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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.

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  • $\begingroup$ also asked in math.stackexchange.com (24 days ago, with no answer) $\endgroup$
    – user 1
    Commented Mar 6, 2015 at 13:23

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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 $m$.

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    $\begingroup$ Oh. Well, if $\phi$ is the endomorphism of multiplication by $x$ on $M$, then $\psi := \phi^n + \sum a_j \phi^{n-j}$ is the endomorphism given by multiplication by $y := x^n +\sum a_j x^{n-j}$. on $M$. And so, to say that $\psi$ is the zero map on $M$ (which it is, by the C-H theorem) is the same as saying that $y$ annihilates $M$. But ``$M$ is faithful'' means that only $0$ annihilates $M$. Hence $y=0$. $\endgroup$
    – Neil Epstein
    Commented Mar 19, 2015 at 0:24

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