A theory is categorical if it has a unique model up to isomorphism. First-order Peano arithmetic is not categorical, but second-order Peano arithmetic is categorical, with the natural numbers as its unique model. The first-order theory of real closed fields is not categorical, but the second-order theory of Dedekind-complete ordered fields is categorical, with the real numbers as its unique model. ZFC is not categorical, but Morse-Kelley Set Theory with an appropriate axiom about inaccessible cardinals is categorical.

My question is, what theory of the field of rational numbers is categorical? Clearly we can’t characterize $\mathbb{Q}$ as the unique countable ordered field whose order is a dense order without endpoints, because the field of algebraic real numbers also satisfies all that. So what else is required?


1 Answer 1


Take the first-order theory of fields of characteristic zero (in the language of rings) and add to it the following second-order axiom:

  • The only subfield is the whole field.

The only field of characteristic zero with this property is $\mathbb{Q}$, so we have the desired categorical theory of rational numbers.

  • $\begingroup$ Does such a field have a natural ordering? I expect $\mathbb{Q}$ to be an ordered field, but I don't see how these properties define an ordering. $\endgroup$
    – Filippo
    Oct 10, 2021 at 16:05
  • $\begingroup$ @Filippo yes, by Lagrange's four-square theorem a rational number is positive if and only if it is a sum of four squares; this gives a first-order definition of the ordering in the rationals $\endgroup$ Jun 24, 2023 at 22:10
  • $\begingroup$ @AtticusStonestrom Alternatively, couldn't we argue that there is a unique ordering of fields s.t. $0\leq 1$? $\endgroup$
    – Filippo
    Jun 25, 2023 at 9:17
  • $\begingroup$ @Filippo well, in every ordered field it must be the case that $0<1$, but it's not true in general that ordered fields have unique orderings. $\endgroup$ Jun 25, 2023 at 11:24
  • $\begingroup$ @AtticusStonestrom I see, thank you for the comment! Okay, but does this happen to be true for $\mathbb Q$? I.e. is there a unique ordering $\leq$ on $\mathbb Q$ s.t. $(\mathbb Q,\leq)$ is an ordered field? $\endgroup$
    – Filippo
    Jun 25, 2023 at 15:20

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