TITLE: Galois group of irreducible cubic equation
QUESTION [7 upvotes]: The polynomial $f(x) = x^3 -3x + 1$  is irreducible, and I'm trying to find the splitting field of the polynomial.
We've been given the hint by our lecturer to allow an arbitrary $\alpha$ to be a root and then to check that $\alpha$ is a root, then $2-\alpha^2$ is also root. I've done this with no problem.
My issue is since this is a polynomial of degree $3$, there must be another root that is somehow related to $\alpha$ so that I can create the splitting field and in turn the Galois group of my polynomial but I have no idea what this last root could be.

REPLY [6 votes]: I'm fairly sure that Kaj Hansen's answer (+1) is the intended solution. If you want to say more, you can observe that if $\zeta=e^{2\pi i/9}$, and $u=\zeta+\zeta^{-1}=2\cos(2\pi/9)$, then
$$
u^3-3u=(\zeta+\frac1\zeta)^3-3(\zeta+\frac1\zeta)=\zeta^3+\zeta^{-3}=-1.
$$
By the doubling formula for cosines
$$
u^2-2=2(2\cos^2(2\pi/9)-1)=2\cos(4\pi/9).
$$
You can probably guess at this point that the third zero of your cubic is $2\cos(8\pi/9)$, doubling the angle again. Anyway, the splitting field is the field $\Bbb{Q}(u)=\Bbb{Q}(\zeta)\cap\Bbb{R}$.

This fits into a grander scheme. You saw that the splitting field $K$ is an abelian extension of $\Bbb{Q}$, and also a subfield of the ninth cyclotomic field $\Bbb{Q}(\zeta)$. Clearly all subfields of cyclotomic fields are abelian extensions of the rationals. The highly non-trivial converse, known as the Kronecker-Weber theorem, is that any abelian extension $K$ of $\Bbb{Q}$ is a subfield of some cyclotomic field. The smallest $f$ such that $K\subseteq \Bbb{Q}(e^{2\pi i/f})$ is called the conductor.
Therefore, given a polynomial with rational coefficients and an abelian Galois group, we are guaranteed the ability to write the roots in terms of roots of unity alone. Which roots of unity? The discriminant $d=81$ is a huge tip off.
This is because the discriminant of a cyclotomic field $\Bbb{Q}(e^{2\pi i/p})$
has $p$ as its only prime factor. The formulas tying the discriminant of a subfield to that of its extensions then tell us that $p=3$ would be the only prime factor of the conductor.