I've found a theorem in the book "Linear Groups" (Dickson, 1901, p.16.): "In $\mathbb{Z}_2$, the degrees of the irreducible divisors of $x^{2^m}-x$ are divisors of $m$."
I've read the prove in this book but i don't understand the proof. Can anybody help me?
Sorry for my bad English.
The idea of the proof is the following.
Let $p$ be any prime, and $f = x^{p^{m}} - x \in \mathbb{Z}_{p}[x]$.
Then the splitting field of $f$ over $\mathbb{Z}_{p}$ is a field $E$ with $p^{m}$ elements. For the degree we have $\lvert E : \mathbb{Z}_{p} \rvert = m$.
Let $g$ be a monic, irreducible divisor of $f$, of degree $k$. Since $E$ is the splitting field of $f$ over $\mathbb{Z}_{p}$, the polynomial $g$ will have a root (actually, all of its roots) $\alpha \in E$.
Since $g$ is irreducible in $\mathbb{Z}_{p}[x]$, $g$ will be the minimal polynomial of $\alpha$ over $\mathbb{Z}_{p}$. Hence $\lvert \mathbb{Z}_{p}[\alpha] : \mathbb{Z}_{p} \rvert = k$.
By the multiplicativity formula for degrees $$ m = \lvert E : \mathbb{Z}_{p} \rvert = \lvert E : \mathbb{Z}_{p}[\alpha] \rvert \cdot \lvert \mathbb{Z}_{p}[\alpha] : \mathbb{Z}_{p} \rvert = \lvert E : \mathbb{Z}_{p}[\alpha] \rvert \cdot k, $$ so that $k$ divides $m$.