Theorem 5

— Let (E_i)_{i \in I}, be a family of extensions of a field K and \Omega an algebraically closed extension of K. Suppose that the inequality

holds. Then there exists an algebraically free family (F_i)_{i \in I}, of subextensions of \Omega such that F_i is K-isomorphic to E_i for all i \in I.

For each $i \in I$ let B_i be a transcendence basis of E_i over K. Let B be a transcendence basis of \Omega over K. By (2) we have Card B \geq \sum_{i \in I} \text{Card} B_i; hence there exists a family (B'_i)_{i \in I}, of pairwise disjoint subsets of B and bijections u_i : B_i \to B'_i for i \in I. By Prop. 2 of V, p. 106, u, extends to a K-isomorphism v_i of K(B_i) onto K(B'_i); since \Omega is algebraically closed and E_i algebraic over

K(B_i), the Cor. (V, p. 23) shows that $v_i$ extends to a K-isomorphism of $E_i$ onto a subextension $F_i$ of $\Omega$. By construction $B'_i$ is a transcendence basis of $F_i$ over K, and Prop. 15 (V, p. 115) shows that the family $(F_i)_{i \in I}$ of subextensions of $\Omega$ is algebraically free over K.