This is a question from Cohn's Basic Algebra.
Question- Let $E|k$ be a normal extension. Show that an element of $E$, of degree $r$ over $k$, has at most $r$ conjugates over $k$, with equality iff it is separable.
Try- Let $\alpha\in E$ has degree $r$ over $k$, then so does its minimal polynomial $p(x)$. Let $\beta$ be a conjugate of $\alpha$. Then there exist a $k$-automorphism $\sigma$ such that $\sigma(\alpha)=\beta$. Then $\beta=\sigma(\alpha)$ is a zero of $p(x)$ because $\sigma$ fixes $k$ elementwise and since deg$(p(x))=r$, it has atmost $r$ zeroes in $E$. Thus, has at most $r$ conjugates in $E$.
To prove equality, we would like to extend $k$-isomorphism $\sigma_{1} : k(\alpha) \to k(\beta)$ in which $\alpha \to \beta$ to $k$-automorphism $\sigma : E \to E$ (to this we need to use induction on finite many steps which only can be true if $E$ is normal as well as finite.), and as $\sigma(\alpha)=\sigma_{1}(\alpha)=\beta$, $\beta$ is a conjugate of $\alpha$ and thus conjugates are precisely the zeroes.
My Doubt is that we cannot do this unless $E/k$ is normal and finite (thus splitting field of some polynomial) which is not given in the question. Where am I going wrong?
The proof of $\leq$ is OK. To prove $\geq$, simply note that the conjugates of $\alpha$ are exactly the roots of its minimal polynomial, which by definition has degree $r$. Now the roots are $r$ if and only if they are all distinct, i.e. if and only if the minimal polynomial is separable.