Exercise 3.25

Proof. $\alpha \equiv 1(\mathit{mod}\lambda )$, so $\alpha =1+\mathit{\beta \lambda },\beta \in \mathbb{Z}[\omega ]$.

$\bar{\lambda }=1-{\omega }^2=(1-\omega )(1+\omega )=-{\omega }^2(1-\omega )=-{\omega }^2\lambda$ (so $\bar{\lambda }$ and $\lambda$ are associate).

Moreover,

since $\beta \equiv 0,1,-1(\mathit{mod}\lambda )$ (see Ex. 3.24).

Therefore ${\lambda }^4\mid {\alpha }^3-1$.

As $\lambda \bar{\lambda }=(1-\omega )(1-{\omega }^2)=1-\omega -{\omega }^2+{\omega }^3=3$, then $\lambda \bar{\lambda }=-{\omega }^2{\lambda }^2=3$, so ${\lambda }^2$ and $3$ are associate : ${\lambda }^2=-\omega 3$. Thus $9={(-{\omega }^2{\lambda }^2)}^2=\omega {\lambda }^4$, so $9\mid {\omega }^{2}9={\lambda }^4\mid {\alpha }^3-1$.

For all $\alpha \in \mathbb{Z}[\omega ]$,

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