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)?
  • $\begingroup$ 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. $\endgroup$
    – Blazej
    Jun 7, 2022 at 10:27


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.