Let R be a commutative ring, and let a⊆R be an ideal. For an R-module we consider the sub-R-modules Γa(M)={x∈M∣∃n∈N:an⊆(0:Rx)} and ˜Γa(M)={x∈M∣a⊆√(0:Rx)} of M. These definitions give rise to left exact subfunctors Γa and ˜Γa of the identity functor on the category of R-modules, and Γa is a subfunctor of ˜Γa.

Recall that a subfunctor F of the identity functor is called a

radicalif 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 ˜Γa is a radical, while Γa need not be so. I conjecture but am unable to prove the following:

Conjecture:˜Γa is the smallest radical containing Γa.Is anything known about this problem?

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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.)

**Answer**

The conjecture is not true.

By Quý’s comment, the conjecture implies that Γm is not a radical if R is a 0-dimensional local ring whose maximal ideal m is idempotent but not nilpotent. This contradicts the following result.

**Lemma** If R is a ring and a⊆R is an idempotent ideal, then Γa is a radical.

*Proof:* Let M be an R-module, and let x∈M be such that x+Γa(M)∈Γa(M/Γa(M)). Then, ax∈Γa(M), and therefore ax=a2x=0, implying x+Γa(M)=0. ◻

**Attribution***Source : Link , Question Author : Fred Rohrer , Answer Author : Fred Rohrer*