Tensor products and isomorphic algebras

Question

I found that $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \simeq \mathbb{C} \oplus\mathbb{C}$ and that $ \mathbb{H} \otimes_{ \mathbb{R}} \mathbb{C} \simeq M_2( \mathbb{C})$. Could anybody hint me how the isomorphisms should look like (these supposed to be algebra isomorphisms)?

Answer 1

For the first, think of one of the copies of $\mathbb{C}$ as $\mathbb{R}[x]/(x^2 + 1)$. For the second, consider $\mathbb{H}$ as a two-dimensional vector space over $\mathbb{C}$ by picking the basis $\{1, j\}$ (so $ij = k$, i.e. we are abusing notation, thinking of the $i$ in $\mathbb{C}$ as the same as the $i$ in $\mathbb{H}$). Now the map sending a quaternion to the endomorphism of this vector space given by right-multiplication is the map that you want.

EDIT: I decided to write this out in better detail because it seemed like the best way to answer your question in the comment. In the process, I realized I was sloppy about something, and I changed left-multiplication to right-multiplication in the above text. You could normalize it either way, but the important thing is that the endomorphism needs to be $\mathbb{C}$-linear, so if you are going to take left-multiplication as the endomorphism, you need to think of $\mathbb{H}$ as a $\mathbb{C}$-vector space via *right*multiplication, which is a little odd, so I changed it to the other way around.

Explicitly, write $\{ 1, \widehat{i}, \widehat{j}, \widehat{k} \}$ for the usual basis of the quaternions. In what follows, we will think of $\mathbb{H}$ as a left $\mathbb{C}$-vector space (i.e. we will declare that $i \in \mathbb{C}$ acts by left-multiplication by $\widehat{i}$ in $\mathbb{H}$.) Now, for $b \in \mathbb{H}$, right-multiplication by $b$ is a $\mathbb{C}$-linear endomorphism of $\mathbb{H}$.

We need to choose a basis for $\mathbb{H}$ as a $\mathbb{C}$-vector space to compute the others; we choose $\{ 1, \widehat{j} \}$. The matrix for $1$ will be the identity, but the others we need to compute. Let's do $\widehat{i}$. Remembering that the endomorphism is given by right-multiplication, we compute $1 \mapsto \widehat{i} = i \cdot 1$, $\widehat{j} \mapsto -\widehat{k} = -\widehat{i}\widehat{j} = i \cdot \widehat{j}$, where the lack of a hat on the $i$ in the last equalities is not a typo but refers to the $\mathbb{C}$-vector space structure. So the matrix corresponding to $\widehat{i}$ is $$ \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}. $$ You can do $\widehat{j}$ and $\widehat{k}$. The point is that we have written down an $\mathbb{R}$-linear map of algebras $\mathbb{H} \to M_2(\mathbb{C})$ and such a thing extends uniquely to a $\mathbb{C}$-linear map of algebras $\mathbb{H} \otimes_\mathbb{R} \mathbb{C} \to M_2(\mathbb{C})$. Our map is obviously injective, and as you've computed in the comments, the dimensions work out, so it is an isomoporphism.