Let $F = \mathbb{Q}(a_1, a_2, . . . , a_n)$ with $a_i^2 \in \mathbb{Q}.$ Prove that $ \sqrt[3]2 \notin F$.

Question

So I tried by claiming each extension $ \mathbb{Q}(a_i) $ was of degree 2 because of the $a_i^2 \in \mathbb{Q}.$ Apparently that wasn't necessarily true. I said that $ \mathbb{Q}(\sqrt[3]2) $ was a degree 3 extension and $F$ had degree $2^n$ so, it wasn't possible. But apparently that isn't necessarily true? SO I'm not too sure how to do it.

Fyi, this was a homework problem but no solutions were given and so, just wanna know how to do it, or get some help.

Answer 1

Hint: $[\mathbb{Q}(a_1,\dots,a_{k+1}:\mathbb{Q}(a_1,\dots,a_k)]=1$ or $2$.