Corollary 3

Let D be a field of finite degree over K with center K. Every element of D is algebraic over K. Let $x$ and $y$ be elements of D; there exists an element $a$ of $D^*$ such that $y=axa^{-1}$ if and only if $x$ and $y$ have the same minimal polynomial over K.

The first assertion follows from Corollary 1 of V, §3, No. 1, p. 17.

Suppose that there exists an element $a$ of $D^*$ such that $y=axa^{-1}$; for every polynomial P of K[X], we have $P(y) =aP(x)a^{-1}$, and, in particular, we have $P(x) = 0$ if and and only if $P(y) = 0$. Consequently, $x$ and $y$ have the same minimal polynomial over K (V, §3, No. 1, p. 16, Theorem 1).

Conversely, suppose that $x$ and $y$ have the same minimal polynomial. By loc. cit., there exists a K-isomorphism $u$ from $K[x]$ to $K[y]$ such that $u(x) =y$, and $K[x]$ is a field. By Corollary 2$,u$ extends to an inner automorphism $\theta :z\mapsto aza^{-1}$ of D, and we therefore have $y=\theta (x) =axa^{-1}$.