108.11 The category of derived complete modules

Let $A$ be a ring and let $I$ be an ideal. Consider the category $\mathcal{C}$ of derived complete modules as defined in More on Algebra, Definition 15.86.4. By More on Algebra, Lemma 15.86.6 we see that $\mathcal{C}$ is abelian.

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 complete module. For a specific example, let $A = \mathbf{Z}_ p$ and $I = (p)$ and $\bigoplus _{n \in \mathbf{N}} \mathbf{Z}_ p$. 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 15.86.3.

Assume $I$ is finitely generated. Let ${}^\wedge : D(A) \to D(A)$ denote the derived completion functor, see More on Algebra, Lemma 15.86.10. We claim that

$M = H^0((\bigoplus M_ t)^\wedge ) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$

is a direct sum of $M_ t$ in the category $\mathcal{C}$. Note that for $E$ a derived complete object of $D(A)$ we have

$\mathop{\mathrm{Hom}}\nolimits _{D(A)}((\bigoplus M_ t)^\wedge , E) = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(\bigoplus M_ t, E) = \prod \mathop{\mathrm{Hom}}\nolimits _{D(A)}(M_ t, E)$

Note that the right hand side is zero if $H^ i(E) = 0$ for $i < 1$. In particular, applying this with $E = \tau _{\geq 1} (\bigoplus M_ t)^\wedge$ which is derived complete by More on Algebra, Lemma 15.86.6 we see that the canonical map $(\bigoplus M_ t)^\wedge \to \tau _{\geq 1}(\bigoplus M_ t)^\wedge$ is zero, in other words, we have $H^ i((\bigoplus M_ t)^\wedge ) = 0$ for $i \geq 1$. Then, for an object $N \in \mathcal{C}$ we see that

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _\mathcal {C}(M, N) & = \mathop{\mathrm{Hom}}\nolimits _{D(A)}((\bigoplus M_ t)^\wedge , N)\\ & = \prod \mathop{\mathrm{Hom}}\nolimits _ A(M_ t, N) \\ & = \prod \mathop{\mathrm{Hom}}\nolimits _\mathcal {C}(M_ t, N) \end{align*}

as desired. This implies that $\mathcal{C}$ has all colimits, see Categories, Lemma 4.14.12. In fact, arguing similarly as above we see that given a system $M_ t$ in $\mathcal{C}$ over a preordered set $T$ the colimit in $\mathcal{C}$ is equal to $H^0((\mathop{\mathrm{colim}}\nolimits M_ t)^\wedge )$ where the inner colimit is the colimit in the category of $A$-modules.

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

$\xymatrix{ \mathbf{Z}_ p/p\mathbf{Z}_ p \ar[r]_{f_ n} \ar[d]^1 & \mathbf{Z}_ p/p^ n\mathbf{Z}_ p \ar[d]_ p \\ \mathbf{Z}_ p/p\mathbf{Z}_ p \ar[r]^{f_{n + 1}} & \mathbf{Z}_ p/p^{n + 1}\mathbf{Z}_ p }$

Now take the colimit of these inclusions in the category $\mathcal{C}$ derived to get $\mathbf{Z}_ p/p\mathbf{Z}_ p \to 0$. Namely, the colimit in $\text{Mod}_ A$ of the system on the right is $\mathbf{Q}_ p/\mathbf{Z}_ p$. The reader can directly compute that $(\mathbf{Q}_ p/\mathbf{Z}_ p)^\wedge = \mathbf{Z}_ p[1]$ in $D(A)$. Thus $H^0 = 0$ which proves our claim.

Lemma 108.11.1. 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.

Proof. See discussion above. $\square$

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