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The question is to compute the isomorphism class of automorphism group of the splitting field over $\mathbb{Q}$ of $x^3 - 6x + 2$.

I know that this polynomial is irreducible over $\mathbb{Q}$ by Eisenstein's Criterion using $p = 2$, so it has no roots in $\mathbb{Q}$ because it's of degree $3$.

I also computed the discriminant of this polynomial: $\Delta = -4(-6)^3 - 27(2^2) > 0$, so it in fact has three distinct real roots.

I know that any such automorphism is determined by how it acts on these roots, and it also has to permute the roots, so there are at most (or exactly?) six automorphisms. My question is how to determine whether this group of order $6$ (if it is even of order $6$ and not order $3$ -- if it is of order $3$ then it must be $Z_3$, but I'd like to know why it would be of order $3$ and not $6$) is $S_3$ or $Z_2 \times Z_3 \simeq Z_6$ in a nice way.

Edit: Please elementary results only! I am just learning about this in my class, and I am not familiar with a lot of results.

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    $\begingroup$ $Z_3$ if $\Delta$ is a square, $S_3$ otherwise. $\endgroup$ – Lord Shark the Unknown Mar 4 at 21:47
  • $\begingroup$ @LordSharktheUnknown hi, thank you so much for the reply! Unfortunately I am not familiar with that result (we just started talking about these in my class). If you happen to know of a slightly more elementary method of solving the problem then I'm all ears :) $\endgroup$ – 0k33 Mar 4 at 21:53
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    $\begingroup$ The Galois group is a group of permutation of the roots so if the splitting field of a degree $m$ polynomial is a degree $m!$ extension then the Galois group is $S_m$. For $m= 3$ if the cubic is irreducible then the only other case is a splitting field of degree $3$ thus with Galois group $\mathbb{Z}/(3)$ $\endgroup$ – reuns Mar 5 at 0:06

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