Lemma 110.11.1. 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.

## 110.11 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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)