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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?

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

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