1- Is there a difference between saying I want to prove this $\Bbb{H\otimes_R C }\cong M_2(\Bbb{C})$ and I want to prove this $\Bbb{H\otimes_R C }\cong_{\mathbb R} M_2(\Bbb{C})$?
I see the OP in the question titled " Why is $\mathbb{H} \otimes \mathbb{C} \cong \text{End}_{\mathbb{C}} (\mathbb{H})$ " here on this site just asked about $\mathbb{H} \otimes \mathbb{C} \cong \text{End}_{\mathbb{C}} (\mathbb{H})$ without specifying the isomorphism he required is of $\mathbb R-$algebras or $\mathbb C-$algebras and he got an isomorphism as complex algebras of $\Bbb{H\otimes_R C }\cong M_2(\Bbb{C})$ .
My second question is there a different isomorphism as $\mathbb R-$ algebras of $\Bbb{H\otimes_R C }\cong M_2(\Bbb{C})$ or can I consider the same isomorphism?
Best Answer
Those two statements are technically different, and the second (an iso as $\mathbb{R}$ algebras) implies the first, so we might expect it to be harder to prove.
However in this instance, any isomorphism you think of will be an isomorphism of $\mathbb{R}$ algebras. It can be helpful to instead think of this kind of thing as:
In this case, once you do step 1, you just need to check whether the complex vector space structure is preserved.
One can give isomorphisms of algebras that dont respect all the potential choices of "underlying scalars" however, eg, consider pointwise complex conjugation on $2\times 2$ complex matrices. This is a morphism of $\mathbb{R}$ algebras, but not of $\mathbb{C}$ algebras.