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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 $F[\alpha] = \{ p(\alpha)\mid p(x)\in F[x]\},$

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How did you choose the title for this question? The body of the question has nothing to do with "quadratic fields". –  KCd Aug 28 '12 at 14:00
    
@KCd, suggest another, please. im nube in it. –  Yola Aug 28 '12 at 14:01
    
I made an edit to the title. I am still curious what you thought "quadratic field" means, such that it would be related to your question. –  KCd Aug 28 '12 at 17:59
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$\alpha = [x]_{q(x)}$ 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)

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you mean $[2x^3]_{q(x)} = 2x^3 + \langle q(x) \rangle$? and why then $q(x)$ is minimal? –  Yola Aug 28 '12 at 13:41
    
$q(x)$ is minimal because it meets the definition of minimal - $q(\alpha)=0$, and $q$ being irreducible no polynomial of smaller degree vanishes at $\alpha$. –  Gerry Myerson Aug 28 '12 at 13:50
    
I edited my answer to include an answer about the minimality. @GerryMyerson see my last note about being monic –  Belgi Aug 28 '12 at 13:51
    
if $q(x) = x^2$ then $\alpha$ may be equal to $x + n x^2$? –  Yola Aug 28 '12 at 13:58
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$q(\alpha)$ = the class of $q(x)$ in the quotient hence it is in the ideal generated by $q(x)$ i.e it is $[0]_{q(x)}$ –  Belgi Aug 28 '12 at 14:01
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