2

Let $R$ be a Noetherian ring, $Q$ a $P$-primary ideal in $R$ for some $P\in \operatorname{Spec}(R)$ and $S$ an arbitrary multiplicative set in $R$ which does not meet $Q$, i.e. $Q\cap S=\varnothing$. What can be said of the contraction: $$Q'=(QR_S)\cap R = \{ x \in R \mid x\cdot y \in Q \text{ for some } y \in S\}?$$

Edit: After some searching, it seems I was confused about the proper symmetric definition of primary, which is $xy \in Q$ if $x \in Q$ or $y \in Q$ or $x\in \sqrt{Q}$ AND $y\in \sqrt{Q}$ which settles the difficulty. Due to this, I've made a dramatic edit to my question and posted most of it as an answer instead just to close the question.

walkar
  • 3,844
  • This is proved in Bourbaki, Commutative Algebra, Ch. 4 Associated Prime Ideals and Primary Decomposition, § 2, n° 1, prop. 3. – Bernard Nov 04 '17 at 17:23

1 Answers1

1

It is known that the preimage of a primary ideal is primary, so I know $Q'$ is primary. Furthermore, note $P\cap S = \varnothing$ as well since $\sqrt{Q}=P$ and if $x \in P \cap S$, then $x^n \in P\cap S$ for all $n$ but also eventually $x^n \in Q$.

Here is my attempt: If $xy \in Q$ for some $y \in S$, then $x\in Q$ or $y \in Q$ or $x\in \sqrt{Q}=P$ and $y \in P$. However, since $Q\cap S = P\cap S = \varnothing$, we see only the first possibility can happen, i.e. $Q'=\{ x \in R \mid xy \in Q \text{ for some } y \in S\}=\{x \in Q\}$, so $Q'=Q$.

walkar
  • 3,844