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I am posting this problem in order to break the problem in my previous post Field Extension problem beyond $\mathbb C$.

Notation: $M(\mathbb C):=$ Field of all meromorphic functions on $\mathbb C$, $K:=$ a proper subfield between $\mathbb C$ and $M(\mathbb C)$

Question 1: Let $\sigma\in Aut(K)$ and $c$ is a constant function. Is it true $\sigma(c)$ is again a constant function ?

I will post subsequent questions after getting a good answer (either by myself or someone else). Let us try how far we can go!

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If $f_n$ converges fo $f$, what about $\sigma(f_n)$? – Blah May 21 '12 at 12:34
@Blah: I don't follow. $\sigma$ is not required to be continuous. – Qiaochu Yuan May 21 '12 at 14:14
true, otherwise the question would make no sense! my mistake. – Blah May 21 '12 at 21:17

Yes, it is true that for $c\in\mathbb{C}$ then $\sigma(c)$ is again in $\mathbb{C}$. This is because the subfield $\mathbb{C}$ of $K$ is distinguished by the following algebraic property.

An element $c\in K$ is in $\mathbb{C}$ if and only if $X^n-c+q=0$ has a root in $K$ for each $n\in\mathbb{N}$ and $q\in\mathbb{Q}$.

Being an automorphism of $K$, $\sigma$ preserves the stated property, so maps elements of $\mathbb{C}$ (or, 'constant functions') into $\mathbb{C}$.

The fact that each $c\in\mathbb{C}$ satisfies the stated property above follows from the fact that $\mathbb{C}$ is algebraically closed. On the other hand, if $c\in K\setminus\mathbb{C}$ is a nonconstant meromorphic function then, by the little Picard theorem, it maps onto $\mathbb{C}$ minus at most two points, so its image will contain (infinitely many) rationals $q$ (or you can use the simpler fact that the image of a nonconstant meromorphic function is a connected dense open subset of $\mathbb{C}$). So, $c-q$ has a zero, which must be of a finite order $m\ge1$. Then, $c-q$ does not have an $n$'th root (in the field of meromorphic functions) for any $n$ not dividing $m$.

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Very ingenious, George! (Which doesn't surprise me from you...) – Georges Elencwajg May 30 '12 at 12:01
@ George Lowther: Sorry I don't follow the answer specifically the algebraic property. I can understand up to $\sigma(x+iy)=x\pm iy$ where $x,y\in \mathbb Q$...would you please elaborate or give me any reference. – users31526 May 30 '12 at 12:12
@Kuashik: I'm not sure what your misunderstanding is. Actually, there was one bit which wasn't quite correct. The image of a meromorphic function is open, but in itself that's not enough to guarantee that it contains a rational. But, still, any meromorphic function maps onto the whole of C minus at most a discrete set. I edited the answer to fix this. – George Lowther May 30 '12 at 20:56

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