Given the polynomial $p(x)=x^6-6x^3+7 \in \mathbb Q$, find its splitting field $\mathbb F\subset \mathbb C$ and the Galois group of the extension $\mathbb F /\mathbb Q$. In fact, the exercise asked to show that $p$ is irreducible, then the degree and a basis of the splitting field as a vector space over $\mathbb Q$, and finally the Galois group of the extension.
The six complex roots are $\zeta_3^k \cdot \sqrt[3]{3\pm \sqrt2}$ with $\zeta_3$ a primitive third root of 1 and $k=0,1,2$.
I've worked out a clumsy solution, so I was thinking to myself:
- Is there a tricky/easy/obvious(!) way to show that $p$ is irreducible over $\mathbb Q$?
- Is there an easy way to show that $[\mathbb Q(\sqrt 2, \sqrt[3]{3\pm\sqrt2}):\mathbb Q(\sqrt 2, \sqrt[3]{3+\sqrt2})]=3$? It seems to me that there's no way to avoid this step or a similar one, like $[\mathbb Q(\sqrt 2, \sqrt[3]{3-\sqrt2}),\mathbb Q(\sqrt 2, \sqrt[3]7)]=3$ or else (you could add $\zeta_3$ just after $\sqrt 2$..). I'm quite sure that $[\mathbb F:\mathbb Q]=36$, but the only non trivial fact in the proof is the one I've just mentioned. Anyway, I could be missing something..
- Claim: the Galois group is $S_3\times S_3$, that is $D_3\times D_3$ (dihedral). Am I right? How can I prove it?
There is no "big hammer" for proving irreducibility. Eisenstein's criterion sometimes works, but does not do so for the problem polynomial nor for the additional polynomial in your comment.
Tricky/obvious solution: $x$ only appears as powers of its cube, so consider the substitution $x^3 \rightarrow y$, yielding $y^2-6y+7$. This has discriminant 36-28 = 8 which is not a square, so $y^2-6y+7$ does not split over $\mathbb{Q}$. Let $r \in \mathbb{C}$ be (either) root of this equation. Then $y^2-6y+7$ does split over $\mathbb{Q}(r)$.
Index = 3: This is easy since you know how to split the cube, yielding three roots in $x$ per root in $y$.
Solved this way, you get a sequence of two extensions from which you should be able to write down your Galois group.