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 $R$.



reduction of an ideal
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 <http://www.math.purdue.edu/~heinzer/preprints/colconf10.pdf>.


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


