Residual finiteness for modules over group rings 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?
 A: Here is another example with different groups.  Formanek showed that the group ring over any field of a free product of non-trivial groups (and not both order 2) is primitive, has a faithful simple module.  That means that $\mathbb F_pG$ is primitive whenever $G=A\ast B$.  And of course $G$ is residually finite if both $A$ and $B$ are.  Obviously a faithful simple $\mathbb F_pG$-module is cyclic as a $\mathbb ZG$-module and since $G$ is infinite, it cannot be finite.  So this gives lots of examples including the free group on two generators.
A: 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.

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.
