## 15.94 Derived completion for Noetherian rings

Let $A$ be a ring and let $I \subset A$ be an ideal. For any $K \in D(A)$ we can consider the derived limit

$K' = R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} A/I^ n)$

This is a functor in $K$, see Remark 15.87.10. The system of maps $A \to A/I^ n$ induces a map $K \to K'$ and $K'$ is derived complete with respect to $I$ (Lemma 15.91.14). This “naive” derived completion construction does not agree with the adjoint of Lemma 15.91.10 in general. For example, if $A = \mathbf{Z}_ p \oplus \mathbf{Q}_ p/\mathbf{Z}_ p$ with the second summand an ideal of square zero, $K = A[0]$, and $I = (p)$, then the naive derived completion gives $\mathbf{Z}_ p[0]$, but the construction of Lemma 15.91.10 gives $K^\wedge \cong \mathbf{Z}_ p[1] \oplus \mathbf{Z}_ p[0]$ (computation omitted). Lemma 15.93.2 characterizes when the two functors agree in the case $I$ is generated by a single element.

The main goal of this section is the show that the naive derived completion is equal to derived completion if $A$ is Noetherian.

Lemma 15.94.1. In Situation 15.91.15. If $A$ is Noetherian, then the pro-objects $\{ K_ n^\bullet \}$ and $\{ A/(f_1^ n, \ldots , f_ r^ n)\}$ of $D(A)$ are isomorphic1.

Proof. We have an inverse system of distinguished triangles

$\tau _{\leq -1}K_ n^\bullet \to K_ n^\bullet \to A/(f_1^ m, \ldots , f_ r^ m) \to (\tau _{\leq -1}K_ n^\bullet )[1]$

See Derived Categories, Remark 13.12.4. By Derived Categories, Lemma 13.42.4 it suffices to show that the inverse system $\tau _{\leq -1}K_ n^\bullet$ is pro-zero. Recall that $K_ n^\bullet$ has nonzero terms only in degrees $i$ with $-r \leq i \leq 0$. Thus by Derived Categories, Lemma 13.42.3 it suffices to show that $H^ p(K_ n^\bullet )$ is pro-zero for $p \leq -1$. In other words, for every $n \in \mathbf{N}$ we have to show there exists an $m \geq n$ such that $H^ p(K_ m^\bullet ) \to H^ p(K_ n^\bullet )$ is zero. Since $A$ is Noetherian, we see that

$H^ p(K_ n^\bullet ) = \frac{\mathop{\mathrm{Ker}}(K_ n^ p \to K_ n^{p + 1})}{\mathop{\mathrm{Im}}(K_ n^{p - 1} \to K_ n^ p)}$

is a finite $A$-module. Moreover, the map $K_ m^ p \to K_ n^ p$ is given by a diagonal matrix whose entries are in the ideal $(f_1^{m - n}, \ldots , f_ r^{m - n})$ as $p < 0$. Note that $H^ p(K_ n^\bullet )$ is annihilated by $J = (f_1^ n, \ldots , f_ r^ n)$, see Lemma 15.28.6. Now $(f_1^{m - n}, \ldots , f_ r^{m - n}) \subset J^ t$ for $m - n \geq tn$. Thus by Algebra, Lemma 10.51.2 (Artin-Rees) applied to the ideal $J$ and the module $M = K_ n^ p$ with submodule $N = \mathop{\mathrm{Ker}}(K_ n^ p \to K_ n^{p + 1})$ for $m$ large enough the image of $K_ m^ p \to K_ n^ p$ intersected with $\mathop{\mathrm{Ker}}(K_ n^ p \to K_ n^{p + 1})$ is contained in $J \mathop{\mathrm{Ker}}(K_ n^ p \to K_ n^{p + 1})$. For such $m$ we get the zero map. $\square$

Proposition 15.94.2. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. The functor which sends $K \in D(A)$ to the derived limit $K' = R\mathop{\mathrm{lim}}\nolimits ( K \otimes _ A^\mathbf {L} A/I^ n )$ is the left adjoint to the inclusion functor $D_{comp}(A) \to D(A)$ constructed in Lemma 15.91.10.

Proof. Say $(f_1, \ldots , f_ r) = I$ and let $K_ n^\bullet$ be the Koszul complex with respect to $f_1^ n, \ldots , f_ r^ n$. By Lemma 15.91.18 it suffices to prove that

$R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} K_ n^\bullet ) = R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} A/(f_1^ n, \ldots , f_ r^ n) ) = R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} A/I^ n ).$

By Lemma 15.94.1 the pro-objects $\{ K_ n^\bullet \}$ and $\{ A/(f_1^ n, \ldots , f_ r^ n)\}$ of $D(A)$ are isomorphic. It is clear that the pro-objects $\{ A/(f_1^ n, \ldots , f_ r^ n)\}$ and $\{ A/I^ n\}$ are isomorphic. Thus the map from left to right is an isomorphism by Lemma 15.87.12. $\square$

Lemma 15.94.3. Let $I$ be an ideal of a Noetherian ring $A$. Let $M$ be an $A$-module with derived completion $M^\wedge$. Then there are short exact sequences

$0 \to R^1\mathop{\mathrm{lim}}\nolimits \text{Tor}_{i + 1}^ A(M, A/I^ n) \to H^{-i}(M^\wedge ) \to \mathop{\mathrm{lim}}\nolimits \text{Tor}_ i^ A(M, A/I^ n) \to 0$

A similar result holds for $M \in D^-(A)$.

Proof. Immediate consequence of Proposition 15.94.2 and Lemma 15.87.4. $\square$

As an application of the proposition above we identify the derived completion in the Noetherian case for pseudo-coherent complexes.

