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Let $A$ be a local noetherian ring with maximal ideal $m$. Let $M$ be an infinitely generated $A$-module and $\hat{M}$ be the $m$-adic completion of $M$. Denote by $\hat{A}$ the $m$-adic completion of $A$. Recall, $\hat{M}$ is an $\hat{A}$-module. My question is: Is $\hat{M}$ infinitely generated as an $\hat{A}$-module?

EDIT Assume that the annihilator of $M$ properly contains the zero prime ideal (the prime ideal containing $0$) and there does not exist an integer $n$ such that $m^n$ is contained in the annihilator of $M$.

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  • $\begingroup$ The $\frak m$-adic completion could even be 0 (for example, when $A$ is domain and $M$ is its fraction field). $\endgroup$
    – Angelo
    Commented Nov 2, 2019 at 9:45
  • $\begingroup$ @Angelo Thank you. I have edited the question slightly with an additional condition on the support. $\endgroup$
    – Ron
    Commented Nov 2, 2019 at 10:08
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    $\begingroup$ Consider $\mathbb Z_p$ as a module over $\mathbb Z_{(p)}$. It is infinitely (even uncountably) generated for cardinality reason, but taking the completions we get $\mathbb Z_p$ over $\mathbb Z_p$. $\endgroup$
    – Wojowu
    Commented Nov 2, 2019 at 10:39
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    $\begingroup$ Your edited condition does not help: take the fraction field of $A/\mathfrak{p}$, where $\mathfrak{p}$ is any prime ideal strictly contained in $\mathfrak{m}$. $\endgroup$
    – abx
    Commented Nov 2, 2019 at 10:42
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    $\begingroup$ Completing a module that is not finitely generated is often very bad for its health. That's why derived completion (a gentler kind of completion) was invented (just do a google search). $\endgroup$
    – Angelo
    Commented Nov 2, 2019 at 11:27

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