Factor $x^8 - x$ into irreducibles in $\mathbb{Z}[x].$

Question

I know $x^8 - x = x(x^7 - 1)$ and the roots of $x^7 -1$ are the 7th roots of unity, but 6 of those roots are complex. I'm pretty sure this can be factored further, but I don't know how to think of it in the integers.

Answer 1

We have $$ x^{n}-1= \prod _{d\mid n}\Phi _{d}(x) $$ where $\Phi _{d}(x)$ is the $d$-th cyclotomic polynomial. Therefore, $$ x^8 - x = x(x^7-1) = x \, \Phi _{1}(x) \, \Phi _{7}(x) = x(x-1)((x^6 + x^5 + x^4 + x^3 + x^2 + x + 1) $$ When $p$ is prime, we have $$ \Phi_p(x) = x^{p-1} + x^{p-2} + \cdots + x + 1 $$ whose irreducibility follows from Eisenstein's criterion applied to $\Phi_p(x+1)$.