Exercise 9.5

Question

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In the text we stated Eisenstein's result $\chi_\gamma(\lambda) = \omega^{2m}$. Show that $\chi_{\gamma}(3) = \omega^{2n}$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
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\begin{proof} Here $\gamma = (3m-1) + 3n\omega$.

Note that $(1-\omega)^2 = -3 \omega$, thus $\chi_\gamma((1-\omega)^2) = \chi_\gamma(-1)\chi_\gamma(3)\chi_\gamma(\omega)$.

Using Eisenstein's result (see a proof in Ex.24-26),
$$\chi_\gamma((1-\omega)^2)= \chi_\gamma(\lambda^2) =\chi_\gamma(\lambda)^2 =\omega^{4m} = \omega^m.$$

As $-1 = (-1)^3, \chi_\gamma(-1) = 1$. Finally  $\chi_\gamma(\omega) = \omega^{m+n}$ by Exercise 9.3. Thus

$$\omega^m = \chi_\gamma(3) \omega^{m+n},\qquad  \chi_\gamma(3) = \omega^{-n} = \omega^{2n}.$$

In conclusion,
$$\chi_\gamma(3) = \omega^{2n}.$$
\end{proof}

Exercise 9.5

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Exercise 9.5

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