Corollary 1

Let K be a commutative field, and let $A_1$ and $A_2$ be K-algebras. Suppose that $A_1$ is absolutely semisimple.

a) If $A_2$ is semisimple, then so is $A_1\otimes_KA_2$.

b) If $A_2$ is absolutely semisimple, then so is $A_1\otimes_KA_2$.

Denote the center of $A_1$ by $Z_1$ and that of $A_2$ by $Z_2$. The center Z of $A_1\otimes_KA_2$ is equal to $Z_1\otimes_KZ_2$ by the corollary of III, §4, No. 4, p. 468. Suppose that $A_2$ is semisimple; then $Z_2$ is a reduced algebra (VIII, p. 137, Propositions 2 and 3). By Theorem 1, $Z_1$ is an étale and therefore separable K-algebra. By Proposition 5 of V, §15, No. 2, p. 120, the ring $Z = Z_1\otimes_KZ_2$ is reduced; since $A_1$ has finite degree over K (Theorem 1), it follows from Proposition 7 of VIII, p. 221 that the ring $A_1\otimes_KA_2$ is semisimple.

Now suppose that $A_2$ is absolutely semisimple. Let L be an extension of K. Then the algebra $A_{1(L)}$ is absolutely semisimple, and the algebra $A_{2(L)}$ is semisimple. Therefore, by a), the algebra $A_1\otimes_KA_{2(L)}$ is semisimple.