Proposition 5

— Let L and M be two extensions of a field K contained in a given extension and algebraically disjoint over K. If M is separable over K, then L(M) is separable over L.

It is enough to show that for every finite subset S of M, $L(S)$ is separable over L (V, p. 122, Prop. 8). Let S be a finite subset of M. Since the field $K(S)$ is separable over K, it has a separating transcendence basis B (Cor. of Th. 5). Since L and M are algebraically disjoint over K, B is a transcendence basis of $L(B)$ over L (V, p. 113, Prop. 12). Further, every element of S is algebraic and separable over $K(B)$, hence over $L(B)$ (V, p. 39, Cor. 2). We deduce (V, p. 39, Prop. 6) that $L(S) = L(B)(S)$ is algebraic and separable over $L(B)$, hence $L(S)$, is separable over L.