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I'm not completely sure what you are asking. Does being $\mu$-critical mean that $\mathrm{Hom}_R(M,R)\neq 0$ (ie canonical dimension $\mu$) but for every proper quotient $M/N$ of $M$, $\mathrm{Hom}_R(M/N,R)\neq 0$? In any case this paper math.washington.edu/~smith/Research/asz6.pdf of Ajitabh, Smith and Zhang is likely to be useful.
If you can prove that (1) the trace of the identity map on $V$ is $\dim V$ (and this is an integer), (2) the trace of the zero map is $0$ and (3) trace is additive in the sense that if $T_1$ acts on $V_1$ and $T_2$ acts on $V_2$ then $\mathrm{tr} (T_1\oplus T_2)=\mathrm{tr} T_1+\mathrm{tr} T_2$ then what you ask for is straightforward since if $E$ is a projector on $V$ then it decomposes canonically as $I_{\ker E}\oplus 0_{\mathrm{Im} E}$. I would imagine that for any sensible definition of trace (1), (2) and (3) should be straightforward.
I'm not even sure that it is easy to extend my strategy to $G=\mathbb{Z}^n$ for $n>1$, although it may be easier than I think. Classifying right ideals in $\mathbb{C}G$ for $G$ a discrete Heisenberg group is probably hard. To get an idea why, see arxiv.org/abs/math/0102190. The first sentence of section 7 of your reference suggests that $Tor^1_{\mathbb{C}G}(\mathbb{C}G/f\mathbb{C}G,l^1(G))=0$ for all non-zero $f\in \mathbb{C}G$ whenever $G$ is torsionfree polycyclic. However, it isn't clear to me how hard this result was to prove already in that generality.
It occurs to me that in the final paragraph one might as well assume that $f$ is irreducible in $R$. Since (by the fundamental theorem of algebra) such an $f$ is a unit times $(x-\lambda)$ for $\lambda\in \mathbb{C}$ non-zero it should be very easy to complete the case $G=\mathbb{Z}$ by hand.
