Degree and Galois group of $\mathbb{Q}(\sqrt[3]{2},\sqrt{-3})/\mathbb{Q}(\sqrt{-3})$.

Question

I would like to find the degree and Galois group of $\mathbb{Q}(\sqrt[3]{2},\sqrt{-3})/\mathbb{Q}(\sqrt{-3})$.

I usually can compute the degree of an extension over $\mathbb{Q}$ as one only needs to find the minimal polynomial over $\mathbb{Q}$ which is relatively easy to compute. On the other hand, whenever I have had to compute the minimal polynomial over some other extension of $\mathbb{Q}$ (for example, $\mathbb{Q}(\zeta_i,\sqrt{2})/\mathbb{Q}(\sqrt{2})$), the calculation has still been easy as it is an imaginary field over a real field. However, in the example I have proposed above, I do not seem to be able to proceed with the extension being real but base field being imaginary. Surely, $x^3-2$ is one of the polynomials, but how do I know if its the minimal polynomial over $\mathbb{Q}(\sqrt{-3})$?

Answer 1

Since the degree of the polynomial $x^3-2$ is $3,$ to prove that it is irreducible over $\Bbb Q(\sqrt{-3}),$ you only need to check that $∀y∈\Bbb Q(\sqrt{-3}),y^3≠2.$ Write $y=a+b\sqrt{-3}$ with $a,b∈\Bbb Q$ and solve.

Now, this proves that the degree of the extension is $3.$ To derive the Galois group, check that your extension is normal. Hint: the three cubic roots of $1$ belong to $\Bbb Q(\sqrt{-3}).$