Degree of extension of fixed field by infinite set of automorphisms.

Question

If $G$ is a finite group of automorphism $E \rightarrow E$, then Dedekind-Artin theorem tells us that $[E:E^G]=\; \mid G \mid$ where $E^G$ is the subfield of $E$ fixed by the automorphisms of $G$. Is this still true if $G$ is infinite or are they simple counter-examples?

Answer 1

No. If $E$ is the algebraic closure of the field $\Bbb{F}_p$, then we can think of $E$ as the nested union $$K_0\subset K_1\subset K_2\subset \cdots,$$ where the field $K_\ell$ is th unique (up to isomorphism) field of $p^{\ell!}$ elements. Then $E=\bigcup_{i=0}^\infty K_i$ is a countably infinite set, and hence $[E:\Bbb{F}_p]$ is countably infinite.

But the group of automorphisms $Gal(E/\Bbb{F}_p)$ is uncountable.

Answer 2

An example never to be ignored is the case when $k$ is a field of characteristic zero, and $E=k(t)$, $t$ being an indeterminate, so that the extension $E\supset k$ is simple transcendental. Now consider the group $G$ of all automorphisms of $E$ of form $t\mapsto t+n$, where $n\in\Bbb Z$. That is, if $f(t)$ is a $k$-rational expression in $t$ the image of $f$ is to be $f(t+n)$. You see that this is a good automorphism of $E$, and certainly the group it generates is countable infinite.

Now, the fixed field of $G$ is $k$ itself, ’cause only a constant rational function is unchanged by the substitution $t\mapsto t+n$, there being infinitely many different such substitutions.

And what is the dimension of $k(t)=E$ as a $k$-vector space? It is at least the cardinality of $k$ (!), since you see that the functions $\{\frac1{t-\alpha}\}$ are all $k$-linearly independent.

Thus you have a situation where the original group $G\cong\Bbb Z$ is countable, and even its completion $\hat{\Bbb Z}$ is merely of continuum cardinality, whereas $k$ itself might have any cardinality whatever that’s at least countable.