Let $R$ be commutative ring with identity, $M$ an $R$-module, and $I$ an ideal of $R$ . One defines $I$-torsion functor $Γ\_I$ as: $\Gamma\_I(M)=\bigcup\_{n\in N} (0:\_MI^n).$ When $R$ is Noetherian, it's known that $$\color{brown}{\Gamma\_I(M)=\Gamma\_{\sqrt I}(M)}.$$ 
When $R$ is non-Noetherian, there is a [counterexample](https://math.stackexchange.com/questions/1315574/local-cohomology-and-radical-of-ideal) for it. 



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> What conditions can be posed on ring (or ideals), to have $\Gamma\_I(R)=\Gamma\_{\sqrt I}(R),$ for **every** ideal of $R$? 
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Thank you.



local cohomology and radical of ideal
This problem is studied in [this article](http://dx.doi.org/10.1007/s40306-013-0020-5) (see [here](http://arxiv.org/pdf/1207.0610.pdf) for the arxiv version), mainly in case the supporting ideal is a monomial ideal in a polynomial ring.


**ADDENDUM (2017):** A completely different condition that ensures $\Gamma\_I=\Gamma\_{\sqrt{I}}$ for every ideal $I$ is that $R$ is absolutely flat.


