Show that a given family of extensions of the rationals does not contain a cube root of 2

Suppose F = \mathbb{Q}(\alpha_1,\ldots,\alpha_k), where \alpha_i^2 \in \mathbb{Q} for each i. Prove that \sqrt[3]{2} \notin F.

We can see, by an inductive argument, that F has degree 2^k over \mathbb{Q} for some k. (In the chain of fields \mathbb{Q} \subseteq \mathbb{Q}(\alpha_1) \subseteq \ldots F, each entry has degree 1 or 2 over its predecessor, and Theorem 14 in D&F applies.)

Suppose now that \sqrt[3]{2} \in F. Then \mathbb{Q}(\sqrt[3]{2}) \subseteq F, and we have the following diagram of field extensions.

A field diagram

Again using Theorem 14, we have that 3 divides 2^k, a contradiction. So in fact \sqrt[3]{2} \notin F.

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