What is the intermediate field of the splitting filed of $x^6+3$ over $\mathbb Q$?

Question

I have worked out that the Galois group of $x^6+3$ should be $S_3$, and let $\alpha$ be a root of $x^6+3$, $w = \frac{1+\alpha^3}{2}$, then $\alpha, w \alpha, w^2 \alpha, -\alpha, -w \alpha, -w^2 \alpha$ are roots of $x^6+3$. The Galois group of it should be $\mathbb Q(\alpha)$. Then an obvious intermeidate field is $\mathbb Q(w)$ which is of degree $2$. How should I locate other intermeidate fields?

Answer 1

There will be three fields of degree $3$, the fixed fields of the subgroups of order $2$ in the Galois group $G\cong S_3$. Let $\sigma$ be the Galois group element mapping $\alpha$ to $-\alpha$. It must have order $2$. Then $$\sigma(\omega)=\sigma\left(\frac12(1+\alpha^3) \right)=\frac12(1-\alpha^3)=\omega^2.$$ An element of the fixed field of $\left<\sigma\right>$ is then $$\omega\alpha+\sigma(\omega\alpha)=\omega\alpha-\omega^2\alpha=\beta_1$$ say. If we choose $\alpha=\exp(\pi i/6)\sqrt[6]3$ then $$\beta_1=\sqrt[6]3(\exp(5\pi i/6)-\exp(-\pi i/2))$$ etc. Likewise, the other two cubic extensions will be generated by $\beta_2=\alpha-\omega\alpha$ and $\beta_3=\alpha-\omega^2\alpha$.