Multiplicative order in field extension

Question

Let $F/K$ be some field extension (both are finite fields) and $u$ be some element in $F$.

I want to know if $u^{|K|} = u$ implies $u \in K$. And why?

Answer 1

As $K^*$ is a group of order $|K|-1$ (under multiplication) every element there satisfies the equation $x^{|K|-1}=1$. So multiplying by $x$, these elements along with $0$, satisfy the equation $x^{|K|}-x=0$. As equations over fields cannot have more roots than the degree it is not possible for elements of the extension field $F$ (not in $K$) to be the root of that equation. This is the argument Hurkyl had in her/his mind and did not have the time write up.