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:



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



On the relation between two definitions of torsion functors
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 $x+\Gamma\_{\mathfrak{a}}(M)=0$. $\square$


