Let denote the Frobenius map on . Prove that is an automorphism and compute its order in .

Recall that is a homomorphism. Moreover, if , then . Since fields contain no nontrivial zero divisors, we have (using induction if you want). So the kernel of is trivial, and thus is injective. Since is finite, is surjective, and so is a field isomorphism.

Next, we claim that for all and all , and will show this by induction. The base case certainly holds, and if , then as desired.

Now , since the elements of are precisely the roots of . So we have .

If , then we have for all , so that each is a root of . So divides , and so divides (by this previous exercise) and then divides (by this previous exercise). In particular, .

So is the order of in .