Let $l$ be a prime number, $n\in \mathbb{Z}$. Is it true that any finitely generated $\mathbb{Z}/l^n\mathbb{Z}$-module has a finite (left) resolution by free finitely generated $\mathbb{Z}/l^n\mathbb{Z}$-modules?


I am not an expert in the field, so the question might not be on the research level. Sorry about that.



Projective resolutions of torsion modules
No. In fact, this is as far from true as possible: a finitely generated $\mathbb{Z}/l^n\mathbb{Z}$-module has a finite free resolution iff it is free. To see this, note that a finitely generated $\mathbb{Z}/l^n\mathbb{Z}$ is a direct sum of modules of the form $\mathbb{Z}/l^m\mathbb{Z}$ for $m\leq n$ (this follows from the classification of finitely generated $\mathbb{Z}$-modules, for instance). We can construct an infinite periodic free resolution of $\mathbb{Z}/l^m\mathbb{Z}$ like this: $$\dots\stackrel{l^{n-m}}\to\mathbb{Z}/l^n\mathbb{Z}\stackrel{l^m}\to\mathbb{Z}/l^n\mathbb{Z}\stackrel{l^{n-m}}\to\mathbb{Z}/l^n\mathbb{Z}\stackrel{l^m}\to\mathbb{Z}/l^n\mathbb{Z}\to\mathbb{Z}/l^m\mathbb{Z}\to 0.$$


If $0<m<n$ (i.e., if $\mathbb{Z}/l^m\mathbb{Z}$ is not free), then when you Hom this resolution into $\mathbb{Z}/l\mathbb{Z}$ all the maps become $0$, so you find that $\operatorname{Ext}^k(\mathbb{Z}/l^m\mathbb{Z},\mathbb{Z}/l\mathbb{Z})=\mathbb{Z}/l\mathbb{Z}$ for all $k$. Thus $\mathbb{Z}/l^m\mathbb{Z}$ has infinite projective dimension.


