a) If u maps E into E, u induces a K-automorphism of E.

b) There exists a K-automorphism v of $\Omega$ extending u.

Suppose that $u(E) \subset E$; to prove a) it is enough to show that $u(E) = E$. Let x be an element of E, f the minimal polynomial of x over K and $\Phi$ the set of roots of fin E. The set $\Phi$ is finite, the mapping u of E into E is injective and we have $u(\Phi) \subset \Phi$; therefore $u(\Phi) = \Phi$, whence $x \in u(\Phi) \subset u(E)$; this shows that $E = u(E)$.

It is clear that $\Omega$ is an algebraic closure of E and of $u(E)$; therefore (V, p. 23, Cor.), the isomorphism u of E onto $u(E)$ extends to an isomorphism v of $\Omega$ onto $\Omega$.

a) x and y are conjugate over K.

b) There exists a K-isomorphism o of $K(x)$ onto $K(y)$ such that $v(x) = y$.

c) x and y have the same minimal polynomial over K.

Suppose first that x and y are conjugate over K; let u be a K-automorphism of $\Omega$ such that $u(x) = y$ and let f be the minimal polynomial of x over K. We have

and $f$ is a monic irreducible polynomial in $\mathbf{K}[X]$; therefore ($V$, p. 16, Th. 1, c)), $f$ is the minimal polynomial of $y$ over $K$. Thus a) implies c).

Suppose now that $x$ and $y$ have the same minimal polynomial $f$ over $K$. There exists a $K$-isomorphism of the field $\mathbf{K}[X]/(f)$ onto $\mathbf{K}(x)$ (resp. onto $\mathbf{K}(y)$) mapping the residue class of $X$ modulo $(f)$ to $x$ (resp. $y$) ($V$, p. 16, Th. 1, b)); hence there exists a $K$-isomorphism $v$ of $\mathbf{K}(x)$ onto $\mathbf{K}(y)$ such that $v(x) = y$. Therefore c) implies b).

Finally, under the hypothesis b), Prop. 1 implies the existence of a $K$-automorphism $u$ of $\Omega$ extending $v$; we then have $u(x) = y$, hence $x$ and $y$ are conjugate over $K$, so b) implies a).

Let $f$ be the minimal polynomial of $x$ over $K$; its roots are the conjugates of $x$ over $K$, and each of these roots admits $f$ as minimal polynomial over $K$. Now $x$ is separable over $K$ if and only if the polynomial $f$ is separable ($V$, p. 39, Prop. 5), that is, if $f$ has $n$ distinct roots in $\Omega$. Cor. 2 follows from this.*

In other words, an element $x$ of $\Omega$ is p-radical over $K$ if and only if it has no conjugate other than itself.

Let $p$ be the characteristic exponent and let $x \in \Omega$. By $V$, p. 44, Prop. 13, there exists an integer $m \geq 0$ such that $y = x^{p^m}$ is algebraic and separable over $K$. Now $x$ is invariant under $G$ if and only if $y$ is invariant under $G$; by Cor. 2 this is the case if and only if $y$ is of degree 1 over $K$, which amounts to saying that $y \in K$. Now the corollary is an immediate consequence.*

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).

This follows from Prop. 1, a) (V, p. 52) and the equivalence of the conditions a) and c) of Prop. 3.

We may assume that $F \subset \Omega$. Let $u$ be an E-automorphism of $\Omega$. Since $u$ is a K-automorphism of $\Omega$ and F is quasi-Galois over K, we have $u(F) = F$, therefore F is quasi-Galois over E.

Let $w$ be a K-automorphism of $\Omega$ extending $u$ (V, p. 52, Prop. 1); since N is quasi-Galois over K, we have $w(N) = N$, whence $w(E) \subset N$ and so $w$ induces a K-automorphism $v$ of N.

Let $(f_i)_{i \in I}$ be a family of non-constant polynomials in $K[X]$ whose splitting field over $K$ is $E$. Then it is clear that $E'$ is the splitting field of the family $(f_i)_{i \in I}$ over $K'$, so it is quasi-Galois over $K'$.

Let $u$ be a $K$-automorphism of $\Omega$. We have $u(N_i) = N_i$ for all $i \in I$, hence we obtain the equalities $u(N) = N$ and $u(M) = M$; now the proposition follows from Cor. 1 (V, p. 54).

Let $A$ be a set of elements of $\Omega$, and let $B$ be the set of elements of $\Omega$ which are conjugates of elements of $A$; expressed differently, if $G$ is the group of $K$-automorphisms of $\Omega$, we have $B = \bigcup_{u \in G} u(A)$. For each $u \in G$ we have $u(B) = B$ whence $u(K(B)) = K(B)$. Therefore $K(B)$ is a quasi-Galois extension of $K$ containing $A$ and it is immediate that every quasi-Galois extension $N$ of $K$ containing $A$ contains $B$, hence $K(B)$. We shall say that $K(B)$ is the quasi-Galois extension generated by $A$. This definition applies in particular when $A$ is an extension of $K$.

The next proposition follows directly from the preceding remarks.