Question 565385
Firstly, we know that (a+b)^p=a^p+b^p in every field of characteristic p. Therefore s(a+b)=s(a)+s(b). Obviously, s(ab)=s(a)s(b). So s(a) is a homomorphism. The only way for s(a) to equal 0 is if a is 0, so the kernel of the homomorphism is {0} and therefore the map is one to one. Finally, we know that s (a) is onto since the field involved is finite. Thus, the Frobenius map is an automorphism.