If $\omega$ is the cube root of unity then I know it’s minimal polynomial over $\mathbb{Q}$ is $x^2 + x +1$. How do I know this is still the minimal polynomial of $\omega$ over $\mathbb{Q}(7^{1/3})$?
I know the only roots of $x^2+x+1$ are $\omega$ and $\omega ^2$. And so I thought this still wouldn’t split in $\mathbb{Q}(7^{1/3})$ and therefore must still be irreducible in the larger field.
But does this logic work in general? If a polynomial is irreducible over a field and none of its roots are in the field extension does that mean it is still irreducible in the field extension?
For a simple counterexample, let $f(x)=x^4-2$.
Then $f$ is irreducible over $\mathbb{Q}$, but over the field $\mathbb{Q}(\sqrt{2})$, $f$ factors as $(x^2+\sqrt{2})(x^2-\sqrt{2})$.
However none of the roots of $x^4-2$ are elements of $\mathbb{Q}(\sqrt{2})$.