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Let $M$ and $N$ be $R$-modules. Suppose we complete them with respect to an ideal $\frak{m}$ of $R$. If we have $$M^\wedge_\mathfrak{m} \simeq N^\wedge_\mathfrak{m}$$must if be the case that $M \simeq N$?

It would appear, from something I am reading, that the answer is no. If so, is there a simple counter example? (You can assume $R$ is nice, e.g. Noetherian, local).

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There are lots of counterexamples. Let $R$ be noetherian ring and $M$ be an $R$-module which is not complete with respect to the $\mathfrak{m}$-adic topology, and let $N$ be its completion. Since completions are idempotent over noetherian rings, the completions of $M$ and $N$ agree, but $M$ and $N$ are not isomorphic.

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Great - thanks (to both of you!) –  Juan S Feb 14 '13 at 2:54

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