Let $G$ be a finitely generated residually finite group and let $M$ be a finitely generated $\mathbb{Z}[G]$-module.


**Question**: Must $M$ be residually finite in the sense that for all nonzero $x \in M$, there exists some submodule $N$ of $M$ such that $x \notin N$ and $M/N$ is finite?


If this is not true in general, is it true if $G$ is also assumed to be nilpotent?



Residual finiteness for modules over group rings
It's true, and due to Ph. Hall, when $G$ is virtually nilpotent, and more generally (Roseblade) when $G$ is virtually polycyclic.


When $G=\mathbf{Z}\wr\mathbf{Z}$ there exists an infinite simple $\mathbf{Z}G$-module, so it's not residually finite.




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Added: The counterexample is due to Ph. Hall. Since I already mentioned the original one for other purposes on this site, let me provide an immediate variant, which entails the result.


Notation: $L\_n=(\mathbf{Z}/n\mathbf{Z})\wr\mathbf{Z}$, $\mathbf{F}\_p=\mathbf{Z}/p\mathbf{Z}$ (viewed as field).


**Proposition** Let $p$, $q$ be primes such that $p$ divides $q-1$. Then there exists an infinite simple $\mathbf{F}\_q L\_p$-module $V$ (which is therefore a simple $\mathbf{Z} L\_p$-module, thus not residually finite).


Proof: Fix an element $x$ of order $p$ in the multiplicative group $\mathbf{F}\_q^\*$. Let $(w\_n)\_{n\in\mathbf{Z}}$
be valued in $\{1,x\}$, with the property that for every $n$ there exists $m$ such that $w\_m=x$ and $w\_i=1$ for all $i$ such that $0<|i|\le n$.


Let $V\_q=\mathbf{F}\_q^{(\mathbf{Z})}$ be the abelian group of finitely supported sequences $\mathbf{Z}\to\mathbf{F}\_q$, with basis $(e\_m)\_{m\in\mathbf{Z}}$. Let $d$ be the diagonal automorphism of $V$: $(x\_n)\mapsto (w\_nx\_n)$. Let $\tau$ be the shift $(x\_n)\mapsto (x\_{n+1})$. Hence $\tau^nf\tau^{-n}$ is diagonal for all $n$ and $f^p=\mathrm{id}$, so that $\tau$, $d$ define a representation $L\_p\to\operatorname{Aut}(V)$. Thus $V\_q$ is a $\mathbf{F}\_qL\_p$-module.


I claim it is a simple $\mathbf{F}\_qL\_p$-module. Indeed, start from a nonzero $v\in V\_q$, and let $W$ be the $L\_p$-submodule generated. Let $S$ be the (finite) support of $v$ and fix $m\in S$. Then there exists a translate of $w$ that equals $1$ on $S\smallsetminus\{m\}$ and equals $x$ on $n$. This corresponds to $\tau^n d\tau^{-n}$ for some $n$. Hence $\tau^nd\tau^{-n}v-v$ is a nonzero scalar multiple of $e\_m$. So $e\_m\in W$, and using $\tau$ it follows that $W$ contains all basis elements. Hence $W=V$, proving simplicity.


