By Eisenstein's criterion this is irreducible over Q, and I know it has only 2 real roots. Is this actually solvable by radicals? I know the Galois group isomorphic to some subgroup of $S_4$, which is a solvable group, which means the Galois group is solvable, but I can't figure out how to do it.
[Math] Determine the Galois group of the splitting $x^4 – 4x +2$ over $\mathbb{Q}$
galois-theory
Related Solutions
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 $\sqrt{-23}$.
It's a general fact that if $f(x)$ is an irreducible polynomial over $\mathbb{Q}$ of prime degree $p\geq 5$ having exactly $p-2$ real roots, then the Galois group of $f$ is $S_p$.
In this case $f(x)=24x^5-30x^4+5$ is irreducible by Eisenstein's criterion and Gauss's lemma. Also $f(-1)<0$, $f(0)>0$, $f(1)<0$, and $f(2)>0$, so $f$ has at least three real roots by the intermediate value theorem.
On the other hand, $f^{\prime}(x)=120x^3(x-1)$, which has exactly two roots. Hence $f(x)$ has at most three real roots by Rolle's theorem.
Thus $f(x)$ has exactly three real roots, hence the Galois group of $f$ over $\mathbb{Q}$ is $S_5$. And $S_5$ is not a solvable group, so $f$ isn't solvable by radicals.
Best Answer
Well, when you derive your polynomial $f$ you get only one real root $1$ for $f'$. Then you see that $f$ is decreasing from $-\infty$ to $1$ and increasing from $1$ to $+\infty$. Now since $f(1)=-1$. We see that $f$ has two real roots and two non-real conjugate complexes.
Now it follows that the coomplex conjugation will fix two roots (the real one) and exchange the two non-real ones. That is the Galois group induces a transposition of the set of roots. It follows that your Galois group as a subgroup of the permutations of the roots $\mathfrak{S}_4$ will contain a transposition. Up to conjugation there are only two transitive subgroups which contain a transposition :
$$\mathfrak{S}_4\text{ and } D_4 $$
Where $D_4$ is one $2$-Sylow of $\mathfrak{S}_4$.
Reducing your polynomial mod $5$ you get $x^4+x+2$. Mod $5$ this factorizes :
$$(x-2)(x^3+2x^2-x-1) $$
Remarking that both polynomials in the factorization are irreducible mod $5$, you have a theorem in your course that say this implies that you have a $3$-cycle in your Galois group. Because of the alternative we already highlighted the Galois group cannot but be $\mathfrak{S}_4$.