Exercise 8.21

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}}

Suppose that $p\equiv 1 \pmod d$, $\zeta = e^{2\pi i/p}$, and consider $\sum_x \zeta^{ax^d}$. Show that $\sum_x \zeta^{ax^d} = \sum_r m(r)\zeta^{ar}$, where $m(r) = N(x^d = r)$.

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}
Let $A_r = \{x \in \mathbb{F}_p \ \vert \ x^d=r\}$

Then  $\mathbb{F}_p = \bigcup_r A_r$, thus

$$\sum\limits_{x\in \mathbb{F}_p} \zeta^{ax^d} = \sum\limits_{r \in \mathbb{F}_p} \sum\limits_{x \in A_r} \zeta^{a x^d} = \sum\limits_{r \in \mathbb{F}_p} \vert A_r\vert \zeta^{ar}= \sum\limits_{r \in \mathbb{F}_p} m(r)\zeta^{ar},$$

where $m(r) = \vert A_r\vert = N(x^d=r)$.
\end{proof}

Exercise 8.21

Answers

Comments

Exercise 8.21

Answers

Comments

Add answer