Let $\pi \in \mathbb{Z}[\omega]$ be a prime. Then $\frac{\mathbb{Z}[\omega]}{\pi \mathbb{Z}[\omega]}$ has $N(\pi)$ elements

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I am currently reading cubic and biquadratic reciprocity from Kenneth Ireland and Michael Rosen's, "A Classical Introduction to Modern Number Theory" have a doubt in proposition 9.2.1.

We take three cases to prove the second part of this proposition: (i) assume $\pi$ is a rational prime congruent to 2 modulo 3, (ii) assume that $p \equiv1\pmod3$ and $\pi \overline{\pi} = p$ , (iii) assume 1- $\omega$ is prime

I am unable to figure out how to begin to prove the third case. Can anyone give me a hint or some idea about how to proceed please? I also want to prove that the number of elements of $\frac{\mathbb{Z}[\omega]}{\pi \mathbb{Z}[\omega]}$ in the first case is not ${{0,1,...,p-1}}$.