I was reading Lemma 2 of Daniel Lascar' The group of automorphisms of the field of complex numbers leaving fixed the algebraic numbers is simple whose statement is the following:
Assume $g \in \text{Aut}(\mathbb{C})$ and $K\subseteq\mathbb{C}$ algebraically closed subfield of cardinality strictly less than the continuum are such that for all $a\in\mathbb{C}$, $g(a)$ is algebraic over $K(a)$. Then $g$ is the identity on $\mathbb{C}$.
In the proof he proves that, under the hypotheses, $g$ is the identity on $\mathbb{C}\setminus K$ and immediately concludes that it needs to be the identity on the whole $\mathbb{C}$.
My question is: Why can he conclude that? Given that $g$ is the identity outside $K$, why does it need to be the identity also on $K$?
Thanks!
Quite generally, suppose $K$ is a proper subfield of $L$ and an automorphism $g$ of $L$ is the identity outside $K$. Fix some $a\in L$ such that $a\notin K$. For any $x\in K$, we have that $x+a\notin K$, so $g$ fixes both $x+a$ and $a$. Being an automorphism, $g$ must therefore also fix their difference $x$.
Note that this proof would work equally well if we just had additive groups $K\subsetneq L$; the rest of the field structure is irrelevant.