Proposition 14

— Let L be an extension of a field K, E and F two subextensions of L and B a transcendence basis of E over K. For E and F to be algebraically disjoint over K it is necessary and sufficient that K(B) and F should be linearly disjoint over K.

For E and F to be algebraically disjoint over K it is necessary and sufficient for B to be algebraically free over F (Prop. 12), that is, for the monomials in the elements of B to be linearly independent over F. Since these monomials form a basis of the vector K-space K[B], it comes to the same to say that K[B] and F are linearly disjoint over K. Finally, since K(B) is the field of fractions of K[B], Prop. 6 of V, p. 14 shows that K[B] and F are linearly disjoint if and only if this is so for K(B) and F.