## 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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