# Can one show that the real field is not interpretable in the complex field without the axiom of choice?

We all know that the complex field structure $$\langle\mathbb{C},+,\cdot,0,1\rangle$$ is interpretable in the real field $$\langle\mathbb{R},+,\cdot,0,1\rangle$$, by encoding $$a+bi$$ with the real-number pair $$(a,b)$$. The corresponding complex field operations are expressible entirely within the real field.

Meanwhile, many mathematicians are surprised to learn that the converse is not true — we cannot define a copy of the real field inside the complex field. (Of course, the reals $$\mathbb{R}$$ are a subfield of $$\mathbb{C}$$, but this subfield is not a definable subset of $$\mathbb{C}$$, and the surprising fact is that there is no definable copy of $$\mathbb{R}$$ in $$\mathbb{C}$$.) Model theorists often prove this using the core ideas of stability theory, but I made a blog post last year providing a comparatively accessible argument:

The argument there makes use in part of the abundance of automorphisms of the complex field.

In a comment on that blog post, Ali Enayat pointed out that the argument therefore uses the axiom of choice, since one requires AC to get these automorphisms of the complex field. I pointed out in a reply comment that the conclusion can be made in ZF+DC, simply by going to a forcing extension, without adding reals, where the real numbers are well-orderable.

My question is whether one can eliminate all choice principles, getting it all the way down to ZF.

Question. Does ZF prove that the real field is not interpretable in the complex field?

I would find it incredible if the answer were negative, for then there would be a model of ZF in which the real number field was interpretable in its complex numbers.

• The completeness of the theories of real-closed fields and algebraically closed fields of characteristic $0$ are provable by elementary syntactic arguments (and in any case, they are arithmetical statements, hence automatically provable without AC). Thus, the problem is equivalent to showing that the field of algebraic numbers does not interpret a real-closed field. Since this field is countable, this should avoid issues with well orderability of $\mathbb R$. Jul 13 at 13:35
• Actually, why am I making it so complicated: the non-interpretability is itself an arithmetic statement, hence it is automatically provable without AC. Jul 13 at 13:37
• Emil, please post an answer! It may be good to explain in detail why the interpretation of those two structures amounts to the interpretability of their theories. I was worried about a model of ZF whose real numbers are very strange (e.g. countable union of countable sets etc.) and you are saying I don't need to worry about that at all. Jul 13 at 13:41
• I don't know about "much more", it's also reliant on absoluteness results. Mine is definitely cooler, though, since it involves forcing and dancing in two weddings with one tuchess. Jul 13 at 13:58
• Related, apparently. Jul 16 at 18:18

An interpretation of $$(\mathbb R,+,\cdot)$$ in $$(\mathbb C,+,\cdot)$$ in particular provides an interpretation of $$\DeclareMathOperator\Th{Th}\Th(\mathbb R,+,\cdot)$$ in $$\Th(\mathbb C,+,\cdot)$$. To see that the latter cannot exist in ZF:

• The completeness of the theory $$\def\rcf{\mathrm{RCF}}\rcf$$ of real-closed fields is an arithmetical ($$\Pi_2$$) statement, and provable in ZFC, hence provable in ZF. Its axioms are clearly true in $$(\mathbb R,+,\cdot)$$, hence $$\Th(\mathbb R,+,\cdot)=\rcf$$.

• Similarly, ZF proves completeness of the theory $$\def\acfo{\mathrm{ACF_0}}\acfo$$ of algebraically closed fields of characteristic $$0$$, hence $$\Th(\mathbb C,+,\cdot)=\acfo$$.

• The non-interpretability of $$\rcf$$ in $$\acfo$$ is again an arithmetical statement ($$\Pi_2$$, using the completeness of $$\acfo$$), hence its provability in ZFC automatically implies its provability in ZF.

Of course, common proofs of some or all the results above may already work directly in ZF (e.g., if you take syntactic proofs of completeness, or if you make it work using countable models, etc.). My point is that it is not necessary to check the proofs, as the results transfer from ZFC to ZF automatically due to their low complexity.

