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.

  • 2
    $\begingroup$ $x^6-1$ is the difference of two squares $\endgroup$ – 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. $


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