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In 2009, Jochen Koenigsmann showed that $\mathbb{Z}$ is universally definable in the field $\mathbb{Q}$. And in 2012, Jennifer Park proved a result which implies that $\mathbb{Z}$ is $\exists\forall$-definable in many if not all finite extensions of $\mathbb{Q}$. And Jan Denef and others showed that $\mathbb{Z}$ is existentially definable in several transcendental extensions of $\mathbb{Q}$, like $\mathbb{R}(t)$ and others. But I'm interested in infinite algebraic extensions.

Now in this paper, Alexandra Shlapentokh discusses $\mathbb{Z}$ being first-order definable in various infinite algebraic extensions of $\mathbb{Q}$. But I'm not sure what the complexities of these first-order definitions are. So my question is, what is the lowest complexity definition of $\mathbb{Z}$ in some infinite algebraic extension of $\mathbb{Q}$?

Note: This is a follow-up to my question here.

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    $\begingroup$ One can try the following strategy to give an existential definition - Choose some equation $f(x,y)$ of high degree, which has no solutions over $\mathbb Q$ but, of course, many solutions over $\overline{\mathbb Q}$. Then adjoin to $\mathbb Q$ a root $y$ of $f(x,y)$ for each $x \in \mathbb Z$. If you're lucky, you haven't accidentally adjoined any roots for any $x \notin \mathbb Z$, and then $\exists y f(x,y)=0$ does the trick. I'm guessing you can find heuristics showing that you will be lucky for some equation, but proving this might be a challenge. $\endgroup$
    – Will Sawin
    Commented Feb 6, 2021 at 22:14

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I do not know what the lowest known complexity of such a definition is, but in principle one can get low complexity definitions as follows: As in my answer to this question, one decomposes the problem into getting a definition of the ring of integers $\mathcal{O}_K$ in $K$, and then a definition of $\mathbb{Z}$ in $\mathcal{O}_K$.

For defining $\mathbb{Z}$ in $\mathcal{O}_K$ one knows cases where this can be done existentially. There is for example the result of Denef that this works if $K$ is totally real and one has an elliptic curve $E/\mathbb{Q}$ with $E(\mathbb{Q})$ infinite and $E(K)/E(\mathbb{Q})$ finite, see the discussion on p. 235 of J. Denef, Diophantine Sets Over Algebraic Integer Rings. II, Trans. Amer. Math. Soc 257, 1980 (link). This elliptic curve method was later exploited and generalized in many ways by people like Poonen, Cornelissen and Shlapentokh, but I'm not sure what the state of the art is.

As for defining $\mathcal{O}_K$ in $K$, the best result in terms of quantifiers I'm aware of right now is Theorem 2.3.7 in this thesis of Philip Dittmann, which gives for every prime number $l>2$ an $\forall\exists$-definition of $\mathcal{O}_K$ in every Galois extension $K/\mathbb{Q}$ which has no finite subextensions of degree divisible by $l$ (and even uniform across such $K$).

So assuming that one finds some $K$ for which both steps work, one would get an $\exists\forall\exists$-definition of $\mathbb{Z}$ in $K$, but maybe someone else knows of better results.

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