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 $R$.