Proposition 9

— Let K be a field and $\Omega$ an algebraically closed extension of K. Suppose that $\Omega$ is not algebraic over K; then the set of elements of $\Omega$ which are transcendental over K is infinite. Moreover, if x and y are two elements of $\Omega$ which are transcendental over K, then there exists an automorphism u of $\Omega$ over K such that $u(x) = y$.

Since $\Omega$ is not algebraic over K, there exists an element x of $\Omega$ transcendental over K; then the elements $x^n$ ($n \in \mathbf{N}$) are distinct and transcendental over K. Suppose that x and y are distinct and transcendental over K; by Prop. 2 (V, p. 106) there exists a K-isomorphism $\tilde{u}$ of $K(x)$ onto $K(y)$ such that $\tilde{u}(x) = y$; since $K(x)$ is of transcendence degree 1 over K, Cor. 2 of Prop. 8 shows that $\tilde{u}$ extends to a K-automorphism u of $\Omega$.