Exercise 9.20

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}}

If $\gamma$ and $ho$ are primary, show that $\chi_\gamma(ho) = \chi_ho(\gamma)$.

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

Important note: The following solution assumes that $\chi_{\pi_2}(\pi_1) = \chi_{\pi_1}(\pi_2)$ for any pair $\pi_1,\pi_2$ of primary primes. But Theorem 1 uses the hypothesis $N(\pi_1) 
e N(\pi_2)$ to prove Cubic Reciprocity.

We can complete the proof in the case where $N(\pi_1) = N(\pi_2)$. Since $\pi_1,\pi_2$ are primary primes, then $\pi_1 = \pi_2$ or $\pi_1 = \overline{\pi_2}$.

In the case $\pi_1 = \pi_2$, then $\chi_{\pi_2}(\pi_1) = 0 =  \chi_{\pi_1}(\pi_2)$.

To prove that $\legendre{\overline{\pi}}{\pi} = \legendre{\pi}{\overline{\pi}}$, we begin with a particular case of the proposition:

\medskip
{\bf Lemma.} {\it Let $n\in \Z$ be a primary element in $A$, and let $\pi$ be a primary prime such that $N(\pi) = p \equiv 1 \pmod 3$. Then
$$\legendre{n}{\pi}_3 = \legendre{\pi}{n}_3.$$
}

\begin{proof}
If $p \mid n$, then $\legendre{n}{\pi}_3 =0 =  \legendre{\pi}{n}_3.$ Now we assume that $p \wedge n = 1$.

The decomposition of $n$ is of the form
\begin{align*} n &= \pm p_1 \cdots p_s q_1 \cdots q_r\ (p_i \equiv 1 \ [3], q_j \equiv -1 \ [3])\
&= \pm \pi_1\overline{\pi_1} \cdots \pi_s \overline{\pi}_s q_1 \cdots q_r,
\end{align*}
where $\pi_i, \overline{\pi_i} (1\leq i \leq s)$ and $q_j (1 \leq j \leq r)$ are primary prime.

Since $N(\pi_i) = p_i 
e p$ and $N(\pi) = p 
e N(q_j) =q_j^2$, Theorem 1 shows that
\begin{align*}
\legendre{n}{\pi}_3 &=\legendre{\pi_1}{\pi}_3\legendre{\overline{\pi_1}}{\pi}_3\cdots \legendre{\pi_s}{\pi}_3\legendre{\overline{\pi_s}}{\pi}_3 \legendre{q_1}{\pi}_3  \cdots \legendre{q_r}{\pi}_3 \
&=\legendre{\pi}{\pi_1}_3\legendre{\pi}{\overline{\pi_1}}_3\cdots \legendre{\pi}{\pi_s}_3\legendre{\pi}{\overline{\pi_s}}_3 \legendre{\pi}{q_1}_3  \cdots \legendre{\pi}{q_r}_3 \
&= \legendre{\pi}{n}_3.
\end{align*}
\end{proof}

We can now remove the useless hypothesis $N(\pi_1) 
e N(\pi_2)$ in Theorem 1.

\medskip

{\bf Proposition.}
Let $\pi_1,\pi_2$ be primary primes. Then
$$\legendre{\pi_2}{\pi_1}_3 = \legendre{\pi_1}{\pi_2}_3.$$

\begin{proof}
By theorem 1, it remains only the case where $N(\pi_1) = N(\pi_2)$.

If $\pi_1 = \pi_2$, then $\legendre{\pi_2}{\pi_1}_3 = \legendre{\pi_1}{\pi_2}_3 =0$.

If $\pi_1 
e \pi_2$, since $\pi_1$ et $\pi_2$ are primary, then $\pi_1, \pi_2$ are primes such that $N(\pi_1) = N(\pi_2) = p \equiv 1 \pmod 3$, and $\pi_2 = \overline{\pi_1}$. Writing $\pi = \pi_1$, it is sufficient to prove
$$\legendre{\overline{\pi}}{\pi}_3 =  \legendre{\pi}{\overline{\pi}}_3.$$
We use the ``Evans' trick'' (see [Lemmermayer, Reciprocity Laws p. 215]). The element $n =-( \pi + \overline{\pi})$ is a  rationnal integer, which is primary. The Lemma gives then
\begin{align*}
\legendre{\overline{\pi}}{\pi}_3 &= \legendre{\pi + \overline{\pi}}{\pi}_3\
&= \legendre{-\pi - \overline{\pi}}{\pi}_3\
&= \legendre{\pi}{-\pi - \overline{\pi}}_3\
&= \legendre{-\overline{\pi}}{-\pi - \overline{\pi}}_3\
&=\legendre{\overline{\pi}}{-\pi - \overline{\pi}}_3\
&=\legendre{-\pi  -\overline{\pi}}{\overline{\pi}}_3\
&=\legendre{-\pi}{\overline{\pi}}_3\
&=\legendre{\pi}{\overline{\pi}}_3
\end{align*}
\end{proof}

We can now give the solution of Exercise 9.20.
\begin{proof}

$ho, \gamma$ are written
\begin{align*}
ho&=\pmho_1ho_2\cdotsho_l,\
\gamma &= \pm \gamma_1 \gamma_2\cdots \gamma_m, 
\end{align*}
 where $ho_i,\gamma_i$ are primary primes.
By the law of Cubic Reciprocity, we obtain
\begin{align*}
\chi_\gamma(ho) &= \prod_{j=1}^m \chi_{\gamma_j}(ho)\
& = \prod_{j=1}^m\prod_{i=1}^l \chi_{\gamma_j}(ho_{i})\
& = \prod_{i=1}^l\prod_{j=1}^m \chi_{\gamma_j}(ho_{i})\
& = \prod_{i=1}^l\prod_{j=1}^m \chi_{ho_i}(\gamma_j)\
&=\prod_{i=1}^l \chi_{ho_i}(\gamma)\
&=\chi_{ho}( \gamma).
\end{align*}

(if $\gamma = -1$, or $ho = -1$, some products are empty, but the result remains true: $\chi_{-1}(ho) = 1 = \chi_{ho}(-1)$.)

\end{proof}

Exercise 9.20

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Exercise 9.20

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