# Galois Group of $x^4+x-1$ over $\mathbb{F}_3$

Consider the finite field $\mathbb{F}_3$ and define the polynomial $f(x)=x^4+x-1$ over $\mathbb{F}_3$.

I want to find its Galois Group.

I observe that $f$ has no root over $\mathbb{F}_3$, so if it is reducible, it must be as quadratic factors. Is there any way to show that this isn't possible?

Assuming it isn't (which is the case), we adjoin a root of $f$, $\alpha$ say to form $F=\mathbb{F}_3(\alpha)$.

This is a degree $4$ extension (since $f$ is irreducible) and so $\mathbb{F}_3(\alpha)\cong \mathbb{F}_{81}$.

Now we need to determine if the other roots of $f$ are in this field - if so then we know our Galois Group has size $4$ and we can work out if it's $C_4$ or $V_4$.

How can I work out whether these roots are in $\mathbb{F}_3 (\alpha)$?

To show that $f$ is irreducible over $\mathbb{F}_{3}$, you may try dividing it by the monic irreducible polynomials of degree $2$. Actually, you need check for all but one, as $f$ has no multiple roots, and thus it is not a square.
So check that $f(x)$ is not divisible by $x^{2} + 1$ and $x^{2} + x - 1$, and you are done with the irreducibility.
You should have seen if $E$ is a finite field, which is an extension of the (necessarily finite) field $F$, then $E/F$ is Galois. Thus $\mathbb{F}_{81}/\mathbb{F}_{3}$ is Galois, and its Galois group is cyclic, generated by the Frobenius automorphism $\sigma : z \mapsto z^ {3}$.
So if $\alpha$ is a root, the four roots are $$\alpha, \alpha^{3}, \alpha^{9}, \alpha^{27}.$$ You can use the fact that $f(\alpha) = 0$ to express these roots in the form $$a_{0} + a_{1} \alpha + a_{2} \alpha^{2} + a_{3} \alpha^{3} \in \mathbb{F}_{3}[\alpha],$$ with $a_{i} \in \mathbb{F}_{3}$.
• Thanks, how about determining whether $\mathbb{F}_3(\alpha)$ is in fact the splitting field? – Mathmo May 14 '14 at 11:17