Proposition 16

— Let (E_i)_{i \in I} be a family of subextensions of an extension of a field K, and let E be the field generated by $\bigcup_{i \in I} E_i$.

a) We have tr . deg_K E \leq \sum_{i \in I} \text{tr . deg}_K E_i, with equality when the family (E_i)_{i \in I} is algebraically free over K.

b) Conversely, assume that tr . deg_K E = \sum_{i \in I} \text{tr . deg}_K E_i and that tr . deg_K E is finite; then the family (E_i)_{i \in I}, is algebraically free over K.

For each $i \in I$ let B_i be a transcendence basis of E_i over K and put $B = \bigcup_{i \in I} B_i$. For each $i \in I$, each element of E_i is algebraic over K(B_i), hence over K(B). Now Cor. 1 of V, p. 18 shows that E is algebraic over K(B); by V, p. 109, Th. 2, B thus contains a transcendence basis of E over K. If moreover the family (E_i)_{i \in I} is algebraically free over K, then the B_i are disjoint and the set B is algebraically free over K. This establishes a) (Set Theory, III, p. 160, Cor. of Prop. 4).

Under the hypotheses of b), E is algebraic over K(B) and of finite transcendence degree over K, and we have Card(B) \leq \text{tr . deg}_K E. By Cor. 1 of V, p. 110, B is algebraically free over K and the B_i are disjoint. Now Prop. 15 shows that the family (E_i)_{i \in I} is algebraically free over K.

Before stating the next theorem let us remark that there exist algebraically closed extensions of K of arbitrary transcendence degree, for example an algebraic closure of an appropriate field of rational fractions.