Find splitting field of $x^5-4x+2$ over $\mathbb{Q}$

I know how to get the splitting field from the roots but I just can't come up with any factorisation that gives me roots. Any suggestions?

Solution 1:

It is quite difficult to find explicitly the splitting field. The Extension is not solvable.

In fact the Galois group of the splitting field of $$ f(x)=x^5-4x+2$$ is isomorphic to $S_5$.

This is because the polynomial has exactly two complex roots (Study the graph of the function to determine it), hence the complex conjugation acts on the set of roots as a transposition.

Moreover, since $5$ divides the order of the group, there exist a $5$-cycle.

So the Galois group $G$ is a subgroup of $S_5$ generated by a transposition and a $5$-cycle. Since $5$ is prime this means that $G=S_5$ (If $p$ is a prime number, a transposition and a $p$-cycle always generate $S_p$).

So you can not express the roots of the polynomial $f$ by radicals, but at least you can say that the degree of the extension is $5!$.