Exercise 8.9

Question

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

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}{\mathrm{N}}

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Suppose that $p\equiv 1 \pmod 3$ and that $\chi$ is a character of order 3. Prove (using Exercise 5) that $g(\chi)^3 = p\pi$, where $\pi = \chi(2)J(\chi,ho)$.

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}}

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

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

\begin{proof}
As $\chi$ is a character of order 3, $\chi^2
eq \varepsilon$. From Exercise 5, we know that
$$g(\chi)^2 = \chi(2)^{-2} J(\chi,ho) g(\chi^2).$$
So
$$g(\chi)^3 = \chi(2)^{-2} J(\chi,ho) g(\chi^2)g(\chi).$$
Recall  (¤8.2) that
$$\overline{g(\chi)} = \sum_t \overline{\chi(t)} \zeta^{-t} = \chi(-1) \sum_t \overline{\chi(-t)} \zeta^{-t} = \chi(-1) g(\overline{\chi}),$$
Here $\chi(-1) = 1$, because $\chi(-1) = \chi((-1)^3) = \chi^3(-1) = \varepsilon(-1) = 1$.
Hence
$$g(\chi^2) g(\chi) = g(\bar{\chi}) g(\chi) =\overline{g(\chi)} g(\chi) = \vert g(\chi) \vert^2 =p.$$
Moreover $\chi(2)^3 = \chi^3(2) = 1$, so $\chi(2)^{-2} = \chi(2)$.

\medskip

Conclusion : if $\chi$ is a character of order 3,
 $$g(\chi)^3 = p \pi, \ \mathrm{where}\ \pi = \chi(2) J(\chi,ho).$$
\end{proof}

Exercise 8.9

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Exercise 8.9

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