How to prove the tensor product of two copies of $\mathbb{H}$ is isomorphic to $M_4 (\mathbb{R})$? 
How to prove the tensor product over $\mathbb{R}$ of two copies of the quaternions is isomorphic to the matrix algebra $M_4 (\mathbb{R})$ as algebras over $\mathbb{R}$? More precisely, the problem is to show the isomorphism $\mathbb{H} \otimes_\mathbb{R} \mathbb{H} \cong M_4 (\mathbb{R})$.

On the book "Spin Geometry" by Lawson and Michelsohn, page 27, there is an isomorphism defined by sending $q_1 \otimes q_2$ to the real endomorphism of $\mathbb{H}$ which is given by $x \mapsto q_1 x \bar{q_2}$, but I don't know how to deduce that this real algebra homomorphism is in fact an isomorphism.
 A: The Brauer group of $\mathbb R$ is $Br(\mathbb R)\cong\mathbb Z/2\mathbb Z$; one way to see this is to note that there are exactly two (iso)classes of central simple f.d. real algebras, in view of Frobenius' theorem which classifies them, 
so the there is no choice for the group structure. It follows that the tensor square of every central simple real algebra is isomorphic to a matrix algebra. In particular, $\mathbb H\otimes_{\mathbb R}\mathbb H$ is a real matrix algebra of dimension $16$, so must be isomorphic to $M_{4}(\mathbb R)$.
A: It's a basic fact (here's a proof in the second proposition on page 157) that the tensor product of two central simple algebras is another central simple algebra. A proof should be available wherever central simple algebras are discussed.

*

*Another location in Jacobson's Basic Algebra II on page 218-219.


*Another location in Rowen's Ring theory Theorem 1.7.27.


*Another location in notes by Morandi.
By simplicity of the ring, the kernel of your (nonzero) ring homomorphism is automatically $\{0\}$, showing it is injective.
Finally, since the image and codomain are both 16-$\Bbb R$-dimensional, they are equal, showing the must also be surjective.
A: I am extremely late to the party, but here is another more direct approach to this problem.
Consider the map $$N:\mathbb{H}\rightarrow \mathbb{H}:q=x+yi+zj+tij\mapsto \overline{q}=x-yi-zj-tij.$$ It's easy to see that $N$ is an anti-algebra automorphism of $\mathbb{H}$ of order two, i.e. $N$ is an involution on $\mathbb{H}$. Recall that if $A$ is an algebra, $A^{\text{op}}$ is an algebra by reversing the multiplication. Using the map $N$ we get an algebra isomorphism $\mathbb{H}\cong \mathbb{H}^{\text{op}}$. 
It's well known that $A\otimes_k A^{\text{op}}\cong \text{End}_k(A)\cong M_n(k)$ for finite-dimensional simple algebras $A$, where $n=\dim_k(A)$. Thus $\mathbb{H}\otimes_{\mathbb R} \mathbb{H}\cong \mathbb{H}\otimes_{\mathbb R} \mathbb{H}^{\text{op}}\cong \text{End}_{\mathbb{R}}(\mathbb{H})=M_4(\mathbb{R})$. This shows that the quaternions have order two in the Brauer group of $\mathbb{R}$.
The above approach is very natural and in fact this technique shows that the inverse of an element in a Brauer group is given by the opposite algebra. For the quaternions the opposite happens to be isomorphic to the original algebra. This approach is also very natural since the map $N$ is simply the norm on $\mathbb{H}$. It can be used to calculate inverses, indeed, the inverse of $q$ is $\frac{\overline{q}}{N(q)}$.
A: Another approach:


*

*$\mathbb H\otimes\mathbb H \cong \mathbb P\otimes\mathbb P$, where $\mathbb P$ is the split quaternions. Explicitly, you can take $i, j, k, i', j', k'$ to $i, ji', ki', i', ij', ik'$. (This works in either direction.)

*$\mathbb P \cong M_2(\mathbb R)$. Explicitly, you can take $i = \bigl(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\bigr), j = \bigl(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\bigr)$.

*$M_2(\mathbb R)\otimes M_2(\mathbb R) \cong M_4(\mathbb R)$.
