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Reduction of an ideal is an important method on commutative algebra. Let $R$ be a Noetherian ring, $J \subseteq I, J \neq I$ two ideals of $R$. Then $J$ is a reduction of $I$ is there exists $k$ such that $I^{k+1} = I^kJ$. Now I interested in a similar condition $J^{k+1} = J^kI$.

Discussion: if $\mathrm{ht}(J)>0$ then the condition $J^{k+1} = J^kI$ (rewrite in form $J J^k = I J^k$) implies that $I$ is integral over $J$ and it is equivalent to $J$ is a reduction of $I$ (see, Huneke, Swanson: Integral closure, Chapter 1). By choice $k$ big enough we can assume that $J^{k+1} = J^kI$ and $I^{k+1} = I^kJ$. Thus for all $n \geq 2k$ we have $$I^n = I^kJ^{n-1} = I^{n-k}J^k = J^n$$ This is a very special property.

Question 1: Let $(R, \mathfrak{m})$ be a local ring of dimension $d>0$. Do there exist two $\mathfrak{m}$-primary ideals $J \subseteq I, J \neq I$ such that $J^{k+1} = J^kI$ for some $k$.

By discussion we know that if $J^{k+1} = J^kI$, then the Hilbert polynomial with respect to $J$ and $I$ are same.

Question 2: Consider question 1 in the case $J = \mathfrak{q}$ be a parameter ideal of $R$.

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up vote 2 down vote accepted

The idea you are flirting with is the ``Ratliff-Rush operation''. Given a regular ideal $J$ (which is a special case of an ideal with positive height), Ratliff and Rush define $\tilde{J} := \bigcup_{n=0}^\infty (J^{n+1} : J^n)$.
Then they show that $\tilde{J}$ is the largest ideal among ideals $I$ such that $I^n = J^n$ for $n \gg 0$.

Later authors have called $\tilde{J}$ the Ratliff-Rush ideal associated with $J$ (or sometimes [misleadingly] the Ratliff-Rush closure of $J$ -- misleading because the operation does not preserve containments of ideals, whereas a true closure operation should). In general, a Ratliff-Rush ideal is an ideal $K$ such that $K = \tilde{K}$.

Your condition can be restated as: $J \subseteq I \subseteq \tilde{J}$. You are right that the condition is stronger than $J$ being a reduction of $I$. For instance, if $R=k[x,y]$ and $J = (x^2, y^2)$, then $J = \tilde{J}$, even though $J$ is a reduction of $(x,y)^2$.

Question 2 (as well as question 1) has a negative answer whenever $R$ is a Dedekind domain, since every ideal there is integrally closed, hence a Ratliff-Rush ideal. I expect that both questions should have a positive answer in most cases, however.

A paper of Heinzer, Lantz and Shah proves the following: Let $R$ be a one-dimensional local domain. Then every ideal of $R$ is Ratliff-Rush if and only if every ideal either has principal reduction or reduction number $1$. This theorem should have bearing on your question as well. (In this situation, of course, the notions of parameter ideal and principal ideal coincide.)

For what I have said here along with other stuff you will likely find interesting, see the survey article

So, not necessarily a complete answer, but this should be enough information to lead you to as complete an answer as is available.

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Thanks you Neil. For Question 2, by regular sequence property we have every parameter ideal in a Cohen-Macaulay ring is Ratliff-Rush. – Pham Hung Quy Jan 17 '13 at 4:35
You're welcome! – Neil Epstein Jan 17 '13 at 18:22

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