Exercise 3.24

Proof. Same definition of congrence in $\mathbb{Z}[\omega ]$ as in Ex. 3.23.

$\omega \equiv 1(\mathit{mod}1-\omega )$, so $a+\mathit{b\omega }\equiv a+b(\mathit{mod}1-\omega )$.

$0=1-{\omega }^3=(1-\omega )(1+\omega +{\omega }^2)$, with $1-\omega \ne 0$, so $1+\omega +{\omega }^2=0$. Hence $3\equiv 0(\mathit{mod}1-\omega )$.

$a+b\equiv 0,1,-1(\mathit{mod}3)$, so $a+b\equiv 0,1,-1(\mathit{mod}1-\omega )$

For all $z\in \mathbb{Z}[\omega ],z\equiv 0,1,-1(\mathit{mod}1-\omega )$. □