• Very nice, thank you. It seems to me that the ZFC argument shows that $\mathbb{R}$ is not interpretable in $\mathbb{C}$ even when parameters are allowed. Does this argument also generalize to the parametric case? It would seem to take us out of the arithmetic realm. Jul 13 at 14:49
• @Joel A parametric interpretation of $\mathbb R$ in $\mathbb C$ gives a parametric interpretation of $\mathrm{Th}(\mathbb R)$ in $\mathrm{Th}(\mathbb C)$. This does not change the complexity of the resulting statement. Jul 13 at 15:00
• @Joel No, sorry, it’s actually a bit weaker, as the usual definition of parametric interpretations of theories requires the set of “good” parameters to be definable. But what it gives is a translation of the language of RCF to the language of ACF_0 with extra free parameters such that for any sentence provable in RCF, the existential closure (over the parameters) of the corresponding translation is provable in ACF_0. This is an arithmetical statement, and in ZFC, it implies that a model of RCF is parametrically interpretable in some model of ACF_0, which is still false. Jul 13 at 15:09

(1) $$\mathbb{C}$$ is stable, (2) $$\mathbb{R}$$ is unstable, and (3) stability is preserved under interpretations. Of course the usual development of stability uses a lot of choice, but (2) and (3) are elementary. (1) follows from QE for algebraically closed fields. One of the reasons why model theorists care about the whole alphabet soup of classification-theoretic properties is that they enable this kind of argument.

Furthermore we have a complete understanding of fields interpretable in $$\mathbb{C}$$: If $$K$$ is an algebraically closed field then any infinite field interpretable in $$K$$ is definably isomorphic to $$K$$. This is in Poizat's book on stable groups, but I forget who it is due to. I don't think this proof requires choice either.

Edit: I just read Joel's post more carefully and realized that it contains "Model theorists often prove this using the core ideas of stability theory". But I think it's still worth pointing out that this argument does not require choice.

Another edit: I thought about this a bit more today and I think that the following is the simplest argument I can think of. It uses ideas from stability, but no actual stability theory, just elementary model theory and algebra. Recall that a theory $$T$$ is $$\omega$$-stable if for any $$M \models T$$ and countable $$A \subseteq M$$, there are only countably many types in $$M$$ over $$A$$. It's easy to see that $$\omega$$-stability is preserved under interpretations - if $$N$$ is interpretable in $$M$$ then a witness of non-$$\omega$$-stability in $$N$$ lifts to one in $$M$$. It's also easy to see that a real closed field is not $$\omega$$-stable - there are uncountably many types over $$\mathbb{Q}$$. So it comes down to showing that algebraically closed fields are $$\omega$$-stable. Quantifier elimination for algebraically closed fields gives a bijection between $$n$$ types over $$A \subseteq \mathbb{C}$$ and prime ideals in $$K [x_1,\ldots,x_n]$$ where $$K$$ is the subfield generated by $$A$$, and every prime ideal in $$K[x_1,\ldots,x_n]$$ is finitely generated, so there are only countably many prime ideals.

Yet another edit: Emil pointed out that the argument above actually uses choice - in the proof of preservation of $$\omega$$-stability under interpretations. He suggests replacing $$\omega$$-stability with the following property: there are only countably many types over any finite set of parameters. When you do that you only need to make finitely many choices in the preservation argument.

Yet another another edit It turns out that this whole question had already been asked and answered on stack exchange. In particular Alex gave a worked out elementary version of the stability argument.

• I see. I think one can probably get the stability argument down to something pretty basic that doesn't actually require the stability-theoretic machinery. Let's use the order property definition of stability - that's the most elementary. It's obvious that $\mathbb{R}$ has the order property, and the absence of the order property is easily seen to be preserved under interpretations - if $M$ interprets $N$ then a witness of the order property in $N$ lifts to one in $M$. So it should just come down to showing that $\mathbb{C}$ does not have the order property. Jul 13 at 22:01
• It should be possible to do this using QE for algebraically closed fields + some algebraic geometry. At this point it would help to know that a boolean combination of stable formulas is stable. I forgot the proof of this but it should just be finite combinatorics. Now we just need to show that $f(x,y) = 0$ does not have the order property, where $x = (x_1,\ldots,x_n)$, $y = (y_1,\ldots,y_m)$, and $f \in \mathbb{C}[x,y]$. Let $\delta(x,y)$ be this formula. So each instance $\delta(a,x)$ of $\delta$ defines a Zariski closed set. Jul 13 at 22:04
• it should follows from Noetherianness of the Zariski topology that there is $n$ such that any finite conjunction of instances of $\delta$ is equivalent to a conjunction of $n$ instances. It should be easy to see that this contracts the order property. Now that I've written this down it seems a lot more involved then the argument on your blog! Jul 13 at 22:10
• Now, one could also apply the fact that a theory is stable if every formula $\delta(x,y)$ is stable where $y$ is a single variable, this reduces to one-variable equations, which are easy. But this would involve using more stability. Jul 13 at 22:12
• @tomasz I gave a detailed argument directly from QE that $\mathbb{C}$ does not have the strict order property in this Math Stackexchange answer (points 5 and 6). I had totally forgotten about it until Asaf linked to it in the comments above. Jul 18 at 14:28