Theorem 1

— Let N be an algebraic extension of K and $\Gamma$ the group of K-automorphisms of N. Then the following assertions are equivalent: a) Every element of N invariant under $\Gamma$ belongs to the image of K in N. b) N is a separable quasi-Galois extension of K. c) For each $x \in N$ the minimal polynomial of x over K splits in $N[X]$ into a product of distinct polynomials of degree 1. The equivalence of b) and c) follows from the Cor. of Prop. 6 (V, p. 40) and the definition of quasi-Galois extension (V, p. 53, Def. 2). Let us identify K with its canonical image in N. $a) \Rightarrow c)$: Suppose that K is the field of invariants of $\Gamma$. Let $x \in N$, with minimal polynomial $f$ over K and let A be the set of all roots of f in N. Put

Every automorphism $a \in \Gamma$ induces a permutation of A, and hence leaves invariant the coefficients of the polynomial $g \in N[X]$. We thus have $g \in K[X]$ and since $g(x) = 0$, the polynomial g is a multiple of $f$ in $K[X]$ (V, p. 16, Th. 1). Further, $f$ and g are monic and g divides $f$ (IV, p. 16, Prop. 5); thus we have $f = g$, which means that the minimal polynomial $f$ of $x$ over $K$ is a product in $N[X]$ of distinct polynomials of degree 1.

c) ⇒ a): let x be an element of N not belonging to K. Denote by Ω an algebraic closure of K containing N as subextension (V, p. 23, Th. 2). Let f be the minimal polynomial of x over K, which is of degree ≥ 2 by hypothesis and let A be the set of roots of $f(X)$ in N. If condition c) holds, we have $f(X) = \prod_{y \in A} (X - y)$ and so (V, p. 53, Cor. 1) A is the set of conjugates of x in Ω. Since f has degree ≥ 2, there exists in A an element $y \neq x$, hence a K-automorphism u of Ω such that $u(x) = y$. Now under the hypothesis c), the extension N of K is quasi-Galois, hence $u(N) = N$ (V, p. 54, Cor. 1); it follows that u induces a K-automorphism a of N such that $\sigma(x) = y \neq x$, so K is the field of invariants of Γ.