Determine the Galois group of the splitting $x^4 - 4x +2$ over $\mathbb{Q}$

Question

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.

Answer 1

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 $\mathfrak{S}_4$.

Answer 2

It is of degree 4, so if all else fails (guessing roots, other tricks) you can always use the quartic equation. http://mathworld.wolfram.com/QuarticEquation.html