Let $E/K$ be an extension and $L_1,L_2$ intermediate fields of $E/K$ with $L_i:K$ finite. Then necessarily $L_1L_2:L_2$ $\le$ $L_1:K$. Prove $L_1L_2:L_2$ is not necessarily a factor of $L_1:K$. Hint: Consider $L_1= \mathbb{Q}(2^{\frac{1}{3}})$ and $L_2= \mathbb{Q}(e^{\frac{2\pi i}{3}}2^{\frac{1}{3}})$
My Progress:
Let $K=\mathbb{Q}$. Then $L_1:K=3$
Let $f(X)= X^2+ e^{\frac{2\pi i}{3}}2^{\frac{1}{3}}X+(e^{\frac{2\pi i}{3}}2^{\frac{1}{3}})^{2}$
Then $f(X)$ is an element of $L_2[X]$ and $f(2^{\frac{1}{3}})=0$. Now I want to show that $f(X)$ is the minimal polynomial of $2^{\frac{1}{3}}$ over $L_2$. So I suppose it’s not i.e. there exists a $g(X)$ element of $L_2[X]$ such that
$g$ | $f$, $g\not=0$ and most importantly $g(2^{\frac{1}{3}})=0$.Implying $g$ is of degree 1. And this is where I get stuck, how do I show that $2^{\frac{1}{3}}$ is not an element of $L_2$?