Corollary

— If L and M are separable extensions, algebraically disjoint over K, then the field $K(L \cup M)$ is separable over K.

For $K(L \cup M) = L(M)$ is separable over L by Prop. 5 (because M is separable over K) and L is separable over K, whence the corollary (V, p. 122, Prop. 9).

The hypothesis that the extensions L and M are algebraically disjoint is indispensable in Prop. 5 and its corollary. For let K be an imperfect field of characteristic $p \neq 0$ and E an extension of the form $K(x, a)$ with x transcendental over K and a p-radical of height 1 over K ; put $L = K(x)$ and $M = K(x + a)$. Then $x + a$ is transcendental over K (if not, $x = (x + a) - a$ would be algebraic over K) and the fields L and M are separable over K. However, $K(L \cup M) = K(x, a)$ is p-radical of degree $p$ over $L = K(x)$ and is not separable over L, nor over K.