TITLE: Find roots or splitting field of a polynomial given its Galois group
QUESTION [8 upvotes]: A well known result from Galois theory is that the roots of a polynomial can be expressed by a formula using field operations and taking $k$-th roots if and only if the Galois group of the polynomial is solvable. 
How would you actually find such a formula given a solution of the Galois group?
For example I tried to find a root of the polynomial $f(x) = x^3-x-1$ (I know that formulae exist for the roots of polynomials with degree $< 5$, but I am hoping for a more general approach). 
Observing that $f$ is irreducible, and that $f(x)$ has only one real root, it is not difficult to show that $Gal(E/\mathbb{Q}) \simeq S_3$ (where $E$ is the splitting field of $f$ ), because it is a subgroup of $S_3$ and it has at subgroup of order 2 (complex conjugation) and it acts transitively on the set of roots. Since $S_3$ is solvable we know that $E$ is contained in a radical extension of $\mathbb{Q}$.
How would you go from here to find the roots? Is there a general approach or some tricks that can be used? I can't just follow the proof of the theorem that guarantees the existence of a radical extension, because i would need to know the automorphisms.

REPLY [3 votes]: I seem to be giving lots of answers that depend on very special properties of the supplied example. Here’s an argument, tailored to your polynomial $f(x)=x^3-x-1$. Not the general method you were hoping for at all.
First set $\alpha$ to ba a root of your polynomial, which we all know is irreducible over $\Bbb Q$. It’s not too hard to calculate the discriminant of the ring $\Bbb Z[\alpha]$ as the norm down to $\Bbb Q$ of $f'(\alpha)=3\alpha^2-1$; this number is $23$, surprisingly small for a cubic extension. The fact that it’s square-free implies that $\Bbb Z[\alpha]$ is the ring of integers in the field $k=\Bbb Q(\alpha)$.
Our field $k$ clearly is not totally real, since $f$ has only one real root. So in the jargon of algebraic number theory, $r_1=r_2=1$, one real and one (pair of) complex embedding(s). We can apply the Minkowski Bound
$$
M_k=\sqrt{|\Delta_k|}\left(\frac4\pi\right)^{r_2}\frac{n!}{n^n}\,,
$$
which for $n=3$ gives a bound less than $2$, so that $\Bbb Z[\alpha]$ is automatically a principal ideal domain.
Let’s factor the number $23$ there: we certainly know that it’s not prime, since $23$ is ramified.
Now, we already know a number of norm $23$, necessarily a prime divisor of the integer $23$, it’s $3\alpha^2-1$, and indeed $23/(3\alpha^2-1)=4 + 9\alpha - 6\alpha^2$. But better than that, $23/(3\alpha^2-1)^2=3\alpha^2-4$. This number has norm $23$ (because the norm of $23$ itself is $23^3$). So we’ve found the complete factorization of $23$.
Now let’s look more closely at $f(x)=(x-\alpha)g(x)$, for a polynomial $g$ that we can discover by Euclidean Division to be $g(x)=x^2+\alpha x+\alpha^2-1$. And the roots of $g$ are the other roots of $f$; the Quadratic Formula tells you what they are, and the discriminant of $g$ is $\alpha^2-4(\alpha^2-1)=4-3\alpha^2$. which we already know as $-23/(3\alpha^2-1)^2$. Going back to the Quadratic Formula, our other roots are
$$
\rho,\rho'=\frac{\alpha\pm\sqrt\delta}2\>,\>\delta=\frac{-23}{(3\alpha^2-1)^2}\>,\>\sqrt\delta=\frac{\sqrt{-23}}{3\alpha^2-1}\>.
$$
And that gives you your roots of this one very special cubic polynomial in terms of one root $\alpha$ and $\sqrt{-23}$.