Saturation of a power of an ideal

Question

Let $k$ be a field and let $R=k[x,y,z]$ and $\mathfrak m=(x,y,z)$. Let $I$ be a graded ideal of $R$. For all $n\in \mathbb{N}$ on has $$ (I^{\rm sat} )^n\subset (I^n)^{\rm sat},$$ where $$I^{\rm sat}= I:_R \mathfrak m^\infty= \{f\in R:\exists s\in \mathbb{N}, f\mathfrak m^s\subset I\}.$$

Question: What about the reverse inclusion? Is there $(I^n)^{\rm sat}=(I^{\rm sat})^n$?

Thank you so much!

Answer 1

I don't think it is true in general.

Let $$M = \begin{pmatrix} x & y & z \\ y & z & x^2 \end{pmatrix}$$ and $p = I_2(M)$. Then $p$ is a prime ideal of height $2$. Hence, $p$ is saturated. However the associated prime of $p^2$ has an embedded component.

For an explicit computation, you can refer to the following Macaulay2 code:

R = QQ[x,y,z]
M = matrix {{x,y,z},{y,z,x^2}}
p = minors(2,M)
isPrime p 
associatedPrimes p^2
primaryDecomposition p^2