Show that as roots of an irreducible polynomial in $ \mathbb{F}_q $ are $a,a^q \ldots a^{q^{n-1}}$

Question

Consider $ a $ a root of f in some extent of $ \mathbb{F}_q $, I can show that in fact $ f(a) = f(a^q) ... = f (a^{q^{n-1}} )= 0$, but I can not show that they are all distinct.

Answer 1

Let $E=\mathbb{F}_q(a)$. Since $f$ is the minimal polynomial of $a$ and has degree $n$, $[E:\mathbb{F}_q]=n$ and so $|E|=q^n$. Now suppose $a^{q^i}=a^{q^j}$ for some $0\leq i<j<n$. Taking the $q^i$th root of both sides (since raising to the $q$th power is injective), we get $a=a^{q^m}$ where $m=j-i$ is a positive integer less than $n$.

Now observe that $K=\{b\in E:b^{q^m}=b\}$ is a subfield of $E$ which contains $\mathbb{F}_q$ and $a$. Since $E=\mathbb{F}_q(a)$, this means $K$ is all of $E$. But $|K|\leq q^m<q^n=|E|$, since the polynomial $x^{q^m}-x$ can have at most $q^m$ roots in $E$. This is a contradiction.