Proposition 7

Let E be an extension of a field K. Every transcendence basis of E is a maximal element of the set (ordered by inclusion) of subsets of E algebraically free over K. Conversely, if S is a subset of E such that E is algebraic over $K(S)$, every maximal algebraically free subset of S is a transcendence basis of E.

Let B be a transcendence basis of E over K and $x \in E - B$. Then $x$ is algebraic over $K(B)$; by V, p. 107, Prop. 4, the subset $B \cup \{x\}$ of E is not algebraically free over K, whence the first part of the proposition. Secondly, if E is algebraic over $K(S)$ and B is a maximal algebraically free part of S, it follows from the Cor. of Prop. 4 that every $x \in S$ is algebraic over $K(B)$; hence (V, p. 18, Cor. 1) $K(S)$ is algebraic over $K(B)$, and so (V, p. 19, Prop. 3), E is algebraic over $K(B)$.