The Frobenius Automorphism If a finite field $F$ has characteristic $p$, prove that every element $a\in F$ can be expressed as $a=b^{p}$ for some $b\in F$?
Hint: Frobenius Automorphism.
Isn't the Frob Automorphism about $a$? Why does it have to do with $b$?
Thanks!!! 
 A: If you say Frobenius "Automorphism" then you should be done...
If you just want to say Frobenius  map then,
Hint :


*

*Prove that map is injective

*any injective map between two finite sets of same cardinality is ???

A: The Frobenius map is just a map: $$\phi: F \rightarrow F$$
where $\phi(a) = a^p$ and char($F$) $=$ $p$
So if you know what an automorphism is, it just follows.
A: There is no constructive reason; it's just the pigeonhole principle. The Frobenius map is injective since the polynomial $x^p-1=(x-1)^p$, so since it acts on a finite set, it must be surjective.
Note that there can be no constructive reason, since there exist non-perfect fields: those for which the Frobenius map is not surjective. An example is $\mathbb{F}_p(t)$, where by degree, $t$ is not a value of the Frobenius map.
A: Supppose that $F=\mathbb F_{q}$ where $q=p^r$ for some positive integer $r$, then elements of $F$ will be the roots of the polynomial $x^q-x$. Thus for every $a\in F$ we have $a=a^q=a^{p^r}=\left(a^{p^{r-1}}\right)^p$, so $b=a^{p^{r-1}}$ is the required solution such that $a=b^p$.
