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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.

Is a (small) filtered colimit (in the category of $R$-modules) of complete modules complete ?

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.

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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 $M/h^nM=0$).

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The answer should be negative. For example, just take $R=k[[t]]$ in your case. We can choose a series of indeterminates $x_i$, and let $k_i=k(x_1,x_2,...,x_i)$, $M_n=k_n[[t]]$, and take the filtered system as inclusion. Then each $M_i$ is complete for the $t$-topology, but the element $x=\sum_{i=1}^\infty x_it^i\in\hat{M}$ is not belong to $M$, here $M$ is your inductive limit.

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  • $\begingroup$ Similar things happens in number theory, that is $\overline{\mathbb{Q}_p}$ is not complete relative to the p-adic topology. $\endgroup$ – Bonbon May 7 '18 at 16:21

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