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J. Robinson has proved that exists a formula $\psi(x)$ in the language of rings which,applied to the rational numbers, defines the the ring integers (making the theory of $\mathbb{Q}$ undecidable, due to Godel's theorem). However, due to Tarski's results, the theory of the real field is complete and therefore the ring of integers $\mathbb{Z}$ is not definable in $\mathbb{R}$.

Let $K/\mathbb{Q}$ be an algebraic extension of the rational numbers. I will ask two related questions:

  1. Without any further assumptions on the extension, is $\mathbb{Z}$ necessarily a definable subset in $\langle K,+,\times,0,1\rangle$?
  2. What purely algebraic properties of the extension might yield a more conclusive answer to the above questions? Specifically: separability, dimension, normality, finiteness, simplicity (maybe even Galois group structure). Moreover, is there a deeper side to this? Do there exist two extensions $K_{1}/\mathbb{Q},K_{2}/\mathbb{Q}$ that satisfy exactly the same algebraic properties denoted above, yet in one of them $\mathbb Z$ is a definable subset, and in the other it is not?

These are not questions on which I have pondered for long, but I am curious whether there exist any interesting examples or logical tools needed in order to tackle these questions.

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  • $\begingroup$ The 1st § sounds a bit vague. Do you mean Robinson proved that there is a formula $F(x)$ in the language of fields such that for $x\in\mathbf{Q}$ we have $F(x)$ $\Leftrightarrow$ $x\in\mathbf{Z}$? In the question 1, by "the integers" do you mean $\mathbf{Z}$ or the ring of integers in $K$? $\endgroup$ – YCor Jun 1 at 14:15
  • $\begingroup$ Also "undecidable", does it mean here that the there's no Turing machine whose input is a free formula in the language of fields and whose output is yes/no according to whether it's true in $\mathbf{Q}$? $\endgroup$ – YCor Jun 1 at 14:21
  • $\begingroup$ @YCor Yes, that is what I mean $\endgroup$ – George Peterzil Jun 1 at 14:24
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    $\begingroup$ Without further assumption on the algebraic extension, $\mathbf{Z}$ is not definable (even with constants) in the real-closed field $\mathbf{R}\cap\bar{\mathbf{Q}}$. $\endgroup$ – YCor Jun 1 at 15:24
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    $\begingroup$ @JoelDavidHamkins because when there's such an immediate answer, I expect from the OP a reaction such as "oh, I meant a finite algebraic extension". $\endgroup$ – YCor Jun 2 at 8:34
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If $K$ is a finite extension of $\mathbb{Q}$, then, yes, $\mathbb{Z}$ is definable in $\langle K, +, \times, 0, 1 \rangle$. See R. Rumely, Undecidability and definability for the theory of global fields, Trans. AMS, 262 (1980), no.1, 195 - 217 http://www.ams.org/journals/tran/1980-262-01/S0002-9947-1980-0583852-6/home.html

As YCor notes, the situation for infinite algebraic extensions is more complicated. There are some such extensions which are known to have decidable theories. For example, the algebraic numbers themselves, $\mathbb{Q}^{\text{alg}}$, form an algebraically closed field and have a decidable theory. Likewise, the field of real algebraic numbers is a real closed field and is decidable by Tarski's theorem. On the other hand, there are natural infinite algebraic extensions in which $\mathbb{Z}$ is definable and others which might not be so natural whose theories are strictly more complicated than that of $\mathbb{Z}$.

There are some interesting cases for which it is not known (at least to me) whether the theory is decidable. For example, we do not know whether the theory of $\mathbb{Q}^{\text{ab}} = \mathbb{Q}( \{ \zeta ~:~ \zeta^n = 1 \text{ for some } n \in \mathbb{Z}_+ \})$ is decidable.

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  • $\begingroup$ What does "strictly more complicated" mean here? Do you just mean that their theories are not interpretable in $\mathbb{Z}$? $\endgroup$ – Alex Kruckman Jun 4 at 15:25
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    $\begingroup$ The precise meaning I had in mind was with respect to Turing reducibility. Other forms of reducibility, such as interpretability, would make sense, too. $\endgroup$ – Thomas Scanlon Jun 4 at 20:25

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