Proposition 5

— Let E be an extension of the field K. For a subset S of E to be algebraically free over K it is necessary and sufficient that for each x \in S, the element x should be transcendental over the field K(S - {x}).

The condition is necessary by Prop. 4.

To prove the sufficiency it is enough (Prop. 3) to show that every finite sequence $(x_1, ..., x_n)$ of distinct elements of $S$ is algebraically free. Now by hypothesis $x_i$ is transcendental over $K(x_1, ..., x_{i-1})$ for $1 \leq i \leq n$, and now our assertion follows by induction on $n$ from the Cor. to Prop. 4.