Let $R$ be a Noetherian semi-local ring, let $\mathfrak m_1,...,\mathfrak m_n$ be the finitely many maximal ideals. Let $J=\mathfrak m_1\cap ...\cap \mathfrak m_n$ denote the Jacobson radical of $R$ . Also assume $R$ is $J$-adically complete.
If $\{I_n\}_{n\ge 0}$ is a descending chain of ideals in $R$ such that $\cap_{n\ge 0} I_n=(0)$ , then how to prove that there exists a function $f: \mathbb N \to \mathbb N$ such that $\lim_{n\to \infty} f(n)=\infty$ and $I_n \subseteq J^{f(n)}, \forall n\ge 1$ ?
Here $\mathbb N$ denotes the set of non-negative integers.
Define $f(n)$ to be the maximum $t$ so that $I_n$ is contained in $J^t$ (with the convention that if $I_n$ is not contained in $J$, we have $f(n)=0$). $f$ has domain the non-negative integers and is clearly non-decreasing by the condition that $I_n\supset I_{n+1}$. This means that the limit of $f(n)$ as $n$ goes to infinity is either infinity or some nonnegative integer $c$. Our goal is to show the latter cannot happen - that is, $f$ can't "get stuck" somewhere.
Suppose $\lim_{n\to \infty} f(n)=c$. Then what this says is that there exists some $a>0$ so that for all $n>a$, we have $I_n\subset J^c$ but $I_n\not\subset J^{c+1}$. We will derive a contradiction.
Let's look at $J^c/J^{c+1}=J^c\otimes_R (R/J)$. First, $J^c/J^{c+1}$ is a finitely-generated $R/J$-module, because $J^c$ is finitely generated (as $R$ is noetherian) and the tensor product of finitely generated modules is finitely generated. As $R/J$ is semisimple (by definition of the Jacobson radical), this means $J^c/J^{c+1}$ is a direct sum of a finite number of nonzero simple submodules.
Now consider $I_n/J^{c+1} \subset J^c/J^{c+1}$. By our assumption that $I_n\subset J^c$ but $I_n\not\subset J^{c+1}$ for all sufficiently large $n$, the decreasing family $I_n/J^{c+1}$ cannot ever be zero. Thus there must exist a nonzero simple $R/J$ submodule, call it $S$, so that $S\subset I_n/J^{c+1}$ for all $n$. By the correspondence theorem, $S=S'/J^{c+1}$ for some nonzero submodule $S'\subset R$. Then $S'\subset I_n$ for all $n$, which implies that $0\neq S'\subset \bigcap I_n = 0$, a contradiction.
Thus $\lim_{n\to\infty} f(n)\neq c$ for any integer $c$, so it must be $\infty$. There might be a slicker solution using properties of topological rings (for instance, this answer gives some nice properties of the $J$-adic topology that one could perhaps use), but I am not super familiar with that area and the above solution was one I was more comfortable writing.