((p+3)/4)}=1$, and so

$$\chi_{\pi}(\overline{\pi})=\chi_\pi(-2) (-1)^{(a^2-1)/8}.  $$
\end{proof}

Exercise 9.30

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Exercise 9.30

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