Exercise 9.10

Proof. $|{(D∕5D)}^{\text{∗}}|=N(5)-1=24$, thus $x^{24}-1=\underset{\alpha \in {(D∕5D)}^{\text{∗}}}{\prod <!--\; FUNCTION\; APPLICATION\; -->}(x-\alpha )$.

(where the $\alpha \in {(D∕5D)}^{\text{∗}}$ are of the form $\alpha =a+b[\omega ],0\le a<5,0\le b<5,(a,b)\ne (0,0))$. □