# I recently saw a proof that the real number field

I recently saw a proof that the real number field is not interpretable in the complex number field. But this required the axiom of choice, namely the existence of wild automorphisms of the complex numbers. Is there a way to prove it in ZF alone?
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billyfcash5n
In the comments, I mentioned the following argument that R is not interpretable in C: C is stable, R is unstable, and stability is preserved under interpretations.
The word "stable" may look a bit scary here - indeed, stability theory is a rather technical subject - but the argument above actually doesn't need any complicated ideas from stability theory. It just boils down to this: the order on the real field is definable, but no infinite linear order is interpretable in the complex field.
I'm taking your question as a challenge to produce as elementary a proof of this as possible, in particular without using the word "stable" and without using any choice. For those in the know, I'm trading in the order property (a theory is stable if it does not have the order property) for the strict order property, to make the argument a bit more transparent.
Suppose for contradiction that the real field R is interpretable in the complex field C.
First, note that the standard order $x\le y$ on R is definable by the formula $\phi \left(x,y\right):\mathrm{\exists }z\phantom{\rule{thinmathspace}{0ex}}\left(x+{z}^{2}=y\right)$.
Since the complex field interprets the real field, and the real field interprets the real order, we can compose these interpretations to conclude that the complex field interprets the real order. More precisely: As part of the data of the given interpretation, we have a definable set $X\subseteq {\mathbb{C}}^{n}$ and a surjective map $\pi :X\to \mathbb{R}$. Each real number $r\in \mathbb{R}$ is represented by an equivalence class ${X}_{r}={\pi }^{-1}\left(\left\{r\right\}\right)$ for a definable equivalence relation on X. Pulling back the formula $\phi$ to C, there is a formula $\psi \left(x,y\right)$ (where now x and y are tuples of length n) such that for all $a\in {X}_{r}$ and $b\in {X}_{s}$, $\mathbb{C}\models \psi \left(a,b\right)$ if and only if $r\le s$.
3. In particular, if we write ${Y}_{b}$ for the subset of X defined by $\psi \left(x,b\right)$, then $\left({Y}_{b}{\right)}_{b\in X}$ is a family of definable sets which is linearly preordered by $\subseteq$, and such that the quotient linear order is isomorphic to the standard order on R. To get a contradiction, we would like to show that the complex field does not admit any such family of definable sets.
4. To understand the definable sets in C, we use quantifier elimination. Now the easiest proofs of quantifier elimination for the complex field use the compactness theorem, which might get you worried that we're using choice. But don't worry: quantifier elimination for C can be proven constructively.
Now there are probably many ways to see that the complex field does not admit any quantifier-free definable family of definable sets which is linearly preordered by $\subseteq$ with the order type of R. Here's the most elementary way that occurred to me - remember, I'm trying to avoid appealing to any more advanced results in algebraic geometry or model theory.
5. First, let's assume $X\subseteq {\mathbb{C}}^{1}$, i.e. x is a single variable, not a tuple of variables. Let $\psi$(x,y) be the formula defining the family of definable sets. By quantifier elimination, we may assume $\psi$ is quantifier-free. Then for any b, $\psi$(x,b) is equivalent to a Boolean combination of polynomial equations p(x)=0 and inequations $p\left(x\right)\ne 0$, with each $p\in \mathbb{C}\left[x\right]$. When $p\ne 0$, the formula $p\left(x\right)\ne 0$ defines a finite set of size at most deg(p), $p\left(x\right)\ne 0$ defines a cofinite set whose complement has size at most deg(p), so letting N be the sum of the degrees (in x) of all the polynomials involved in $\psi \left(x,y\right)$, we have that $\psi \left(x,b\right)$ defines a finite set of size at most N or a cofinite set whose complement has size at most N. Hence, a $\subseteq$chain of definable sets defined by instances of $\psi$ can have length at most 2N+2, and in particular every such chain has a $\subseteq$minimal element.
6. Now let's prove by induction on n, where n is the length of the tuple of variables x, that for any formula $\psi \left(x,y\right)$, there is no family $\left({X}_{b}{\right)}_{b\in Y}$ of definable sets defined by $\psi$ which is linearly preordered by $\subseteq$ and has no minimal element. We've established the base case n=1. So let $\mathrm{\forall }x\phantom{\rule{thinmathspace}{0ex}}\left(\psi \left(x,b\right)\to \psi \left(x,{b}^{\prime }\right)\right)$. Let's write $b\le {b}^{\prime }$ when Xb⊆Xb′, and note that this relation is definable (by ∀x(ψ(x,b)→ψ(x,b′))). For any b and any $a\in {\mathbb{C}}^{n}$, we can look at the set ${Z}_{a,b}$ defined by $\psi \left(a,{x}_{n+1},b\right)$. Since ${Z}_{a,b}\subseteq {\mathbb{C}}^{1}$ is the fiber over a of ${X}_{b}$, we have ${Z}_{a,b}\subseteq {Z}_{a,{b}^{\prime }}$ whenever $b\le {b}^{\prime }$. For fixed a, since $\left({Z}_{a,b}{\right)}_{b\in Y}$ is a definable family of subsets of ${\mathbb{C}}^{1}$, it has a $\subseteq$least element, i.e. ${Z}_{a,b}$ is constant for a downwards-closed set of bs. Let's call this downwards-closed set ${Y}_{a}$. We can use this observation to definably linearly preorder the n-tuples a: $a\le {a}^{\prime }$′ if ${Y}_{a}\subseteq {Y}_{{a}^{\prime }}$′. Applying induction to the family of downwards-closed sets in this order, by induction the order has a minimal element a∗. But now for any $b\in {Y}_{{a}^{\ast }}$, I claim that ${X}_{b}$ is minimal in the original family of sets. Indeed, if b′ such that that ${X}_{{b}^{\prime }}⊊{X}_{b}$, then there is some a such that ${Z}_{a,{b}^{\prime }}⊊{Z}_{a,b}$. But then $b\notin {Y}_{a}$, so ${Y}_{a}⊊{Y}_{{a}^{\ast }}$, contradicting minimality of a∗.