Frobenius Homomorphism and p-root

Question

I want to know how the Frobenius homomorphism helps me to prove that for all $a$ exist $b$ such that $b^{p} = a$.

\begin{align*} f: &F \rightarrow F\\ &a \rightarrow a^{p} \end{align*}

(where $F$ is a finite field with characteristic $p$)

For the Homomorphism, I get that $f(a)=a^{p}$, I tried to prove it with the fact that it is a biyection so I have that all element in $F$ has an image in $F$, but I am stuck in this, I would like to know if I am correct or maybe exist other way to do it.

Thanks in advice :).

Answer 1

Assuming $F$ is a finite field of characteristic $p$, then let $F=\mathbb{F}_{p^r}$. Then $F^*$ is a cyclic group under multiplication, of order $p^r-1$. In particular, for any $x\in F$: $$x=x^{p^r-1}x=x^{p^r}=f^r(x).$$

In particular: $$x=f(f^{r-1}(x))=(f^{r-1}(x))^p.$$