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Let $Aut(\mathbb{C}/\mathbb{Q})$ be the field automorphisms of $\mathbb{C}$, and $\mathbb{C}^{Aut(\mathbb{C}/\mathbb{Q})}$ the subfield of $\mathbb{C}$ fixed by this group. I supsect that it is equal to $\mathbb{Q}$ but I have difficulty proving it.

Here is what I do : let $x \notin \mathbb{Q}$. We have to find an automorphism not fixing $x$. It is easy to find an embeding $\mathbb{Q}(x) \rightarrow \mathbb{C}$ not sending $x$ on $x$, and use Zorn's lemma to extend it to a maximal subfield (of $\mathbb{C}$) $K \supset \mathbb{Q}(x)$. But it is not true that $K=\mathbb{C}$. So how to handle it ?

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What's stopping you from extending until you get all the way to $\mathbb{C}$? –  Qiaochu Yuan May 25 '11 at 15:34
You might want to take a look in here: mathdl.maa.org/images/upload_library/22/Ford/PaulBYale.pdf –  user9413 May 25 '11 at 15:37
You are asserting that $K$ is never equal to $\mathbb{C}$? On what grounds? –  Arturo Magidin May 25 '11 at 18:17
The problem is that if I apply Zorn's lemma like I do, then the map $K \rightarrow \C$ can be an isomophism without having $K=C$. –  user10676 May 25 '11 at 21:17
Instead of looking for embeddings of $K$ into ${\mathbb C}$, can't you simply apply Zorn Lemma for automorphisms of $K$? The hard part is the first step: for $x$ irrational, find $x \in K$ and an automorphism of $K$ not sending $x$ to $x$, and the Zorn it. –  N. S. May 26 '11 at 18:46
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2 Answers

up vote 8 down vote accepted

In this work in progress I state the following Fundamental Theorem of Galois Theory:

Let $K/F$ be any field extension. The following are equivalent:
(i) For all subextensions $L$ of $K/F$, $K^{\operatorname{Aut}(K/L)} = L$.
(ii) At least one of the following holds:
(a) $K/F$ is algebraic, normal and separable (i.e., a Galois extension in the usual sense), or
(b) $K$ is algebraically closed of characteristic zero.

That's the good news. The bad news is that this "Theorem", which I wrote down several years ago, is actually only a conjecture (it has remained as an open question on MathOverflow for more than a year, which is some indication of its nontriviality, at least). Okay, but there is more good news: the implication (ii) $\implies$ (i) is proven, and is a fairly routine application of basic field theory. Your question is a special case of (ii) $\implies$ (i), so there you go.

Added: Mea culpa, the proof of (ii) $\implies$ (i) in the linked to notes is "$\ldots$". (When you can't prove the big theorem you announce on the first page, you lose some motivation to fill in the other details, it seems.) Instead, please see $\S 10.1$ of my field theory notes in which there is a complete proof of (even a result slightly more general than) (ii) $\implies$ (i). Really, I promise.

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It is always wonderful when open research questions reach this site, thanks for sharing yours :-) –  Asaf Karagila May 25 '11 at 21:14
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I think that the article Chandru posted indicates that you can interchange any two irrational algebraic numbers with the same minimal polynomial or any two transcendental numbers with an automorphism of C. Thus the fixed field of Aut(C/Q) is exactly Q. The trick here is not to consider the poset of extension maps of Q(x) into C, but rather to start with an automorphism of the field and then look at the poset of automorphisms of some field extension. The punchline is towards the end of the article.

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