How do I factor $x^8-x$ over $\mathbb{Z}_2$?

Question

I am trying to factor the polynomial $x^8-x$ over $\mathbb{Z}_2$ to get a splitting field for it.

I got that it is equal to $x(x-1)(x^3-x-1)(x^3+x^2+1)$ over $\mathbb{Z}_2$ but cannot proceed any further.

Answer 1

Well, it's a general (and obvious) fact that if $X-a$ divides a polynomial $P(X)\in k[X]$ for $a\in k$ ($k$ a field), then $a$ is a root of the polynomial. Since neither of your polynomials have $0$ or $1$ as a root, it must be that these cubic polynomials do not have linear factors in $k[X]$. But then this means that one cannot reduce them non-trivially anyway, so the reduction you have done so far is sufficient to complete the problem.