I got confused when I read the question titled "Why is $\mathbb{H} \otimes \mathbb{C} \cong \text{End}_{\mathbb{C}} (\mathbb{H})$" here on this site.
I know that the OP did not specify the tensor product over which field in the main question (but he specified it in the comments of one answer), so my question, just to clarify, if we want to prove that $$\mathbb{C} \otimes_{\mathbb R} \mathbb H \cong_{\mathbb R} M_2(C),$$
which is an isomorphism of $R$-algebras for the field of real numbers.
1- Do we have to take into account $\mathbb H$ as a left $\mathbb C$-vector space or $\mathbb H$ as a left $\mathbb C$-vector space as far as I understand there is no distinction between left and right vector spaces so why here there is a distinction?
2- Do we have to prove first the isomorphism $\mathbb{H} \otimes \mathbb{C} \cong \text{End}_{\mathbb{C}} (\mathbb{H})$ and then prove the required isomorphism $\mathbb{C} \otimes_{\mathbb R} \mathbb H \cong_{\mathbb R} M_2(C)$ as mentioned in the answer of @reuns?
Here is the answer of @reuns:
$\Bbb{H=C+Cj}$ 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 $a\pmatrix{1&0\\0&1}+b\pmatrix{i&0\\0&-i}+c\pmatrix{0&1\\-1&0}+d\pmatrix{0&i\\i&0}$
3-How can we show $End_\Bbb{C}(\Bbb{H})\cong M_2(\Bbb{C})$?
Any clarification will be appreciated!
EDIT:
Also, I was looking at that question: Why is $\mathbb{H}\otimes\mathbb{H}\cong\text{End}_\mathbb{R}\mathbb{H}$?
Why is $\mathbb{H}\otimes\mathbb{H}\cong\text{End}_\mathbb{R}\mathbb{H}$?
Best Answer
It is unclear what is unclear to you.
I am saying that $\Bbb{H}$ is just a 2-dimensional (left) vector space so we can replace $End_\Bbb{C}(\Bbb{H})$ by $End_\Bbb{C}(\Bbb{C}^2)$.
It means that I am interpreting $End_\Bbb{C}(\Bbb{H})$ as $$End_{left\ \Bbb{C}\ vector\ space}(\Bbb{H})$$
Then $1\otimes a+i\otimes b+j\otimes c+ij\otimes d\mapsto a\pmatrix{1&0\\0&1}+b\pmatrix{i&0\\0&-i}+c\pmatrix{0&1\\-1&0}+d\pmatrix{0&i\\i&0}$ gives the isomorphism $\Bbb{H\otimes_R C }\to End_\Bbb{C}(\Bbb{C}^2)$.
The $i$ of $i\otimes b$, the $i$ hidden in $a=A+iB$, and the $i$ in $\pmatrix{i&0\\0&-i}$ are not the same, that's part of the game to understand what we identity the latter two.
The distinction between left and right vector space is because $\Bbb{C}$ is a subring of $\Bbb{H}$ but which doesn't commute with $\Bbb{H}$.