Let $\mathfrak a$ be a decomposable ideal in a (commutative ring with unity) $A$, let $\Sigma$ be an isolated set of prime ideals belonging to $\mathfrak a$, and let $\mathfrak q_\Sigma$ be the intersection of the corresponding primary components. Let $f$ be an element of $A$ such that, for each prime ideal $\mathfrak p$ belonging to $\mathfrak a$, we have $f\in\mathfrak p\Leftrightarrow\mathfrak p\not\in\Sigma$, and let $S_f$ be the set of all powers of $f$. Show that $\mathfrak q_\Sigma=S_f(\mathfrak a)=(\mathfrak a:f^n)$ for all large $n$.
It is easy enough to see that $\mathfrak q_\Sigma=S_f(\mathfrak a)=\bigcup_{n>0}(\mathfrak a:f^n)$. (If $\mathfrak a=\cap_{i=1}^n\mathfrak q_i$ is a primary decomposition, then $S_f(\mathfrak a):=\mathfrak a A[S_f^{-1}]\cap A$, where $\cap A$ is an abuse of notation standing in for the preimage under the natural map $A\rightarrow A[S_f^{-1}]$, and we do have $S_f(\mathfrak a)=\cap_iS_f(\mathfrak q_i)$. Now $S_f(\mathfrak q_i)=(1)$ if $f\in\mathfrak p_i$, $\mathfrak q_i$ if $f\not\in\mathfrak p_i$. At the same time, it is easily seen that $S_f(\mathfrak a)=\bigcup_{n>0}(\mathfrak a:f^n)$, since if $\alpha\in S_f(\mathfrak a)$, then $\alpha/f^k\in A$ f.s. $k\geq0$, providing one inclusion. The other inclusion follows since if $\beta\in(\mathfrak a:f^k)$ f.s. $k$, $f^k\beta\in\mathfrak a$, so $\beta\in S_f(\mathfrak a)$.)
But now why must the chain $$(\mathfrak a:f)\subset\cdots\subset(\mathfrak a:f^k)\subset(\mathfrak a:f^{k+1})\subset\cdots$$ stabilize?
Let $\Upsilon$ denote the (finite) set of primes belonging to $\mathfrak a$. Then $\bigcap_{\sigma\in\Sigma}\mathfrak q_\sigma=:\mathfrak q_\Sigma=\bigcup_{n=1}^\infty(\mathfrak a:f^n)$, $\mathfrak a=\bigcap_{y\in\Upsilon}\mathfrak q_y$, $f\in\bigcap_{y\in\Upsilon-\Sigma}\mathfrak p_y$, hence $f\in\sqrt{\mathfrak q_{\Upsilon-\Sigma}}$. Suppose $f^n\in\mathfrak q_{\Upsilon-\Sigma}$. Then of course $f^n\mathfrak q_\Sigma\subset\mathfrak a$. $\blacksquare$