# On inequality between number of generators of ideals

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.