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Edit: According to the comments of abx and Yemon Choi I revise the question as follows:

Let $G$ be a group and $\mathcal{A_G}$ be the category of $G$-module commutative algebras, that is the category of commutative algebras which are acted by $G$ and the action preserves the algebra structures.

We define the functor $Nil_G:\mathcal{A_G}\to \mathcal{A_G} $ as follows:

$$Nil_G (A)=\{a\in A \mid g.a-a\; \text{is a nilpotent elementof $A$},\;\forall g \in G\} $$

This functor can be defined on the category of commutative rings or comuutative Banach algebras. For the later case we consider $Nil_G(A):= \overline{Nil_G(A)}$.

A motivation for construction of such functor comes from the classical "Group Cohomology". In the group Cohomology we start with the functor which maps every $G$-module $M$ to $\{m| gm=m\}$ and the corresponding derived functor of this functor generats "group cohomology". On the other hand $g.m=m$ is equivalent to $g.m-m \in \{0\}$. Now we replace $\{0\}$ by the ideal of nilpotent elements or the torsion subgroup. in the group case.

Is this functor identical to a well known functor in the corresponding categories? Does it posses some non trivial properties? In particular, is it always a full functor? If not, for which kind of groups $G$ this functor is a full functor?In this way, can we get a non trivial invariant in group theory?

The Group Cohomology Version

If we consider the same construction on the category of Abelian groups, we obtain a left exact functor $Nil:\mathcal{Gr}_G \to \mathcal{Gr}$ with $$Nil_G (H)=\{a\in H \mid g.a-a\; \text{is a torsion element of $H$},\;\forall g \in G\} $$

Here $\mathcal{Gr}_G$ and $\mathcal{Gr}$ are category of $G$-modules and abelian group, respectively.

Does the corresponding derived functor gives us a kind of useful cohomology theory? What would be the precise formulation of corresponding chain complex and coboundary maps?

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    $\begingroup$ Are you aware that your categories are not abelian? $\endgroup$ – abx Apr 12 '18 at 20:22
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    $\begingroup$ You have to be careful with what you mean by "left exact" in categories like that of commutative Banach algebras. This is not just because the category is non-additive, but because there can be morphisms which are monic and epic yet are not isomorphisms $\endgroup$ – Yemon Choi Apr 12 '18 at 20:30
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    $\begingroup$ @abx I am sorry that I did not pay attention to this fact. $\endgroup$ – Ali Taghavi Apr 12 '18 at 20:37
  • $\begingroup$ And I thank you and @YemonChoi for your comments. $\endgroup$ – Ali Taghavi Apr 12 '18 at 20:37
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    $\begingroup$ About the last question: what is a "nilpotent element" in an abelian group? $\endgroup$ – Laurent Moret-Bailly Apr 20 '18 at 12:03

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