SOLUTION: Prove: Let F be a finite field of characteristic p. Then the Frobenius map [s(a)=a^p] is an automorphism.

Question 565385: Prove: Let F be a finite field of characteristic p. Then the Frobenius map [s(a)=a^p] is an automorphism.
Answer by ad_alta(240) (Show Source): You can put this solution on YOUR website!
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.
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