Let $L$ be a first-order language and $M$ be an $L$-structure. Let $D \subseteq M^n$ . Let us say $D$ is definable in $M$ if for some finite set (possibly empty) $A=\{a_1,...,a_m\} \subseteq M$ and some formula $\psi[x_1,...,x_n,y_1,...,y_m]$ , $D=\{(b_1,...,b_n)\in M^n : M\vDash \psi[b_1,...,b_n,a_1,...,a_m] \}$.

(i.e. $D$ is definable by a finite set of parameters according to this definition https://en.wikipedia.org/wiki/Definable_set )

Now consider the first-order theory of fields. Let $\bar {\mathbb Q}$ be the algebraic closure of $\mathbb Q$ in $\mathbb C$. Take $M=\bar {\mathbb Q}$.

My question is : Is $\mathbb Q$ definable in $\bar {\mathbb Q}$ ?


No. The most basic way to see this is to use the fact that ACF$_0$ (the theory of algebraically closed fields of characteristic $0$, of which $\overline{\mathbb{Q}}$ is a model) has quantifier elimination. So every formula $\varphi(x,\overline{a})$ (with parameters $\overline{a}$, in one free variable $x$) is equivalent to a quantifier-free formula. But any quantifier-free formula is equivalent to a boolean combination of polynomial equalities of the form $p(x) = 0$, where $p\in \mathbb{Q}[\overline{a}]$. Since a polynomial has at most finitely many solutions, such a formula defines a finite or cofinite set. And $\mathbb{Q}$ is infinite and coinfinite in $\overline{\mathbb{Q}}$.

At a higher level: What we've really shown above is that ACF$_0$ is strongly minimal (every definable set in one variable is finite or cofinite), from which it follows that it is uncountably categorical, in particular $\omega$-stable, i.e. the tamest of the tame kind of theory. On the other hand, any theory which interprets $\mathbb{Q}$ (as a field) is the wildest of the wild kind of the theory: it can define arithmetic and all recursively enumerable sets and do Gödelian tricks, and it's definitely definitely not $\omega$-stable, much less strongly minimal.

So this is stronger: not only is the "standard" copy of $\mathbb{Q}$ not definable in a model of ACF$_0$, but ACF$_0$ doesn't interpret $\text{Th}(\mathbb{Q})$, by which I mean there is no definable subset $Q\subseteq (\overline{\mathbb{Q}})^n$ and definable functions $\hat{+}\colon Q^2\to Q$ and $\hat{*}\colon Q^2\to Q$ such that $(Q,\hat{+},\hat{*})\models \text{Th}(\mathbb{Q},+,*)$.

  • $\begingroup$ +1. Note for the OP: the fact that $\mathbb{Q}$ (as a field) is "wild" - in particular, $\mathbb{Z}$ is definable - is very surprising, and as far as I know all definitions of $\mathbb{Z}$ in $\mathbb{Q}$ are quite complicated. $\endgroup$ – Noah Schweber Feb 26 '18 at 15:44
  • 2
    $\begingroup$ @Noah: I copied Robinson's definition in my answer here math.stackexchange.com/a/2644566/630 - it is not so bad. $\endgroup$ – Carl Mummert Feb 26 '18 at 15:46
  • $\begingroup$ Thanks for your answer ... I will have to study about the fact that $ACF_0$ has quantifier elimination ( I am new to these things ) $\endgroup$ – user Feb 26 '18 at 15:48
  • $\begingroup$ @CarlMummert Ah, right - I was thinking of definitions shooting for optimal quantifier complexity (like this one). $\endgroup$ – Noah Schweber Feb 26 '18 at 15:49
  • $\begingroup$ @Noah yes, much worse! $\endgroup$ – Carl Mummert Feb 26 '18 at 15:50

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