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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?

blancket
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    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. – Gaussian Apr 22 '20 at 07:30
  • @Gaussian I didn't, but I want to know how that theorem implies the result! – blancket Apr 22 '20 at 07:34
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    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 $S_5$ ? – Peter Apr 22 '20 at 07:51
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    @Peter Here the complex conjugate yields a product of two transpositions (it interchanges both pairs of conjugate complex roots) – Ewan Delanoy Apr 22 '20 at 08:11
  • @Peter Does a $5$-cycle and a product of two transpositions generate $S_5$? I know that a $5$-cycle and a transposition generates $S_5$, but I'm not sure in this case. – blancket Apr 22 '20 at 08:12
  • @Troaro I am sceptical in this case, but as said, I am not an expert of galois-groups. – Peter Apr 22 '20 at 08:45
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    @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 $G=S_5$. – Ewan Delanoy Apr 22 '20 at 08:48
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    @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 $S_5$. – Matt B Apr 22 '20 at 09:16
  • @MattB That would be the missing piece of the puzzle ; but how do you show that a product of two disjoint transpos and a $5$-cycle generate $A_5$ ? – Ewan Delanoy Apr 22 '20 at 09:20
  • @EwanDelanoy I just used a computer to check and not sure of a direct proof as the only standard presentation I know is that alternating groups are generated by all $3$-cycles. You can certainly move the fixed point in the double transpos by conjugating by some power of the $5$-cycle and conjugation of the $5$-cycle by these should then yield all $5$-cycles... – Matt B Apr 22 '20 at 09:32
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    In group-props it is stated that a 5-cycle and a 3-cycle generates $A_5$. It is not difficult to show that a 5-cycle and the product of two transpositions can generate a 3-cycle. – Peter Apr 22 '20 at 09:40
  • @MattB Wait a minute, group-props also says that (1,2,3,4,5) with (2,5)(4,3) generates "$D_{10}$ in $A_5$" which is a strict subgroup of $A_5$. Too bad ... – Ewan Delanoy Apr 22 '20 at 09:59
  • ah thanks. I did originally think it might only be dihedral, but my one check with (1,2,3,4,5) and (1,2)(3,4) gave me $A_5$ and (wrongly) made me think this was always true – Matt B Apr 22 '20 at 10:06
  • @EwanDelanoy This confuses me even more. How can we then create a 3-cycle ? $D_{10}$ cannot have an element of order $3$, what am I missing ? – Peter Apr 22 '20 at 10:07
  • @Peter Well, obviously this $D_{10}$ counterexample shows that we need yet another idea to rule it out. – Ewan Delanoy Apr 22 '20 at 10:10
  • What I mean : Does the possibility to create a 3-cycle not contradict that $D_{10}$ is created by the two generators ? – Peter Apr 22 '20 at 10:29
  • @Peter $D_{10}$ contains 5-cycles but no 3-cycles, so I don't see any contradiction – Ewan Delanoy Apr 22 '20 at 11:09
  • @EwanDelanoy I think, this is because we do not have all possible products, but only that every element can be generated this way. It depends on how we define "generated by". I am only confused because usually we consider all possible products. – Peter Apr 22 '20 at 11:11
  • @Troaro: I didn't read the comments given above, but you already now that the galois group contains a 5-cycle. Furthermore, if you use Dedekind's Theorem mod $2$, you'll get a $2$-cycle in the Galois group. Now, enough to show that a subgroup of $\mathfrak{S}_5$ containing a $5$-cycle and a $2$-cycle is equal to $\mathfrak{S}_5$. – Gaussian Apr 22 '20 at 14:59
  • @Gaussian No, what you just said is wrong. See my first comment above – Ewan Delanoy Apr 22 '20 at 15:02
  • Please consider taking this discussion to chat. It will be faster for you and cleaner for the site. – robjohn Apr 22 '20 at 18:38
  • @EwanDelanoy: No, I don't see anyhting contradicting what I said in my comment. It is indeed true that $\mathfrak{S}_5$ is generated by any combination of a transposition (what I call a $2$-cycle) and a $5$-cycle. See this question for a proof: https://math.stackexchange.com/questions/355516/why-is-s-5-generated-by-any-combination-of-a-transposition-and-a-5-cycle . Let me be more precise about Dedekind's theorem: it gives you a $(2,3)$ permutation $\sigma$ in $G$, hence a transposition (which is $\sigma^3$). – Gaussian Apr 22 '20 at 20:30
  • See Example 3.33. (A simpler argument is the following: the Galois group contains a product of two disjoint cycles, one of length 2 and the other of length 3. This shows that it contains a transposition. Since it also contains a 5-cycle, the Galois group is $S_5$.) – user26857 Apr 22 '20 at 22:31
  • @Gaussian Got it. I didn't realize that if $\sigma$ is a (2,3)-permutation, then $\sigma^3$ is a transposition. Thanks for the explanation – Ewan Delanoy Apr 23 '20 at 05:36

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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 $S_5$.