**Question**. Let $(R,m)$ be a Noetherian local ring and $M$ be a finite *faithful* $R$-module. Let $I$ be an ideal of $R$ such that $IM=mM$. Can we say that $I$ is a *reduction ideal* of $m$? Recall that $I$ is a reduction ideal of $m$ if $Im^n=m^{n+1}$ for some (or equivalently all) sufficiently large $n$. 


If not what conditions can we add (on $I$ or $R$ or...) to be able to say $I$ is a *reduction ideal* of $m$? 


Thank you.



$IM=mM$. can we say that $I$ is a reduction ideal of $m$?
The answer is always yes. It's a basic tool in studying the integral closures of ideals.


Let $x \in m$. Let $\phi: M \rightarrow M$ be the endomorphism of $M$ given by multiplication by $x$. Then since $\phi(M) \subseteq IM$, the Cayley-Hamilton theorem shows that there exist $a\_j \in I^j$ and $n \in \mathbb N$ such that $$
\phi^n + \sum\_{j=1}^n a\_j \phi^{n-j}=0
$$
as an element of the endomorphism ring of $M$. But since $M$ is faithful, it follows that $x^n + \sum\_{j=1}^n a\_j x^{n-j}=0 \in R$. Thus, $x$ is in the integral closure of the ideal $I$.


Since this holds for an arbitrary element of $m$, it follows that $I$ is a reduction of $m$.


