Do you know what they mean by a primitive root? That means the set is the set of all -th roots of unity. Additionally, . So, the -th roots of unity are solutions to . Obviously, 1 is a solution. But the other solutions can not be real so long as . We know that the multiplicative group of is cyclic. This is why they use primes. If we used an arbitrary value, then it might not have any primitive roots of unity. For example, there are no primitive 8th-roots of unity. The 8th-roots of unity (as a group) are isomorphic to (the product of a two-cycle and a four-cycle). Hence they cannot be generated by a single element. I don't have time to go into any more detail at the moment. I hope this helps clarify at least why they are using a prime number for the example. Also, . The second factor is irreducible over the rationals, and it is also irreducible over . I hope this helps a little. If I have more time, I will try to write up a bit more explanation.