For primes p,q I am trying to find the number of irreducible factors of $f(x)=x^{p^q}-x\in \mathbb{F}_p$. My first thought was that since q is prime there is no proper subfield $$\mathbb{F}_p \subset E \subset \mathbb{F}_{p^q}$$ therefore for every root $\alpha$ of $f$, $F(\alpha)=\mathbb{F}_p$ or $F(\alpha)=\mathbb{F}_{p^q}$. Hence if $m_\alpha(x)$ is the minimal polynomial of $\alpha$ in $\mathbb{F}_p[x]$ then $deg(m_\alpha)=1$ or $deg(m_\alpha)=q$ since $[\mathbb{F}_{p^q}:\mathbb{F}_p]=q$. But, this means that all irreducible polynomials of $$\tilde{f}(x)=f(x)/(x^p-x)$$ are of degree $q$ which means that $q\vert (p^q-p)$ which is usually not the case. So my argument is incorrect and i dont know the correct one. Will be glad for a hint
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1They can also be of degree $1$, such as $x$ or $x-1$. For a concrete example, $x^{3^2}-x = x(x+1)(x+2)(x^2 + 1)(x^2 + x + 2)(x^2 + 2x + 2)$ in $\Bbb F_3[x]$. – Arthur Jul 25 '17 at 10:25
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You can find your answer at the accepted answer of this question – Amin235 Jul 25 '17 at 10:33
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Fermat’s little theorem tells you that it is usually (ehm, always) the case that $$ q\mid (p^q-p) $$
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