Exercise 11.18

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}}

Let the notation be as in Exercise 16. Use the Hasse-Davenport relation to show that $J(\chi'_1,\chi'_2,\ldots,\chi'_n) = (-1)^{(s-1)(n-1)} J(\chi_1,\chi_2,\ldots,\chi_n)^s$, where the $\chi_i$ are non trivial characters of $F$ and $\chi_1\chi_2\cdots \chi_n 
e \varepsilon$.

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} Note that $(\chi ho)' = \chi' ho'$, thus $(\chi_1 \ldots \chi_n)'= \chi'_1 \chi'_2\cdots \chi'_n$.

The conditions on the characters, and Exercise 16,  show that $\chi'_i 
e \varepsilon$ and $\chi'_1\chi'_2\cdots \chi'_n 
e \varepsilon$. By Theorem 3 of Chapter 8,
$$J(\chi'_1,\ldots,\chi'_n) = \frac{g(\chi'_1)g(\chi'_2) \cdots g(\chi'_n)}{g(\chi'_1 \chi'_2\cdots \chi'_n)}.$$
Then the Hasse-Davenport relation gives
\begin{align*}
J(\chi'_1,\ldots,\chi'_n) &= \frac{[-(-g(\chi_1))^s] [-(-g(\chi_2))^s] \cdots[- (-g(\chi_n))]^s]}{-(-g(\chi_1 \chi_2 \cdots \chi_n))^s}\
&=(-1)^{n-1}(-1)^{s(n-1)} \left(\frac{g(\chi_1)g(\chi_2) \cdots g(\chi_n)}{g(\chi_1 \chi_2\cdots \chi_n)}ight)^s\
&=(-1)^{(s+1)(n-1)}  J(\chi_1,\ldots,\chi_n)^s\
&=(-1)^{(s-1)(n-1)}  J(\chi_1,\ldots,\chi_n)^s.\
\end{align*}
\end{proof}

Exercise 11.18

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Exercise 11.18

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