Remark 1

Let $F$ be an extension of $K$ and $E$ a subextension of $F$, and suppose that $E$ is quasi-Galois over $K$. Let us show that every $K$-automorphism $u$ of $F$ leaves $E$ invariant. For let $x \in E$ and $f$ the minimal polynomial of $x$ over $K$. Since $E$ is quasi-Galois over $K$, there exist $a_1, \ldots, a_n \in E$ such that $f(X) = \Pi(X - a_i)$; we have $f(u(x)) = u(f(x)) = 0$, hence $u(x)$ is one of the $a_i$ and so belongs to $E$. We have proved $u(E) \subset E$, and now $u(E) = E$ follows by V, p. 52, Prop. 1.