Does $x^p=x$ for any finite field with characteristic $p$?

Question

I know it's true when the field has $p$ elements (an easy application of Lagranges theorem to the group of units), but how about when the field has $p^k$ elements?

Answer 1

(Counterexample given in comments by David Hill) (Remark given in comments by Ted Shifrin)

Consider $\bar{x} \in \Bbb{F}_2[x]/(x^2+x+1) \cong \Bbb{F}_4$. Then $\overline{x}^2 = \overline{x+1} \neq \overline{x}$

As a remark, any $x \in \Bbb{F}_{p^k}\backslash \Bbb{F}_p$ will be a counterexample. To see this, notice that the polynomial $x^p-x$ will have at most $p$ distinct roots, but all of the elements of $\Bbb{F}_p$ are roots of it. So they are the only roots.

However, you can get a result that says that if $\text{char}(\Bbb{F})=p$, then for any $x,y \in \Bbb{F}$, $(x+y)^p = x^p+y^p$.