Exercise 9.17

Proof. If $\gamma \equiv 2,\rho \equiv 2(\mathit{mod}3)$, then $-\mathit{\gamma \rho }\equiv -2\times 2\equiv 2(\mathit{mod}3)$, so $-\mathit{\gamma \rho }$ is primary.

By Ex. 9.2, $\gamma$ can be written

where ${\pi }_i\equiv 2(\mathit{mod}3),a\in \{0,1\},b\in \{0,1,2\}$.

As ${\pi }_i\equiv -1(\mathit{mod}3)$, and $\gamma \equiv -1(\mathit{mod}3)$, we obtain ${\omega }^b{\lambda }^c\equiv \pm 1(\mathit{mod}3)$. We prove that $b=c=0$.

Note that ${\lambda }^2={(1-\omega )}^2=-3\omega \equiv 0(\mathit{mod}3)$. If $c\ge 2$, we would obtain $\gamma \equiv 0(\mathit{mod}3)$, in contradiction with the hypothesis, thus $c=0$ or $c=1$.

If $c=1$,

Since $1-\omega \not\equiv \pm 1,1+2\omega \not\equiv \pm 1,-2-\omega \not\equiv \pm 1(\mathit{mod}3)$, this is impossible, so $c=0$.

Then ${\omega }^b\equiv \pm 1(\mathit{mod}3)$, where ${\omega }^b\in \{1,\omega ,-1-\omega \}$. Since $\omega \not\equiv \pm 1(\mathit{mod}3)$, and $-1-\omega \not\equiv \pm 1(\mathit{mod}3)$, then ${\omega }^b=1,0\le b\le 2$, thus $b=0$.

Finally, $\gamma ={(-1)}^a{\pi }_1^{a_1}\cdots {\pi }_t^{a_t}$.

Conclusion: Every primary $\gamma \in D$ is under the form

where the ${\gamma }_i$ are primary primes. □