How to show that $\mathbb{Q} \subset \mathbb{Q}(i, \sqrt{3}, \sqrt[3]{2})$ is a normal extension?
I tried:
The minimal polynomial of $i$ over $\mathbb{Q}$ is $x^2+1$. Irreducible with Eisenstein $p=2$.
The minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}$ is $x^2-3$. Irreducible with Eisenstein $p=3$.
The minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}$ is $x^3-2$. Irreducible with Eisenstein $p=2$.
But the other roots of $\sqrt[3]{2}$ are not real:
$x_2=\sqrt[3]{2} \omega$ and $ x_3=\sqrt[3]{2} \omega^2$, with $\omega={\frac{-1+i\sqrt{3}}{2}}$.
So it doesn't split in $\mathbb{Q}$. But then it can't be normal?
How to show it in the right way that it is normal?
Note that $\omega \in \mathbb{Q}(i, \sqrt{3},\sqrt[3]{2})$. So the field $\mathbb{Q}(i,\sqrt{3},\sqrt[3]{2})$ is the splitting field of $(x^2+1)(x^2-3)(x^3-2)$, a nice separable polynomial. Notice that the roots being real or not does not matter here!(Just like $i$ is not real, but $\mathbb{Q}(i)/\mathbb{Q}$ is a Galois extension!).