Exercise 9.1

Proof. Let $\lambda =1-\omega$, and $\alpha =a+\mathit{b\omega }\in D=\mathbb{Z}[\omega ],a,b\in \mathbb{Z}$.

$\omega \equiv 1(\mathit{mod}\lambda )$, so $\alpha \equiv a+b(\mathit{mod}\lambda )$, $\alpha \equiv c$ with $c=a+b\in \mathbb{Z}$.

$c\equiv 0,1,-1(\mathit{mod}3)$, and since $\lambda \mid 3$, $c\equiv 0,1,-1(\mathit{mod}\lambda )$.

Every $\alpha \in D$ is congruent to either $0,1$, or $-1$ modulo $\lambda =1-\omega$.

The classes of $0,1,-1$ in $D∕\mathit{\lambda D}$ are distinct. Indeed, $1\not\equiv -1(\mathit{mod}\lambda )$, if not $\lambda \mid 2$, so $2=\lambda {\lambda }^{\prime }$, $N(2)=N(\lambda )N({\lambda }^{\prime })$, thus $4=3N({\lambda }^{\prime })$, so $3\mid 4$, which is nonsense.

$\pm 1\equiv 0(\mathit{mod}\lambda )$ implies $\lambda \mid 1$, so $\lambda$ would be a unit, in contradiction with $\lambda$ prime.

So there exist exactly three classes modulo $\lambda$ in $D$ : $|D∕\mathit{\lambda D}|=3=N(\lambda )$. □