Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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")

share|improve this question
1  
Do you seek first-order or second-order axiomatizations? –  Asaf Karagila Jun 21 '12 at 20:12
1  
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
6  
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
3  
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

5 Answers 5

up vote 2 down vote accepted

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.

share|improve this answer
    
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.

share|improve this answer
    
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.

share|improve this answer
    
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.

share|improve this answer
    
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

You could first axiomize $\mathbb{N}$ using peano axioms. This will lead you to the group $\langle \mathbb{N},+ \rangle$ This group has a neutral element which is $0$.

Then you can define the additive inverse of $n\in\mathbb{N}$ and name it $(-n)$. Define $(-n)$ as the element for which $n+(-n)=0$.

Note that $(-n)\not \in \mathbb{N}$. Except for $n=0$.

Now $\mathbb{Z}$ is the collection which contains $\{x | x\in\mathbb{N} \vee (-x)\in\mathbb{N} \}$.

PS: (I'm not totally sure it is correct)

share|improve this answer
3  
$(\mathbb N,+)$ is not a group. It's a monoid. –  user23211 Jun 21 '12 at 20:41
    
Thanks for the response, and welcome to the community! (I say that like I'm not new :) But as I said, I'm not really interested in building $\mathbb{Z}$ from $\mathbb{N}$ for this purpose. –  Eric Stucky Jun 21 '12 at 20:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.