Remark 3

Let $E$ be an algebraic extension of $K$ and $x, y$ two elements of $E$. If there exists a $K$-automorphism of $E$ which maps $x$ to $y$ then $x$ and $y$ have the same minimal polynomial over $K$ and so are conjugate over $K$ by Prop. 2 (V, p. 52). The converse holds if $E$ is quasi-Galois, because every $K$-automorphism of $\Omega$ induces a $K$-automorphism of $E$. The hypothesis that $E$ be quasi-Galois is not superfluous as the following example shows: \* let $K = \mathbf{Q}$ and $\Omega$ the field of all algebraic numbers; put $E = \Omega \cap \mathbf{R}$. It may be shown (V, p. 153, Ex. 2) that every automorphism of $E$ is the identity mapping and that $\sqrt{2}$ and $-\sqrt{2}$ are elements of $E$ conjugate over $\mathbf{Q}$. \*