Proposition 12

— Let L be an extension of a field K and E, F subextensions of L. Then the following conditions are equivalent:

a) E and F are algebraically disjoint;

b) there exists a transcendence basis of E over K which is algebraically free over F;

c) every subset of E which is algebraically free over K is algebraically free over F.

Let us introduce the following conditions:

$b'$') there exists a transcendence basis of F over K which is algebraically free over E;

$c'$') every subset of F which is algebraically free over K is algebraically free over E.

$a) \Rightarrow b'$') : Suppose that E and F are algebraically disjoint. Let B (resp. C) be a transcendence basis of E (resp. $F$) over K. Then $B \cap C = \varnothing$ and $B \cup C$ is algebraically free over K, hence C is algebraically free over $K(B)$ (V, p. 107, Prop. 4); since E is algebraic over $K(B)$, Prop. 6 of V, p. 108 shows C to be algebraically free over E.

$b'$') $\Rightarrow$ c) : Suppose that there exists a transcendence basis C of F over K which is algebraically free over E. Let A be a subset of E algebraically free over K. Then C is algebraically free over $K(A)$, hence A is algebraically free over $K(C)$ (V, p. 107, Prop. 4) and therefore over $F$ (V, p. 108, Prop. 6), because F is algebraic over $K(C)$.

c) $\Rightarrow$ a) : This follows at once from Prop. 4 (V, p. 107).

The implications $a) \Rightarrow b) \Rightarrow c' ) \Rightarrow a)$ may be proved in the same way.