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Let $(R, \mathfrak m,k )$ be a regular local ring of dimension $3$ with infinite residue field $k$. Let $I$ be an $\mathfrak m$-primary ideal such that for every ideal $J$ containing $I$, it holds that $\mu(J)\le \mu(I)$.

Then, is it necessarily true that $(I: \mathfrak m)=(I:x)$ for some $x\in \mathfrak m\setminus \mathfrak m^2$ ? If this is not true in general, what if we assumed the stronger condition that for every ideal $J$ strictly containing $I$, it holds that $\mu(J)< \mu(I)$ ?

(Here, $(I:J):=\{r\in R: rJ\subseteq I\}$ and $(I:x):=(I: xR)$ )

For some motivations regarding the question, see https://doi.org/10.1080/00927870902747340 and https://www.semanticscholar.org/paper/The-strong-Rees-property-of-powers-of-the-maximal-J.Puthenpurakal-Watanabe/0d3f621f19df02fadf583065c0f857ed22d968cd . An $\mathfrak m$-primary ideal $I$ is said to satisfy the Rees property (resp. Strong Rees property) if $\mu(J)\le \mu(I)$ (resp. $\mu(J)< \mu(I)$ ) holds for every ideal $J$ containing (resp. strictly containing) $I$.

An ideal $I$ is called full iff $(I: \mathfrak m)=(I:x)$ for some $x\in \mathfrak m\setminus \mathfrak m^2$ . It is known that in a regular local ring of dimension $2$ with infinite residue field, an $\mathfrak m$-primary ideal is full if and only if it has the Rees property.

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