Consider $f=x^4-2\in \mathbb{F}_3[x]$, the field with three elements. I want to find the Galois group of this polynomial.
Is there an easy or slick way to factor such a polynomial over a finite field?
Consider $f=x^4-2\in \mathbb{F}_3[x]$, the field with three elements. I want to find the Galois group of this polynomial.
Is there an easy or slick way to factor such a polynomial over a finite field?
Recall that over $\mathbb{F}_q$, the polynomial $x^{q^n} - x$ is precisely the product of all irreducible polynomials of degree dividing $n$. The following then gives a straightforward algorithm to determine the degrees of the irreducible factors of a polynomial $f(x)$ over $\mathbb{F}_q$:
In this case by inspection $f$ has no linear factors so we only have to check for quadratic factors, hence we only have to compute $\gcd(x^4 - 2, x^9 - x)$. But again by inspection, $(x^4 - 2)(x^4 + 2) = x^8 - 1$
so in fact $x^4 - 2$ must be a product of two (distinct) irreducible quadratics.
In this case, there is a straightforward, mindless thing to do:
And in all cases, you can refine the argument to find all of the roots, and/or to find the factors of $f$.
We can get a bit of a shortcut by observing all of the fourth roots of unity have to be in the splitting field of $f$, so $\mathbb{F}_9$ has to be involved.
We can get even more of a shortcut by observing, for $x \in \overline{\mathbb{F}_3} \setminus 0$:
If we just want the Galois group, we can stop here. :)
The coefficients are reduced modulo 3, so $ x^4-2=x^4-3x^2+1=(x^4-2x^2+1)-x^2=(x^2-1)^2-x^2=(x^2+x-1)(x^2-x-1). $
It is easy to see that neither $x^2+x-1$ nor $x^2-x-1$ have any roots any $F_3$. As they are both quadratic, the roots are in $F_9$. Therefore the Galois group is $Gal(F_9/F_3)$, i.e. cyclic of order two.