Factoring $x^{15}−1$ into irreducible polynomials over $\mathrm{GF}(2)$

Question

Factorize $x^{15}−1$ into irreducible polynomials over $\mathrm{GF}(2)$

The answer is $$(x+1)(x^2+x+1)(x^4+x+1)(x^4+x^3+1)(x^4+x^3+x^2+x+1)$$ but how would I find this? Please help.

Answer 1

The polynomial $x^{16}-x\in\mathbb{F}_2[x]$ has roots precisely equal to the elements of $\mathbb{F}_{16}$, and the subfields of $\mathbb{F}_{16}$ are $\mathbb{F}_{16}$, $\mathbb{F}_{4}$, and $\mathbb{F}_{2}$, which have degrees $$[\mathbb{F}_{16}:\mathbb{F}_2]=4\qquad [\mathbb{F}_{4}:\mathbb{F}_2]=2\qquad [\mathbb{F}_{2}:\mathbb{F}_2]=1$$ Therefore $x^{16}-x$ is the product of the irreducible polynomials in $\mathbb{F}_2[x]$ of degrees $1$, $2$, and $4$.

Therefore $x^{15}-1$ is the product of the irreducible polynomials in $\mathbb{F}_2[x]$ of degrees $1$, $2$, and $4$, other than the irreducible polynomial $x$.

To find these irreducible polynomials, you can use Sieve of Eratosthenes (Wikipedia link), adapted to the ring $\mathbb{F}_2[x]$.