invariants of a representation over a local ring from the residual representation

Question

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.

Answer 1

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.