Suppose $F = \mathbb{Q}(\alpha_1,...,\alpha_n)$, where $\alpha_i^2 \in \mathbb{Q}$, for $i = 1,...,n$. Prove that $\sqrt[3]2 \not\in F$.

Question

Suppose $F = \mathbb{Q}(\alpha_1,...,\alpha_n)$, where $\alpha_i^2 \in \mathbb{Q}$, for $i = 1,...,n$. Prove that $\sqrt[3]2 \not\in F$.

Attempt: Suppose $F/\mathbb{Q}$ is a finte extension and let $\mathbb{Q}(\sqrt[3]2)$ be a subfield of F containing $\mathbb{Q}$.
Then $[\mathbb{Q}(\sqrt[3]2): \mathbb{Q}] = 3$

And $[F : \mathbb{Q}]$, depending if $i $ is even or odd will have a different value.

I was hoping to find a contradiction in finding $[F : \mathbb{Q}]$, so that $[\mathbb{Q}(\sqrt[3]2): \mathbb{Q}] = 3$ does not divide $[F : \mathbb{Q}]$. But I am having trouble in finding $[F : \mathbb{Q}]$ , so that $\sqrt[3]2 \not\in F$.

Can someone please help? Thank you!

Answer 1

You can adjoin each $\alpha_i$ one at a time, and each extension will have degree $2$ or $1$, so by the tower law, the whole extension has degree $2^k$. In other words, you can write $F$ as $$F=K_{n−1}(\alpha_n),$$ where $K_{n−1}= \mathbb{Q}(\alpha_1,…,\alpha_{n−1})$. And similarly, you can write $K_{n−1} = K_{n−2}(\alpha_{n−1})$, where $K_{n−2} = (\alpha_1,…,\alpha_{n−2})$,... and so on and so forth, and each extension $K_i$ has degree $1$ or $2$ over $K_{i−1}$. Then apply the tower law to find $[F : \mathbb{Q}] = 2^k$.