Factorization of $x^8-x$ over $F_2$ and $F_4$

Solution 1:

This is an answer to "Is there a strategy to find this out?" referring to the factorization shown in Jack D'Aurizio's answer.

In this case, the factorization is easy to discover by simple calculations. First, we note that there are no repeated roots since $x^8-x$ and its formal derivative $8x^7-1 = -1$ are relatively prime polynomials. Next, we have the obvious factorization $$\begin{align}x^8-x &= x(x-1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)\\ &= x(x+1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1) \end{align}$$ where the factors of $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ (if any) must be irreducible polynomials of degree $2$ or more; we have already accounted for all the degree-$1$ factors of $x^8-x$. Now, the only irreducible quadratic is $x^2+x+1$ (all other quadratic polynomials have $x$ or $x+1$ as factors), and it is easy to verify (by inspection, almost!) that $x^2+x+1$ does not divide $x^6 + x^5 +\cdots+ x + 1$. Turning to cubic polynomials, we can discard $x^3+1$ and $x^3+x^2+x+1$ as possible candidates since they both have $x+1$ as a factor. So we are left with irreducible cubic $x^3+x^2+1$ and $x^3+x+1$ as possibilities, and it is with much joy and satisfaction that we verify that their product does indeed equal $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ giving us that $$x^8-x = x(x-1)(x^3+x^2+1)(x^3+x+1).\tag{1}$$