Existence of an automorphism of $\Bbb C$ that fixes a finite set pointwise but does not fix $\Bbb R$ setwise

Question

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.

Answer 1

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).