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Assume a finite field extension $\mathbb{C}/K$ such that $[\mathbb{C}:K]>2$. Let $\varphi \in \text{Aut}(\mathbb{C}/K)$, so $\varphi \in \text{Aut}(\mathbb{C})$ and $\varphi\vert_K=\text{id}\vert_K$.

Is it possible to show one of the three:

i. that the image $\varphi(\mathbb{R})$ is a subset of $\mathbb{R}$.

ii. that $\varphi$ is also a $\mathbb{R}$-Automorphism (which would follow from i.).

iii. that $\varphi$ is continuous on $\mathbb{R}$ (which follows from either i. or ii.).

We know that $(\varphi(i))^2=\varphi(i^2)=\varphi(-1)=-1$ as $\varphi\vert_{\mathbb{Q}}=\text{id}_{\mathbb{Q}}$ and we therefore have either $\varphi(i)=i$ or $\varphi(i)=-i$.

We also know that $\varphi$ fixes $K\cap\mathbb{R}$ pointwise and due to Artin-Schreier we have that $i$ is not in either $K$ or $K\cap\mathbb{R}$ as $K \cap \mathbb{R}$ is a subfield of $K$ and $\mathbb{R}$.

As $\mathbb{C}$ has characteristic $0$, so does $K$ and $K\cap\mathbb{R}$, and therfore $\mathbb{Q} \subseteq K\cap\mathbb{R} \subseteq K \subset \mathbb{C}$.

I tried to experiment with $K\cap\mathbb{R}$ but quite unsuccesful, as it could be equal to $\mathbb{Q}$ which isn#t of much use. It's also to note, that $\mathbb{C}/K$ is a Galois-Extension.

In the first version of the post I forgot to mention, that the degree of $\mathbb{C}/K$ is supposed to be finite!

Thanks for checking in :)

Cedric Brendel
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2 Answers2

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All three of your statements are true, but only vacuously so! There are no subfields $K\subseteq \mathbb{C}$ such that $\mathbb{C}$ is finite over $K$ and $[\mathbb{C}:K]>2$. More generally, by a theorem of Artin and Schreier, if any field $K$ is not algebraically closed but has an algebraic closure $\overline{K}$ that is finite over $K$, then $[\overline{K}:K]=2$, $K$ is a real-closed field, and $\overline{K}=K(\sqrt{-1})$.

Eric Wofsey
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  • That's indeed what I'm hinting at :) But I wan't to go the other way around! If a $\mathbb{C}$-automorphism that fixes $K$ pointwise turns out to be continuous on $\mathbb{R}$ it will also be continuous on $\mathbb{C}$ and then turns out to be the identity or the conjugation. We have exactly two elements in $\text{Aut}(\mathbb{C}/K)$. But since the extension is galois, we have that $\vert\text{Aut}(\mathbb{C}/K)\vert=[\mathbb{C}:K]>2$. A contradiction, there is no such $K$. – Cedric Brendel Mar 23 '19 at 07:00
  • I don't think this is a fruitful line of pursuit. Note that if you did not assume that $[\mathbb{C}:K]>2$, then it would not be true: there are automorphisms of $\mathbb{C}$ which fix $K\subset\mathbb{C}$ with $[\mathbb{C}:K]=2$ which are not continuous on $\mathbb{R}$. So you would somehow need to use the assumption $[\mathbb{C}:K]>2$ to prove it. – Eric Wofsey Mar 23 '19 at 07:04
  • If $[\mathbb{C}/K]=2$ isn't $K=\mathbb{R}$? So all $\mathbb{C}$-automorpisms that fix $K=\mathbb{R}$ are continuous... Or are there any other such subfields of $\mathbb{C}$ that I don't know of? And yes: to use $[\mathbb{C}/K]>2$ what I was going for :) – Cedric Brendel Mar 23 '19 at 07:18
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    No. There are lots and lots of index 2 subfields of $\mathbb{C}$ besides $\mathbb{R}$. It is rather difficult for me to imagine a natural way to use the hypothesis $[\mathbb{C}:K]>2$ to prove $\varphi$ is continuous on $\mathbb{R}$. The usual proof takes a very different route, first reducing to the case that $[\mathbb{C}:K]$ is prime and then investigating how $\varphi$ behaves on roots of unity. – Eric Wofsey Mar 23 '19 at 07:37
  • Ah I see... pretty hard stuff :P It's quite hard to connect $K$ to $\varphi$ besides of the pointwise fixation... If I have an idea I'll let you know – Cedric Brendel Mar 23 '19 at 08:19
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As linked to in a comment, $\mathbb{C}$ has "wild" automorphisms over $\mathbb{Q}$ or really any subfield of $\mathbb{C}$ of transcendence degree at most countable. These are basically impossible to describe, but they definitely don't satisfy any of these properties you mentioned.

