Suppose that F is a field contained in an algebraically closed field A. Prove that every automorphism of F can be extended to an automorphism of A.
Let $\sigma$ be an automophism of $F$. Let $S$ be a transcendence basis of $A$ over $F$. There exists a unique automorphism $\tau$ of $F(S)$ such that $\tau|F = \sigma$ and $\tau(x) = x$ for every $x \in S$. Since $A$ is an algebraic closure of $F(S)$, there exists an automorphism $\rho$ of $A$ extending $\tau$. Then $\rho$ extends $\sigma$.