Can anybody explain to me how to find such a factorization? I have yet to find some kind of "algorithm" to do it properly. I know that
$$x^7 - 1 = (x - 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)$$
with $(x-1)$ being irreducible. But how to I factorize the second polynomial further? Plus, are those factorizations always unique or are there several factorizations for the same polynomial?
On
You don't have lots of ways of doing it, right? Your polynomial is a $6$th degree polynomial without roots in $\mathbb{F}_2$. Therefore, if it is reducible, it will have irreducible factors of degree $2$ or $3$. There is only one quadratic irreducible polynomial in $\mathbb{Z}_2[x]$, which is $x^2+x+1$. And there are only two cubic irreducible polynomials in $\mathbb{Z}_2[x]$, which are $x^3+x+1$ and $x^3+x^2+1$.
Using the Berlekamp Algorithm we obtain $$ x^7-1=(x^3 + x^2 + 1)(x^3 + x + 1)(x + 1). $$ The factorization is unique, up to permutations and units, because the polynomial ring $\Bbb{F_2}[x]$ is a UFD.