## A sufficient condition for the conjugates over a field of an algebraic element to be distinct

Let $F$ be a field and $\alpha$ be algebraic of degree $n$ over $F$. Suppose there exists an element $\lambda \in F(\alpha)$ such that $\lambda \alpha = \beta \in F$. Show that the conjugates of $\alpha$ over $F(\alpha)$ are distinct.

Certainly we have $F(\lambda) \subseteq F(\alpha)$. Now $\lambda^{-1} \in F(\lambda)$, so $\alpha = \beta\lambda^{-1} \in F(\lambda)$. (Recall that $\beta \in F$, so that $\beta\lambda^{-1}$ can be written uniquely as a polynomial in $\lambda$.) So $\alpha \in F(\lambda)$, and we have $F(\alpha) \subseteq F(\lambda)$. Hence $F(\alpha) = F(\lambda)$.

By Theorem 5.10 in TAN, because $F(\alpha) = F(\lambda)$, the conjugates of $\lambda$ over $F(\alpha)$ are distinct.