Small elementary extensions of $\langle\mathbb Q, 0,1,+,-,×,<\rangle$?

Question

Is it possibile to describe the elementary extensions of $\mathbb Q$, in the language of ordered rings, that have no infinitesimals? ( Subquestion: can they all be embedded in $\mathbb R$? )

Answer 1

Yes: $\mathbb{Q}$ is the only elementary extension of $\mathbb{Q}$ that has no infinitesimals.

First, it's a famous theorem of Julia Robinson that $\mathbb{N}$ is definable in $\mathbb{Q}$. Let's say $\varphi(x)$ is the formula defining $\mathbb{N}$.

Second, there is a definable function $f\colon (n,m)\mapsto \frac{n}{m}$, defined by $f(n,m) = q \iff q\cdot m = n$ which is surjective from the pairs of positive natural numbers to the positive rational numbers.

Now let $\mathbb{Q}\prec \mathcal{Q}$ be a proper elementary extension. Since $a\mapsto -a$ is a bijection between the positive rational numbers and the negative rational numbers, $\mathcal{Q}$ contains new positive rational numbers which are not in $\mathbb{Q}$. Since $f$ is surjective onto the positive rational numbers, this implies that $\mathcal{Q}$ also contains new "natural numbers": $\varphi(\mathbb{Q})\subsetneq \varphi(\mathcal{Q})$. Now for any standard natural number $n$, the theory of $\mathbb{Q}$ asserts that there are exactly $n$ elements less than $n$ which satisfy $\varphi(x)$. So a new element of $\varphi(\mathcal{Q})$ cannot be less than $n$ for any $n\in \mathbb{N}$. Thus $\mathcal{Q}$ contains an infinite element $\omega$ satisfying $\varphi(x)$, and hence an infinitesimal element: the multiplicative inverse of $\omega$.

The answer to your subquestion is then (trivially) yes. But it's also not hard to show that every (not necessarily elementary) ordered ring extension of $\mathbb{Q}$ which has no infinitesimals can be embedded in $\mathbb{R}$: just note that every element fills a proper Dedekind cut in $\mathbb{Q}$, and map it to the corresponding real number.