# Axiomatization of $\mathbb{Z}$

Though I've seen several cool axiomizations of $\mathbb{R}$, I've never seen any at all for $\mathbb{Z}$.

My initial guess was that $\mathbb{Z}$ would be a ordered ring which is "weakly" well-ordered in the sense that any subset with a lower bound has a minimal element.

However, after seeing this definition of a discrete ordered ring, I'm less sure. I made that guess under the impression that the fundamental characteristic of $\mathbb{Z}$ is that every nonzero element has exactly one representation of the form $\pm (1+1+\dots+1)$, but that seems to be shared by every DOR.

Presumably, this definition wouldn't exist if every instance of it was isomorphic to $\mathbb{Z}$, so can someone give me an example of another discrete ordered ring? More to the point, what is a sufficient characterization of $\mathbb{Z}$? (and a proof sketch of uniqueness would be nice)

I'm aware that $\mathbb{Z}$ is pretty easily constructible from $\mathbb{N}$, but I want to use this for a seminar and given the audience I am expecting, I would rather not deal with Peano. (And I guess it feels like cheating to say "$\mathbb{N}$ is a well-ordered rig")

-
Do you seek first-order or second-order axiomatizations? – Asaf Karagila Jun 21 '12 at 20:12
Isn't the word "axiomatization"? I've never seen "axiomization" before, though it has some Google hits. – user23211 Jun 21 '12 at 20:12
Also, what sort of structure? $\mathbb Z$ as a countable set? As an ordered set? As an ordered group? As an ordered ring? As an ordered Euclidean ring? As an ordered PID? As an ordered integral domain? etc. etc. – Asaf Karagila Jun 21 '12 at 20:20
There is a very easy higher-order axiomatisation of $\mathbb{Z}$ that does not invoke order structure: it is the unique (up to unique isomorphism) free ring on no generators, i.e. it is the initial object in the category of rings. – Zhen Lin Jun 21 '12 at 20:20
Any ordered ring R whose positive elements are well-ordered in R is isomorphic to $\mathbb Z$ as an ordered ring. Does that help? – Bill Dubuque Jun 21 '12 at 20:32

Second-order quantification allows us to talk about properties of subsets of the ring, much like the completeness axiom of the real numbers (which is too a second-order statement).

We can adjoin the usual theory of ordered rings the following axiom:

$$\forall A(A\neq\varnothing\land\exists x\forall a(a\in A\rightarrow x<a)\rightarrow\exists y(y\in A\land\forall x(x\in A\rightarrow y\leq x)))$$

Saying that for non-empty every set $A$, if there is a lower bound for $A$ then $A$ has a minimal element.

We can also follow Zhen Lin's suggestion in the comments. Notice that $\mathbb Z$ is the unique free additive group which has only one generator. That is: $$\exists x(x\neq 0\land\forall A(x\in A\land\forall a\forall b(a\in A\land b\in A\rightarrow a+b\in A)\land\forall a(a\in A\rightarrow -a\in A)\rightarrow\forall y(y\in A)$$

This is a very complicated way of saying that there exists some $x$ which is non-zero and every $A$ in which $x$ is an element, and $A$ is closed under addition and negation imply that $A$ is everything.

In $\mathbb Z$ this is true because $x=1$. However this is not true for any other ordered ring.

-
I love the smell of downvotes in the morning... – Asaf Karagila Jun 22 '12 at 7:36

Any ordered ring R whose positives P are well-ordered in R is isomorphic to $\mathbb Z$ as an ordered ring. The proof is easy. Hint:  the natural image of $\,\mathbb Z\,$ in R is an order mononomorphism, so it remains to show it is onto. If not, R has a positive element $\rm\:w\not\in \mathbb Z.\:$ $\rm w$ is not infinite $\rm (w\! >\! n,\, \forall\, n\in\mathbb N)\,$ else $\rm\,w > w\!-\!1 > w\!-\!2,\ldots\,$ is an infinite descending chain in P, contra P well-ordered. Therefore $\rm\:w\:$ must lie between two naturals $\rm\:n < w < n\!+\!1,\:$ hence $\rm\ 0 < \epsilon < 1\:$ for $\rm\:\epsilon = w\!-\!n,\:$ therefore $\rm\: \epsilon > \epsilon^2 > \epsilon^3 > \ldots\,$ is an infinite descending chain in P,  contra P is well-ordered.  QED

You ask for another example of a discrete ordered ring. Order the ring $\rm\,\mathbb Z[x]\,$ of integer coefficient polynomials by declaring $\rm\:f > 0\:$ iff it has leading coefficient $> 0,\,$ i.e. iff $\rm\:f\:$ is positive at $+\infty,$ and $\rm\:f > g\:$ iff $\rm\,f\!-\!g > 0.\:$ Here, as above, $\rm\:x > x\!-\!1 > x\!-\!2 > \ldots\,$ so its positives are not well-ordered.

-
But then is $x>1$? – Eric Stucky Jun 22 '12 at 18:52
@Eric Yes, $\rm\:x > n\:$ since $\rm\:x-n\:$ has leading coefficient $1 > 0$ or, equivalently, it is eventually positive. – Bill Dubuque Jun 22 '12 at 19:06

Any ultrapower of $\mathbb{Z}$ will be a discrete ordered ring.

-
Suppose that $Z$ is an ultrapower of the integers by a free ultrafilter, it will have a non-standard part (i.e. transfinite numbers), and now if you take the standard part you will have a bounded set without a minimal element. – Asaf Karagila Jun 21 '12 at 20:18
@AsafKaragila: Yes, but a small part of the question, which is all I intended to answer, was about an example of a discrete ordered ring other than $\mathbb{Z}$, no further restrictions. I suppose I could have made it a comment, though. Too late now. – Harald Hanche-Olsen Jun 21 '12 at 20:23

When you talk about isomorphism you should indicate the structure. If you consider $\mathbb{Z}$ with the only the order structure, the statement

"fundamental characteristic of $\mathbb{Z}$ is that every nonzero element has exactly one representation of the form ±(1+1+⋯+1)"

is not true. $\mathbb{N}$ also has this property. However if you add to your property above that in the linear ordering every element has an element smaller than it, then I believe you do get $(\mathbb{Z}, <)$. The idea is to make define an equivalence relation based on the property above and show that any linear ordering with these properties has a single equivalence class and each equivalence class is isomorphic to $\mathbb{Z}$ as linear ordering.

Also I don't think the "weakly" well-ordered property you stated above unique characterizes the the linear ordering $\mathbb{Z}$. If $W$ is any well ordered set, $\omega^* + W$, where $\omega^*$ is the backward $\mathbb{N}$, would also have what you called the "weakly" well ordered property.

-
I understood the question as looking for a ring structure axioms for $\mathbb Z$ and not just an ordered set. – Asaf Karagila Jun 21 '12 at 20:19
Thank you very much for the counterexamples! However, Asaf was correct in that I would like to consider the ring structure as well. – Eric Stucky Jun 21 '12 at 20:40