# Primary decomposition of ideal saturation

This is a follow-up to my question here: Primary decomposition and contraction

Let $R$ be a Noetherian ring, $I\subset R$ ideals, and $S \subset R$ a multiplicative set. Define: $$I:\langle S\rangle := (IR_S)\cap R = \{ x \in R \mid xy \in I \text{ for some } y \in S\}.$$ Then we can get an explicit primary decomposition of this new saturation" ideal in terms of a primary decomposition of $I$.

In particular, let $I = \cap_{i=1}^n Q_i$ be a primary decomposition with $Q_i$ primary to $P_i\in \operatorname{Spec}(R)$ for each $i$. Without loss of generality assume $Q_i \cap S = \varnothing$ for $1 \le i \le m$ and $Q_i \cap S \neq \varnothing$ for $m+1 \le i \le n$. Then $I:\langle S \rangle = \cap_{i=1}^m Q_i$. Intuitively, the operation $\bullet : \langle S\rangle$ of saturating at $S$ smooths away" the primary components which meet $S$.

I have heard of another notion of "saturating" an ideal $I$ at another ideal $J$. Define: $$I:\langle J \rangle := \bigcup_{n \ge 1} I:J^n = \{ x \in R \mid xJ^n \subset I \text{ for some } n \ge 1\}.$$ A paper I am reading implies that these two notions are equivalent. In particular, the authors make two claims to compare the two different kinds of saturations.

First, the authors claim that given $I$ and $J$, let: $$\mathcal{S} = \{ P \in \operatorname{Spec}(R) \mid J \not \subset P \text{ and } P \in \operatorname{Ass}(R/I^n) \text{ for some } n\}.$$ Then let $x \in J \setminus \cup_{P \in \mathcal{S}} P$, nonempty by Prime Avoidance. Then $I:\langle S\rangle = I:\langle J \rangle$ where $S= \{ 1, x,x^2,\cdots\}$.

I can show one direction of this, not using that $x$ is special: if $y J^n \subset I$ then $yx^n \in I$ so $I:\langle J \rangle \subset I:\langle S\rangle$. To show the other direction, I want to try and use the smoothing components" idea above. Is there a similar sort of lemma one can prove about a primary decomposition of $I:\langle J \rangle$ in terms of primes somehow connected to $I$ and $J$?

If the claim is to be believed (or if I understand it correctly), it seems it should involve primes in the set $A(I) = \cup_{n \ge 1} \operatorname{Ass}(R/I^n)$ (known to be a finite set by work of Brodmann) and either $V(J) = \{ P \in \operatorname{Spec}(R) \mid J \subset P\}$ or its complement $D(J) = \{ P \in \operatorname{Spec}(R) \mid J \not \subset P\}$.

I think if I could understand the primary decomposition of $I:\langle J \rangle$ I could also prove their second claim, which is stated more vaguely: symbolic powers are saturations at a suitable fixed ideal," where the symbolic power $I^{(n)}$ is $I:\langle W_I\rangle$, for $W_I = R-\cup_{P \in \operatorname{Ass}(R/I)} P$. I believe they claim that $I^n : \langle W_I\rangle = I^n:\langle J\rangle$ for some ideal $J$ but I can't produce one. I also think any construction of a $J$ that would work for $W_I$ would work for an arbitrary multiplicative set $S$ to show that in fact both of these notions of saturation are always equivalent.

Let $$I=\bigcap_{i=1}^n\mathfrak q_i$$ be a primary decomposition of $$I$$ and suppose that $$\newcommand{\qq}{\mathfrak q} \newcommand{\pp}{\mathfrak p}J\not\subseteq\sqrt{\qq_i}:=\pp_i$$ for $$1\leq i\leq m$$ and $$J\subseteq \pp_i$$ for $$m+1\leq i\leq n$$. We claim that $$(I:J^\infty) := \bigcup_{n=1}^\infty (I:J^n)=\bigcap_{i=1}^m\qq_i$$.
Suppose the claim true. Choose $$s\in J$$ such that $$s\notin \pp_i$$ for $$1\leq i\leq m$$. Let $$S$$ denote the multiplicatively closed subset $$\{1,s,s^2,\ldots\}$$. Then $$I:\langle S\rangle$$, the saturation of $$I$$ with respect to $$S$$, is also given by $$\bigcap_{i=1}^m\qq_i$$, hence $$I:\langle S\rangle=(I:J^\infty)$$.
Proof of the claim: Let $$x\in\bigcap_{i=1}^m\qq_i$$. Pick $$t$$ such that $$\pp_i^t\subseteq \qq_i$$ for $$m+1\leq i\leq n$$. Note that since $$J\subseteq \pp_i$$ for $$m+1\leq i\leq n$$, $$J^t\subseteq \qq_i$$ for $$m+1\leq i\leq n$$. Thus, $$xJ^t\subseteq \bigcap_{i=1}^n\qq_i=I$$, and hence $$x\in (I:J^\infty)$$. Conversely, suppose that $$x\in (I:J^\infty)$$. We then have $$xJ\subseteq \qq_i$$ for all $$i$$, and since $$J\not\subseteq \pp_i$$ for $$1\leq i\leq m$$, we must have $$x\in \bigcap_{i=1}^m\qq_i$$ (recall that $$\qq_i$$ is $$\pp_i$$-primary). This completes the proof of the claim.