# Theory with non archimedean models

Is there a consistent theory in the language of ordered groups (or ordered rings) whose models are non archimedean ordered groups (or rings or fields)?

(note: I am not asking for the existence of an axiomatization of the quality of being a non achimedean ordered group.)

If the underlying order of a nontrivial ordered group is discrete (i.e. every element has a successor and a predecessor), then either the group is isomorphic to $\mathbb{Z}$ or it is non-archimedean. Indeed, if $1$ (the successor of $0$) does not generate the whole group, then there is some element which is greater than $n\cdot 1$ for all $n\in \mathbb{N}$. The same statement holds for ordered rings, since the only ordered ring structure on $\mathbb{Z}$ compatible with the ordered group structure is the standard one.

So for a theory of ordered groups or rings all of whose models are non-archimedean, we can take $\mathrm{Th}(M)$, where $M$ is any discretely ordered group or ring which is not elementarily equivalent to $\mathbb{Z}$. At this point, you might worry that the theory of discretely ordered groups or discretely ordered rings is complete, so every model is elementarily equivalent to $\mathbb{Z}$. Fortunately, that's not the case.

For example, consider $T = \text{Th}(\mathbb{Z}\times\mathbb{Z},0,+,-, <)$, where $\mathbb{Z}\times \mathbb{Z}$ is ordered lexicographically. The sentence asserting that for all $x$, either $x$ or its successor is divisible by $2$ is true in $\mathbb{Z}$ but false in $\mathbb{Z}\times\mathbb{Z}$ (neither $(1,0)$ or $(1,1)$ are divisible by $2$).

• @Noah Schweber Those are actually both great answers! Jan 8, 2017 at 23:43

Here's a silly example, coming from proof theory:

Consider the theory $T=PA+\neg Con(PA)$. Since $PA$ is consistent, the standard natural numbers don't satisfy $T$; but by Goedel's Second Incompleteness Theorem, $T$ is consistent. So any model of $T$ has to be a non-Archimedean ordered semiring.

Now moving this example from semirings to rings is an easy exercise.

More generally, we can get an example of this type from any sentence $\varphi$ in the language of ordered semirings which is false in $\mathbb{N}$ but not disprovable in $PA$.

This also lets us get an example in fields. It turns out that $\mathbb{Z}$ is definable in $\mathbb{Q}$; this gives us a recursively axiomatizable theory $T$ of fields which is strong enough to do Goedel coding (by moving things onto the integer part), and such that $\mathbb{Q}\models T$. Now consider the theory $S=T+\neg Con(T)$. This theory will be consistent, so it will have models (which in particular are ordered fields) - but those models will need to have nonstandard integers since $T$ is actually consistent, and hence any model of $S$ will be non-Archimedean.