Exercise 9.33

Question

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

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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}}

Let $q$ be a positive prime, $q\equiv 3 \pmod 4$. Show $\chi_q(1+i) = i^{(q+1)/4}$. [Hint : $(1+i)^{q-1} \equiv -i \pmod q$.]

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}}

\bigskip

The sentence is false and must be replaced by
$$ \chi_q(1+i) = (-i)^{(q+1)/4 } = i^{-(q+1)/4} .$$
We verify this on the example $q=11$ :
\begin{align*}
\chi_{q}(1+i) &\equiv (1+i)^{(q^2-1)/4}\
& \equiv (1+i)^{30}\
&\equiv -2^{15} i \equiv -32 i \equiv i \pmod {11},
\end{align*}
so $\chi_{11}(1+i) = i$, and $i^{(-q-1)/4} = i^{-3} = i$ (but $i^{(q+1)/4} = -i$).

\begin{proof}

Write $q = 4k + 3, k \in \N$.

As $(1+i)^2 = 2i$, $(1+i)^{q-1} =  (2i)^{(q-1)/2}$.

$2^{(q-1)/2} \equiv \legendre{2}{q} \ [q]$ and $ \legendre{2}{q} = (-1)^{(q^2-1)/8} = (-1)^{2k^2+3k+1} = (-1)^{k+1}$

$i^{(q-1)/2} = i^{2k+1} = (-1)^k i$.

So
$$ (1+i)^{q-1} \equiv -i\ [q].$$
$N(q) = q^2$, so $\chi_q(1+i) \equiv (1+i)^{(q^2-1)/4} = [(1+i)^{q-1}]^{(q+1)/4}\equiv (-i)^{(q+1)/4} \ [q]$ :

$$\chi_q(1+i) =(-i)^{(q+1)/4} = i^{(-q-1)/4}.$$
\end{proof}

Exercise 9.33

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Exercise 9.33

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