Proposition 2

— Let x and y be two elements of $\Omega$. Then the following conditions are equivalent:

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).