Let $ \mathbb{H} $ be the ring of quaternions and make the vector space $A = \mathbb{H} \otimes \mathbb{C}$ into a ring by defining
$$(a \otimes w)(b \otimes z) = (ab \otimes wz) $$
for $a,b \in \mathbb{H}$ and $w,z \in \mathbb{C}$Show that $A \simeq \mathrm{M}_2(\mathbb{C})$
the ring of complex $2\times 2$ matrices.
I would normally post my solution attempt but i haven't really gotten anywhere with this problem.
Update
Haven't managed to solve this yet, could use some more help.
I admit, i don't quite understand the quaternions or the tensor product.
Does any of this make sense?
$ \mathbb{H} \otimes \mathbb{C}$ = ($ \mathbb{R}1 \oplus \mathbb{R}i \oplus \mathbb{R}j \oplus \mathbb{R}k$) $\otimes$ $\mathbb{C} $ = ($ \mathbb{R}1 \otimes \mathbb{C} $) $\oplus$ ($\mathbb{R}i \otimes \mathbb{C} $) $\oplus$ ($\mathbb{R}j \otimes \mathbb{C} $) $\oplus$ ($\mathbb{R}k \otimes \mathbb{C} $) $\simeq $ $\mathbb{C} \oplus \mathbb{C} \oplus \mathbb{C} \oplus \mathbb{C}$ $ \simeq \mathrm{M}_2(\mathbb{C})$. The last isomorphism is as additive groups by sending (a,b,c,d) to \begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix} a,b,c,d $\in \mathbb{C}$
But I don't know how to make it into a ring isomorphism.
Perhaps I could use that $ \mathbb{C} \simeq \begin{pmatrix}
a & -b \\
b & a \\
\end{pmatrix} $ $ a,b \in \mathbb{R}$
Best Answer
The canonical isomorphism $\Phi:\mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}\to \mathrm{M}_2(\mathbb{C})$ is constructed as follows:
$\mathbb{H}\simeq \mathbb{C}^2$ as vector spaces over $\mathbb{C}$. Explicitly, $\gamma: \mathbb{C}^2\to \mathbb{H}$ is $\gamma(z_1,z_2)=z_1 + z_2j$ [Here $i$ is the imaginary unit of $\mathbb{C}$, and $i,j,ij=k$ are the three imaginary units of $\mathbb{H}$]. Given $\mathbb{H}\ni u =z+wj$, we have $$\bar{u}=\bar{z}+\overline{wj}=\bar{z}+\bar{j}\bar{w}=\bar{z}-j\bar{w}=\bar{z}-wj$$
We define the $\mathbb{C}$-algebra homomorphism $\Phi:\mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}\to \mathrm{M}_2(\mathbb{C})$ via $$\Phi(1\otimes u)V=\gamma^{-1}(\gamma(V)\bar{u}), \qquad (z\in \mathbb{C}, u\in \mathbb{H}, V\in \mathbb{C}^2)$$ Explicitly, if $u=z+w j$, then $$ \phi(u):=\Phi(1\otimes u)=\begin{pmatrix} \bar{z} & \bar{w}\\ -w& z \end{pmatrix} $$ To finish, note that $\dim_{\mathbb{C}}\mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}=\dim \mathrm{M}_2(\mathbb{C})=4$, so it suffices to check that $\Phi$ is surjective. Define $I=\phi(i)$, $J=\phi(j)$, $K=\phi(ij)$, then it is easy to check that $$\mathrm{M}_2(\mathbb{C})\ni \begin{pmatrix} a & b\\ c & d \end{pmatrix}= \frac{a+d}{2}+\frac{i(a-d)}{2}I+\frac{b-c}{2}J+\frac{i(b+c)}{2}K $$ I want to also mention that the isomorphism $\phi$ is the natural isomorphism $\mathrm{Sp}(1)\to \mathrm{SU}(2)$, where $\mathrm{Sp}(1)$ is the group of unit quaternions.