Any field automorphism must preserve $\mathbb{Q}$ pointwise, so if it were continuous on $\mathbb{R}$ it would have to be the identity on $\mathbb{R}$ and therefore the automorphism of $\mathbb{C}$ would have to be either the identity map or complex conjugation.

Nate
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  • Of course. But remember I am talking about a extra condition that the automorphism fixes $K$ pointwise, not just any automorphism. I'm failry convinced, that this extra condition rules out any wild isomorphisms and suffices to show continuity. That's what I'm looking for and that's also why the linked comment doesn't quite help :( – Cedric Brendel Mar 22 '19 at 19:54
  • I'm not quite sure what you are asking then. As I said in the first paragraph, if $K = \mathbb{Q}$ (or any subfield of $\mathbb{C}$ of at most countable transcendence degree) there are wild automorphisms of $\mathbb{C}$ over $K$, meaning they fix $K$ pointwise. – Nate Mar 22 '19 at 20:06
  • Do you want some additional condition on $K$? – Nate Mar 22 '19 at 20:14
  • Isn't that just false? $\mathbb{C}/\mathbb{R}$ for example is algebraic and therefore has transcendence degree $0$. Since $\mathbb{C}/\mathbb{R}$ is an Galois-Extension we have $\vert \text{Aut}(\mathbb{C}/\mathbb{R})\vert=[\mathbb{C}:\mathbb{R}]=2$ (both $\text{id}:z \rightarrow z$ and $\text{conj}:z \rightarrow \bar{z}$ are in $\text{Aut}(\mathbb{C}/\mathbb{R})$) and there are no wild automorphisms left that fix $\mathbb{R}$ pointwise. – Cedric Brendel Mar 22 '19 at 20:24
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    Sorry when I said the transcendence degree of $K$ I meant transcendence degree of $K$ over $\mathbb{Q}$ – Nate Mar 22 '19 at 20:38
  • Ok, I now get the problem! I forgot a little word that meant all the trouble: $[\mathbb{C}:K]$ was supposed to be finite. That's embarassing... It now follows that $K/\mathbb{Q}$ has trancendence degree $\mathfrak{c}$. I'll edit the post... – Cedric Brendel Mar 22 '19 at 20:51
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    Ah ok that makes a big difference. Anyway doesn't the Artin-Schreier theorem say that any $K$ such that $\mathbb{C}/K$ is a finite extension must have $[\mathbb{C} : K] = 2$? – Nate Mar 22 '19 at 21:46
  • Yeah, but I wantbto go the other way around. If you show that $\varphi$ is continuous on $\mathbb{R}$ it's also continuous on $\mathbb{C}$ and either the identity or the conjugation. But $\mathbb{C}/K$ is a Galois-Extension and therefore $\vert \text{Aut}(\mathbb{C}/K)\vert=[\mathbb{C} :K]>2$. Contradiction. There is no such $K$ – Cedric Brendel Mar 23 '19 at 06:31