I was wondering how one computes the Galois group of $x^4-x-1$ over $\mathbb{Q}$. The roots are fairly messy, but going about this just by looking at them indicates that the group should be $S_4$, but I was wondering if someone could provide me with a nicer method of determining the Galois group?
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Did you try the resolvent cubic (which have the same discriminant)? What is the discriminant of it? And, did you try reducing the poly. to mod p to see n-cycles? – offret Apr 08 '18 at 21:13
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1Last time I checked the field $\Bbb{Q}$ has infinitely many elements so I don't see any finite fields here. Retagged a bit. – Jyrki Lahtonen Apr 08 '18 at 21:19
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Are you familiar with Dedekind's theorem? See e.g. here for a proof. That will make light work of this (but isn't really necessary for quartics). – Jyrki Lahtonen Apr 08 '18 at 21:31
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There are some general ways of determining the Galois Group of a quartic. If you have $\textbf{Dummit & Foote}$, there's a section in there in chapter 14 called Galois Groups of Polynomials where he gives an algorithm for determining the Galois group of a quartic $g(x) = x^4 + px^2 + qx + r$ by examining its resolvent cubic:
$h(x) = x^3 - 2px^2 + (p^2 - 4r)x + q^2$
The resolvent cubic is the polynomial whose roots are the differences between the roots of the original polynomial, so you can use it to find information about your original polynomial's Galois group. In particular, the Galois group of $h(x)$ is a subgroup of the Galois group of $g(x)$.
If $g(x) = x^4 - x - 1$ then $p = 0, q = r = -1$, so your resolvent is
$h(x) = x^3 + 4x + 1$. If you look up and follow the algorithm, you can use that to find your Galois group. It'll involve factoring $h(x)$ into irreducibles.
Edit: I changed chapter 11 to chapter 14.
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