algebraic number theory - In what sense (if any) does the cohomology of profinite groups commute with projective limits?

Background:

Let $G$ be a profinite group. If $M$ is a discrete $G$-module, then $M=varinjlim_U M^U$, where the direct limit is taken with respect to inclusions over all open normal subgroups of $G$, and one naturally has $H^n(G,M)simeqvarinjlim H^n(G/U,M^U)$, where the cohomology groups on the right can be regarded as the usual abstract cohomology groups of the finite groups $G/U$ (this is sometimes, as in Serre's Local Fields, taken as the definition of $H^n(G,M)$).

More generally if one has a projective system of profinite groups $(G_i,varphi_{ij})$ and a direct system of abelian groups $(M_i,psi_{ij})$ such that $M_i$ is a discrete $G_i$-module and the pair $(varphi_{ij},psi_{ij})$ is compatible in the sense of group cohomology for all $i,j$, then $varinjlim M_i$ is canonically a discrete $varprojlim G_i$-module, the groups $H^n(G_i,M_i)$ form a direct system, and one has $H^n(varprojlim G_i,varinjlim M_i)simeqvarinjlim H^n(G_i,M_i)$. The statement and straightforward proof of this more general result can be found, for instance, in Shatz' book on profinite groups.

Question:

In general, I'm wondering if there are, under appropriate hypotheses, any similar formulae for projective limits of discrete $G$-modules. Now, given a projective system of discrete $G$-modules $(M_i,psi_{ij})$, it isn't even obvious to me that the limit will again be a discrete $G$-module, and at any rate, while each $M_i$ is discrete, the limit (in its natural topology) will be discrete if and only if it is finite. So, for the sake of specificity, I'll give a particular situation in which I'm interested. If $R$ is a complete, Noetherian local ring with maximal ideal $mathfrak{m}$ and finite residue field and $M$ is a finite, free $R$-module as well as a discrete $G$-module such that the $G$-action is $R$-linear, then the canonical isomorphism of $R$-modules $Msimeqvarprojlim M/mathfrak{m}^iM$ is also a $G$-module isomorphism (each $M/mathfrak{m}^iM$ is a discrete $G$-module with action induced from that of $M$). Moreover, in this case, one can see that the limit is a discrete $G$-module (because it is isomorphic to one as an abstract $G$-module!). There is a natural homomorphism $C^n(G,M)rightarrowvarprojlim C^n(G,M/mathfrak{m}^iM)$ where the projective limit is taken with respect to the maps induced by the projections $M/mathfrak{m}^jMrightarrow M/mathfrak{m}^iM$, and this induces similar map on cohomology. I initially thought the map at the level of cochains was trivially surjective, just because of the universal property of projective limits. However, given a ``coherent sequence" of cochains $f_i:Grightarrow M/mathfrak{m}^iM$, the property gives me a map $f:Grightarrow M$ that is continuous when $M$ is regarded in its natural profinite topology, which is, as I noted above, most likely coarser than the discrete topology, so this might not be a cochain. So, what I'd really like to know is whether or not the map on cohomology is an isomorphism.

Why I Care: The reason I'd like to know that the map described above is an isomorphism is to apply it to the particular case of $G=hat{mathbb{Z}}$. It is well known (and can be found, for instance, in Serre's Local Fields) that $H^2(hat{mathbb{Z}},A)=0$ for $A$ a torsion abelian group. In particular the higher cohomology of a finite $hat{mathbb{Z}}$-module vanishes, and I'd like to be able to conclude that the same is true for my $M$ above, being a projective limit of finite abelian groups.

Thanks!

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