First-order properties and models of $\mathbb{Q}$

Question

It's an open question whether there exists an existential first-order definition of $\mathbb{Z}$ in $\mathbb{Q}$ in the language of rings $\mathcal{L}_\text{ring} = (+, \cdot, 0, 1)$, i.e. a formula of the form $t \in \mathbb{Z} \Leftrightarrow \exists x_1, \ldots, x_n : \phi(t, x_1, \ldots, x_n)$ where $\phi$ is a quantor-free formula. Koenigsmann noted in his 2016 article "Defining $\mathbb{Z}$ in $\mathbb{Q}$" that it might be able to obtain an answer to this question by studying models of the theory of $\mathbb{Q}$ and their "rings of integers" (defined by transferring a first-order definition of $\mathbb{Z}$ in $\mathbb{Q}$).

A quick internet search for "models of the rationals" yielded no current research about this topic, although I cannot imagine it has gone unexplored. So my question: what is known about the first-order theory of $\mathbb{Q}$ and have meaningful non-standard models of $\mathbb{Q}$ been constructed/studied?

Some examples (again due to Koenigsmann) of facts about the theory of $\mathbb{Q}$ which would lead to results about the original question:

• A necessary condition for $\mathbb{Z}$ to be existentially definable in $\mathbb{Q}$, is that, for any two models $Q_1$ and $Q_2$ of the theory of $\mathbb{Q}$, $Q_1 \subseteq Q_2$ implies that $Q_1$ is relatively algebraically closed in $Q_2$.
• A necessary and sufficient condition for $\mathbb{Z}$ to be definable in $\mathbb{Q}$ is that, for any two models $Q_1, Q_2$ of the theory of $\mathbb{Q}$ with "rings of integers" $Z_1$ and $Z_2$, $Q_1 \subseteq Q_2$ implies $Z_1 \subseteq Z_2$.

Answer 1

A field which is elementarily equivalent to $\mathbb{Q}$ is sometimes called a Peano field. Some structural results about these fields are proven in Chapter 4 of the book Model Theoretic Algebra, by Jensen and Lenzing.

Answer 2

By the classic back-and-forth argument the following facts characterise $\mathbb{Q}$ as an ordered space:

Let $(X,<)$ be a countable ordered space. If

1. $\forall x \in X: \exists y \in X: y < x$ (no minimum)

2. $\forall x \in X: \exists y \in X: x < y$ (no maximum)

3. $\forall x.y \in X: (x < y) \implies \left( \exists z \in X: (x < z \land z < y)\right)$ dense order.

Then there is an order isomorphism $f: (\mathbb{Q}, <) \to (X,<)$. So $\mathbb{Q}$ is $\omega$-categorical in the theory of linearly ordered spaces.