Theorem 3

— Let E be an extension of a field K. All transcendence bases of E over K have the same cardinal.

It is enough to prove the inequality Card(B) ≥ Card(B'), when B and B' are two transcendence bases of E over K, and we may suppose that B' is not empty. We first suppose that B is finite and use induction on its cardinal n; for n = 0, E is algebraic over K and B' is empty. Suppose now that n ≥ 1; given x ∈ B', the exchange theorem provides a subset C of B such that x ∉ C and {x} U C is a transcendence basis of E over K; we have C ≠ B by Prop. 7 whence Card C < n. Put K_1 = K(x) and C' = B' - {x}; then C and C' are algebraically free over the field K_1 (V, p. 107, Prop. 4) and E is algebraic both over K_1(C) = K(C U {x})) and K_1(C') = K(B'). In other words, C and C' are two transcendence bases of E over K_1. Since Card(C) < n, the induction hypothesis implies the inequality Card(C') ≤ Card(C) ≤ n - 1, whence Card(B') ≤ n = Card(B).

Suppose now that B is infinite. Every x ∈ B is algebraic over K(B') and so there exists a finite subset S(x) of B' such that x is algebraic over K(S(x)). Write S = U_{x ∈ B} S(x), then S ⊂ B' and since B is infinite, we have Card(S) ≤ Card(B) (E, III, p. 49, Cor. 3). But since every element of B is algebraic over K(S) and E is algebraic over K(B), we conclude that E is algebraic over K(S) (V, p. 19, Prop. 3). Now Prop. 7 implies that S = B', whence the desired inequality Card(B') ≤ Card(B).