Does $\mathbb Q(\alpha )\mathbb Q(\beta )=\mathbb Q(\alpha ,\beta )$? Does $\mathbb Q(\alpha )\mathbb Q(\beta )=\mathbb Q(\alpha ,\beta )$ ?
I recall that $EF=\{ef\mid e\in E, f\in F\}$. It's clear that $\mathbb Q(\alpha )\mathbb Q(\beta )\subset \mathbb Q(\alpha ,\beta )$, but the reverse inclusion looks correct too. I tryied with concret example, and it looks to work, so may be it's equal... But if we make the distinction, it can't be the case. So do you have an example where it doesn't work ? And under which condition the equality hold ?

$EF$ as you've defined it is not a field in general - it is not necessarily closed under addition. 
For example, is $\sqrt{2}+\sqrt{3}$ in $\mathbb Q(\sqrt{2})\mathbb Q(\sqrt{3})$? 
Solve: $$\sqrt{2}+\sqrt{3}=(a+b\sqrt{2})(c+d\sqrt{3})=ac + bc\sqrt{2}+ad\sqrt{3}+bd\sqrt{6},$$ we know $1,\sqrt{2},\sqrt{3},\sqrt{6}$ are linearly independent over $\mathbb Q$, so this means $ac=bd=0$ and $bc=ad=1$. Show that is not possible.
