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Let $A$ be a commutative noetherian ring, and suppose that $A$ is $I$-adically complete with respect to some ideal $I\subseteq A$. Is it true that for any ideal $J\subseteq I$, the ring $A$ is also $J$-adically complete?

Edit. Recall that a ring $A$ is $I$-adically complete if the canonical morphism $A\to \varprojlim A/I^n$ is an isomorphism.

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Can you expand on the definition of $I$-adically complete? I'm not sure if $\bigcap_nI^n=0$ is necessary, which would make a difference to the counterexample, if any. –  George Lowther Jan 16 '11 at 14:41
@George: Dear George, Usually $I$-adically complete is taken to mean "complete and separated", i.e. that the intersection you ask about does equal $0$. –  Matt E Jan 16 '11 at 15:24

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up vote 4 down vote accepted

The answer is "yes".

Since $A$ is Noetherian, for any $m$ the finitely generated $A$-module $A/J^m$ is $I$-adically complete, and so $A/J^m$ is the inverse limit over $n$ of $A/(I^n + J^m)$. Now $J^m \subset I^m,$ and so $I^n \subset I^n + J^m \subset I^m$ when $n \geq m$. Thus the inverse limit (over $m$) of $A/J^m$ is the same as the inverse limit (over $n$) of $A/I^n$, and we see that $A$ is $J$-adically complete.

Another way to think about it is that $A$ is $I$-adically complete (and separated, which is part of the requirement of "complete") if and only if any $I$-adic Cauchy sequence of elements of $A$ has a unique $I$-adic limit. Since a $J$-adic Cauchy sequence is also an $I$-adic sequence, a $J$-adic Cauchy $(a_n)$ sequence also has a unique $I$-adic limit, say $a$.

Now if we choose $n_0$ so that $a_m - a_n \in J^k$ if $m,n \geq n_0$, then we see that $a - a_m = a - a_{n} + a_{n} - a_m \in J^k + I^l,$ where $l$ can be made arbitrarily large by choosing $n$ large enough (since $a_n$ converges to $a$ in the $I$-adic topology). Thus $a - a_m \in \cap_l J^k + I^l.$ This intersection is equal to $J^k$ (by $I$-adic completeness of $A/J^k$) and so $a-a_m \in J^k$. Thus in fact $(a_n)$ converges to $a$ in the $J$-adic topology, and so $A$ is $J$-adically complete.

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Matt: can you expand on that? –  George Lowther Jan 16 '11 at 15:34
@Matt: The bit I found tricky when thinking about this is showing that $A/J^m$ is $I$-adically complete. Why is that? Does it follow from being finitely generated? –  George Lowther Jan 16 '11 at 17:24
Actually, it does not follow. I have a counterexample in mind. –  George Lowther Jan 16 '11 at 17:38
I'm not sure that the ring in my counterexample is noetherian. Do you have a reference or proof that $A/J^m$ is $I$-adically complete? –  George Lowther Jan 16 '11 at 18:01
@George: Dear George, The Artin--Rees lemma implies that any f.g. module of an $I$-adically complete Noetherian ring is itself $I$-adically complete. Regards, –  Matt E Jan 16 '11 at 19:21

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