If $F$ is a finite field of size $q=p^n$ and $b\in K$ is algebraic over $F$ then $b^{q^{m}} = b$ for some $m > 0$

Question

I want to apply a similar type of argument to show that $\alpha^{q} = \alpha$ for all $\alpha\in F$. When we know that the characteristic of $F$ is $p$ and $|F| = q = p^{n}$. But I dont know how to incorporate the condition that $b\in K$ is algebraic over $F$. I know there is some $f\in F[t]$ such that $f(b) = 0$. Also, $F(b)/F$ is finite... I don't know how to proceed. Any suggestions?

Answer 1

If $b=0$ then you're done. Otherwise it is a unit and the multiplicative order of a unit $b$ in the group of units divides the order of said group. Given that $F(b)$ is a vector space over a field of size $q$, we find that $|F(b)|=q^m$ for $m=\dim_{\,F}F(b)$. What is the size of the group of units in $F(b)$ then?