Exercise 1.34

Proof. As ${\omega }^3=1,\bar{\omega }={\omega }^2$, and $1+\omega +{\omega }^2=0$, so

$|1-\omega {\text{|}}^2=(1-\omega )(1-{\omega }^2)=1+{\omega }^3-\omega -{\omega }^2=3$, therefore

Consequently,

3 is divisible by ${(1-\omega )}^2$ in $\mathbb{Z}[\omega ]$.

Note: As $-{\omega }^2$ is an unit, $3$ and ${(1-\omega )}^2$ are associated. This shows that $3$ is not irreducible in $\mathbb{Z}[\omega ]$. □