TITLE: If $I$ and $J$ are ideals and I know an algorithm to find $I : J$, can I use this to find the saturation of $I$ by $J$?
QUESTION [5 upvotes]: I know a pair of ideals $I$ and $J$ in a Noetherian ring and have an algoritm to find a generating set of the quotient ideal $I : J$. If I keep applying this algorithm successively finding generating sets for

$I : J$
$(I : J) : J$
$((I : J) : J) : J$
$(((I : J) : J) : J) : J$

and so on then this will at some point stabilise by the ascending chain condition. My question is whether it is guaranteed to stabilise at the saturation $I : J ^\infty$?

REPLY [2 votes]: First, a lemma:
Lemma: Let $I,J,$ and $K$ be ideals of some commutative ring $R.$ Then $((I:J):K) = (I:JK).$
Proof: Let $f\in R.$ Then we have
\begin{align*}
f\in ((I:J):K)&\iff fK\subseteq (I:J)\\
&\iff fk\in I:J,&\forall k\in K\\
&\iff fkJ\subseteq I,&\forall k\in K\\
&\iff fkj\in I,&\forall j\in J,k\in K\\
&\iff fJK\subseteq I,\\
&\iff f\in (I:JK).&\square
\end{align*}
(We use the fact that $JK$ is generated by products of the form $jk$ with $j\in J$ and $k\in K$ at the penultimate step.)
By the lemma, your chain of ideals
$$
(I:J)\subseteq ((I:J):J)\subseteq(((I:J):J):J)\subseteq\cdots
$$
is nothing more than the chain of ideals
$$
(I:J)\subseteq (I:J^2)\subseteq (I:J^3)\subseteq\cdots.
$$
So, you want to show that the chain $(I:J)\subseteq (I:J^2)\subseteq (I:J^3)\subseteq\cdots$ stabilizes at the stablization $(I:J^\infty).$ This should be easier to prove! If you get stuck proving this, see below.

 Say the chain stabilizes at the step $(I:J^N)$. Almost by definition, we have $(I:J^N)\subseteq (I:J^\infty).$ Conversely, if $f\in (I:J^\infty),$ then $fJ^M\subseteq I$ for all $M\gg 0$. But since $$fJ^N\subseteq I\quad\textrm{ if and only if }\quad fJ^{N+1}\subseteq I\quad\textrm{ if and only if }\quad fJ^{N+2}\subseteq I,$$ and so on, it follows that $fJ^M\subseteq I$ for all $M\gg 0$ if and only if $fJ^N\subseteq I$ if and only if $f\in (I:J^N).$