Nonaxiomatizability of lattice-orderable rings

Abstract

Two elementarily equivalent rings, one of which is lattice-orderable and the other is not lattice-crderable, are constructed.Hence follows the elementary nonclosedness and the nonaxiomatizability of the class of all lattice-orderable rings. This example shows that the class of all lattice-orderable rings is nonaxiomatizable in the class of directedly orderable rings.It is shown in [i] that the class of all linearly orderable rings can be axiomatized. It will be proved here. that the lattice-orderable rings are nonaxiomatizable.For proving this we will construc~ "an example of two elementarily equivalent rings, one of which is lattice-orderable and the other is not lattice-orderable.Let us consider the set of numbers which can be represented in the form k/2n, where k and n are integers.This set is a ring with respect to addition and m~itiplication; we denote it by K:. Let us take the set N of natural numbers and an ultrafilter D on it. We denote the ultrapower of the ring K~ by K2 --KN/D.The usual linear order of the ring KI induces a linear order in the ultrapower K2.Let us consider the algebra over the ring K2 with basis elements a, b, c, and the defining relations :This is an associative and anticommutative algebra. Therefore, by [2, Corollary 9.11], A, and A2 are elementarily equivalent. THEOREM.The ring AI is not lattice-orderable.Proof. Let us assume that ring A~ has a lattice order P. We take an element ~b~N1, where 15 ~ Kz is that element which, under factorization with respect to the ultrafilter, coincides with an element of Cartesian power K N having r-th component 3 r. Then B is "divided" in K2 by an arbitrary power 3 k, where k is a fixed integer, since almost all the components of 13 satisfy this condition, i.e., for each k there exists a ~ (k)~ K~, such that 3k~ (k) = ~. Let l~b[ = ~la %-~,b %-7~c J-k I be the modulus of the element Bb; here ~i, ~i, ~'i K~ and k i~K x. Let us assume that k I ~:0. We now find a power k of the number 3 such that k~ is not divisible by 3 k.. There does not exist any element in K~ which would give kx on multiplication by 3 k. Therefore, the element has no modulus, which is impossible.Thus, k, = 0 and o~1a%-~ib%-y~c~P.Let al = BI = 0.Then ~b~ --~b----?tc.