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On my homework I have been asked to compute the Galois group of a quintic. I have no idea how to do this, except

(a) I calculated that it was irreducible (brute-force)

(b) Since it is irreducible, its splitting field must have degree divisible by $5$

(c) The Galois group must be a subgroup of $S_5$.

There is also a fact that looks helpful, about $S_n$ only having one normal subgroup for $n \geq 5$. Does this mean there's only one possible Galois extension, or is that only if $S_n$ was already some sort of Galois group?

Aside from that, I have no idea how to do the problem, so can you help me?

Thanks!

P.S. The quintic in question is $x^5 + x - 1 \in \mathbb{Q}[x]$

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    $\begingroup$ $x^5 + x - 1 = (x^2-x+1) (x^3+x^2-1)$ is reducible. $\endgroup$ – lhf May 13 '11 at 15:02
  • $\begingroup$ Well, that is very good to know! Thanks. $\endgroup$ – badatmath May 13 '11 at 15:10
  • $\begingroup$ Related: math.stackexchange.com/questions/1375747 $\endgroup$ – Watson Dec 26 '16 at 13:07
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Here are some general facts about Galois groups of irreducible quintics.

  • There are five transitive subgroups of $S_5$ up to conjugacy: $S_5, A_5, D_5, C_5, F_{20}$. All should be familiar to you except possibly the last one, which is the Frobenius group of invertible affine linear transformations $x \mapsto ax + b$ on $\mathbb{F}_5$.
  • Of these five, only $S_5$ and $F_{20}$ lie outside $A_5$. Thus if the discriminant is a square, the Galois group must be one of $A_5, D_5, C_5$, and otherwise the Galois group must be one of $S_5, F_{20}$.
  • $S_5$ is the only one of these groups containing a transposition. Indeed, a more general statement is true: if a transitive subgroup of $S_n$ contains a transposition and a $p$-cycle for some prime $p > \frac{n}{2}$, then it must be $S_n$. See, for example, these notes by Keith Conrad.

Edit: Ah, I see in the comments that the polynomial is reducible. That explains it, then.

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  • $\begingroup$ Well, we covered Galois groups of cubics and quartics. But now that I know it is reducible I can reduce it to those :D Interesting facts though! $\endgroup$ – badatmath May 13 '11 at 15:18
  • $\begingroup$ I have a similar problem as above except my polynomial is also not solvable by radicals. I am to show that there is at most one subextension of splitting field of degree 2. What are the possible galois groups? $\endgroup$ – Yunus Syed Apr 28 '17 at 0:57
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Here's a general theorem that may help: If $f$ is an irreducible rational polynomial of prime degree $p$ with exactly two non-real roots, then the Galois group of $f$ is the full symmetric group $S_p$.

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  • $\begingroup$ Thanks! I don't think we covered that though... $\endgroup$ – badatmath May 13 '11 at 14:55
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    $\begingroup$ you could try proving it... your teacher would like that! $\endgroup$ – yoyo May 13 '11 at 20:38
  • $\begingroup$ Is it because complex conjugation acts as the transposition that exchanges the two non-real roots and fixes all the rest? $\endgroup$ – Nicolas Miari Jun 6 '16 at 8:34

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