Suppose $E = \mathbb{Q}(\alpha_1,\alpha_2,...,\alpha_n)$ where $\alpha_i^2\in \mathbb{Q}$ for $i=1,2,...,n.$ Prove that $2^{1/3} \not\in E$.
I thought I could prove it by contradiction but I have made no progress. Can someone share some insights? Thanks a lot!
This is terse, but hopefully helpful: The minimal polynomial of $2^{1/3}$ over $\mathbb{Q}$ will be $x^3-2$. This is irreducible over $\mathbb{Q}$, so $3$ must divide the degree of the extension $\mathbb{Q} \to \mathbb{Q}(2^{1/3})$. But the degree of the extension $\mathbb{Q} \to E$ will be a power of $2$, so $\mathbb{Q}(2^{1/3})$ cannot be an intermediate field.
If you have any question about this, please edit your post to include those questions there. :)