Prove that for a finite field $R$ of characteristic $p$, the mapping $f(a)=a^p$ for every element $a$ of $R$ is an isomorphism.

Question

$R$ is a finite field having characteristic $p(>0)$. Let $f(a)=a^p$ where $a\in R$ be a mapping from $R$ to $R$. I have to show that it is an isomorphism.

I am done with the homomorphism part. But when it comes to injectivity part,I am having problems. I only know that as $p>0$, it has to be some prime integer as $R$ is a finite field. Also R has only improper ideals i.e. ${0}$ and $R$ itself.

How can I show that for any $a$,$b\in R$, $f(a)=f(b)\implies a=b$ i.e. $a^p=b^p\implies a=b$.

Please suggest how to proceed.

Answer 1

If $a^p = 0$ for some $a\in R$, then (since $R$ is a field) $a = 0$. Thus $\mathrm{Ker}(f) = 0$ and hence $f$ is injective.