Proposition 5

— Let $K$ be a field and $A, B$ commutative $K$-algebras. If $A$ is reduced and $B$ separable then $A \otimes_K B$ is reduced.

By the Cor. of Prop. 2 (V, p. 119) $A$ is isomorphic to a subalgebra of a product $\prod_{i \in I} L_i$ where $L_i$ is an extension of $K$ for $i \in I$. Therefore $A \otimes_K B$ is isomorphic to a subring of $\left( \prod_{i \in I} L_i \right) \otimes_K B$ and this latter ring is isomorphic to a subring of $\prod_{i \in I} (L_i \otimes_K B)$ (II, p. 306, Prop. 15). Since $B$ is separable, each of the rings $L_i \otimes_K B$ is reduced, and so the same is true of $\prod_{i \in I} (L_i \otimes_K B)$, and a fortiori of $A \otimes_K B$.