Lemma 109.11.1. Let $A$ be a ring and let $I \subset A$ be an ideal. The category $\mathcal{C}$ of derived complete modules is abelian and the inclusion functor $F : \mathcal{C} \to \text{Mod}_ A$ is exact and commutes with arbitrary limits. If $I$ is finitely generated, then $\mathcal{C}$ has arbitrary direct sums and colimits, but $F$ does not commute with these in general. Finally, filtered colimits are not exact in $\mathcal{C}$ in general, hence $\mathcal{C}$ is not a Grothendieck abelian category.

## 109.11 The category of derived complete modules

Please read More on Algebra, Section 15.92 before reading this section.

Let $A$ be a ring, let $I$ be an ideal of $A$, and denote $\mathcal{C}$ the category of derived complete modules as defined in More on Algebra, Definition 15.91.4.

Let $T$ be a set and let $M_ t$, $t \in T$ be a family of derived complete modules. We claim that in general $\bigoplus M_ t$ is not a derived complete module. For a specific example, let $A = \mathbf{Z}_ p$ and $I = (p)$ and consider $\bigoplus _{n \in \mathbf{N}} \mathbf{Z}_ p$. The map from $\bigoplus _{n \in \mathbf{N}} \mathbf{Z}_ p$ to its $p$-adic completion isn't surjective. This means that $\bigoplus _{n \in \mathbf{N}} \mathbf{Z}_ p$ cannot be derived complete as this would imply otherwise, see More on Algebra, Lemma 15.91.3. Hence the inclusion functor $\mathcal{C} \to \text{Mod}_ A$ does not commute with either direct sums or (filtered) colimits.

Assume $I$ is finitely generated. By the discussion in More on Algebra, Section 15.92 the category $\mathcal{C}$ has arbitrary colimits. However, we claim that filtered colimits are not exact in the category $\mathcal{C}$. Namely, suppose that $A = \mathbf{Z}_ p$ and $I = (p)$. One has inclusions $f_ n : \mathbf{Z}_ p/p\mathbf{Z}_ p \to \mathbf{Z}_ p/p^ n\mathbf{Z}_ p$ of $p$-adically complete $A$-modules given by multiplication by $p^{n - 1}$. There are commutative diagrams

We claim: the colimit of these inclusions in the category $\mathcal{C}$ gives the map $\mathbf{Z}_ p/p\mathbf{Z}_ p \to 0$. Namely, the colimit in $\text{Mod}_ A$ of the system on the right is $\mathbf{Q}_ p/\mathbf{Z}_ p$. Thus the colimit in $\mathcal{C}$ is

by More on Algebra, Section 15.92 where ${}^\wedge $ is derived completion. This proves our claim.

**Proof.**
See More on Algebra, Lemma 15.92.1 and discussion above.
$\square$

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