Proposition 15

— Let $(E_i)_{i \in I}$ be a family of subextensions of a given extension $L$ of a field $K$. Then the following conditions are equivalent :

a) The family $(E_i)_{i \in I}$ is algebraically free.

b) For each $i \in I$ the extension $E_i$ is algebraically disjoint over $K$ from the extension $F_i$ generated by the $E_j$ for $j \neq i$.

c) There exists a family $(B_i)_{i \in I}$ of disjoint subsets of $L$, such that $B_i$ is a transcendence basis of $E_i$ over $K$ for each $i \in I$, and $B = \bigcup_{i \in I} B_i$ is algebraically free over $K$.

It is clear that a) implies c).

Assuming c), let us choose $i$ in $I$; put $C_i = \bigcup_{j \neq i} B_j$. For each $j \neq i$ every element of $E_j$ is algebraic over $K(B_j)$ and a fortiori over $K(C_i)$. By Cor. 1 of $V$, p. 18, the field $F_i$ is therefore algebraic over $K(C_i)$. Further, we have $B_i \cap C_i = \varnothing$ and $B - B_i \cup C_i$ is algebraically free over $K$; therefore $B_i$ is algebraically free over $K(C_i)$ ($V$, p. 107, Prop. 4), hence also over $F_i$ (which is algebraic over $K(C_i)$) by Prop. 6 of $V$, p. 108. Thus we have proved that $E_i$ is algebraically disjoint from $F_i$ over $K$ ($V$, p. 113, Prop. 12), hence c) implies b).

Let us now assume b) and prove a). It is enough to show that if $i_1, \ldots, i_n$ are distinct elements of $I$, then the family of extensions $(E_{i_1}, \ldots, E_{i_n})$ is algebraically free; we argue by induction on $n$, the case $n = 1$ being trivial. Suppose then that $n > 1$ and that the family $(E_{i_1}, \ldots, E_{i_{n-1}})$ is algebraically free; for $1 \leq k \leq n$ choose a subset $A_k$ of $E_{i_k}$ algebraically free over $K$ and put

B = A, U ... U A_{n-1}. By the induction hypothesis the subsets A_1, ..., A_{n-1} are pairwise disjoint and B is algebraically free over K; by b) E_{i_n} is algebraically disjoint from F_i and since B is contained in F_{i_n} we have B \cap A_i = \varnothing and B \cup A_i = A, U ... U A_i is algebraically free over K. We have thus shown that the family (E_{i_1}, ..., E_{i_n}) is algebraically free.

The following proposition generalizes part b) of Prop. 13 (V, p. 113).