$I$-adic completion 
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.
 A: 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.
