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Is any finitely generated module $M$ over a complete commutative ring $A$ complete?

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.

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  • $\begingroup$ The theorem is for Noetherian rings. I just wonder about the general case. $\endgroup$ – wuzx May 2 '18 at 19:35
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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.

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

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.

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    $\begingroup$ Just a comment regarding your terminology: "The $I$-adic completion of the $A$-module $A$ does not exist." I think one should say: "The $I$-adic completion of the $A$-module $A$ is not $I$-adically complete" instead of "does not exist". Compare with Example 1.8 on page 4225 of: math.bgu.ac.il/~amyekut/publications/adic_free/journal.pdf $\endgroup$ – Mahdi Majidi-Zolbanin May 2 '18 at 23:59
  • $\begingroup$ This is a point that bogged me a number of times. I long ago decided to never call "completion" (of a metrizable topological space, hence linearly topologized group/ring/module) a non-complete object. This is standard terminology in EGAI and III and in other references. I believe in Fujiwara and Kato's book too. $\endgroup$ – user87684 May 3 '18 at 0:03
  • $\begingroup$ I think it becomes confusing because when you say the $I$-adic completion of $A$ does not exist, it sounds like the inverse limit$\ $ $A/I^k$ does not exist. $\endgroup$ – Mahdi Majidi-Zolbanin May 3 '18 at 0:14
  • $\begingroup$ @Merlin: There is still more confusion to be had: Since in your example one has $\cap I^n=0$, the topology on $A$ is a topology of the metric, given by $d(f, g)=2^{-k}$ where $f-g \in I^k \setminus I^{k+1}$. Then one can consider the Cauchy sequence construction of completion, which is isomorphic to $A'$, and it is complete with respect to the extended metric. I guess that the only issue is that the metric topology now does not coincide with the $IA'$-adic one. $\endgroup$ – Pavel Čoupek May 3 '18 at 2:08
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    $\begingroup$ Correct. Such topology is not the $IA’$-adic. $\endgroup$ – user87684 May 3 '18 at 3:55

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