Example. $\mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right) / \mathbb{Q}$ [2305A]

In (2) of Example 1505, we have seen that $\mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right)$ is the splitting field of $x^{3} - 2$ over $\mathbb{Q}$, so $\mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right)/ \mathbb{Q}$ is Galois. By (5) of Example 2106, we know that $\mathrm{Gal}\left( \mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right) / \mathbb{Q} \right) = \left< \sigma_{1}, \sigma_{2} \right> \cong S_{3}$ with $$\begin{aligned} \sigma_1 : \sqrt[3]{2} &\longmapsto \sqrt[3]{2}\frac{-1 + \sqrt{-3}}{2}, &\qquad \sigma_2 : \sqrt[3]{2} &\longmapsto \sqrt[3]{2}, \\ \sqrt{-3} &\longmapsto \sqrt{-3}, &\qquad \sqrt{-3} &\longmapsto -\sqrt{-3}. \end{aligned}$$ The nontrivial proper subgroups of $\mathrm{Gal}\left( \mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right) / \mathbb{Q} \right)$ are $\left< \sigma_{1} \right>, \left< \sigma_{2} \right>, \left< \sigma_{1}\sigma_{2} \right>, \left< \sigma_{1}^{2}\sigma_{2} \right>$, and $\left< \sigma_{1} \right>$ is normal of order $3$, $\left< \sigma_{2} \right>, \left< \sigma_{1} \sigma_{2} \right>, \left< \sigma_{1}^{2} \sigma_{2} \right>$ are not normal and of order $2$. $$\begin{aligned} \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})^{\langle \sigma_1 \rangle} = \mathbb{Q}(\sqrt{-3}), &\qquad \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})^{\langle \sigma_2 \rangle} = \mathbb{Q}(\sqrt[3]{2}), \\ \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})^{\langle \sigma_1\sigma_2 \rangle} = \mathbb{Q}\left(\sqrt[3]{2}\frac{-1-\sqrt{-3}}{2}\right), &\qquad \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})^{\langle \sigma_1\sigma_2^2 \rangle} = \left(\sqrt[3]{2}\frac{-1+\sqrt{-3}}{2}\right), \end{aligned}$$ and these are all the proper nontrivial sub-extensions of $\mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right) / \mathbb{Q}$.