Exercise 8.25

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}}

Deduce from Exercise 8.24 that $$|N-p^{n-1}| \leq (p-1)(d_1-1)\cdots(d_n-1)p^{(n/2)-1}.$$

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}
As $\vert g_{aa_i}(\chi_i)  \vert= \sqrt{p}$,

$$\left\vert\sum\limits_{\chi_i^{d_i} = \varepsilon,\,\chi_i 
eq \varepsilon} g_{aa_i}(\chi_i)ight\vert \leq \sqrt{p} \ n_i,$$ where $n_i = \mathrm{Card}\,\{\chi_i 
eq \varepsilon \ \vert\  \chi_i^{d_i} = \varepsilon \}$.

As $d_i \mid p-1$, there exists exactly $d_i$ characters of order dividing $d_i$, so $n_i = d_i-1$ :

$$\left | \sum\limits_{\chi_i^{d_i} = \varepsilon,\,\chi_i 
eq \varepsilon} g_{aa_i}(\chi_i)ight | \leq \sqrt{p} (d_i-1).$$

By Exercise 8.24,
 $$N = p^{n-1}+p^{-1} \sum\limits_{a\in \mathbb{F}_p^*}  \prod\limits_{i=1}^n \sum\limits_{\chi_i^{d_i} = \varepsilon,\chi_i 
eq \varepsilon} g_{aa_i}(\chi_i)$$
 
so $$\vert N - p^{n-1}\vert  \leq p^{-1} (p-1) \sqrt{p}(d_1-1)\cdots \sqrt{p}(d_n-1),$$ that is
 
$$\vert N - p^{n-1}\vert  \leq (p-1)(d_1-1)\cdots(d_n-1) p^{\frac{n}{2}-1}.$$
\end{proof}

Exercise 8.25

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Exercise 8.25

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