Let $F = \mathbb{Q}(\sqrt{2}, \sqrt{-3})$ be a biquadratic extension. I want to know about the ring $F \otimes \mathbb{R}$ is it equivalent to the quaternions? I think the notation is ambiguous so there could be three possible tensor products:

* $ \mathbb{Q}(\sqrt{2}, \sqrt{-3}) \otimes\_{\mathbb{Z}} \mathbb{R}$
* $ \mathbb{Q}(\sqrt{2}, \sqrt{-3}) \otimes\_{\mathbb{Q}} \mathbb{R}$
* $ \mathbb{Q}(\sqrt{2}, \sqrt{-3}) \otimes\_{\mathbb{R}} \mathbb{R} \stackrel{?}{\simeq} \mathbb{R}^4$   

however... what about the multiplication. LHS is an *algebra* while RHS is a vector space

I still have trouble grasping the tensor product here, so there three separate questions.