$m$ contains a non-zero-divisor on $R/(I:m^{\infty}) $

Question

Let $(R,m)$ be a noetherian local ring with the maximal ideal $m$ and $I$ an $R$-ideal such that $\sqrt{I}\neq m$. Then, is it true that $m$ contains a non-zero-divisor on $R/(I:m^{\infty})$ . The notation $I:J^{\infty}$ is the notation used to indicate the least $n$ such that $IJ^{n+1}=IJ^n$, also called the saturation of $J$ with respect to $I$. The notation $(I:J^{\infty})$ is used for $\cap_nI:J^n$.

I know that there exists an $R-$linear map from $R/I\to R/I.a^n$, for some element $a\in R$. I also know that $I:J=\cap_{i=1}^nI:a_i^{\infty}$, where the ideal $J$ is generated by $\{a_1,a_2\ldots,a_n\}$. How do we proceed using these facts? Suppose $m=(a_1,a_2,\ldots,a_n)$, then $R/(I:m^{\infty})=\cap_{i=1}^n R/(I:a_i^{\infty})$. What is the signifiance of the condition $\sqrt{I}\neq m$? Any hints? thanks beforehand.

Answer 1

Yes. Here’s a proof.

First, note that since $\sqrt{I} \neq m$, we have $I : m^\infty \neq R$, so that $R/(I : m^\infty) \neq 0$.

Now suppose every element of $m$ is a zero-divisor on $R/(I : m^\infty)$. Then $m \subseteq \bigcup \operatorname{Ass} (R/(I : m^\infty))$, so by prime avoidance $m \in \operatorname{Ass} (R/(I: m^\infty))$. Thus there is some $x\in R \setminus (I : m^\infty)$ with $m x \subseteq I : m^\infty$. That is, there is some $t$ with $mx \subseteq I : m^t$. Thus, $x \in ((I : m^t) : m) = (I : m^{t+1}) \subseteq (I : m^\infty)$, which is a contradiction.