Let $p\equiv -1 \pmod 9$ be a prime. Determine the factorization of $f(x)=x^9-1$ over $\mathbb{F}_p$.
My proof: First of all we observe that $f(x)=(x-1)(x^2+x+1)(x^6+x^3+1)$. The degree of the splitting field of $f(x)$ over $\mathbb{F}_p$ is $\text{ord}_{(\mathbb{Z}/9\mathbb{Z})^*}(p)=2$. At the same way we can determine the degree of the splitting field of $x^3-1$ over $\mathbb{F}_p$ that is $2$, hence $x^2+x+1$ is an irreducible polynomial over $\mathbb{F}_p$. Now we have to determine the degrees of the factors of $g(x):=x^6+x^3+1$ over $\mathbb{F}_p$: for sure the are only factors with degree $1$ or $2$. Suppose $g(x)$ has a linear factor over $\mathbb{F}_p$, then it has also a linear factor over the splitting field of $f(x)$ that is $\mathbb{F}_{p^2}$. But $\mathbb{F}_{p^2}\cong \mathbb{F}[x]/(x^2+x+1):=L$ and $g(x)=-(x+1)^3+x^3+1=-3x(x+1)$ over $L$. Therefore, if $\alpha$ is a root of $g(x)$ over $\mathbb{F}_p$ then $\alpha=0, -1$, absurd. We conclude that $g(x)$ splits in three factors of degree $2$ over $\mathbb{F}_p$.
Can you check my proof, please? I have some doubts about the last part...
Yes, $x^9-1$ splits into irreducible factors $x-1$, $x^2+x+1$ and another three irreducible quadratic polynomials over $\mathbb{F}_p$ for $p\equiv -1(9)$, which are the irreducible factors of $\Phi_9(x)=x^6+x^3+1$: $$ x^9-1=\Phi_1(x)\Phi_3(x)\Phi_9(x)=(x-1)(x^2+x+1)(x^2+ax+1)(x^2+bx+1)(x^2+cx+1). $$