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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?

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    $\begingroup$ 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. $\endgroup$
    – Gaussian
    Apr 22, 2020 at 7:30
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    $\begingroup$ 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 $S_5$ ? $\endgroup$
    – Peter
    Apr 22, 2020 at 7:51
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    $\begingroup$ @Peter Here the complex conjugate yields a product of two transpositions (it interchanges both pairs of conjugate complex roots) $\endgroup$ Apr 22, 2020 at 8:11
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    $\begingroup$ @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 $G=S_5$. $\endgroup$ Apr 22, 2020 at 8:48
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    $\begingroup$ @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 $S_5$. $\endgroup$
    – Matt B
    Apr 22, 2020 at 9:16

1 Answer 1

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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 $S_5$.

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