Consider a field extension $\mathbb{Q}(\alpha,\beta)/\mathbb{Q}$ with $\alpha = 2^{\frac{1}{3}}$ and $\beta = e^{\frac{2\pi i}{3}}$.
g is the minimal polynomial of $\alpha$ over $\mathbb{Q}$, which is $X^3 - 2$
How to determine the minimal polynomial of $a$ over $\mathbb{Q}(\beta)$? What is the factorization of $g$ into irreducibles in $\mathbb{Q}(\beta)[X]$, $\mathbb{Q}(\alpha)[X]$, and $\mathbb{Q}(\alpha,\beta)[X]$? thank you!
The polynomial $\;x^3-2\in\Bbb Q(b)[x]\;$ is irreducible since otherwise it'd have a root in that field. But its roots (in some algebraic closure of the rationals) are $\;\sqrt[3]2\,,\,\,\sqrt[3]2\,b\,,\,\sqrt[3]2\,b^2\;$, and none of these belongs to $\;\Bbb Q(b)\;$, because:
$$[\Bbb Q(b):\Bbb Q]=2\;\;,\;\;\text{since}\;\;x^2+x+1\in\Bbb Q[x]$$
is the minimal polynomial of $\;b\;$ over $\;\Bbb Q\;$. Try now to develop these ideas.