Skip to main content
MathOverflow

Return to Question

edited title
Link
Sidney Raffer
Sidney Raffer
  • 6.2k
  • 1
  • 27
  • 42

Discrete orderings on polynomial rings$\mathbb{Z}[x,y]$ that violate the universal theory of the integers

added 32 characters in body
Source Link
Sidney Raffer
Sidney Raffer
  • 6.2k
  • 1
  • 27
  • 42

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.

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 $\mathbb{Z}[x,y]$ that 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.

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.

Source Link
Sidney Raffer
Sidney Raffer
  • 6.2k
  • 1
  • 27
  • 42

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 $\mathbb{Z}[x,y]$ that 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.