Corollary 2

Let $A_1$ and $A_2$ be simple K-algebras; suppose that $A_1$ is finite-dimensional over K. If the center of $A_1$ or $A_2$ is equal to K, then the algebra $A_1\otimes A_2$ is simple. This is, in particular, the case when K is algebraically closed.

The centers $Z_1$ and $Z_2$ of $A_1$ and $A_2$, respectively, are fields; if one of the rings $Z_1$ and $Z_2$ is equal to K, then the ring $Z_1\otimes Z_2$ is a field. It therefore suffices to apply Proposition 7.

If the field K is algebraically closed, then the center of $A_1$ is equal to K.