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

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?

• Do you know Dedekind's theorem on Galois groups? If so, I claim that this theorem is enough to compute the Galois group of your polynomial. Apr 22, 2020 at 7:30
• I am not an expert in galois-groups, but isn't the complex conjugate a transposition , and does the fact that we have non-real roots not imply that the group contains a transposition which would complete the proof that the galois group is $S_5$ ? Apr 22, 2020 at 7:51
• @Peter Here the complex conjugate yields a product of two transpositions (it interchanges both pairs of conjugate complex roots) Apr 22, 2020 at 8:11
• @Gaussian The only primes $p\leq 31$ for which $P=X^5-X-1$ is not irreducible are $p=2$ (indeed $P=(X^2+X+1)(X^3+X^2+1)$ mod $2$) and $p=7$ (indeed $P=(X^2-X+3)(X^3+X^2-2X+2)$ mod $7$). So Dedekind's theorem tells us that $G$ contains a permutation of type (2,3), and nothing more. Further, the discriminant of $P$ is $2869=19\times 151$, a non-square, so we know that $G \not\subseteq A_5$. It is not obvious (at least to me) that this suffices to show that $G=S_5$. Apr 22, 2020 at 8:48
• @Peter as Ewan points out, complex conjugation is a product of two (disjoint) transpositions, and with the 5-cycle, this generates the alternating group $A_5$ (note both generators are even permutations so we definitely have containment). However, when you combine this with the discriminant in Ewan's argument above, you do then get the whole of $S_5$. Apr 22, 2020 at 9:16

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 $$S_5$$.