Exercise 8.24

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) Let $f(x_1,x_2,\ldots,x_n) = a_1x_1^{m_1}+a_2x_2^{m_2}+\cdots+a_n x_n^{m_n}$. Let $d_i = (m_i,p-1)$. Show that $N = p^{n-1} + p^{-1} \sum_{a
eq 0} \prod_{i=1}^n \sum_{\chi_i} g_{aa_i}(\chi_i)$ where $\chi_i$ runs over all characters such that $\chi_i^{d_i} = \varepsilon$ and $\chi_i 
eq \varepsilon$.

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 8.2, $$N = N(a_1 x_1^{m_1}+\cdots a_n x_n^{m_n} = 0) = N(a_1 x_1^{d_1}+\cdots a_n x_n^{d_n} = 0),$$where $d_i = m_i \wedge (p-1)$ divides $p-1$.

By Exercise 8.23, 
$$N = p^{n-1}+p^{-1} \sum\limits_{a\in \mathbb{F}_p^*} \sum\limits_{(x_1,x_2,\cdots,x_n) \in \mathbb{F}_p^n} \zeta^{a(a_1 x_1^{d_1}+\cdots a_n x_n^{d_n} )}$$

By Exercise 8.22, since $p 
mid a, p
mid a_i$,
\begin{align*}
\sum\limits_{(x_1,x_2,\cdots,x_n) \in \mathbb{F}_p^n} \zeta^{a(a_1 x_1^{d_1}+\cdots a_n x_n^{d_n} )} &=\left( \sum\limits_{x_1 \in \mathbb{F}_p} \zeta^{aa_1 x_1^{d_1}}ight)\cdots\left(\sum\limits_{x_n \in \mathbb{F}_p} \zeta^{aa_n x_n^{d_n}}ight)\
&=\left(\sum\limits_{\chi_1^{d_1} = \varepsilon,\, \chi_1
eq \varepsilon} g_{aa_1}(\chi_1)ight)\cdots\left(\sum\limits_{\chi_n^{d_n}= \varepsilon,\,\chi_n 
eq \varepsilon} g_{aa_n}(\chi_n)ight)\
&=\prod\limits_{i=1}^n \sum\limits_{\chi_i^{d_i}= \varepsilon,\, \chi_i 
eq \varepsilon} g_{aa_i}(\chi_i)
\end{align*}
In conclusion,

$$N = p^{n-1}+p^{-1} \sum\limits_{a\in \mathbb{F}_p^*}  \prod\limits_{i=1}^n \sum\limits_{\chi_i^{d_i} = \varepsilon,\chi_i 
eq \varepsilon} g_{aa_i}(\chi_i).$$
\end{proof}

Exercise 8.24

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Exercise 8.24

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