Does there exist an irreducible factor of $x^{32}-x \in\mathbb{Z_3}[x]$ with degree 4?

Question

I would like to find whether there exist an irreducible factor of $x^{32}-x \in\mathbb{Z_3}[x]$ with degree $4$.

Firstly, this polynomial is expressed as the product $x(x-1)(x^{30}+x^{29}+...+x+1)$ over $\mathbb{Z}_3$

Then, how can i check whether the 3rd term is irreducible ? And is this statement true?

Answer 1

Compute the gcd of this polynomial with $x^{3^4}-x$. The Euclidean algorithm will help you compute the gcd. We know that all irreducibles of degree $4$ are a factor of $x^{3^4}-x$, so you need only work from there. This is also relatively easy to do

$$x^{81}-x = (x^{49}+x^{18})(x^{32}-x) +x^{19}-x=q(x)(x^{32}-x)+r(x)$$

iterating gives

$$x^{32}-x = x^{13}(x^{19}-x) + x^{14}-x$$

$$x^{19}-x = x^5(x^{14}-x)+x^6-x$$

$$x^{14}-x = (x^8+x^3)(x^6-x)+x^4-x$$

$$x^6-x = x^2(x^4-x) + x^3-x$$

$$x^4-x = x(x^3-x)+x^2-x$$

$$x^3-x = (x+1)(x^2-x)+0$$

So the gcd is of degree $2<4$, making it impossible to have a factor of degree $4$. The nice thing about this approach is it doesn't require you to figure out if the extra factor is reducible or not!