Galois group of $x^5-x-1$ over $\Bbb Q$

Question

I am trying to compute the Galois group of $x^5-x-1$ over $ \Bbb Q$. I've shown that this polynomial is irreducible over $\Bbb Q$, by showing that it is irreducible over $\Bbb Z_5$. Let $F$ be the splitting field of $x^5-x-1$ over $\Bbb Q$. This polynomial has $1$ real root and $4$ complex (non-real) roots. If $\alpha \in F$ is the real root of $x^5-x-1$, then $[\Bbb Q(\alpha):\Bbb Q]=5$, and $\Bbb Q(\alpha)\subset \Bbb R$. Since $F \not\subset \Bbb R$, from this we conclude that $[F:\Bbb Q]$ is strictly bigger than $5$, and that the Galois group $G$ has a subgroup of order $5$, i.e., contains a $5$-cycle. But I got stuck here. Any hints?

Answer 1

We know the Galois group will be a transitive subgroup of $S_5$.

The discriminant is 2869, a non-square. So the Galois group is not contained in $A_5$. It will either be $S_5$ or $F_5$: the Frobenius group of order 20. It contains a 5 cycle and two transpositions so we need to know something more to differentiate between the two.

One such tool is the sextic resolvent: The sextic resolvent has a rational root iff the Galois group is conjugate to a subgroup of $F_5$. David Cox - Galois Theory Theorem 13.2.6. In our case this is:

$$y^6 - 8y^5 + 40y^4 - 160y^3 + 400y^2 - 3637y + 9631$$

This has no rational roots therefore the Galois group is $S_5$.