Well behaved colon between powers of ideals when the associated graded ring is Cohen-Macaulay.

Question

I'm reading a paper: A formula for the core of an ideal, by Claudia Polini and Bernd Ulrich and I'm in trouble with the following problem:

Let $R$ be a Cohen-Macaulay ring and $I$ be an ideal of $R$ with $\mathrm{ht}(I)>0$. If $\mathrm {gr}_I(R)$ is Cohen-Macaulay, then $I^{m+n}:_RI^n = I^m$ for all $n,m \in \mathbb{N}$.

Well, I only proved this result with the following additional hypothesis: "R Cohen-Macaulay local ring with infinite residue field", because I was able to use some classical theorems that allowed me to find a homogeneous $\mathrm{gr}_I(R)$-regular element of degree 1.

Can someone help me in this problem?

In the article: Reductions Numbers and Balanced Ideals, by Louiza Fouli, this result is apparently used at the end of proposition 2.1. She says the cancellation law is clear, because $grade(gr_I(R)_+)>0$, since $gr_I(R)$ is Cohen-Macaulay and $ht(I)>0$.

Answer 1

Let $G = gr_I(R)$ and $G_n$ denote the n-th graded piece $I^n/I^{n+1}$ of $G$. We will show that if the grade of $G_+$ is positive, then $I^{n+i} : I^n = I^i$.

For $l > 0$, we will use the following.

(1) For any non-zero element $Z \in G_n$, there exists a lift $z \in I^n \setminus I^{n+1}$ of $Z$. That is $Z = z + I^{n+1} \in I^n/I^{n+1} = G_n$.
(2) If $Z \in G_l$ is a nonzerodivisor and $z \in I^l \setminus I^{l+1}$ is a lift of $Z$, then $I^{n+l}:z = I^n$ for $n > l$.
(3) If grade $G_+ > 0$, then there exists a homogeneous nonzerodivor in $G_+$.
(4) For any $m \ge n$, $I^{n+i} : I^n \subset I^{m+i} : I^m$.

In any case, $I^i \subset I^{n+i}:I^n$. Let $y \in I^{n+i}:I^n$. Let $Z \in G_l$ be a nonzerodivisor and $z$ a lift of $Z$ in $I^l$. Choose $m$ such that $m \ge n$ and $l \mid m$. Then $y \in I^{m+i}:I^m$ and $$ yz^{m/l} \subset yI^m \subset I^{m+i}. $$ Thus, $y \in I^{m+i}:z^{m/l} = I^i$.