$[\mathbb{Q}(\omega \cdot 2^{\frac{1}{3}}) : \mathbb{Q}] = 3$ and $[\mathbb{Q}(\sqrt{2}) : \mathbb{Q}] = 2$

Question

We know that $[\mathbb{Q}(\omega \cdot 2^{\frac{1}{3}}) : \mathbb{Q}] = 3$ and $[\mathbb{Q}(\sqrt{2}) : \mathbb{Q}] = 2$, where $\omega$ is a primitive cube root of unity.

My algebra instructor then made the claim that this implies that $\omega \cdot 2^{\frac{1}{3}} \notin \mathbb{Q}(\sqrt{2})$. Why does this claim follow ?

Thanks!

Answer 1

If that was the case, then surely we'd have $\Bbb Q(\omega\cdot 2^{\frac{1}{3}}) \subset \Bbb Q(\sqrt 2)$. But by considering the degrees you've already computed, that's impossible.

Answer 2

Hint:

If $K\subset E\subset F$ are field extensions, we have $$[F:K]=[F:E]\cdot[E:K].$$