I want to understand, how to deal with following task:
Which of the following extensions are normal?
- $\mathbb{Q}(i \sqrt[6]{3}) / \mathbb{Q}$;
- $\mathbb{C}(t) / \mathbb{C}\left(t^4\right)$;
- $\mathbb{R}(t) / \mathbb{R}\left(t^4\right)$;
The first one is pretty obvious, if i am right. I know, that extension is finite and normal iff it is splitting field for some polynomial. In first example polynomial is $x^6+3$, and it is easy to calculate, that every other root of that polynomial is algebraic over $\mathbb{Q}(i \sqrt[6]{3})$. It means that $\mathbb{Q}(i \sqrt[6]{3}) / \mathbb{Q}$ is splitting field, i.e. extension is normal.
But i don't know how to approach 2nd and 3rd cases. I think that both extensions are finite, but i can't think of some polynomials like i did in 1st example. Can i use an exact same logic or i need something different?
Upd: as @RobertShore mentioned i didn't wrote proof of first example correct. I meant, that every other root of $x^6+3$ is contained in $\mathbb{Q}(i \sqrt[6]{3})$, which leads us to $\mathbb{Q}(i \sqrt[6]{3}) / \mathbb{Q}$ being normal, as it is splitting field for $x^6+3$.
The minimal polynomial of $t$ in $\Bbb C(t^4)$ or in $\Bbb R(t^4)$ is simply $p(x)=x^4-t^4$. Moreover, the roots of $p(x)$ (in a suitable algebraic extension) are simply $\pm t, \pm it$.
This tells you that $\Bbb C(t)$ is normal over $\Bbb C(t^4)$, but since $i \notin \Bbb R(t^4)$, we also find that $\Bbb R(t)$ is not normal over $\Bbb R(t^4)$.
And yes, since $(i \sqrt[6]{3})^3=-i\sqrt 3$, you are correct that the sixth roots of unity are in $\Bbb Q(i \sqrt[6]{3})$ and that field is normal over $\Bbb Q$.