Exercise 8.22

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}}

(continuation) Prove that $\sum_x \zeta^{ax^d} = \sum_\chi g_a(\chi)$, where the sum is over all $\chi$ such that $\chi^d=\varepsilon, \chi 
e \varepsilon$. Assume that $p 
mid a$.

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 Exercise 8.21,
$$S = \sum_{x \in \F_p} \zeta^{ax^d} = \sum_{ r \in \F_p} m(r) \zeta^{ar}.$$

As  $d \mid p-1$, by Proposition 8.1.5,
$$m(r) = N(x^d=r) = \sum_{\chi^d = \varepsilon} \chi(r).$$

Therefore
 $$S = \sum_{r\in \F_p}\, \sum_{\chi^d = \varepsilon} \chi(r) \zeta^{ar} =\sum_{\chi^d = \varepsilon} \sum_{r\in \F_p} \chi(r) \zeta^{ar}.$$

If $\chi = \varepsilon, \sum\limits_{r\in \F_p} \chi(r) \zeta^{ar} = \sum\limits_{r\in \F_p} \zeta^{ar} = 0$, since $a
ot \equiv 0 \pmod p$.

By definition $g_a(\chi) =  \sum_r \chi(r) \zeta^{ar}$, so, if $d \mid p-1$, $p 
mid a$,

$$\sum_{x\in \mathbb{F}_p} \zeta^{a x^d} = \sum_{\chi^d=\varepsilon,\, \chi
eq \varepsilon} g_a(\chi).$$
\end{proof}

Exercise 8.22

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Exercise 8.22

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