Example of associated ideal in primary decomposition

Question

Let $I$ be a decomposable ideal of a commutative ring $R$ with minimal primary decomposition $I=\bigcap_{i=1}^n\mathfrak q_i$.

The first uniqueness theorem shows that $\{\sqrt {\mathfrak q_i}:1\le i\le n\}=\{\sqrt{(I:x)}\in \operatorname{Spec} R:x\notin I\}$

Is there any example imply that we could take off the radical, i.e., for some primary ideal $\mathfrak q_i$ of $I$, $\sqrt {\mathfrak q_i}$ can not be written of the form $(I:x)$?

Answer 1

In particular, you are asking the following:

If $\mathfrak q$ is a $\mathfrak p$-primary ideal do we have $\mathfrak p=(\mathfrak q:x)$, that is, $\mathfrak p\in \mathrm{Ass}(R/\mathfrak q)$?

This holds if $R$ is noetherian; see How to prove Ass$(R/Q)=\{P\}$ if and only if $Q$ is $P$-primary when $R$ is Noetherian?
Otherwise, let $R=K[X_1,\dots,X_n,\dots]$ and $\mathfrak q=(X_1^2,\dots,X_n^2,\dots)$. Now notice that $\mathrm{Ass}(R/\mathfrak q)=\emptyset$.