Proposition 13

— Let L be an extension of a field K and E, F two subextensions of L.

a) We have $\operatorname{tr. deg}_L F(E) \leq \operatorname{tr. deg}_K E$. When E and F are algebraically disjoint over K, then every transcendence basis of E over K is a transcendence basis of

F(E) over F and we have tr . deg_F F(E) = tr . deg_K E. Conversely, this equality implies that E and F are algebraically disjoint over K when tr . deg_K E is finite.

b) We have tr . deg_K K(E U F) ≤ tr . deg_K E + tr . deg_K F. When E and F are algebraically disjoint over K, we have tr . deg_K K(E U F) = tr . deg_K E + tr . deg_K F. Conversely, this equality implies that E and F are algebraically disjoint over K when E and F are of finite transcendence degree over K.

a) Let B be a transcendence basis of E over K; then E is algebraic over K(B), and Cor. 2 of V, p. 18 shows that F(E) is algebraic over F(K(B)) = F(B). By Th. 2 (V, p. 109) B contains a transcendence basis of F(E) over F; when E is algebraically disjoint from F over K, B is algebraically free over F (Prop. 12) and this is then a transcendence basis of F(E) over F. The three first assertions of a) follow from this. Suppose now that E is of finite transcendence degree over K, equal to that of F(E) over F; since F(E) is algebraic over F(B) and Card B = tr . deg_F F(E), Cor. 1 of V, p. 110 shows B to be algebraically free over F, so E is algebraically disjoint from F over K (Prop. 12).

b) We have K(E U F) = F(E) and so the Cor. of V, p. 110 implies the equality:

Now b) follows at once from a) and this equality.