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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?

user26857
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roi_saumon
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    To make sure you have heard of this. The Galois correspondence for infinite extensions takes a slightly different form. We need to give the Galois group the structure of a topological group. That way the correspondence survives in the form that the intermediate field are the fixed fields of closed subgroups. The fixed fields of a subgroup and its closure are the same. You need to consult a textbook for those pieces of theory. – Jyrki Lahtonen Jan 03 '19 at 18:37
  • Oh, this is already Galois theory level 16. – roi_saumon Jan 03 '19 at 18:54
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    Of course, @JyrkiLahtonen, you’re assuming that the total extension is algebraic, which OP did not do. – Lubin Jan 03 '19 at 23:02
  • @Lubin Correct. That was an oversight. – Jyrki Lahtonen Jan 04 '19 at 17:05

2 Answers2

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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.

Kenny Lau
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Jyrki Lahtonen
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  • The same holds for $E=$ the algebraic closure of $\Bbb{Q}$ in $\Bbb{C}$. – Jyrki Lahtonen Jan 02 '19 at 20:49
  • Oh, I get the idea, thanks! And what does the inclusion for the finite fields mean? What would be the embedding of $\mathbb{F}9$ into $\mathbb{F}{27}$. Also could you point to a proof that $Gal(E/\mathbb{F}_p)$ is uncountable? – roi_saumon Jan 02 '19 at 23:30
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    @roi_saumon $\Bbb{F}{27}$ has no subfield isomorphic to $\Bbb{F}_9$. Remember that $\Bbb{F}{p^n}$ is a subfield of $\Bbb{F}{p^m}$ only when $n\mid m$. That is why I use the factorial $\ell!$ as the exponent in the cardinality of $K\ell$. – Jyrki Lahtonen Jan 02 '19 at 23:33
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    See Wikipedia for a description of the Galois group of $E/\Bbb{F}_p$. Over $\Bbb{Q}$ we have a very complicated group. – Jyrki Lahtonen Jan 02 '19 at 23:42
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    Also here for a nice but brief account of why $\hat{\Bbb{Z}}$ really parametrizes the elements of the Galois group – Jyrki Lahtonen Jan 02 '19 at 23:46
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    This thread describes $\hat{\Bbb{Z}}$ in terms of $p$-adic rings of integers. I have written an answer explaining why the union $F$ of $\Bbb{F}_{p^{2^\ell}}$ has uncountably many automorphisms (in that case the automorphism group is the ring of 2-adic integers), but may be I shouldn't link to it. A) self-advertising is bad, B) I skipped why automorphisms of $F$ can be extended to the algebraic closure. – Jyrki Lahtonen Jan 03 '19 at 08:18
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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.

Lubin
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  • Very nice example! I'm wondering if there is a form of Dedekind-Artin theorem for infinite automorphism groups, namely: If is a group of automorphisms →, is $E/E^G$ still Galois? And if so, how is it related to $G$ (maybe profinite completion)? Of course, in your example, $E/E^G$ is even transcendental, so we at least need to assume that $E/E^G$ is algebraic. – Lao-tzu Feb 06 '22 at 09:32