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I am doing this question:

Show that $f(x)=x^3-3x-1 $ is an irreducible element of $\mathbb{Z}[x]$. Compute the Galois group of the splitting field of $f$ over $\mathbb{Q}$ and over $\mathbb{R}$.

First by Eisenstein's criteria, let $y=x-1$ and substitute it to get $y^3+3y^2-3$, with prime 3 I know it has no root in its fraction field, $\mathbb{Q}$, hence irreducible in $\mathbb{Z}[x]$.

Next I observe that this polynomial happens to have 3 real roots. Taking first derivative, $3x^2-3$ has $\pm1$. Plug in $f(-1)>0, f(1)<0$. So it should be separable over $\mathbb{R}$. Furthermore, such Galois group is isomorphic $\{1\}$.

Last step is to factor it over $\mathbb{Q}$. First introduce a root $\theta$, $f$ is its minimal polynomial, of degree 3. $f(x)=(x-\theta)(x^2+x\theta+\theta^2-3)$. Check that $\theta$ is not a root of the second factor, as otherwise $\theta=\pm1$. So now introduce another root $\alpha$, to $g(x)=x^2+x\theta+\theta^2-3$. Factor that again, $g(x)=(x-\alpha)(x+\alpha +\theta)$. The last root is $-\alpha-\theta$.

My question is that what is the Galois group of that over $\mathbb{Q}$. $[\mathbb{Q}(\theta)(\alpha):\mathbb{Q}]=[\mathbb{Q}(\theta)(\alpha):\mathbb{Q}(\theta)][\mathbb{Q}(\theta):\mathbb{Q}]=2\cdot 3 = 6$ I think is clear at this moment.

Do I consider $\operatorname{Gal}(\mathbb{Q}(\theta)(\alpha)/\mathbb{Q})$ isomorphic to $S_3$ or $S_2$?

Since $\alpha$ depends on the choice of $\theta$, if I consider that in general, it seems that they all should be roots to $f$ originally, and thus any permutation would be counted. However at a moment I feel that $\alpha$ itself has a minimal polynomial of degree 2, and then it can be switched with $-\alpha-\theta$. Then only $S_2$ would satisfy.

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    $\begingroup$ See this question for references. In Conrad's article your question is answered, see here. The Galois group over $\Bbb{Q}$ is $A_3$, see Example $2.2$. $\endgroup$ – Dietrich Burde Aug 21 '18 at 20:37
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    $\begingroup$ ummm; the roots are $-2 \cos \frac{2 \pi}{9} \; , \; $ $-2 \cos \frac{4 \pi}{9} \; , \; $ $-2 \cos \frac{8 \pi}{9} \; . \; $ Just one of those things. $\endgroup$ – Will Jagy Aug 21 '18 at 21:26
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    $\begingroup$ And if you can't be bothered with cosines you can just prove that if $\theta$ is one of the zeros $\alpha=2-\theta^2$ is another. $\endgroup$ – Jyrki Lahtonen Aug 24 '18 at 9:39

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