Grothendieck rings of ordered subgroups of $\mathbb{Q}$

TL;DR

This work computes the model-theoretic Grothendieck ring of the ordered group $(G;+,<)$ for a proper subgroup $G$ of $\mathbb{Q}$. It shows $K_0(G;+,<)$ is a quotient of $(\mathbb{Z}/q\mathbb{Z})[T]/(T+T^2)$, where $q$ is the largest odd integer dividing $p-1$ for all primes $p\notin S_G$, with $S_G$ the set of primes such that $G$ is $p$-divisible. The authors derive explicit values for unary definable sets, in particular for $\mathbb{Q}$-intervals, and identify torsion contributions from primes outside $S_G$, leading to criteria for when the ring is trivial. They then show the Grothendieck ring is generated by unary definable sets (via a Presburger-style quantifier elimination and a cell-decomposition transfer from $\mathbb{Q}$ to $G$), implying a reduction to unary, quasi-cell data. The results provide a concrete algebraic description of $K_0(G;+,<)$ and connect model-theoretic definable geometry in ordered subgroups of $\mathbb{Q}$ to explicit finite-type quotient rings, with implications for motivic-style invariants in this setting.

Abstract

Let $G$ be a proper subgroup of $\mathbb{Q}$ and $S_G$ be the set of primes $p$ for which $G$ is $p$-divisible. We show that the model-theoretic Grothendieck ring of the ordered abelian group $(G;+,<)$ is a quotient of $(\mathbb{Z}/q\mathbb{Z})[T]/(T+T^2)$, where $q$ is the largest odd integer that divides $p-1$ for all $p \notin S_G$. This implies that the Grothendieck ring of $(G;+,<)$ is trivial in various salient cases, for example when $S_G$ is finite, or when $S_G$ does not contain some prime of the form $2^n+1$, $n\in \mathbb{N}$.

Theorems & Definitions (17)

• Theorem 1.1
• Lemma 2.1
• proof
• Lemma 2.2
• proof
• Lemma 2.3
• proof
• proof
• Lemma 2.5
• proof