Describing all models of the theory of $(\mathbb{Z}, =, S, 0)$

Question

Consider the theory of integers with equality, a constant 0 and the successor operation. In other words, that's a theory with axioms

• { axioms of equality },
• $\forall x \forall y (S(x) = S(y) \rightarrow x = y)$,
• $\forall x \exists y (S(y) = x)$,
• $\forall x \lnot (x = S(x))$,
• $\forall x \lnot (x = S(S(x)))$,
• $\forall x \lnot (x = S(S(S(x))))$,
• and countably infinite number of similar axioms each stating that $k$ applications of $S$ don't cycle.

What would be the models of this theory?

Clearly, one example is $\mathbb{R}$ with the natural interpretation. It just decomposes to $\mathbb{Z} \times \mathbb{R}$, so we have $\mathfrak{c}$ copies of $\mathbb{Z}$ that don't care about each other. This suggests that for every uncountable cardinality $\alpha$ we can do something similar and decompose that into $\mathbb{Z} \times \alpha$ (and similarly for countable models of the form $\mathbb{Z} \times k$, where $k$ is finite or countably infinite). But is there anything else?

If there is, how does it look? If there isn't, does it mean that this theory is uncountably categorical? The latter would be interesting, since this theory clearly isn't countably categorical.

Answer 1

Yes, a decomposition like this is always possible. Just look at the equivalence classes under the relation $x\sim y$ iff there is an $n$ such that $S^n(x) = y$ or vice versa. So all models can be expressed as disjoint copies of $\mathbb Z.$ It follows that the theory is $\kappa$-categorical for any uncountable $\kappa,$ but not $\aleph_0$-categorical (since the models with finitely many copies will not be isomorphic to each other or to models with countably infinite numbers of copies).

Answer 2

Partial answer: Let's say $M=(\Bbb Z,=, S, 0)$ to save typing. Note first it seems likely that you don't actually mean to include "$=$" in the signature, it seems more likely that you''re talking about the theory of $(\Bbb Z, S, 0)$ in first-order logic with equality. Assuming that:

It's not at all clear to me that the theory of $M$ is in fact characterized by those axioms, although it seems right. In any case, the models of those axioms are exactly $\Bbb Z\times A$ for some set $A$, with the obvious interpretation of $S$ (and with $0$ any element of the structure; note $0$ is never mentioned in the axioms).

This is clear. In any model $M'$ the interpretation of $S$ is a bijection such that no $S^k$ has a fixed point. Say $$O(x)=\{S^k(x):k\in\Bbb Z\}.$$Then the sets $O(x)$ give a partition of $M'$, and each $O(x)$ is isomorphic to $\Bbb Z$.