Tensor completions of 2-nilpotent finitely generated torsion-free groups

Abstract

In this paper, we study tensor completions $G \otimes_{\mathcal{N}_{2,R}} R$ of finitely generated torsion-free nilpotent groups $G$ of class $2$ in the quasivariety $\mathcal{N}_{2,R}$ of $R$-exponential 2-nilpotent groups over a binomial integral domain $R$. We show that the classical Hall completion $G\otimes_{\mathcal{H}} R$ embeds as an abstract group (the embedding is not an $R$-homomorphism) into $G \otimes_{\mathcal{N}_{2,R}} R$, such that $G\otimes_{\mathcal{N}_{2,R}} R \simeq (G \otimes_{\mathcal{H}} R) \times D$, where $D$ is an $R$-module and the direct product is a product of abstract groups (not $R$-groups!). In particular, the canonical $R$-epimorphism $\mu: G \otimes_{\mathcal{N}_{2,R}} R \to G \otimes_{\mathcal{H}} R$ is a retract on $G \otimes_{\mathcal{H}} R$ with abelian kernel $D$. Moreover, in addition to the algebraic structure, we describe precisely how raising to an $R$-exponent works in the group $G \otimes_{\mathcal{N}_{2,R}} R$. To do this, we introduce a new type of commutators, the so-called c-commutators, which are interesting in their own right. These results answer an old question of Remeslennikov about the algebraic structure of free 2-nilpotent R-groups in the quasivariety $\mathcal{N}_{2,R}$. Indeed, it was shown in \cite{AMN} that if $G$ is a free 2-nilpotent group with basis $X$ (in the variety of abstract 2-nilpotent groups), then $G \otimes_{\mathbb{N}_{2,R}} R$ is a free 2-nilpotent R-group in $\mathcal{N}_{2,R}$ with basis $X$. Note that in this case $G \otimes_{\mathcal{H}} R$ is a free 2-nilpotent Hall $R$-group with basis $X$. As an illustration, for a free 2-nilpotent group $G$ of rank 2, we describe the group $G \otimes_{\mathcal{N}_{2,R}} R$, the action of $R$ on $G \otimes_{\mathcal{H}} R$, and the module $D$ in the case where $R$ is either the polynomial ring $\mathbb{Q}[t]$ or the field of rational functions $\mathbb{Q}(t)$ with coefficients in the field of rational numbers $\mathbb{Q}$.