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?

• Consider $\mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$. – dxiv Apr 3 '17 at 23:38
• No: $\mathbf Q(\pi,\sqrt 2)$ is not even finite-dimensional. – Bernard Apr 3 '17 at 23:38
• @Bernard This is a good one, thanks. – Sam Y. Apr 3 '17 at 23:43

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