Lemma 15.92.1. Let A be a ring and let I \subset A be an ideal. The category \mathcal{C} of derived complete modules is abelian, has arbitrary limits, and the inclusion functor F : \mathcal{C} \to \text{Mod}_ A is exact and commutes with limits. If I is finitely generated, then \mathcal{C} has arbitrary colimits and F has a left adjoint
15.92 The category of derived complete modules
Let A be a ring and let I be an ideal. Denote \mathcal{C} the category of derived complete modules, see Definition 15.91.4. In this section we discuss some properties of this category. In Examples, Section 110.11 we show that \mathcal{C} isn't a Grothendieck abelian category in general.
By Lemma 15.91.6 the category \mathcal{C} is abelian and the inclusion functor \mathcal{C} \to \text{Mod}_ A is exact.
Since D_{comp}(A) \subset D(A) is closed under products (see discussion following Definition 15.91.4) and since products in D(A) are computed on the level of complexes, we see that \mathcal{C} has products which agree with products in \text{Mod}_ A. Thus \mathcal{C} in fact has arbitrary limits and the inclusion functor \mathcal{C} \to \text{Mod}_ A commutes with them, see Categories, Lemma 4.14.11.
Assume I is finitely generated. Let {}^\wedge : D(A) \to D(A) denote the derived completion functor of Lemma 15.91.10. Let us show the functor
is a left adjoint to the inclusion functor \mathcal{C} \to \text{Mod}_ A. Note that H^ i(M^\wedge ) = 0 for i > 0 for example by Lemma 15.91.20. Hence, if N is a derived complete A-module, then we have
as desired.
Let T be a preordered set and let t \mapsto M_ t be a system of derived complete A-modules, i.e., a system over T in \mathcal{C}, see Categories, Section 4.21. Denote \mathop{\mathrm{colim}}\nolimits _{t \in T} M_ t the colimit of the system in \text{Mod}_ A. It follows formally from the above that
is the colimit of the system in \mathcal{C}. In this way we see that \mathcal{C} has all colimits. In general the inclusion functor \mathcal{C} \to \text{Mod}_ A will not commute with colimits, see Examples, Section 110.11.
Proof. This summarizes the discussion above. \square
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