Exercise 5.33

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}}

Use Proposition 5.3.2 to derive the quadratic character of $-1$.

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}
Let $f(z) = e^{2\pi i z}- e^{-2\pi i z}$. If $p$ is an odd prime, $a \in \Z$, and $p
mid a$, we know from Prop. 5.3.2 that
$$\prod_{l=1}^{(p-1)/2} f\left(\frac{la}{p} ight) = \legendre{a}{p} \prod_{l=1}^{(p-1)/2} f\left( \frac{l}{p} ight).$$
For $a = -1$, as $f(-z) = -f(z)$, 
\begin{align*}
\legendre{-1}{p} \prod_{l=1}^{(p-1)/2} f\left( \frac{l}{p} ight) &= \prod_{l=1}^{(p-1)/2} f\left(\frac{-l}{p} ight)\
&=(-1)^{(p-1)/2} \prod_{l=1}^{(p-1)/2} f\left( \frac{l}{p} ight)
\end{align*}
Moreover $f(z) = 0 \iff e^{4\pi i z} = 1 \iff 4\pi i z = 2 k i \pi, k \in \Z \iff z = k/2, k\in \Z$, so, if $l \in \Z$,
$f\left( \frac{l}{p} ight) = 0 \iff l/p = k/2, k \in \Z \iff p \mid 2l \iff p \mid l$. For $1\leq l <p$, this is impossible, so $\prod_{l=1}^{(p-1)/2} f\left( \frac{l}{p} ight) 
e 0$. Consequently,
$$\legendre{-1}{p} = (-1)^{(p-1)/2}.$$
\end{proof}

Exercise 5.33

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Exercise 5.33

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