$I$-torsion functor and $\sqrt{I}$

Question

Let $R$ be a Noetherian ring and $I,J\subseteq R$ ideals. For any $R$-module $M$, define $\Gamma_I(M)=\{x\in M\mid I^n x=0\text{ for some }n\in\mathbb{N}\}$. 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 $r\in \sqrt{I}$, I think I want some quotient $R/A$ (where $A$ is an ideal involving $I$, or maybe $r$) such that $r\in\Gamma_I(R/A)=\Gamma_J(R/A)$, i.e. such that $(r+A)J^e\subseteq A$ for some $e$.

Answer 1

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.

Answer 2

1. If $I\subset J$, then $\Gamma_I(M)\supset \Gamma_J(M)$. As $A$ is Noetherian, $\Gamma_I=\Gamma_{\sqrt{I}}$.
2. We assume that $I,J$ are radical, then take $M=(I+J)/I$.

Exercise II.5.6(d) in Hartshorne's Algebraic Geometry is linked to this.