Ordered field of rationals axiomatizable

Question

Is there a set of sentences in the language of ordered fields whose models are precisely the rationals and any ordered field isomorphic to them?

Answer 1

The answer is no. For by the) Lowenheim-Skolem theorem, if the set $T$ of statements has an infinite model, then $T$ has models of arbitrarily high cardinality.

Answer 2

No, for obvious reasons.

There is exactly one field isomorphic to the rational numbers. The rational numbers. Even if you mean order isomorphic, and not ordered-field isomorphic, these are all countable fields.

And of course there is no set of axioms which have an infinite model only of one cardinality in first-order logic. So this is impossible.

But even if you limit yourself to countable fields this is not possible. Take any uncountable model of this set of axioms, then it has at least one number which is irrational and not algebraic; call it $t$. Now look at the least elementary submodel containing $t$, this is a countable field which has an irrational number. So it's not isomorphic to $\Bbb Q$.

Answer 3

The existence of such a set of sentences is precluded by the compactness theorem of first-order logic. Consider the sentences \begin{align} a & > 1 \\ a+a & > 1 \\ a+a+a & > 1 \\ a+a+a+a & > 1 \\ & \,\,\, \vdots \end{align} Every finite subset of this set is satisfiable within the rationals. Therefore by the compactness theorem, all of them are simultaneously satisfiable in a structure satisfying all first-order sentences true in $\mathbb Q$. But such a structure clearly cannot be isomorphic to $\mathbb Q$.