What is $ \mathbb{Q}(\sqrt{2}, \sqrt{-3}) \otimes \mathbb{R}$? 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.
 A: The first two of your tensor products are the same. The last does not
make sense, alas. The tensor product is a commutative ring, so certainly
not the quaternions.
If $K=\Bbb Q(\sqrt2,\sqrt{-3})$ then $K$ has two essentially
different embeddings in $\Bbb C$, viz., $\sigma_1:\sqrt2\mapsto\sqrt2$,
$\sqrt{-3}\mapsto i\sqrt3$, and $\sigma_2:\sqrt2\mapsto-\sqrt2$,
$\sqrt{-3}\mapsto i\sqrt3$. This extends to a map $\Sigma:K\to\Bbb C^2$
with $\Sigma(\alpha)=(\sigma_1(\alpha),\sigma_2(\alpha))$
which is a ring homomorphism. This extends to $K\otimes\Bbb R\to
\Bbb C^2$ by $\alpha\otimes x\mapsto(x\sigma_1(\alpha),x\sigma_2(\alpha))$.
This is a ring isomorphism.
A: $\mathbb{Q}(\sqrt{2}, \sqrt{-3}) \simeq \mathbb{Q}[t, u] / \langle t^2 - 2, u^2 + 3 \rangle$.  Therefore, $\mathbb{Q}(\sqrt{2}, \sqrt{-3}) \otimes_{\mathbb{Q}} \mathbb{R} \simeq \mathbb{R}[t, u] / \langle t^2 - 2, u^2 + 3 \rangle$.  Now, the $u$ part $\mathbb{R}[u] / \langle u^2 + 3 \rangle$ is isomorphic to $\mathbb{C}$, so $\mathbb{Q}(\sqrt{2}, \sqrt{-3}) \otimes_{\mathbb{Q}} \mathbb{R} \simeq \mathbb{C}[t] / \langle t^2 - 2 \rangle$.  Now, since $t^2 - 2 = (t - \sqrt{2}) (t + \sqrt{2})$, by the Chinese Remainder Theorem, this is further isomorphic to $\mathbb{C}[t] / \langle t - \sqrt{2} \rangle \oplus \mathbb{C}[t] / \langle t + \sqrt{2} \rangle \simeq \mathbb{C} \oplus \mathbb{C}$.
(Incidentally, $\mathbb{Q}(\sqrt{2}, \sqrt{-3}) \otimes_{\mathbb{Q}} \mathbb{R}$ is automatically a commutative ring, so there is no way it could be isomorphic to the quaternions.)
