Axioms for $\mathbb Z$-groups without named one?

Question

The theory of $(\mathbb Z,+,0,1)$ has been studied as the theory of $\mathbb Z$-groups, and it has been examined as a series of exercises (and I'm sure other places) in David Marker's Model Theory textbook.

But what happens if you lose one? That is, how can we axiomatize $(\mathbb Z,+,0)$?

This theory has intrinsic interest; among other things, it's one of the few natural examples of a complete first order theory without a prime model. There are other somewhat remarkable qualities about it, but this is enough.

How can it be axiomatized? I don't want to reinvent the wheel and Google is turning up nothing useful.

Answer 1

As a part of her work on abelian groups (which includes the famous fact that the theory of abelian groups is decidable), Wanda Szmielew gave a complete classification of all abelian groups up to elementary equivalence. This is Theorem 5.2 in her paper Elementary properties of abelian groups (caveat: this is written in a rather outdated language, so it might be hard to read).

In a modern language (unless I'm reading it wrong myself), it says that two abelian groups $A,B$ are elementarily equivalent if and only if they are both of finite exponent or both of infinite exponent, and they both have the same "dimensions" (which are natural numbers or $\infty$), defined by the maximal number of:

• elements of order $p^k$ and "independent modulo $p^k$" (in the sense that if a linear combination is zero, then the coefficients are zero modulo $p^k$),
• elements "strongly independent" modulo $p^k$ (this means that if a linear combination is zero modulo $p^k$, then each coefficient must be zero modulo $p^k$),
• elements of order $p^k$ strongly independent modulo $p^k$,

all for each prime $p$ and natural $k$.

Integers are torsion-free, so they have infinite exponent and the first and third kinds of dimensions are both zero. The second one is, I believe, $1$ for all $p,k$ (because $A/p^kA\cong {\bf Z}/{p^k{\bf Z}}$ for $A={\bf Z}$, more or less).

This, I think, implies that $A\equiv({\bf Z},+)$ precisely when $A$ is a torsion-free abelian group such that for each $p,k$ we have that $A/p^kA$ has $p^k$ elements (it's not hard to express this in a first-order way). It may be slightly redundant, I haven't given it much thought.

If I recall correctly, adding symbols for divisibility and subtraction gives you q.e. in this theory (which is another way to show that this is an axiomatisation, circumventing Szmielew's paper), and it is about as tame as it gets without being $\omega$-categorical.

As a bonus, this gives immediately the idea on how to show that ${\bf Z}^n$ are not elementarily equivalent for different $n$: if you take $2^n+1$ elements of ${\bf Z}^n$, some two of them will add up to something even.