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.


1 Answer 1


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.


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