Question regarding invariant separable elements of an algebraic closure of a field. Let $\bar k$ be an algebraic closure of $k$ and $\alpha \in \bar k$ be separable over $k$. Suppose that $\sigma(\alpha)=\alpha$ for any non-zero ring homomorphism $\sigma:\bar k\rightarrow \bar k $. Show that $\alpha \in k$.

The roots of the minimum polynomial of $\alpha$ are among $\{\sigma (\alpha ) | \sigma : \overline{k} \rightarrow \overline{k}  \ $, $\sigma \ $ring homomorphism$ \}$.
But $\sigma( \alpha) = \alpha \ \forall  \ \sigma$ and being $\alpha$ separable this forces minpol($\alpha$) $= x- \alpha \Rightarrow \alpha \in k $.
