Exercise 9.18

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}}

(continuation) If $\gamma = \pm \gamma_1 \gamma_2\cdots \gamma_t$ is a primary decomposition of the primary element $\gamma$, define $\chi_{\gamma}(\alpha) = \chi_{\gamma_1}(\alpha)\chi_{\gamma_2}(\alpha)\cdots\chi_{\gamma_t}(\alpha)$. Prove that $\chi_{\gamma}(\alpha) = \chi_{\gamma}(\beta)$ if $\alpha \equiv \beta \pmod \gamma$ and $\chi_{\gamma}(\alpha \beta) = \chi_\gamma(\alpha) \chi_\gamma(\beta)$. If $ho$ is primary, show that $\chi_ho(\alpha) \chi_\gamma(\alpha) = \chi_{-ho \gamma}(\alpha)$.

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}
If $\alpha \equiv \beta\ [\gamma]$, then $\alpha \equiv \beta\pmod {\gamma_i}, 1 \leq i \leq t$, so $\chi_{\gamma_i}(\alpha) = \chi_{\gamma_i}(\beta)$, thus $\chi_\gamma(\alpha) = \chi_\gamma(\beta)$.

By Proposition 9.3.3,
\begin{align*}
\chi_\gamma(\alpha \beta) &= \chi_{\gamma_1}(\alpha \beta)\chi_{\gamma_2}(\alpha \beta)\cdots\chi_{\gamma_t}(\alpha \beta)\
&= \chi_{\gamma_1}(\alpha)\chi_{\gamma_2}(\alpha)\cdots\chi_{\gamma_t}(\alpha)\chi_{\gamma_1}(\beta)\chi_{\gamma_2}(\beta)\cdots\chi_{\gamma_t}(\beta)\
&= \chi_\gamma(\alpha) \chi_{\gamma}(\beta)
\end{align*}
Finally, if $ho = \pm ho_1ho_2\cdotsho_l$ is primary, then $-ho \gamma = \pm ho_1ho_2\cdotsho_l\gamma_1\gamma_2\cdots\gamma_t$ is primary by Ex. 9.17, therefore

$$\chi_{-ho \gamma}(\alpha) = (\chi_{ho_1} \chi_{ho_2}\cdots \chi_{ho_l}\chi_{\gamma_1} \chi_{\gamma_2}\cdots \chi_{\gamma_t} )(\alpha)= \chi_ho(\alpha)\chi_\gamma(\alpha).$$
\end{proof}

Note: The unit $-1$ is primary by definition, and $-1$ is the opposite of the empty product, so for all $\alpha$ in $D$, $\chi_{-1}(\alpha) = 1$ by definition. The result of the exercises remain true if we accept the unit $-1$ as a primary element.

Exercise 9.18

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Exercise 9.18

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