showing $x^4+x^2+x+1$ is reducible in $GF(81)=GF(3^4)$

Question

I am trying to show that in $GF(81)=GF(3^4)$,

$$x^4+x^2+x+1$$ is reducible

I proved that it was irreducible in $\mathbb{Z}_3$

How can I prove this ?

More generally, how to prove if a polynomial is / is not reducible in finite fields ?

Thanks

Answer 1

You know that $f(x)=x^4+x^2+x+1$ has a linear factor if and only if $f(x)$ has a root in $\mathbb{F}_3$. We see $f(0),f(1),f(2) \not = 0$ so $f(x)$ has no linear factor.

Then if it is reducible it must have quadratic factors that are irreducible in $\mathbb{F}_3$. $$x^4+x^2+x+1 = (ax^2+bx+c)(dx^2+ex+f)$$ $$x^4+x^2+x+1 = adx^4+(ae+bd)x^3+(af+cd)x^2+(bf+ce)x+cf$$ See if you can solve from there.