Notation for finite fields What is the meaning of the following:

If $q(x)$ is irreducible polynomial of degree $d$ and $d$ divides $n$, then $q(x)$ divides $x^{p^n}-x$.
Let $F = F_p[x]/(q(x)) = F_p[\alpha]$ where $\alpha = [x]_{q(x)}$. Then $q(x)$ is the minimal polynomial over $F_p$ of $\alpha$.

Whats  $\alpha = [x]_{q(x)}$ mean? And why "Then $q(x)$ is the minimal polynomial over $F_p$ of $\alpha$.".
I know what $F_p[x]/(q(x))$ is a quotient ring, but why its equal to $F_p[\alpha]$, where as i thought $F[\alpha] = \{ p(\alpha)\mid p(x)\in F[x]\},$
 A: $\alpha = [x]_{q(x)}$ means the equivalence class of $x$ in the quotient, i.e $x+\langle q(x) \rangle$.
The reason $q(x)$ is the minimal polynomial of $\alpha$ over $F_p$ is that if for some polynomial $g(x)$ it holds that $g(\alpha)=0$ then by the euclidean algorithm you get $g(x)=q(x)p(x)+r(x)$  hence $r(x)=q(x)p(x)-g(x)$ thus $r(\alpha)=q(\alpha)p(\alpha)-g(\alpha)=0$ (since $\alpha = [x]_{q(x)}$ is a root of $q(x)$ in $F$) so we conclude $r(x)=0$ i.e $g(x)=q(x)p(x)$ i.e $g(x)\in \langle q(x) \rangle$.
So you can see that the set of the polynomials that $\alpha$ is a zero of them is generated by $q(x)$ and it is clear that there is no polynomial in $\langle q(x) \rangle$ with smaller degree.
Note: you did not mention this, but for any $\beta\in F_p$ s.t. $\beta\neq 0$ it holds that $\langle q(x) \rangle=\langle \beta q(x) \rangle$ so it is not clear that your $q(x)$ is monic, but if $a_n$ is the leading coefficient of $q(x)$ then $\frac{q(x)}{a_n}$ is the minimal polynomial of $\alpha$ over $F_p$ (this is because the minimal polynomial is defined to be monic)
