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> Is any finitely generated module $M$ over a complete commutative ring $A$ complete?
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Here complete is with respect to the $I$-adic topology, where $I$ is a given ideal of $A$. 


Edit: as Luc Guyot mentioned, the answer is positive in the noetherian case.



Finitely generated modules over a complete ring
First off, a general comment.


**Remark.**
It is not always the case that for any ring $A$, and any ideal $I$ of $A$, the $I$-adic completion of an $A$-module (no matter if finitely generated) exists. (By "exists" I mean: it is not always the case that for an $A$-module $M$, $\varprojlim\_{k\ge 1} M/I^kM$ is $I$-adically complete. I prefer to not call "completion" a possibly non-complete object).


For instance:


**Example.** Let $A = k[t\_1,t\_2,\ldots ]$ be the polynomial algebra over $k$, a field, in infinitely many indeterminates, and $I$ the ideal generated by $t\_i, i\ge 1$. The $I$-adic completion of the $A$-module $A$, does not exist. If it did, then $A' := \varprojlim\_{k\ge 1}A/I^k$ should be $IA'$-adically complete. However, the infinite sum $\sum\_{k\ge 1}t\_k^k$ does not converge with respect to the $IA'$-adic topology.


When $I$ is **finitely generated**, then indeed one can show the $I$-adic completion of an $A$-module exists.


Usually, finite generation of the ideal of definition of the adic topology on a commutative ring, is assumed to ensure the completion of the ring itself exists. In particular, it is not necessarily the case that such finite generation is needed, to ensure finitely generated modules are complete, as long as the ground ring is **already** assumed to be complete.


Since you do assume, in your question, $A$ is $I$-adically complete, it is not a priori necessary to further assume $I$ is finitely generated. In other words, your question is well posed.


We may show the following:



> 
> **Lemma** Let $A$ be an $I$-adically complete ring, with $I$ an arbitrary ideal. Let $M$ be a finitely generated $A$-module. If $M$ is $I$-adically separated, $M$ is $I$-adically complete.
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> *Proof.* We may write $M \simeq A^{\oplus n}/N$, for some $A$-submodule $N\subset A^{\oplus n}$. Note that $A^{\oplus n}$ is $I$-adically complete, as $A$ is. The $I$-adic completion $M^{\wedge}$ of $M$ is isomorphic to $A^{\oplus n}/\overline{N}$, where $\overline{N}$ is the closure of $N$ in $A^{\oplus n}$ with respect to the $I$-adic topology (check!). Therefore, the canonical map $M\to M^{\wedge}$ is surjective. Since $M$ is $I$-adically separated, it is also injective. QED
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If $A$ is Noetherian and $M$ is finitely generated, then by the Krull Intersection Theorem, $M$ is always $I$-adically separated, which is why, by the Lemma, $M$ is also $I$-adically complete.


**To have a Krull Intersection Theorem**, what you need is:


(a) the $A$-submodule $N := \bigcap\_{k\ge 0} I^kM$ of $M$ to be finitely generated;


(b) $I^kM\cap N \subseteq IN$ for large enough $k\ge 0$.


If so, by Nakayama's Lemma $N=0$ and $M$ is complete by the above Lemma.


The are indeed large classes of non Noetherian $I$-adically complete and separated rings such that finitely generated ideals satisfy property (b), by a version of the Artin-Rees Lemma. In other words and more generally, the Artin-Rees Lemma can be salvaged in a lot of situations, allowing non Noetherian $A$.


The real issue is (a), which much more easily fails.


I leave to you the task of finding an example, following this last comment, showing your question has, in general, negative answer, unless you assume $M$ is $I$-adically separated.


