I'm working on an exercise from Ireland and Rosen, and I want to know if I'm on the right track.
Let $\omega = \frac{ -1 + \sqrt{-3}}{2}$, and consider $\mathbb{Z}[\omega]$. Let $\gamma$ be a primary prime in $\mathbb{Z}[\omega]$, which means $\gamma \equiv 2 \pmod{3}$. I want to show that $(\omega / \gamma)_3$ is equal to $1, \omega, \omega^2$ depending on whether $\gamma \equiv 8, 2, 5 \pmod{3(1- \omega)}$.
We can express $\gamma$ as $\gamma = (3m -1) + (3n)\omega$, where $m,n \in \mathbb{Z}$. From an earlier exercise, we have that $(\omega / \gamma)_3 = \omega^{m+n}$. The current exercise's hint is to observe that $\gamma \equiv -1 + 3(m+n) \pmod{3(1 - \omega)}$, which I can derive.
If $\gamma \equiv 8 \pmod{3(1-\omega)}$, then $9 \equiv 3(m+n) \pmod{3(1- \omega)}$, but I'm not sure where to go from here. I would want to show that the cubic character is equal to $1$. I could say that $3(m+n) - 9 = 3(1 - \omega)k$ for some $k$. If I equate the coefficients of $\omega$, I would get that $k = 0$, and therefore, $m+n = 3$ which would get my desired result. However, I don't believe equating coefficients is valid because $k$ is not necessarily a rational integer.
I have the answer.
Suppose $\gamma \equiv 8 \pmod{3(1- \omega)}$. Then $0 \equiv 3(m+n) \pmod{3(1 - \omega)}$. There exists some $\alpha \in \mathbb{Z}[\omega]$ such that $3(m+n) = 3(1-\omega)\alpha$. Equivalently, $m+n = (1 - \omega) \alpha$. Taking the norm of both sides gives $(m+n)^2 = 3N(\alpha)$. So $3 | (m+n)^2$, which implies $3 | (m+n)$ since $3$ is prime. We conclude $m+n \equiv 0 \pmod{3}$, so $(\omega / \gamma)_3 = \omega^{m+n} = 1$ as desired.
Changing the congruence class of $\gamma$ to $2$ or $5$ gives a character of $\omega$ and $\omega^2$ using identical reasoning.