Proposition 3

— Let E be an extension of K contained in $\Omega$. Then the following conditions are equivalent: a) E is a quasi-Galois extension of K. b) For each $x \in E$ the conjugates of $x$ over K in $\Omega$ all belong to E. c) Every K-automorphism of $\Omega$ maps the field E into itself. d) Every K-homomorphism of E into $\Omega$ maps E into itself. e) E is the splitting extension in $\Omega$ of a family $(f_i)_{i \in I}$ of non-constant polynomials in $K[X]$ (V, p. 24, Remark 3).

The equivalence of c) and d) stems from the fact that every K-homomorphism of E into $\Omega$ is induced by a K-automorphism of $\Omega$ (V, p. 52, Prop. 1).

By definition a quasi-Galois extension is the splitting field of the family of minimal polynomials (over K) of its elements, so a) implies e). Under the hypothesis e) let $u$ be an automorphism of $\Omega$; for each $i \in I$, $u$ permutes the set $R_i$ of roots of $f_i$ and since $E = K \left( \bigcup_{i \in I} R_i \right)$, we have $u(E) = E$; thus e) implies c).

The definition of conjugate elements shows that c) implies b). Finally assume that b) holds; let $f$ be a monic irreducible polynomial in $K[X]$ having at least one root $x$ in E; since $\Omega$ is algebraically closed, there exist elements $a_k$ of $\Omega$ ($1 \leq k \leq n$) such that $f(x) = \prod_{k=1}^n (X - a_k)$ and since the $a_k$ are the conjugates of x over K (V, p. 53, Cor. 1), they belong to E by hypothesis. So we have shown that b) implies a).