Most convenient axioms to reason from when thinking about $\mathbb{Z}$ and $\mathbb{Q}$?

Question

When doing real analysis, one way that we can ground ourselves is by constructing the real numbers (e.g. as Dedekind cuts), and then proving the basic properties of $\mathbb{R}$ from this construction. However, we can also start with the axioms of a complete ordered field, and it turns out these completely characterize $\mathbb{R}$. Even though we still need to construct $\mathbb{R}$ in order to show that a complete ordered field exists, we don't actually need to reason directly from this construction at any point while working with $\mathbb{R}$. We can disregard the construction as soon as we are done with it, working only from the complete ordered field axioms instead.

Similarly, although in set theory we need to construct $\mathbb{N}$, we can work with $\mathbb{N}$ directly assuming nothing but the Peano axioms. This is much nicer than working with a specific construction.

However, when it comes to $\mathbb{Z}$ and $\mathbb{Q}$, I have never seen them presented in this axiomatic way. If I want to prove basic facts about $\mathbb{Z}$ or $\mathbb{Q}$, I would have to do so directly from the usual constructions. This isn't very aesthetically pleasing to me.

I'm sure there are various ways to characterize $\mathbb{Z}$ and $\mathbb{Q}$ "axiomatically" (I don't mean first-order axioms here; neither the axioms of a complete ordered field nor the Peano axioms are first-order). My question is, is there a particular way of defining these structures that is particularly convenient, or particularly conducive to proving basic arithmetic facts? If you were asked to rigorously build up the theory of $\mathbb{Z}$ or $\mathbb{Q}$ from first principles, without reference to their constructions, where would you start?

Answer 1

$\mathbb{Z}$ is characterized as the unique ordered integral domain whose positive elements are well-ordered (i.e., every non-empty subset of positive integers has a least element).

(Sketch of the argument: Take any ordered integral domain with well-ordered positive elements. If $x$ is the least positive element, then $x=1$ because otherwise $0 < x < 1$ implies $0 < x^2 < x$, a contradiction to the minimality of $x$. So the elements 0, 1, 2, 3, ... are distinct and there are no elements in between two consecutive values. If there were an $x$ larger than all of these elements then there would be no least positive element among $x, x-1, x-2, \ldots$. Now it is easy to see that $\mathbb{Z}$ is the only possibility.)

$\mathbb{Q}$ is the quotient field of $\mathbb{Z}$.

Answer 2

Both $\mathbb{Z}$ and $\mathbb{Q}$ are initial in their respective categories:

• For any unital ring $R$ there is exactly one unital ring homomorphism from $\mathbb{Z}$ to $R$. Moreover, any two rings with this property are uniquely isomorphic.

  • The zero ring may look like a counterexample at first, but it's not: unital ring homomorphisms have to preserve the unit!

• For any field $F$ of characteristic zero there is exactly one field homomorphism from $\mathbb{Q}$ to $F$ (remember that field homomorphisms have to be injective). Moreover, any two fields of characteristic zero with this propery are uniquely isomorphic.

Note that $\mathbb{N}$ is similarly initial, namely as a unital semiring (to me $0\in\mathbb{N}$): it has a unique unital semiring homomorphism to any given unital semiring. $\mathbb{R}$, however, is a bit wonkier and is hard to construe as initial in a satisfying sense. To some extent (e.g. downward Lowenheim-Skolem) this is due to the uncountability of $\mathbb{R}$.