Let $R$ be Noetherian ring. Let $I \subseteq J$ be proper ideals of $R$. If $R$ is $J$-adically complete, then is $R$ complete $I$-adically ?
I was proceeding as follows: Let $\{x_n\}$ be an $I$-adic Cauchy sequence in $R$ ; to show $\{x_n\}$ converges $I$-adically in $R$ . Now for every $I^k$, there is some $n_0$ such that $x_{n+1}-x_n \in I^k,\forall n \ge n_0$. Then $x_{n+1}-x_n \in J^k,\forall n \ge n_0$, so $\{x_n\}$ is $J$-adic Cauchy in $ R$, so converges $J$-adically to some $x\in R$ i.e. for every $k \ge 1$, $\exists n_k\ge 1$ such that $x_n-x \in J^k,\forall n \ge n_k$. Then I am stuck. I don't know if it can be done considering the algebraic definition of completion or not.
Please help.
The sequence $(x_n)$, being $I$-adically Cauchy by assumption, tends $I$-adically, and thus $J$-adically, to $y$ for some $y$ in $R$. This implies $y=x$.
Edit 1. I used this answer of Matt E.
EDIT 2. We can also repeat Matt E's nice argument:
In the setting of the question, let $p$ be a positive integer, let $q\in\mathbb N$ be such that $r\ge q$ implies $x_r-x_{r+1}\in I^p$. It suffices to show $x_r-x\in I^p$.
Let's forget the $I$-adic topology, and work only with the $J$-adic topology.
Set $u_n:=x_r-x_{r+n}$. Then $(u_n)$ is a sequence in $I^p$ converging to $x_r-x$. As $I^p$ is closed, we have $x_r-x\in I^p$, as desired.