Proposition 10

— Let K be a field, $\Omega$ an algebraically closed extension of K and G the group of K-automorphisms of $\Omega$. Let $x \in \Omega$.

a) For x to be algebraic over K it is necessary and sufficient for the set of elements $u(x)$ as u runs over G to be finite,

b) For x to be p-radical over K it is necessary and sufficient that $u(x) = x$ for all $u \in G$.

In particular if K is perfect, the set of invariants of the group G is equal to K.

Suppose first that x is transcendental. By Prop. 9 the set T of elements of $\Omega$ transcendental over K is infinite, and for each $y \in T$ there exists $u \in G$ such that $u(x) = y$. Hence the set of elements $u(x)$ as u runs over G is infinite.

Suppose now that x is algebraic over K and denote by fits minimal polynomial over K; the set of roots of f in $\Omega$ is finite and for each $u \in G$ we have $f(u(x)) = u(f(x)) = 0$. Hence the set of elements $u(x)$ as u runs over G is finite. This proves a).

Let L be the set of elements y of $\Omega$ such that $u(y) = y$ for all $u \in G$, and let $\bar{K}$ be the relative algebraic closure of K in $\Omega$. By what has been said, L is a subextension of $\bar{K}$ over K. Further (Cor. 1 to Prop. 8) every K-automorphism of $\bar{K}$ is the restriction to $\bar{K}$ of an element of G. Assertion b) of Prop. 10 now follows from Cor. 3 of V, p. 53.