Basis for $\mathbb{Q}(\alpha, \beta)$ over $\mathbb{Q}$ One can prove that a basis for $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ is the set $\{1, \sqrt{2}, \sqrt{3}, \sqrt{6} \}$.  This got me wondering if the following is true:

Let $\alpha, \beta$ be elements that are (1) not rational and (2) not scalar multiples of each other (where the scalars come from $\mathbb{Q}$).  Then is $\{1, \alpha, \beta, \alpha\beta\}$ a basis for $\mathbb{Q}(\alpha, \beta)$ over $\mathbb{Q}$?  

(Note: I said $\alpha, \beta$ are not rational instead of irrational since I am not assuming $\alpha, \beta \in \mathbb{R}$.)
If this is true, can you provide a proof?  And if not, perhaps give a counterexample?

(Compiled from the comments and posted as CW in order to mark the question as answered.)
The proposition does not hold true in general. Counterexamples:


*

*finite extension: $\;\mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$

*infinite extension: $\;\mathbb{Q}(\pi, \sqrt{2})$
