Why is $\mathbb{H} \otimes \mathbb{C} \cong \text{End}_{\mathbb{C}} (\mathbb{H})$

Question

As the title insinuates, in my readings I can across this isomorphism $\mathbb{H} \otimes \mathbb{C} \cong \text{End}_{\mathbb{C}} (\mathbb{H})$ and i cannot see why this is the case. Can someone help me see why this is an isomorphism. Here $\mathbb{H}$ is referring to the quaternions.

Answer 1

$\Bbb{H=C+Cj}$ is a 2-dimensional complex vector space so $End_\Bbb{C}(\Bbb{H})\cong M_2(\Bbb{C})$ as complex algebras.

Then we are willing to check that $\Bbb{H\otimes_R C }\cong M_2(\Bbb{C})$ as complex algebras, where $\Bbb{H\otimes_R C }$ comes with its own complex algebra structure.

For this, send $1\otimes a+i\otimes b+j\otimes c+ij\otimes d$ to $a\pmatrix{1&0\\0&1}+b\pmatrix{i&0\\0&-i}+c\pmatrix{0&1\\-1&0}+d\pmatrix{0&i\\i&0}$

Answer 2

You should clarify certain what field you are tensoring over and how you view the quaternions as a complex vector space (left or right: they are not in the center of the quaternions). See Example 2.6 and Exercise 5 of Section 2 here for a starting point.