In the ring $R = \mathbb{H} \otimes_{\mathbb{R}} M_{2}(\mathbb{C})$ I have computed the center as $Z(R)= \mathbb{C}$. I am however struggling to show that $R$ is a simple ring and consequently find the division ring $D$ such that $R \cong M_{n}(D)$ for some $n\in \mathbb{N}$.
Any help would be appreciated
It's a well-known lemma that the tensor product of a central simple $k$ algebra with a simple $k$ algebra is a simple $k$ algebra. Since $\Bbb H$ is central simple and $M_2(\Bbb C)$ is simple, the product is a simple $\Bbb R$ algebra.
Now the product is a $\Bbb C$ algebra, and the division ring $D$ must be a finite extension of $\Bbb C$, but since $\Bbb C$ is algebraically closed, $D=\Bbb C$.
By inspection the product is $32$-$\Bbb R$ dimensional, so we can deduce the $\Bbb C$ dimension is $16$, and so it's $M_4(\Bbb C)$.