Let $n \in \mathbb{N}$.
How to compute the splitting field of $x^n-2$ over $\mathbb{Q}$?
I did it like this:
$x^n-2=0 \Leftrightarrow x=\sqrt[n]{2}$
Then $x=\sqrt[n]{2}$ is the root of $x^n-2$.
So the splitting field is $\mathbb{Q}(\sqrt[n]{2})$.
I'm not sure if this is complete. Or can another root be found here?
All the roots of this polynomial have the form $\sqrt[n]{2}\times e^{\frac{2\pi ik}{n}}$ when $0\leq k\leq n-1$ is an integer. Indeed, easy to see that they are all roots, and since a polynomial of degree $n$ over a field can't have more than $n$ distinct roots we know these are all the roots. So the splitting field of the polynomial over $\mathbb{Q}$ is the field which is generated by all these roots. You can also check that this field is equal to $\mathbb{Q}(\sqrt[n]{2},e^{\frac{2\pi i}{n}})$. (all the roots obviously belong to this field and hence $\mathbb{Q}(\{roots\})\subseteq\mathbb{Q}(\sqrt[n]{2},e^{\frac{2\pi i}{n}})$. For the other direction note that if any extension field contains all the roots then it is easy to see that $\sqrt[n]{2}$ and $e^{\frac{2\pi i}{n}}$ also must belong to such a field)