Exercise 8.8

Question

\def\imp{\Rightarrow}
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Generalize Exercise 3 in the following way. Suppose that $p$ is a prime, $\sum_t \chi(1-t^m) = \sum_{\lambda} J(\chi,\lambda)$, where $\lambda$ varies over all characters such that $\lambda^m = \varepsilon$. Conclude that $\left | \sum_t \chi(1-t^m) ight | \leq (m-1)p^{1/2}$.

Richard Ganaye · Accepted Answer

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

ewcommand{\D}{\mathrm{d}}

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

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

\begin{proof}
For all $y \in \F_p$, write  $A_y =\{x \in \mathbb{F}_p\ \vert \ x^m = y\}$. Then $\vert A_y\vert = N(x^m = y)$.

$\mathbb{F}_p = \bigcup_{y \in \mathbb{F}_p} A_y$ is the disjoint union of the $A_y$:

\begin{enumerate}
\item[$\bullet$] if $x \in A_y \cap A_{y'}$, then $x^m = y = y'$. This proves
$$y 
e y' \Rightarrow A_y \cap A_{y'} = \varnothing.$$
\item[$\bullet$] Every $x \in \F_p$ satisfies $x \in A_{x^m}$, thus
$$\bigcup_{y \in \F_p} A_y = \F_p.$$
\end{enumerate}
(Note that some $A_y$ may be empty.)

Therefore
$$\sum_{t\in \mathbb{F}_p} \chi(1-t^m) = \sum_{y\in \mathbb{F}_p} \sum\limits_{t \in A_y} \chi(1-t^m) = \sum_{y\in \mathbb{F}_p} \vert A_y \vert \chi(1-y) =\sum_{y\in \mathbb{F}_p} N(x^m = y) \chi(1-y).$$
Moreover, $N(x^m=y) = \sum\limits_{\lambda^m = \varepsilon} \lambda(y)$ (Prop. 8.1.5), so
\begin{align*}
\sum_{t\in \mathbb{F}_p} \chi(1-t^m) &= \sum_{y\in \mathbb{F}_p}  \sum_{\lambda^m = \varepsilon} \lambda(y)  \chi(1-y)\
& =  \sum\limits_{\lambda^m = \varepsilon}  \sum\limits_{x+y=1} \chi(x) \lambda(y)\
& = \sum\limits_{\lambda^m = \varepsilon} J(\chi,\lambda).
\end{align*}
Conclusion : $$\sum\limits_{t\in \mathbb{F}_p} \chi(1-t^m) = \sum\limits_{\lambda^m = \varepsilon} J(\chi,\lambda).$$

We know that there exist $m$ characters whose order divides $m$. We know that $\chi 
e \varepsilon$, $J(\chi,\varepsilon)=0$, and  $\vert J(\chi,\lambda) \vert = \sqrt{p}$ for every $\lambda 
e \varepsilon,\lambda
e \chi^{-1}$ (Theorem 1 and Corollary).

Moreover, by Theorem 1(c), $|J(\chi,\chi^{-1})|  = 1 \leq \sqrt{p}$, so
$$\left \vert \sum_{t\in \mathbb{F}_p} \chi(1-t^m)ight  \vert \leq \sum_{\lambda^m = \varepsilon, \lambda 
eq \varepsilon} \vert J(\chi,\lambda) \vert \leq  (m-1)\sqrt{p}.$$
\end{proof}

Exercise 8.8

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Exercise 8.8

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