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Let $R$ be a commutative ring, and let $\mathfrak{a}\subseteq R$ be an ideal. For an $R$-module we consider the sub-$R$-modules $$\Gamma_{\mathfrak{a}}(M)=\{x\in M\mid\exists n\in\mathbb{N}:\mathfrak{a}^n\subseteq(0:_Rx)\}$$ and $$\widetilde{\Gamma}_{\mathfrak{a}}(M)=\{x\in M\mid\mathfrak{a}\subseteq\sqrt{(0:_Rx)}\}$$ of $M$. These definitions give rise to left exact subfunctors $\Gamma_{\mathfrak{a}}$ and $\widetilde{\Gamma}_{\mathfrak{a}}$ of the identity functor on the category of $R$-modules, and $\Gamma_{\mathfrak{a}}$ is a subfunctor of $\widetilde{\Gamma}_{\mathfrak{a}}$.

Recall that a subfunctor $F$ of the identity functor is called a radical if $F(M/F(M))=0$ for every $R$-module $M$. Moreover, for every subfunctor $F$ of the identity functor there exists the smallest radical containing $F$.

Now, one can show that $\widetilde{\Gamma}_{\mathfrak{a}}$ is a radical, while $\Gamma_{\mathfrak{a}}$ need not be so. I conjecture but am unable to prove the following:

Conjecture: $\widetilde{\Gamma}_{\mathfrak{a}}$ is the smallest radical containing $\Gamma_{\mathfrak{a}}$.

Is anything known about this problem?

(Motivation: In the literature about torsion functors and their right derived functors (i.e., local cohomology) both definitions are used. Since most authors work over noetherian rings this does not matter: For an ideal of finite type, the two functors coincide. But if we wish to study non-noetherian situations it might be helpful to understand the precise relation between the two definitions.)

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  • $\begingroup$ Could you give a detail example for the non-radical property of $\Gamma_{\mathfrak{a}}$ $\endgroup$
    – Pham Hung Quy
    Commented Feb 9, 2017 at 14:22
  • $\begingroup$ Em Quý, you can take the example in Proposition 2.12 of our paper... $\endgroup$
    – Fred Rohrer
    Commented Feb 9, 2017 at 15:05
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    $\begingroup$ Thanks you very much, i don't remember it. If your conjecture hold true and $\widetilde{\Gamma}_{\mathfrak{a}} \neq \Gamma_{\mathfrak{a}}$, then $\Gamma_{\mathfrak{a}}$ is not radical. Now, I consider local ring $(R, \mathfrak{m})$ of dimension zero with idempotent maximal ideal, that is $\mathfrak{m} = \mathfrak{m}^2$. So we have $\Gamma_{\mathfrak{m}}(M) = \mathrm{Soc}(M) = 0:_M \mathfrak{m}$. But $\widetilde{\Gamma}_{\mathfrak{m}}(M) = M$ since $\dim R = 0$. Could you explain why $\Gamma_{\mathfrak{m}}$ is not radical. $\endgroup$
    – Pham Hung Quy
    Commented Feb 9, 2017 at 16:06

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The conjecture is not true.

By Quý's comment, the conjecture implies that $\Gamma_{\mathfrak{m}}$ is not a radical if $R$ is a $0$-dimensional local ring whose maximal ideal $\mathfrak{m}$ is idempotent but not nilpotent. This contradicts the following result.

Lemma If $R$ is a ring and $\mathfrak{a}\subseteq R$ is an idempotent ideal, then $\Gamma_{\mathfrak{a}}$ is a radical.

Proof: Let $M$ be an $R$-module, and let $x\in M$ be such that $x+\Gamma_{\mathfrak{a}}(M)\in\Gamma_{\mathfrak{a}}(M/\Gamma_{\mathfrak{a}}(M))$. Then, $\mathfrak{a}x\in\Gamma_{\mathfrak{a}}(M)$, and therefore $\mathfrak{a}x=\mathfrak{a}^2x=0$, implying $x+\Gamma_{\mathfrak{a}}(M)=0$. $\square$

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