invariants of a representation over a local ring from the residual representation Let $(R, \mathfrak m)$ be a local ring (not necessarily an integral domain) and $T$ be a free $R$-module of finite rank $n\geq 2$. Let $\rho: G \to \mathrm{Aut}_{R\text{-linear}}(T)$ be a represenation of a group G. Is it true that if the residual representation $\overline \rho: = \rho \textrm{ mod } \mathfrak m $ has no nonzero $G$-invariant, then the representation $\rho$ has no nonzero $G$-invariant? It seems that the answer is no in general. In fact, the short exact sequence 
$0 \to \mathfrak mT \to T \to T/\mathfrak m T \to 0$
gives the exact sequence 
$0 \to (\mathfrak mT)^G \to (T)^G \to (T/\mathfrak m T)^G.$
This gives 
$(\mathfrak mT)^G \simeq (T)^G$ 
as $\overline \rho = T/\mathfrak m T$ has no nonzero $G$-invariant. So my question is equivalent to asking for an example of a representation $T$ over a local ring such that $\mathfrak mT$ has a nonzero $G$-invariant. 

Note that $\mathfrak m^n T/\mathfrak m^{n+1}$ is naturally isomorphic to
$(\mathfrak m^n/\mathfrak m^{n+1})\otimes_k  T/\mathfrak m$ as a $G$-representation,
with $G$-acting through the right-hand factor.  In particular, if $T/\mathfrak m$ 
has trivial $G$-invariants, so does $\mathfrak m^n/\mathfrak m^{n+1}$.  
From this, an easy devissage shows that $T/\mathfrak m^{n+1}$ has trivial $G$-invariants for every $n$, and hence so does the $\mathfrak m$-adic completion of $T$.    If $R$ is Noetherian (so that $T$ embeds into its $\mathfrak m$-adic completion) or complete, we then see that $T$ itself has trivial $G$-invariants.
Any counterexample thus has to be non-Noetherian and not complete.  I don't have enough feeling for those contexts to say for sure whether a counterexample actually exists without thinking more about it.
