Factorise $x^{9}-1$ in $\Bbb{F}_7[x]$

Question

I don't know much about how to solve it since we are just learning the topic, but I tried to answer it, so far I got $$(x-1)(x-2)(x-4)({x}^{6}+{x}^{3}+1)$$ but the answer is $$(x-1)(x-2)(x-4)(x^{3}-2)(x^{3}-4).$$

Answer 1

Mod $7$, $x^6+x^3+1\equiv x^6+x^3-6= (x^3-2)(x^3+3)\equiv (x^3-2)(x^3-4)$.

Answer 2

So, it appears the only issue was that you didn't fully factorise the sextic polynomial. You needed to either factorise it further or prove that it was irreducible (as it turns out, it was reducible).

Note that substituting elements of $\Bbb{F}_7$ into the polynomial will find all linear factors (i.e. factors of the form $x - \alpha$ for $\alpha \in \Bbb{F}_7$), but it's possible for a reducible polynomial of degree $4$ or more to not have any linear factors. In this case, the sextic polynomial has no roots, but it still factorises into two irreducible cubic roots.

So, in this case, I would note that $x^6 + x^3 + 1$ is a quadratic in $x^3$, and use this fact as inspiration to see if we can factorise it into the form $(x^3 - \alpha)(x^3 - \beta)$. There's a few ways we can do this, but a simple way is to simply look at the corresponding quadratic $x^2 + x + 1$, and find its roots in the same way you did for the original polynomial.

Substitution into $x^2 + x + 1$ should reveal that $$x^2 + x + 1 = (x - 2)(x - 4),$$ and hence $$x^6 + x^3 + 1 = (x^3)^2 + (x^3) + 1 = (x^3 - 2)(x^3 - 4).$$ Now, are we done? Can we reduce these cubics any further? If we could, then they would have to factorise into a linear factor and a quadratic factor. The linear factor would mean that the cubic polynomial has a root. But, substituting all of $\Bbb{F}_7$ into these polynomials will yield no roots at all, which means these cubics are irreducible, and we are indeed done.

(Again, remember that the fact that these polynomials are cubics is very important in this step! If we had a quartic, it'd be possible to factorise into two irreducible quadratics, and there'd still be no roots.)