# Find roots or splitting field of a polynomial given its Galois group

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.

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 $\sqrt{-23}$.

• I left out square root (of discriminant) in the fmla for Mink bound. Corrected – Lubin Jan 29 '16 at 21:03

I also came across same question just few months before, and while going through books on Galois Theory, not clear answer found to me (at that time) except some simple illustrations made in the book Galois Theory- Ian Stewart. This may not be a full answer to your question, but the last paragraph quoted here would say this it may be difficult to obtain the roots by Galois Theory.

8.3 How to Use the Galois Group

As an example, consider the polynomial equation $f(t)=t^4−4t^2−5 = 0$ which ... factorizes as $(t^2 + 1)(t^2−5) = 0$, so there are four roots $t = i, −i, \sqrt{5}, -\sqrt{5}$.

Pretend for a moment that we don’t know the explicit zeros $i, −i, \sqrt{5}, -\sqrt{5}$ but that we do know the Galois group G. In fact, consider any quartic polynomial $g(t)$ with the same Galois group as our example $f(t)$ above [isomorphic to $Z_2\times Z_2$]; that way we cannot possibly know the zeros explicitly.

Let them be $\alpha,\beta,\gamma,\delta$. Consider three subfields of $\mathbb{C}$ related to $\alpha,\beta,\gamma,\delta$, namely $$\mathbb{Q} \subseteq \mathbb{Q}(\gamma,\delta) \subseteq \mathbb{Q}(\alpha,\beta,\gamma,\delta)$$ Let $H = \{I, R=(\alpha \beta)(\gamma)(\delta)\} \subseteq G$. Assume that we also know the following two facts:

(1) The numbers fixed by H are precisely those in $\mathbb{Q}(\gamma,\delta)$.

(2) The numbers fixed by G are precisely those in $\mathbb{Q}$.

Then we can work out how to solve the quartic equation $g(t)=0$, as follows. The numbers $\alpha+\beta$ and $\alpha\beta$ are obviously both fixed by $H$. By fact (1) they lie in $\mathbb{Q}(\gamma,\delta)$. But since $$(t-\alpha)(t-\beta)=t^2-(\alpha+\beta)t+\alpha\beta,$$ this means that $\alpha$ and $\beta$ satisfy a quadratic equation whose coefficients are in $\mathbb{Q}(\gamma,\delta)$. That is, we can use the formula for solving a quadratic to express $\alpha,\beta$ in terms of rational functions of $\gamma,\delta$, together with nothing worse than square roots. Thus we obtain $\alpha$, $\beta$ as radical expressions in $\gamma,\delta$.

But we can repeat the trick to find $\gamma,\delta$. The numbers $\gamma+\delta$ and $\gamma\delta$ are fixed by the whole of $G$: they are clearly fixed by $R$, and also by $S=(\alpha)(\beta)(\gamma\delta)$, and these generate $G$. Therefore $\gamma+\delta$ and $\gamma\delta$ belong to $\mathbb{Q}$ by fact (2) above. Therefore $\gamma$ and $\delta$ satisfy a quadratic equation over $\mathbb{Q}$, so they are given by radical expressions in rational numbers. Plugging these into the formulas for $\alpha$ and $\gamma$ we find that all four zeros are radical expressions in rational numbers.

We have not found the formulas explicitly. But we have shown that certain information about the Galois group necessarily implies that they exist. Given more information, we can finish the job completely.