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I would like to construct a (ring-theoretic) automorphism of $\Bbb C$ that fixes a finite set $A$ pointwise but does not fix $\Bbb R$ setwise. Marker's Model Theory, Corollary 1.3.6 does that in this way:

Let $r, s \in \Bbb C$ be algebraically independent over $A$ with $r \in \Bbb R$ and $s \not \in \Bbb R$. There is an automorphism $\sigma$ of $\Bbb C$ such that $\sigma|_A$ is the identity and $\sigma (r) = s$. Thus $\sigma(\Bbb R) \neq \Bbb R$ [...]

(By the way, the existence of such an automorphism implies $\Bbb R$ cannot be definable by a first-order formula in $\Bbb C$, and that's what this part of the book is all about.)

I suppose the proof can be divided into two parts, each of which corresponds to the first and the second sentence, resp. What are the general facts used in this proof? I would also be grateful if you could suggest materials on the field of mathematics that include those facts.

Pteromys
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    One way is to note that there is an irreducible cubic over $\mathbb Q(A)$ with exactly one real root, and then use a modicum of Galois theory: There is an automorphism of the splitting field that permutes the real root with a complex one, and any such automorphism extends to one of $\mathbb C$. See also here. – Andrés E. Caicedo May 17 '14 at 22:58
  • (By the way, I've asked versions of this question in logic courses I've taught in the past. Nice to see that Marker's book also discusses this.) – Andrés E. Caicedo May 17 '14 at 22:59
  • @AndresCaicedo Thank you. Which of the two parts does your solution correspond to? – Pteromys May 17 '14 at 23:16
  • @Pteromys: the second sentrnce follows immediately from the first. – tomasz May 18 '14 at 09:10
  • @tomasz Could you please elaborate on that? – Pteromys May 20 '14 at 08:36
  • @Pteromys: $\sigma[{\bf R}]\ni \sigma(r)\notin {\bf R}$. – tomasz May 21 '14 at 00:17
  • @tomasz tomasz, I've assumed by the second sentence you meant the sentence that begins with "There." Now I know I was wrong. :-) – Pteromys May 21 '14 at 00:25
  • @Pteromys: Well, you referred to the first an second sentence in the statement of the question as things to be proved, and I used the same convention. The literally first sentence is not something to be proved, but rather an assumption. – tomasz May 21 '14 at 00:34
  • @tomasz I didn't think it trivial for such $r$ and $s$ to exist, so I thought the literally first sentence included something to be proved. – Pteromys May 21 '14 at 00:59
  • @Pteromys: I see, that's what you meant... – tomasz May 21 '14 at 01:59
  • @Pteromys: For that you need to see that the intersection of the algebraic closure of ${\bf Q}(A)$ and ${\bf R}$ is countable, so there is a real independent from $A$. The existence of $s$ follows similarly. – tomasz May 21 '14 at 02:07
  • @tomasz I've assumed that the first sentence states that there exist $r$ and $s$ such that for any nonzero $f(X, Y) \in \Bbb Q(A)[X,Y]$, f(r, s) = 0. Is this how you interpreted it? – Pteromys May 21 '14 at 23:50

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There are two (purely algebraic) facts you need:

  • for any field homomorphism $f\colon F\to K$ and $x$ is algebraically independent from $F$, while $x'$ is algebraically independent from $K$, then $f$ extends to a homomorphism $\overline f\colon F(x)\to K(x')$ which takes $x$ to $x'$. Similarly for (arbitrarily large) algebraically independent sets of $x, x'$.
  • any field isomorphism $f\colon F\to K$ extends to an isomorphism of algebraic closures.

You use the first fact twice, for $F={\bf Q}(A)$, $f=\operatorname{id}$, $x=a$, $x'=b$ and again with $x=b,x'=a$. Then you may choose a transcendental basis $(x_i)_{i\in I}$ of ${\bf C}$ over ${\bf Q}(A,a,b)$, and use similar fact to extend $f$ to an automorphism $\overline f$ of ${\bf Q}(A,a,b,x_i)_{i\in I}$ such that $\overline f(x_i)=x_i$.

You use the second fact to extend the resulting isomorphism to an automorphism of ${\bf C}$ (recall that field homomorphisms are automatically injective).

tomasz
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  • Thank you for explaining the details. Is the algebraic closure of $\Bbb Q(A, a, b)$ necessarily $\Bbb C$? Or, can every field isomorphism be extended to one between extension field? I have the feel that such a proposition can be proved by using Zorn's Lemma. – Pteromys May 21 '14 at 01:10
  • @Pteromys: My bad. You actually need to add enough elements to ${\bf Q}(A,a,b)$ to make sure that ${\bf C}$ is algebraic over the resulting field. – tomasz May 21 '14 at 02:01
  • so is it not true that for a general field extension $E/F$ and a field isomorphism $\sigma:F\rightarrow F$ there exists an extension of $\sigma$, namely, $\tau: E\rightarrow E$? – Pteromys May 21 '14 at 03:55
  • @Pteromys: No, for instance if $F={\bf Q}[\sqrt[3]{2},e^{2\pi i/3}\sqrt[3]{2}]$, $E=F[\sqrt[6]{2}]$ and $\sigma$ swaps $\sqrt[3]{2}$ and $e^{2\pi i/3}\sqrt[3]{2}$. I think this is true if $F\subseteq E$ is normal, but don't quote me on that. – tomasz May 21 '14 at 08:43