$I$-adic completion

Solution 1:

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.