Proposition 6

— Let E be an extension of a field K and X an algebraically free subset of E over K. If K' \subset E is an algebraic extension of K then X is algebraically free over K'.

Arguing by contradiction, let us suppose that X is algebraically related over K'. By Prop. 5 there exists an element $x \in X$ which is algebraic over the field $K'(M)$, where $M = X - \{x\}$. Since $K'(M) = K(M)(K')$ and $K'$ is algebraic over $K$, Cor. 2 of V, p. 18 shows that $K'(M)$ is algebraic over $K(M)$; since $x$ is algebraic over $K'(M)$, it is therefore algebraic over $K(M) = K(X - \{x\})$, by Prop. 3 of V, p. 19. Now Prop. 5 shows X to be algebraically related over K and we have a contradiction.