On cellular covers with free kernels

TL;DR

This paper demonstrates that any finite rank cotorsion-free module can be realized as the kernel of a cellular cover of a rank 2 cotorsion-free module, highlighting limitations for rank 1 modules.

Contribution

It establishes the existence of cellular covers with free kernels for finite rank cotorsion-free modules, extending previous results and clarifying rank restrictions.

Findings

Every finite rank cotorsion-free module is a kernel of a cellular cover of a rank 2 module.

Rank 1 cotorsion-free abelian groups do not admit nontrivial cellular covers with free kernels.

The results are motivated by prior examples and recent research in the area.

Abstract

Recall that a homomorphism of $R$-modules $\pi: G\to H$ is called a {\it cellular cover} over $H$ if $\pi$ induces an isomorphism $\pi_*: \Hom_R(G,G)\cong \Hom_R(G,H),$ where $\pi_*(\varphi)= \pi \varphi$ for each $\varphi \in \Hom_R(G,G)$ (where maps are acting on the left). In this paper we show that every cotorsion-free module $K$ of finite rank can be realized as the kernel of a cellular cover of some cotorsion-free module of rank 2. In particular, every free abelian group of any finite rank appears then as the kernel of a cellular cover of a cotorsion-free abelian group of rank 2. This situation is best possible in the sense that cotorsion-free abelian groups of rank 1 do not admit cellular covers with free kernel except for the trivial ones. This work comes motivated by an example due to Buckner and Dugas, and recent results obtained by G\"obel--Rodr\'iguez--Str\"ungmann, and Fuchs--G\"obel.