Exercise 9.8

Proof. $7=N(2+3\omega )$, thus $7=(2+3\omega )(2+3{\omega }^2)=(2+3\omega )(-1-3\omega )$, where $2+3\omega$ and $-1-3\omega$ are primes in $D$, since their norm is a prime integer. Since these primes are primary, they are not associate.

$21=3\times 7=-{\omega }^2{\lambda }^2(2+3\omega )(-1-3\omega )$ since $3=-{\omega }^2{(1-\omega )}^2$.

$45=3^2\times 5=\omega {\lambda }^{4}5$, where $5\equiv 2(\mathit{mod}3)$ is a primary prime in $D$.

$22=2\times 11$ , where 2 and 11 are primes in $D$.

$143=11\times 13=11(-4-3\omega )(-4-3{\omega }^2)=11(-4-3\omega )(-1+3\omega )$. □