## Extensions

Suppose $\displaystyle E \leq \bar{F}$ is an algebraic extension of of $\displaystyle F$ and $\displaystyle \alpha, \beta \in E$ are conjugate over $\displaystyle F$. Then the conjugation isomorphism $\displaystyle \psi_{\alpha, \beta}: F(\alpha) \to F(\beta)$ can be extended to an isomorphism of $\displaystyle E$ onto a subfield of $\displaystyle \bar{F}$.

So is it never always possible to extend it onto $\displaystyle \bar{F}$ itself? Is $\displaystyle \mathbb{C}$ its own algebraic closure? In other words, does $\displaystyle \mathbb{C} = \bar{\mathbb{C}}$?