Timeline for Discrete orderings on $\mathbb{Z}[x,y]$ that violate the universal theory of the integers
Current License: CC BY-SA 3.0
9 events
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| Jun 13, 2016 at 13:42 | answer | added | Sidney Raffer | timeline score: 4 | |
| Jun 13, 2016 at 13:28 | history | edited | Sidney Raffer | CC BY-SA 3.0 |
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| Sep 20, 2012 at 19:03 | comment | added | Gerhard Paseman | 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 | |
| Sep 20, 2012 at 16:47 | comment | added | Sidney Raffer | @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. | |
| Sep 20, 2012 at 16:23 | comment | added | Gerhard Paseman | 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 | |
| Sep 20, 2012 at 15:24 | comment | added | Sidney Raffer | @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. | |
| Sep 20, 2012 at 14:55 | comment | added | David E Speyer | 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 $\alpha \in (0,1)$. | |
| Sep 20, 2012 at 14:02 | history | edited | Sidney Raffer | CC BY-SA 3.0 |
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| Sep 20, 2012 at 12:16 | history | asked | Sidney Raffer | CC BY-SA 3.0 |