According to Milne for each power $q=p^n$ there exists a field $\mathbb{F}_q$ with $q$ elements. I would like to understand what exactly happens with the roots of this.
Let us consider the very simple case where $p=3$, $n=2$ and thus $q=9$. Then the roots of $\mathbb{F}_q$ must be given by $$ x^{9}-x = 0 \Rightarrow x=0 \text{ or } x^8-1=0 $$ Already we see the trivial root $x=0$. Then, another one is $x=1$ since $x^8=1$. After that I am confused with the rest of the roots since it is not very clear to me what are the exact modulo rules for this kind of fields. Any help?
And since I already asked the above let me ask one more question. For the specific example what would be the algebraic closure $\mathbb{F}$ of $\mathbb{F}_9$? What is the usual way to find such a closure?
Thanks a lot.
On the algebraic closure:
1) It is $\mathbb{F}_{3^n} \subset \mathbb{F}_{3^m}$ if and only if $n$ divides $m$.
2) In $\mathbb{F}_{3^n}$, all polynomials of degree $n$ factorize.
3) So in the field $F=\cup_k \mathbb{F}_{3^{k!}}$ (which is infinite of course), all polynomials factorize. This is the algebraic closure of $\mathbb{F_3}$.