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.
