Fermat's little theorem only holds for integers. This root is something that lives in the splitting field of , and is probably not an integer, or even an element of Z/pZ. However, since you know there exists a splitting field for every polynomial f(x), you can assume the existence of a root.

In this instance j is necessary an integer (or more precisely, an element of Z/pZ). Fermat's little theorem holds in this case.

2) Fine, let's accept that [; \alpha ;] is indeed a root, that my objection in (1) is wrong. Then why is j^p-j = 0? Is it because j belongs to k while [; \alpha ;] does not?

Yes.(Bonus question: Determine all intermediate fields of [; Q(\sqrt{2+\sqrt{2}}) ;]. The intermediate fields correspond to the subgroups of the Galois group, in this case the cyclic group on four elements, so wouldn't there be only one intermediate field, corresponding to the (up to isomorphism) unique group of two elements, namely [; Q(\sqrt{2}) ;] ?)