# Previous step of the supplement to the law of cubic reciprocity

Let $\gamma$, $\rho\in\mathbb{Z[\omega]}$ be different primary irreducibles (i.e. $\gamma$, $\rho\equiv 2(3)$), where $\omega=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$. I have to prove that $\chi_{\gamma}(\rho)=\chi_{\rho}(\gamma)$, where $\chi_{\pi}(\alpha)$ is the cubic residue character.

Let $\lambda=1-\omega$, if $\gamma$, $\rho\equiv 2(3)$, then $\lambda$ does not divide $\gamma$ and $\rho$. If $N(\gamma)\neq N(\rho)$ then $\chi_{\gamma}(\rho)=\chi_{\rho}(\gamma)$ is a consequence of the cubic reciprocity law.

Under these conditions, if $N(\gamma)= N(\rho)$, then $\overline{\gamma}=\rho$. In the book of Ireland and Rosen is proved that $\overline{\chi_{\pi}(\alpha)}={\chi_{\overline{\pi}}(\overline{\alpha})}$, so the only thing left is to show that $\chi_{\gamma}(\rho)=1$. It is supposed to be an easy consequence of the properties of the cubic residue character (according to Kenneth S. Williams) but I don't get to prove it.

This question is implicitly on problem 20 in chapter 9 of Ireland and Rosen. Any help will be appreciated. If any of my guesses is wrong, a correction would be appreciated too.

If $\bar\gamma=\rho$ and $\gamma\neq\rho$ then $\gamma\rho$ is some rational prime $p=3k+1$, and $\rho \equiv \gamma+\rho \bmod \gamma$ with $\gamma+\rho = a$ the rational integer such that $a \equiv 1 \bmod 3$ and $4p = a^2 + 27b^2$ for some integer $b$. Now if you already know the neat formula $$a \equiv -\left({3k\atop k, k, k}\right) = \frac{-(3k)!}{k!^3} \bmod p$$ then you're done, because the numerator is $-(p-1)! \equiv 1$ by Wilson's theorem, and the denominator is a cube, so $a$ is a cube mod $p$ and thus modulo the factor $\gamma$ of $p$ as desired.