# On the relation between two definitions of torsion functors

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 and 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 $\Gamma_{\mathfrak{a}}$.

(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.)

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 $x+\Gamma_{\mathfrak{a}}(M)=0$. $\square$