Xrfgtjtk

Let $R$ be a Noetherian ring and $I,Jsubseteq R$ ideals. For any $R$-module $M$, define $Gamma_I(M)={xin Mmid I^n x=0text{ for some }ninmathbb{N}}$. Suppose that $Gamma_I=Gamma_J$. I want to show that $sqrt{I}=sqrt{J}$.

My issue is that I don't know which $R$-module to consider. Picking any $rin sqrt{I}$, I think I want some quotient $R/A$ (where $A$ is an ideal involving $I$, or maybe $r$) such that $rinGamma_I(R/A)=Gamma_J(R/A)$, i.e. such that $(r+A)J^esubseteq A$ for some $e$.

First check that we can use the Noetherian hypothesis to rewrite the definition of the $I$-torsion functor as $Gamma_I(M) = {m in M mid I subseteq sqrt{text{Ann}_R(m)} }$. I think this well illuminates an approach to the problem by associated primes.

Hint about what module to apply the functor to:

I would look at $Gamma_I(R/I) = Gamma_J(R/I)$.

More details, if it's helpful:

In particular, pick a prime $mathfrak{p}$ minimal over $I$, so that $mathfrak{p}$ is an associate prime of $R/I$. There exists an element $m in R / I$ such that $text{Ann}_R(m) = mathfrak{p}$, hence $I subseteq mathfrak{p} = sqrt{text{Ann}_R(m)}$ and $m in Gamma_I(R/I)$. By assumption, $m in Gamma_J(R/I)$, too. Thus $J subseteq mathfrak{p}$ as well. It follows easily that $V(I) subseteq V(J)$. A symmetric argument shows the opposite inclusion.