Fixed field is a field

Question

Let $L/K$ be a algebraic field extension and let $G\subset Aut(L/K)$ be a subgroup. It is easy to check that the set of fixed points of $G$ is a field by checking, that it is closed under multiplication, addition and inverse elements. But is there any more elegant prove of this?

Answer 1

There are some redundancies in your setting.

Let $F$ be a field and $\varphi$ an automorphism of $F$. Then the set of elements of $F$ fixed by $\varphi$ is a subfield of $F$.

This can be proved by showing closure under $0$, $1$, addition, multiplication and inverses. I see no “more elegant” way.

Consequently, if $X$ is a set of automorphisms of $F$, the set $\operatorname{Fix}(X)$ of elements of $F$ fixed by all automorphisms in $X$ is a subfield (being the intersection of subfields).

More interesting is perhaps that $\operatorname{Fix}(X)$ equals $\operatorname{Fix}(\langle X\rangle)$, where $\langle X\rangle$ is the subgroup of $\operatorname{Aut}(F)$ generated by $X$. Indeed, if $\varphi$ and $\psi$ are automorphisms of $F$, then $$ \operatorname{Fix}(\{\varphi^{-1}\})=\operatorname{Fix}(\{\varphi\}), \qquad \operatorname{Fix}(\{\varphi,\psi\})\subseteq\operatorname{Fix}(\{\varphi\psi\}) $$ Since $\langle X\rangle$ consists of finite products of elements in $X$ and inverses thereof, we get $\operatorname{Fix}(X)\subseteq\operatorname{Fix}(\langle X\rangle)$. The converse inclusion is obvious, as $X\subseteq\langle X\rangle$.