Corollary 2

— Let $\Omega$ be an algebraically closed extension of a field $K$, $E$ and $F$ two subextensions of $\Omega$ and $u$ a $K$-isomorphism of $E$ onto $F$. If the transcendence degree of $E$ over $K$ is finite (in particular if $E$ is algebraic over $K$) then there exists a $K$-automorphism of $\Omega$ extending $u$.

Let us denote by $n, d(E)$ and $d(F)$ the transcendence degrees of $E$ over $K$, of $O$ over $E$ and of $C?$ over $F$ respectively. The existence of the $K$-isomorphism $u$ shows that the transcendence degree of $F$ over $K$ is equal to $n$. By the Cor. of Th. 4 the transcendence degree of $\Omega$ over $K$ is equal to $d(E) + n$ and also to $d(F) + n$. Therefore ($E$, III, p. 28, Prop. 8) we have $d(E) = d(F)$ and we can apply Prop. 8.