Let F be a finite field of characteristic $p$. Show $f(a) = a^p$ is a ring homomorphism, injective, and surjective

Question

Let F be a finite field of characteristic $p$. Show that the function $f:F \to F$ defined by $f(a) = a^p$ is a) a ring homomorphism, b) injective and, c) surjective.

I tried to approach this problem by proving $f(a+b) = f(a) + f(b), f(ab) =f(a)f(b) $ for ring homomorphism.

But I don't know how to solve b)and c). Could you please help me to solve it ? Thank you!!!

Answer 1

If we have that $a^p=b^p$, then $(a-b)^p=\sum_i a^ib^{p-i}\binom{p}{i}=a^p-b^p=0$, so that, assuming $a-b\neq 0$, $((a-b)^{-1})^p(a-b)^p=1=(a-b)^{-p}0=0$, a contradiction, so that $a=b$ and the function is injection. Now note that for a finite set $S$, if we have an injection $f:S\to S$, it must be a surjection, since if $|f(S)|<|S|$, then some two must be mapped to the same point by the pigeon-hole principle.

Answer 2

Hint: a field homomorphism is always injective, and an injective map from a finite set to itself is always surjective.