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Let $R$ be a commutative ring with unity and let $I$ be a proper ideal. (I'm not assuming $R$ is Noetherian.) For every $M \in R$-Mod, let $\Gamma_I(M):=\{m \in M : I^n m=0$ for some $n\ge 1\}$.

If $f \in \mathrm{Hom}_R (M,N)$, it can be seen that $f(\Gamma_I(M)) \subseteq \Gamma_I(N)$, giving us a map $\Gamma_I (f):=f|_{\Gamma_I(M)} \in \mathrm{Hom}_R (\Gamma_I(M) , \Gamma_I(N))$. Thus we have a co-variant functor $\Gamma_I : R$-Mod $\to R$-Mod. It can be shown that $\Gamma_I$ is additive, and left exact. So we can consider its right derived functors $R^i\Gamma_I$.

How to show that $R^i\Gamma_I$ is naturally isomorphic to $\varinjlim \mathrm{Ext}^i_R(R/I^n,-)$ ?

[Note that $R^i \mathrm{Hom}_R(R/I^n,-)$ is $\mathrm{Ext}^i_R(R/I^n,-)$]

Relevant related questions:

On the natural isomorphism between $I$-torsion functor and direct limit of $\mathrm{Hom}$ functor

and
Direct limit of directed system of modules commutes with right derived functors of additive, covariant, left exact functor?

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  • $\begingroup$ The first relevant related question gives you a natural isomorphism between $\Gamma_I$ and something in terms of $\hom$s. Are you aware that if $F$ is exact and $G$ is left exact, then $R^i(F\circ G) = F\circ R^iG$ ? $\endgroup$ Commented Mar 15, 2019 at 7:51
  • $\begingroup$ @Max: Wow ... I am not aware of the fact you mention ... that would right away give the claim in question ... could you maybe give a reference or post a proof as an answer please ? thanks .. $\endgroup$
    – user102248
    Commented Mar 15, 2019 at 7:54

3 Answers 3

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Let $G:A\to B, F:B\to C$ be additive functors between abelian categories and assume $G$ is left exact, $F$ is exact, and $A$ has anough injectives so we may define $R^iG$ and $R^i(F\circ G)$.

Then $R^i(F\circ G)$ is naturally isomorphic to $F\circ R^iG$ for all $i$, and as you pointed out in the comments, from this and one of the links you posted + the fact that directed colimits are exact in categories of $R$-modules, the claim immediately follows.

But this natural isomorphism is really easy : it simply comes from the fact that $F$ being exact implies that it commutes with cohomology.

Indeed let $x\in A$ and let $x\to I^\bullet$ be an injective resolution. Then $R^iG(x) = H^i(G(I^\bullet))$ by definition, and $R^i(F\circ G)(x) = H^i(F\circ G(I^\bullet))$.

$F$ being exact, if $C^\bullet$ is any cochain complex, $H^i(F(C^\bullet)) = F(H^i(C^\bullet))$. With $C^\bullet = G(I^\bullet)$ you get $R^i(F\circ G)(x) = FH^i( G(I^\bullet))=FR^iG(x)$ (those equals should really be canonical isomorphisms), hence $R^i(F\circ G) = F\circ R^iG$

Note : this may already have appeared somewhere on the site so I don't know if it's a duplicate, I don't have time to look it up right now

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@Maxime Ramzi's answer is great. Here we can also prove the following main claim by spectral sequence:

Claim: Let $G:A\to B, F:B\to C$ be additive functors between abelian categories and assume $G$ is left exact, $F$ is exact, and $A$ has anough injectives so we may define $R^iG$ and $R^i(F\circ G)$. Then $R^i(F\circ G)$ is naturally isomorphic to $F\circ R^iG$ for all $i > 1$.

Proof: Since $F$ is exact, $R^iF$ are zero for all $i>0$. Hence clearly $G$ sends injective objects of $A$ to $F$-acyclic objects of $B$. Therefore, we may apply the Grothendieck spectral sequence $$ E_2^{p,q} = R^pF(R^qG(A)) \Rightarrow R^{p+q}(F \circ G)(A). $$ As $F$ is exact, the above spectral sequence collapses on the $q$-axis on page 2. Hence $$ R^n(F \circ G)(A) = F(R^n G(A)) \quad (\star) $$ for all $n \geq 1$. The Grothendieck spectral sequence is natural in $A$, which eventually gives the natrality in $A$ of $(\star)$.

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Let me add another answer on how to prove the claim in my post on April 25th, 2022.

Actually we can use the machinery of derived categories. This gives the isomorphisms $$ R(F \circ G) \xrightarrow{\sim} R(F) \circ R(G) \xrightarrow{\sim} F \circ R(G), $$ where the second isomorphism follows from the exactness of $F$. This is the full information we have for the composition of functors. Now taking the cohomology gives $$ R^i(F \circ G) \xrightarrow{\sim} H^i(F \circ R(G)) \xrightarrow{\sim} F \circ R^i G, $$ where the second ismorphism follows from that exact functors commute with cohomology. This is a "section" of information of the derived functors.

Comparing to the previous post I gave in April when I haven't learnt the stuff of derived category, it seems that

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  • $\begingroup$ And one may try to use this argument to prove the Shapiro's lemma (both the cohomology version and the homology version) in the theory of group (co)homology. See Rotman's An Introduction to Homological Algebra, especially in Chapter 10 there, for a proof of the Shapiro's lemma using spectral sequences. Compare that approach with this "derived category" approach. $\endgroup$
    – Hetong Xu
    Commented Aug 22, 2022 at 2:15

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