Tensor product between quaternions and complex numbers.

Let $H$ be the Hamiltonian quaternions, $\mathbb C$ the complex numbers and $\mathbb R$ the real numbers. Identify $H\otimes_{\mathbb R} \mathbb C$ in familiar terms.

This is an exercise of Modern Algebra of Garrett Birkhoff. I think it may have something to do with the Octonions. Any ideas?

any ideas?

Since $i$ and $j$ both satisfy $x^2+1$, we should look for matrices that also satisfy this polynomial. Take $I$ to be the Jordan canonical form and $J$ to be the rational canonical form for $x^2+1$: $$ I = \begin{pmatrix} \sqrt{-1} & 0\\ 0 & -\sqrt{-1} \end{pmatrix} \qquad J = \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} $$ Then $JI = -IJ$ as well, so the map \begin{align*} \mathbb{H} \otimes_\mathbb{R} \mathbb{C} &\to M_2(\mathbb{C})\\ i \otimes 1, j \otimes 1 &\mapsto I,J \end{align*} is a homomorphism, and in fact an isomorphism.

More concretely, this is just $M_2(\mathbb{C})$.

Not sure exactly how Birkhoff wants you to see this, but here's how I did it, using the judo of central simple algebras. $\mathbb{H}$ is a central division algebra (so a CSA) over $\mathbb{R}$. A commonly used fact is that base-extending a CSA gives a CSA, so $\mathbb{H} \mathop{\otimes}\limits_{\mathbb{R}} \mathbb{C}$, viewed as a $\mathbb{C}$ algebra, is a CSA, of the same dimension (4). Another common result is that any CSA over an algebraically closed field is just a matrix algebra over that field, so this is $M_2(\mathbb{C})$.