Corollary

Let L and M be two extensions of a field K. Suppose that the relative separable algebraic closure of K in L is equal to K. Then the set p of nilpotent elements of $L \otimes_K M$ is a prime ideal. If moreover L or M is separable over K, then $L \otimes_K M$ is an integral domain.

We may assume that L and M are algebraically disjoint subextensions of an extension $\Omega$ of K (V, p. 116, Th. 5); then p is a prime ideal by Prop. 1 (V, p. 139). If moreover L or M is separable over K, then $L \otimes_K M$ is a reduced ring by the definition of separable extension (V, p. 119, Def. 1); so $p = 0$ and $L \otimes_K M$ is an integral domain because p was prime.