My professor said the following:
Suppose I know that $\sqrt[3]{3}\notin \mathbb{Q}(\sqrt[3]{2})$ and the two other conjugates of $\sqrt[3]{3}$ over $\mathbb{Q}$ are not real hence also not contained in $\mathbb{Q}(\sqrt[3]{2})$. One immediately deduces that the degree of $\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3})$ over $\mathbb{Q}$ is $9$.
Everything is obvious to me here except how it follows that degree is 9.
I have considered the following tower: $\mathbb{Q}\leq \mathbb{Q}(\sqrt[3]{2})\leq \mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3})$ and $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$. I know that $[\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3}):\mathbb{Q}(\sqrt[3]{2})]\leq 3$ but since $\sqrt[3]{3}\notin \mathbb{Q}(\sqrt[3]{2})$ then it cannot be $1$. How it follows that it cannot be $2$?
Can anyone show this in clear way, please?
If $[\mathbf Q(\sqrt[3]{3}, \sqrt[3]{2}) : \mathbf Q(\sqrt[3]{2})]$ is equal to $2$, then the minimal polynomial $m(X) \in \mathbf Q(\sqrt[3]{2})[X]$ of $\sqrt[3]{3}$ over $\mathbf Q(\sqrt[3]{2})$ will have to be of degree two.
Is this possible? We know that $\sqrt[3]{3}$ is a root of the polynomial $X^3 - 3 \in \mathbf Q(\sqrt[3]{2})[X]$. So its minimal polynomial $m(X)$ must divide $X^3 - 3$.
If $m(X)$ is of degree $2$, then there are two possibilities: either $$m(X) = (X - \sqrt[3]{3})(X - \sqrt[3]{3}e^{2\pi i / 3}),$$ or $$m(X) = (X - \sqrt[3]{3})(X - \sqrt[3]{3}e^{4\pi i / 3}).$$ However, if you expand out these polynomials, you'll see that the coefficients are non-real, and hence are not actually elements of $\mathbf Q(\sqrt[3]{2})$. This is a contradiction.