# Factoring a polynomial in $\mathbb{Z_3}$ into linear factors.

I am trying to factor the polynomial $$x^{7} - x$$ over the field $$\mathbb{Z_3}$$. The solution is: $$x^{7} - x = x(x^6 - 1) = x(x^3 - 1)(x^3 + 1) = x(x - 1)^3(x + 1)^3.$$

I understand that the last step follows from the fact that in a field $$p$$ elements, $$x^p + y^p = (x + y)^p$$, however, my confusion lies with factoring $$x^6 - 1$$ into two polynomials with the same degree. If $$1$$ and $$-1$$ are the roots of $$x^6 - 1$$, is there some obvious way to guarantee that the factors split evenly.

I guess I could use long division each time I factor a root, and I would get the same answer, but is there some intuition, or an obvious fact behind this that I am missing? Thanks.

• $x^6-1$ is the difference of two squares – J. W. Tanner Mar 10 '19 at 17:59

In any field, $$x^6-1$$ is a difference of squares so it will always factor into $$(x^3-1)(x^3+1)$$.
Another way is: $$x^6 - 1 = (x^2 - 1)^3 = ((x - 1)(x + 1))^3 = (x - 1)^3(x + 1)^3.$$