This is probably a textbook question but i haven't been able to find a reference. Let $R$ be a complete commutative Noetherian local ring and $I$ its unique maximal ideal (I'm mostly interested in the case $R=k[[h]]$ for $k$ a field of characteristic 0). Recall that an $R$-module is complete if the canonical map
$$ M\rightarrow \hat M:=\lim M/I^n$$
is an isomorphism.



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> Is a (small) filtered colimit (in the category of $R$-modules) of complete modules complete ?
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This is true for $\kappa$ filtered colimit if $\kappa>\aleph\_0$, since in that case in $R-mod$ $\kappa$-filtered colimits commute with small limit. On the other hand, according to nlab it's still true in that case that small limits distributes over filtered colimits in the sense that if $I$ is a small filtered category the evaluation map
$$ Fun(I,R-mod) \rightarrow R-mod$$
commutes with small limits. It seems to me that it implies the result I want, but i'd be happy to have a reference or a confirmation, or ideally a more direct argument.



Is a filtered colimit of complete module complete?
The answer is no. Here is another example. Let the ring $R=k[[h]]$ and the modules be $M\_n=k[h]/h^n$ with the bounding maps $M\_n\to M\_{n+1}$ sending 1 to $h$.


Then all $M\_n$ are complete (they are $h$-torsion), but $M=\mathrm{colim}\_n M\_n=k[h^{\pm1}]/k[h]$ (including $M\_n$ as the submodule spanned by $h^{-n}$). The colimit is $h$-divisible, hence its $h$-completion is 0 (since $M/h^nM=0$).


