Let E be a splitting field for f(x) = over F of char p. Then for be a root for f(x), then r + 0, r + 1, ..., r + p - 1 are also roots.

I have to make a quick point, that f has no repeated roots.for if, assuming root a in E, f(x) = h(x)(x-a) , then f'(x) = h(x) + h'(x)(x-a), and f'(a) = h(a).

Since f'(x) = f'(a) = h(a) , so no repeated roots.

Since = = 0 . for (Since

so

Take g(x) to be the minimal polynomial such that g(r) = 0.

Then [E : F ] = deg(g).

If f = is the decomposition of f into unique factorization, since g is the min. polynomial s.t g(r) = 0, g divides f. since we made f into a unique factorization g divides one of the , since are monic and irreducible, this means g(x)*1 = , thus deg( ) = deg(g) for all .

So deg(g) divides deg(f).

Since deg(f) = p, either f = g and f is irreducible

or deg(g) = 1, and [E: F] = 1, and so