Calculating the Galois group of an (irreducible) quintic 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 $x^5 + x - 1 \in \mathbb{Q}[x]$
 A: 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 $S_p$.
A: 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. 
