Quantifier Elimination for Lexicographic Products of Ordered Abelian Groups

Abstract

Listen

Translate article

Let $\Lg = \{+,-,0\}$ be the language of the abelian groups, $L$ an expansion of $\Lg(<)$ by relations and constants, and $\Lmod = \Lg \cup \{\equiv_n\}_{n \geq 2}$ where each $\equiv_n$ is defined as follows: $x \equiv_n y$ if and only if $n|x-y$. Let $H$ be a structure for $L$ such that $H|\Lg(<)$ is a totally ordered abelian group and $K$ a totally ordered abelian group. We consider a product interpretation of $H \times K$ with a new predicate $I$ for $\{0\}\times K$ defined by N.~Suzuki \cite{Sz}. Suppose that $H$ admits quantifier elimination in $L$. 1. If $K$ is a Presburger arithmetic with smallest positive element $1_K$ then the product interpretation $G$ of $H \times K$ with a new predicate $I$ admits quantifier elimination in $L(I, 1) \cup \Lmod$ with $1^G = (0^H, 1_K)$.2. If $K$ is dense regular and $K/nK$ is finite for every integer $n \geq 2$ then the product interpretation $G$ of $H \times K$ with a new predicate $I$ admits quantifier elimination in $L(I, D) \cup \Lmod$ for some set $D$ of constant symbols where $G \models I(d)$ for each $d \in D$.3. If $K$ admits quantifier elimination in $\Lmod(<, D)$ for some set $D$ of constant symbols then the product interpretation $G$ of $H \times K$ with a new predicate $I$ admits quantifier elimination in $L(I, D) \cup \Lmod$ unless $K$ is dense regular with $K/nK$ being infinite for some $n$. Conversely, if the product interpretation $G$ of $H \times K$ with a new predicate $I$ admits quantifier elimination in $L(I, D) \cup \Lmod$ for some set $D$ of constant symbols such that $G \models I(d)$ for each $d \in D$ then $H$ admits quantifier elimination in $L \cup \Lmod$, and $K$ admits quantifier elimination in $\Lmod(<, D)$. We also discuss the axiomatization of the theory of the product interpretation of $H \times K$. %For some set $C$ of constants in $K$.