## Analyze an argument about field extensions

Consider the following argument. ($\omega$ is a cube root of 1.)

Because $(\mathbb{Q}(\omega\sqrt[3]{2})/\mathbb{Q}) = (\mathbb{Q}(\sqrt[3]{2})) = 3$, we have $(\mathbb{Q}(\omega\sqrt[3]{2}/\mathbb{Q}(\sqrt[3]{2})) = 1$, and thus $\mathbb{Q}(\omega\sqrt[3]{2}) = \mathbb{Q}(\sqrt[3]{2})$.

Is this reasoning valid?

This argument would only hold if one of $\mathbb{Q}(\omega\sqrt[3]{2})$ and $\mathbb{Q}(\sqrt[3]{2})$ contains the other. A more transparent example is $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$; again these are degree 2 extensions of the rationals, but neither contains the other, so they are not equal.