Let $R$ be a commutative ring, $I \subset R$ is an ideal. The functor of $\Gamma_I(M) = \varinjlim (R/I^t, M)$ as an endofunctor of $\operatorname{Mod}(R)$ is a subfunctor of the identity functor, since $\Gamma_I(M) \subset M$ and idempotent $\Gamma_I^2=\Gamma_I$.
In the theory of torsion pairs functor with such two properties are called idempotent radicals. Set $\mathcal{T} = \{M \in \operatorname{Mod}(R) \mid \Gamma_I(M)=M$ } to be $I$-torsion modules and $\mathcal{F} = \{M \in \operatorname{Mod}(R) \mid \Gamma_I(M)=0$ } to be $I$-torsion-free modules. Do this two classes form a torsion pair $(\mathcal{T}, \mathcal{F})$?
It seems to me that it is correct and even obvious from the point of view of idempotent radicals. We even work with another definition of torsion pair, in which a torsion pair is a pair of classes $(\mathcal{T}, \mathcal{F})$ of modules such that $\operatorname{Hom}(\mathcal{T}, \mathcal{F})=0$ and for any module $m$ there is a short exact sequence $$ 0 \to T \to M \to F \to 0, $$ where $T \in \mathcal{T}$ and $F \in \mathcal{F}$. In the example with $I$-torsion this short exact sequence is $$ 0 \to \Gamma_I(M) \to M \to M/\Gamma_I(M) \to 0 $$ and there are no module maps from $I$-torsion modules to $I$-torsion-free module.
Is the above correct and we indeed getting a torsion pair for any ideal $I$? The well-known example of a torsion pair (torsion modules, torsion-free modules) is not of this type. Is it possible to describe all torsion pairs for a (noetherian) commutative ring?