Computing the center of a ring

Question

if i have the following ring $R = \mathbb{H} \otimes _\mathbb{R} M_2(\mathbb{C}) $

then how would i find the center $Z(R)$? Also is this ring simple, i am sure it is but am struggling to show that is actually is

Any help would be really useful and greatly appreciated

thank you.

Answer 1

Clearly $Z(\Bbb H)\otimes_\Bbb R Z(M_2(\Bbb C))$ is in the center of the tensor product. Now $Z(\Bbb H)=\Bbb R$ and $Z(M_2(\Bbb C))\cong \Bbb C$, so we're looking at $\Bbb R\otimes_\Bbb R \Bbb C\cong \Bbb C$ as $\Bbb R$ algebras. We're looking for a copy of $\Bbb C$ inside this ring.

Clearly $Z(\Bbb H)\otimes_\Bbb R Z(M_2(\Bbb C))=\{1\otimes \begin{bmatrix}c&0\\0&c\end{bmatrix}\mid c\in \Bbb C\}$ is the copy we're looking for.

Answer 2

There is the following general theorem:

Let $A$ be a central simple $k$-algebra and $B$ a simple $k$-algebra with center $T$. Then $A \otimes_k B$ is a simple $k$-algebra with center $1 \otimes_k T$.

You can find the proof in any introductory book of non-commutative ring theory. It is Theorem 3.60 in "Methods of Representation Theory" by Curtis and Reiner.