# Discrete orderings on polynomial rings that violate the universal theory of the integers

Working in the language of ordered rings, which we take to have type $(+ - \times < 0\, 1)$, can anyone give an example of a discrete ordering on the polynomial ring in two variables $\mathbb{Z}[x,y]$ such that the resulting ordered ring does not satisfy the universal theory of the integers?

It is not difficult to show that as a ring, i.e. forgetting the less-than symbol, the ring $\mathbb{Z}[x,y]$ does in fact satisfy the universal theory of the ring of integers. The problem is to find a discrete ordering on $\mathbb{Z}[x,y]$ such that some system of inequalities is solvable in $\mathbb{Z}[x,y]$ but not in the integers, or, on the contrary, to prove that there is no such ordering.

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Since I misread the question at first, let me point out two key provisions. "Universal": The sentence should be of the form $\forall x_1 x_2 x_3 \cdots : \phi(x_1, \ldots, x_n)$ where $\phi$ is a list of inequalities. If we don't have this, order $\mathbb{Z}[x]$ by leading term and consider the sentence "For all $0 < a < b$, there are $p$ and $q>0$ with $a q^2 < p^2 < b q^2$." "Discrete": If you don't have this, consider the sentence $\forall f: f^2 \geq f$ and order $\mathbb{Z}[x]$ by evaluation at $\alpha$ for some irrational $\alpha \in (0,1)$. – David Speyer Sep 20 '12 at 14:55
@David: Yes, discreteness and universality are the key assumptions. More intuitively, the problem is whether the ring structure on Z[x,y] together with discreteness is enough to determine which systems of inequalities are and are not solvable. The axioms for discretely ordered rings are very weak, insofar as what they can prove about number theory, so it seems that the answer is likely to be "no", but I don't see any obvious examples to prove this. – SJR Sep 20 '12 at 15:24
For clarification, can someone explain why some version of lexicographic order might (or might not) work? I am thinking such an order might be where any polynomial that has a monomial containing y is greater than any polynomial that has no y whatsoever. (If on the other hand, all such orders have to respect the order on Z, then I think it unlikely such an order will be found.) Gerhard "Ask Me About System Design" Paseman, 2012.09.20 – Gerhard Paseman Sep 20 '12 at 16:23
@Gerhard: There is a unique ordering on Z[x,y] in which the element x is greater than any integer, and the element y is greater than any element of Z[x]. The resulting ordered ring is indeed discrete, but unfortunately it satisfies the universal theory of the integers, since, e.g., the ring in question can be embedded in any ultrapower of the integers. – SJR Sep 20 '12 at 16:47
If all the axioms of an ordered ring are to be satisfied, then I think there are few choices, as (if I understand correctly) the basic orders that are discrete are determined by the order of x,y and the integers. So I suggest no such order exists that will not satisfy the universal theory. Gerhard "Ask Me About System Design" Paseman, 2012.09.20 – Gerhard Paseman Sep 20 '12 at 19:03