I have problem with finding irreducible factors of $x^{10}-1$ in $\mathbb{Z}_3[x]$. Easily one gets that $$x^{10}-1=(x-1)(x+1)(x^8+x^6+x^4+x^2+1)$$ And now I am stuck. I tried dividing the polynomial $x^8+x^6+x^4+x^2+1)$ by irreducible polynomials of order 2 (only three of them but no luck there). Do I have to try now the irreducible polynomials of order 3,4, etc.? Does anyone has simpler/faster/neater solution? Without using Wolfram ;) :)
Thanks
On
Any complex root of $x^{10}-1$ is a tenth root of unity, hence $$ x^{10}-1 = (x-1)(x+1)\Phi_5(x)\Phi_{10}(x) \tag{1}$$ and the splitting field of $\Phi_n(x)$ (that is the minimal polynomial over $\mathbb{Q}$ of a primitive $n$-th root of unity) over $\mathbb{F}_{p}$ is given by $\mathbb{F}_{p^d}$ with $d$ being the least integer such that $n\mid(p^d-1)$. Since $3$ is a generator for both $\mathbb{Z}/(5\mathbb{Z})^*$ and $\mathbb{Z}/(10\mathbb{Z})^*$, $\Phi_5$ and $\Phi_{10}$ are irreducible over $\mathbb{F}_3$ and $(1)$ is a factorization over $\mathbb{F}_3$.
Certainly one could muse over a faster and neater solution. However, it seems better to apply the Berlekamp algorithm right away, to get $$ x^8+x^6+x^4+x^2+1=(x^4 + 2x^3 + x^2 + 2x + 1)(x^4 + x^3 + x^2 + x + 1). $$ And we do not need Wolfram for it, though it is possible to use it.