Prove $1-ω$ is prime in $\mathbb{Z}[ω]$ Let $p$ be an odd prime and $ω=e^{2\pi i /p}$. Determine if $1-ω$ is prime in $\mathbb{Z}[ω]=\mathbb{Z}+\mathbb{Z}ω+\mathbb{Z}ω^2+...+\mathbb{Z}ω^{p-2}$
My attempt
I have tried using the definition of prime and also tried to show $\langle 1-\omega \rangle$ is maximal but I end up with more unknowns than equations.
I know the norm of the ideal is $N(\langle 1-\omega \rangle)=|1-\omega|^{p-2}$.
 A: One way to show an ideal $I$ is maximal is to show that $R/I$ is a field. This approach avoids norms.
(1) Given $$a=\sum_{i=0}^{p-2}a_i\omega^i,$$ show that $$a-(a_0+a_1+\dots+a_{p-2})\in\langle 1-\omega\rangle.$$
(2) Use this to show that $$\mathbb Z[\omega]/\langle 1-\omega\rangle\cong \mathbb Z/\langle q\rangle,$$ where $\langle q\rangle=\mathbb Z\cap \langle1-\omega\rangle.$
(3) Show that $\langle 1-\omega\rangle\cap\mathbb Z$ contains a prime, so $q$ is either that prime or a unit.
(4) Show that $1-\omega$ is not a unit, so neither is $q,$ and thus $q$ is prime and the quotient ring is a field.

The easiest way to prove the last step is to use norms, but it can be proved directly.
If $$1=(1-\omega)a=a_0+a_{p-2}+\sum_{i=1}^{p-2} (a_i-a_{i-1}+a_{p-2})\omega^i$$
Then $a_{p-2}=1-a_0$ and $a_i=a_{i-1}-a_{p-2}=a_{i-1}+(a_0-1).$ So $a_i=(i+1)a_0-i$ for all $i.$
That gives you $1-a_0=a_{p-2}=(p-1)a_0-(p-2)$ or $a_0=\frac{p-1}{p},$ which is not an integer.
A: The minimal polynomial of $\omega$ is $\frac{X^p-1}{X-1}$, thus the minimal polynomial of $\omega-1$ is $\frac{(X+1)^p-1}{X}=X^{p-1}+pX^{p-2}+\binom{p}{2}X^{p-3}+\dots+p$. The constant term in the minimal polynomial is (up to a sign) the norm, so $\omega-1$ has norm $\pm p$, hence it is prime, because the absolute value of the norm is the order of $\Bbb Z[\omega]/(1-\omega)$ and the only ring of order $p$ is $\Bbb F_p,$ which is a field.
