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Let $L=\{+,\dot{},0,1\}$ be the language of fields. I wish to show that the reals ($N=(\mathbb{R},+,\dot{},0,1)$) are not interpretable in the structure $M = (\mathbb{C},+,\dot{},0,1)$.

I have the following solution to show that $N$ is not definable in $M$: Suppose that the reals were definable in $M$. Then we can write a formula that defines a linear order inside of $M$ exploiting the fact that $<$ is definable in $N$. This gives a contradiction since $Th(M)$ is $\aleph_{1}$- categorical and hence $\omega$-stable but the above implies its unstable.

I believe the same proof holds with definability replaced with interpretability.

My questions is: Is there a more elementary proof of this fact?

Edit1: I saw a lot of comments below. I'm sorry about the typo. It has been corrected but it probably read $Th(N)$ where it should have been $Th(M)$.

  • What does it mean to say a structure is interpretable in another structure? (Did you maybe mean you want to show the theory of the reals is not interpretable in the theory of $\Bbb C$?) – David C. Ullrich Feb 13 '16 at 15:58
  • @DavidC.Ullrich See page 32 of http://library.msri.org/books/Book39/files/marker.pdf –  Feb 13 '16 at 16:01
  • How do you get definable linear order on $M$ from the data required to interpret $N$ in $M$ according to the definition you cite? I.e., how do you get a linear order on $M$ from a definable linear order on some definable quotient of $M$? – Rob Arthan Feb 13 '16 at 17:09
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    Note that, since $M$ has elimination of imaginaries, being able to interpret something is no easier than being able to define something. –  Feb 13 '16 at 18:59
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    @MikeHaskel: are you saying that the question boils down to "is $\Bbb{R}$ isomorphic to a definable subfield of $\Bbb{C}$?"? If so then that's easy to answer algebraically: if such a subfield were definable, then a particular root of $z^3 = 2$ would be definable, but the group of automorphisms of $\Bbb{C}$ acts transitively on the roots of that equation. – Rob Arthan Feb 13 '16 at 19:05
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    @RobArthan No, the question boils down to whether there's a definable subset $R$ of $\mathbb{C}^n$ for some $n$ which carries a definable field structure (not necessarily having anything to do with the field structure on $\mathbb{C}$), making $R$ a real closed field. – Alex Kruckman Feb 13 '16 at 19:11
  • If we had that, we'd have a definable linear order on the definable set $R$ (not on all of $\mathbb{C}$ - that's why kav writes "linear order inside of $M$" instead of "linear order on $M$"). The point about elimination of imaginaries is that we can take $R$ to be a definable subset, instead of a definable quotient of a definable subset - this is the difference between definability and interpretability. – Alex Kruckman Feb 13 '16 at 19:16
  • @AlexKruckman: you are getting closer to the point of my original comment (which would have been clearer if I'd written "subquotient" rather than "quotient"). What requirements on the linear order are needed to make kav's argument go through? – Rob Arthan Feb 13 '16 at 19:35
  • Well, kav's argument goes through as stated. Any theory which interprets an infinite linear order is unstable - all you need for instability is that some infinite sequence of tuples is linearly ordered by some formula. – Alex Kruckman Feb 13 '16 at 20:29
  • @AlexKruckman: thanks, the magic word was "infinite". I am sorry if my blundering attempts to understand the question have created a distraction, but there is one more thing I don't understand: is the first-order theory of the reals really $\aleph_1$-categorical? Isn't the field of Puiseux series over a countable real closed field, a real closed field with the same cardinality as $\Bbb{R}$ that is not isomorphic to $\Bbb{R}$? – Rob Arthan Feb 13 '16 at 21:35
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    The theory of the reals is not uncountably categorical - in fact, it has the maximal number $2^{\kappa}$ of models of size $\kappa$ for all uncountable $\kappa$. This is not obvious, but it follows from the fact that the theory is unstable, by Shelah's work. I should say that despite all these comments I don't have an answer to kav's question, ie I don't know how to see this in a more elementary way. – Alex Kruckman Feb 13 '16 at 21:46
  • @AlexKruckman: I am now confused. The question proposes an argument (for the statement in the title of the question) based on the claim that the theory of the reals is $\aleph_1$-categorical and asks for a more elementary argument. We've just agreed the claim is false, so we don't have any proof on the table for the statement in the title of the question. So surely the question now needs to be corrected. – Rob Arthan Feb 13 '16 at 22:20
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    Oh, now I understand the confusion - there is a typo in the question! Of course it's the theory of $M$ (the complex numbers) that is $\aleph_1$-categorical. – Alex Kruckman Feb 13 '16 at 22:26

1 Answers1

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OK, here's an argument that avoids stability theory:

Suppose we could interpret $\mathbb{R}$ in a field $K\models \text{ACF}_0$. Note that $K$ must be uncountable. By elimination of imaginaries in $\text{ACF}_0$, we may assume that the domain of the interpretation is a definable subset of $K$, defined by $\varphi(\overline{x})$, rather than a quotient of a definable set.

Let $A$ be the finite set of parameters used in the interpretation (i.e. in $\varphi$, as well as in the formulas defining the field operations on the interpreted copy of $\mathbb{R}$), and let $F$ be the algebraic closure in $K$ of the subfield of $K$ generated by $A$. Note that $F$ is countable, and for any $b\in K\setminus F$, there is an automorphism of $K$ which fixes $F$ pointwise but moves $b$ (extend $\{b\}$ to a transcendence basis for $K$ over $F$, and permute the basis).

Now since $\mathbb{R}$ is uncountable, there is a tuple $\overline{a}$ satisfying $\varphi(\overline{x})$ such that one of the entries $a_i$ of the tuple is not in $F$. Let $\sigma\in \text{Aut}(K)$ fix $F$ pointwise but move $a_i$. Then $\sigma$ induces an automorphism of the interpreted copy of $\mathbb{R}$ which moves $\overline{a}$. This contradicts the fact that $\mathbb{R}$ has no nontrivial automorphisms, QED.

Note that this argument does not show that $\text{Th}(\mathbb{R})$ is not interpretable in $\text{Th}(\mathbb{C})$, just that no uncountable rigid model of this theory (e.g. $\mathbb{R}$ or any uncountable subfield) is in the image of any such interpretation. Hence it also shows that $\text{Th}(\mathbb{R})$ and $\text{Th}(\mathbb{C})$ are not bi-interpretable.

As explained in the question, $\text{Th}(\mathbb{R})$ is not in fact interpretable in $\text{Th}(\mathbb{C})$ - it would be interesting to see a proof of this which avoids stability theory.

Alex Kruckman
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  • Thank you Alex. This is what I was looking for. But as you said it would be interesting to see a proof of Th$\mathbb{R}$ is not interpretable in Th$\mathbb{C}$. Please post if you do figure it out :) –  Feb 15 '16 at 14:48