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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 for it.

What conditions can be posed on ring (or ideals), to have $\Gamma_I(R)=\Gamma_{\sqrt I}(R),$ for every ideal of $R$?

Thank you.

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This problem is studied in this article (see here 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.

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  • $\begingroup$ Thank you @Fred. Does the article suggest conditions that the equality holds for every ideal of $R$? $\endgroup$
    – user 1
    Commented Jun 16, 2015 at 7:42
  • $\begingroup$ No, not beside the obvious ones (cf. Corollary 1.3 and the following remark). $\endgroup$
    – Fred Rohrer
    Commented Jun 16, 2015 at 12:04

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