# $\mathbb Z_n[x_1^\pm,\dots,x_D^\pm]$-modules extended from $\mathbb Z_n$

Let $$n$$ be a positive integer and let $$\mathbb Z_n=\mathbb Z/n \mathbb Z$$. Consider the ring of Laurent polynomials $$R=\mathbb Z_n[x_1^\pm,\dots,x_D^\pm]$$. $$R$$-modules of the form $$M=M_0 \otimes_{\mathbb Z_n} R$$, where $$M_0$$ is a $$\mathbb Z_n$$-module, will be called quasi-free. I'm happy to restrict attention to finitely generated quasi-free modules whenever that makes a difference. I'm interested in answers to several questions about quasi-free modules (or pointers to literature).

1. Is there some characterization of quasi-free modules, e.g. in homological terms or expressed by subquotients $$p^k M / p^{k+1}M$$? Here is a working conjecture that I came up with: if $$n=p^t$$ for a prime number $$p$$, then $$M$$ is quasi-free if and only if $$p^k M/p^{k+1}M$$ is free over $$R/(p)$$ for every $$k$$.
2. One source of my interest in descriptions as above is to address questions such as: (a) are summands of quasi-free modules quasi-free? (b) for a short exact sequence $$0 \to A \to B \to C \to 0$$ with $$C$$ quasi-free, is it true that $$A$$ quasi-free $$\iff$$ $$B$$ quasi-free (if not, do we have at least a one-sided implication)?
• In the first version of this question I mentioned that quasi-free modules $M$ satisfy $\mathrm{Ext}^i_R(M,N)=0$ for every $i>0$ and every $N$ free over $\mathbb Z_n$ (equivalently, $N$ of finite projective dimension) and asked if the converse is true. Then I realized that module $M = \mathrm{im} \begin{pmatrix} 2 & 0 \\ 1-x & 2 \end{pmatrix}$ with $n=4$ and $D=1$ provides a counterexample to this converse. Jun 7, 2022 at 10:27