Is the theory of the rational ordered field countably categorical?

Question

Consider the structure $(\mathbb{Q},+,-,\times,0,1,<).$ Now certainly, by Lowenheim-Skolem, we can't have a set of sentences in first-order logic whose models are precisely that structure and any structure isomorphic to it. But what if we require the model to be countable? That is, if a countably infinite ordered field satisfies the theory of the rational ordered field, is it isomorphic to the rational ordered field?

Edit: Also, if the answer is no, can someone exhibit an ordered field not isomorphic to $\mathbb{Q}$ which satisfies its complete theory?

Answer 1

By compactness, $\mathbb{Q}$ has a non-Archimedean elementary extension $\mathcal{A}$. Of course $\mathcal{A}$ may be uncountable, but we can get around that: let $a\in\mathcal{A}$ be an infinite element, and apply downward Lowenheim-Skolem to get a countable $\mathcal{B}\preccurlyeq\mathcal{A}$ with $a\in \mathcal{B}$.

Since $\mathcal{B}\preccurlyeq\mathcal{A}\equiv\mathbb{Q}$ we have $\mathcal{B}\equiv\mathbb{Q}$. But by choice of $a$, $\mathcal{B}$ must be non-Archimedean - and so not isomorphic to $\mathbb{Q}$.

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EDIT: Whoops, I was really wrong initially - here's a correct statement:

It turns out in fact that there's a very useful general principle available to us: if $T$ is a (complete, countable language) theory which interprets a non-$\aleph_0$-categorical theory, then $T$ itself is not $\aleph_0$-categorical. So we can prove non-categoricity just by finding "bad configurations." This is a consequence of the Ryll-Nardzewski theorem, that a (countable complete) first-order theory is $\aleph_0$-categorical iff it has finitely many $n$-types for each $n$. See Alex Kruckman's comment below.

Answer 2

Your question admits a very strong negative answer: No infinite integral domain has a countably categorical theory.

Further, you can take "infinite integral domain" to mean any infinite structure $R$ in a countable language $L$ containing the language of rings, such that the reduct of $R$ to the language of rings is an integral domain. So, for example, this applies to the ordered field $\mathbb{Q}$.

Let $R$ be an infinite integral domain. For every $n$, the polynomial $x^n - 1$ has at most $n$ roots in $R$, so since $R$ is infinite, there is no finite upper bound on the multiplicative orders of non-zero elements of $R$. Then the formulas $y = x^n$ are pairwise inequivalent for all $n$ (given $m < n$, if $a\in R$ is a non-zero element of multiplicative order greater than $n$, and $b = a^n$, then $R\models b = a^n$, but $R\not\models b = a^m$). By the Ryll-Nardzewski theorem, $\text{Th}(R)$ is not countably categorical.