definable set with parameters in $\mathbb{Z}$

Question

I'm facing some difficulties in understanding the relation between definable set with parameters and without parameters.

Roughly speaking, a definable set with parameters allows us to add parameters in the sentence determining the definable set. That is $A\subset S\iff S\models\varphi(a,p)$ where $p$ is a parameter in $S$.

$\mathbb{N}$ is obviously not definable without parameters in $(\mathbb{Z},+)$. (Use the automorphism $x\mapsto -x$.)

My question is:

Is $\mathbb{N}$ definable with parameters in $(\mathbb{Z},+)$?

I guess that the answer is negative but I have no idea of the proof.

Any help will be appreciated.

Answer 1

The answer is indeed negative. There are a couple ways to see this; here is my favorite.

First, note that the only parameter we could possibly need is $1$: every element is parameter-freely-definable in $(\mathbb{Z};+,1)$. (In fact, any nonzero integer will do in place of $1$ here.) So we can shift attention to parameter-free definability over $\mathcal{Z}=(\mathbb{Z};+,1)$.

Now if $\mathbb{N}$ were definable in $\mathcal{Z}$, we could in turn define an ordering on $\mathbb{Z}$ via $x\le y$ iff there is some $n$ in $\mathbb{N}$ such that $x+n=y$ (I consider $0\in\mathbb{N}$). This means that we'll be done if we can show that there is a model of $Th(\mathcal{Z})$ with no parameter-freely-definable ordering (the property of a formula defining a linear order on the whole structure is preserved under elementary equivalence).

This is a problem we can attack with automorphisms (even though $\mathcal{Z}$ itself is rigid). Specifically, we can produce a model of $Th(\mathcal{Z})$ which has an automorphism swapping a pair of elements; any such structure cannot have a parameter-freely-definable ordering. This is a fun exercise, so I've hidden the answer below:

Consider the additive subgroup $\mathcal{G}$ of $\mathbb{Q}[X,Y]$ consisting of elements with integral constant term, with a constant symbol naming $1$. This structure has an obvious automorphism swapping $X$ and $Y$.

It's not hard to check using EF-games that this structure does indeed satisfy $Th(\mathcal{Z})$. However, this raises the question of whether this problem can be solved without EF-games, quantifier elimination, or similar tools. I'm going to tentatively claim that the answer is no - because the base structure $\mathcal{Z}$ is rigid and $\mathbb{N}$ is "almost-definable" over it (in the sense of structure-specific implicit definability), genuinely technical arguments will be needed at some point here. But this is a rather subjective claim of course.

Answer 2

As an alternative to Noah's answer, we can solve the problem by classifying the definable sets in $(\mathbb{Z},+)$.

The complete theory of the structure $\mathcal{Z} = (\mathbb{Z}; 0,+,-,(n|)_{n\in \mathbb{Z}})$ has quantifier elimination. Here $n\mid$ is a unary relation symbol defining the multiples of $n$, so $n\mid x$ means $x\equiv 0 \text{ (mod $n$)}$. I previously wrote a longer answer about this, see here. Since $(\mathbb{Z}; 0,+,-,(n|)_{n\in \mathbb{Z}})$ is a definitional expansion of $(\mathbb{Z};+)$, they have the same definable sets.

A term in variables $x_1,\dots,x_k$ is equivalent in $\mathcal{Z}$ to one of the form $\sum_{i=1}^k a_ix_i$, with each $a_i\in \mathbb{Z}$. So every atomic formula (without parameters) is equivalent in $\mathcal{Z}$ to one of the form $\sum_{i=1}^k a_ix_i = 0$ or $\sum_{i=1}^k a_ix_i\equiv 0\text{ (mod $n$)}$. If we allow parameters, every atomic formula is equivalent to one of the form $\sum_{i=1}^k a_ix_i = b$ or to $\sum_{i=1}^k a_ix_i\equiv b\text{ (mod $n$)}$. By quantifier elimination, every formula is equivalent to a boolean combination of formulas of these forms. This gives a useful description of all definable subsets of $\mathbb{Z}^k$.

Specializing to the case $k = 1$, we find that every atomic formula with parameters has the form $ax = b$ or $ax\equiv b\text{ (mod $n$)}$.

Case 1: $ax = b$. This formula defines $\varnothing$, $\mathbb{Z}$, or the singleton $\{b/a\}$, depending on the values of $a$ and $b$.

Case 2: $ax\equiv b \text{ (mod $n$)}$. This formula defines a finite union of bi-infinite (infinite in both directions) arithmetic progressions of common difference $n$, obtained by lifting the solutions to $ax = b$ in $\mathbb{Z}/n\mathbb{Z}$ back to $\mathbb{Z}$.

Thus every definable (with parameters) subset of $\mathbb{Z}$ is a finite boolean combination of singletons and bi-infinite arithmetic progressions. Conversely, every singleton and every bi-infinite arithmetic progression is definable in $\mathbb{Z}$, so the finite boolean combinations of these are exactly the definable sets.

One consequence is that if $D\subseteq \mathbb{Z}$ is definable and contains infinitely many positive integers, it also contains infinitely many negative integers. In particular, $\mathbb{N}$ is not definable in $(\mathbb{Z},+)$, even with parameters.