Lemma 15.94.4. Let $A$ be a Noetherian ring and $I \subset A$ an ideal. Let $K$ be an object of $D(A)$ such that $H^ n(K)$ a finite $A$-module for all $n \in \mathbf{Z}$. Then the cohomology modules $H^ n(K^\wedge )$ of the derived completion are the $I$-adic completions of the cohomology modules $H^ n(K)$.

Proof. The complex $\tau _{\leq m}K$ is pseudo-coherent for all $m$ by Lemma 15.64.17. Thus $\tau _{\leq m}K$ is represented by a bounded above complex $P^\bullet$ of finite free $A$-modules. Then $\tau _{\leq m}K \otimes _ A^\mathbf {L} A/I^ n = P^\bullet /I^ nP^\bullet$. Hence $(\tau _{\leq m}K)^\wedge = R\mathop{\mathrm{lim}}\nolimits P^\bullet /I^ nP^\bullet$ (Proposition 15.94.2) and since the $R\mathop{\mathrm{lim}}\nolimits$ is just given by termwise $\mathop{\mathrm{lim}}\nolimits$ (Lemma 15.87.1) and since $I$-adic completion is an exact functor on finite $A$-modules (Algebra, Lemma 10.97.2) we conclude the result holds for $\tau _{\leq m}K$. Hence the result holds for $K$ as derived completion has finite cohomological dimension, see Lemma 15.91.20. $\square$

Lemma 15.94.5. Let $I$ be an ideal of a Noetherian ring $A$. Let $M$ be a derived complete $A$-module. If $M/IM$ is a finite $A/I$-module, then $M = \mathop{\mathrm{lim}}\nolimits M/I^ nM$ and $M$ is a finite $A^\wedge$-module.

Proof. Assume $M/IM$ is finite. Pick $x_1, \ldots , x_ t \in M$ which map to generators of $M/IM$. We obtain a map $A^{\oplus t} \to M$ mapping the $i$th basis vector to $x_ i$. By Proposition 15.94.2 the derived completion of $A$ is $A^\wedge = \mathop{\mathrm{lim}}\nolimits A/I^ n$. As $M$ is derived complete, we see that our map factors through a map $q : (A^\wedge )^{\oplus t} \to M$. The module $\mathop{\mathrm{Coker}}(q)$ is zero by Lemma 15.91.7. Thus $M$ is a finite $A^\wedge$-module. Since $A^\wedge$ is Noetherian and complete with respect to $IA^\wedge$, it follows that $M$ is $I$-adically complete (use Algebra, Lemmas 10.97.5, 10.96.11, and 10.51.2). $\square$

Lemma 15.94.6. Let $I$ be an ideal in a Noetherian ring $A$.

1. If $M$ is a finite $A$-module and $N$ is a flat $A$-module, then the derived $I$-adic completion of $M \otimes _ A N$ is the usual $I$-adic completion of $M \otimes _ A N$.

2. If $M$ is a finite $A$-module and $f \in A$, then the derived $I$-adic completion of $M_ f$ is the usual $I$-adic completion of $M_ f$.

Proof. For an $A$-module $M$ denote $M^\wedge$ the derived completion and $\mathop{\mathrm{lim}}\nolimits M/I^ nM$ the usual completion. Assume $M$ is finite. The system $\text{Tor}^ A_ i(M, A/I^ n)$ is pro-zero for $i > 0$, see Lemma 15.27.3. Since $\text{Tor}_ i^ A(M \otimes _ A N, A/I^ n) = \text{Tor}_ i^ A(M, A/I^ n) \otimes _ A N$ as $N$ is flat, the same is true for the system $\text{Tor}^ A_ i(M \otimes _ A N, A/I^ n)$. By Lemma 15.94.3 we conclude $R\mathop{\mathrm{lim}}\nolimits (M \otimes _ A N) \otimes _ A^\mathbf {L} A/I^ n$ only has cohomology in degree $0$ given by the usual completion $\mathop{\mathrm{lim}}\nolimits M \otimes _ A N/ I^ n(M \otimes _ A N)$. This proves (1). Part (2) follows from (1) and the fact that $M_ f = M \otimes _ A A_ f$. $\square$

Lemma 15.94.7. Let $I$ be an ideal in a Noetherian ring $A$. Let ${}^\wedge$ denote derived completion with respect to $I$. Let $K \in D^-(A)$.

1. If $M$ is a finite $A$-module, then $(K \otimes _ A^\mathbf {L} M)^\wedge = K^\wedge \otimes _ A^\mathbf {L} M$.

2. If $L \in D(A)$ is pseudo-coherent, then $(K \otimes _ A^\mathbf {L} L)^\wedge = K^\wedge \otimes _ A^\mathbf {L} L$.

Proof. Let $L$ be as in (2). We may represent $K$ by a bounded above complex $P^\bullet$ of free $A$-modules. We may represent $L$ by a bounded above complex $F^\bullet$ of finite free $A$-modules. Since $\text{Tot}(P^\bullet \otimes _ A F^\bullet )$ represents $K \otimes _ A^\mathbf {L} L$ we see that $(K \otimes _ A^\mathbf {L} L)^\wedge$ is represented by

$\text{Tot}((P^\bullet )^\wedge \otimes _ A F^\bullet )$

where $(P^\bullet )^\wedge$ is the complex whose terms are the usual $=$ derived completions $(P^ n)^\wedge$, see for example Proposition 15.94.2 and Lemma 15.94.6. This proves (2). Part (1) is a special case of (2). $\square$

[1] In particular, for every $n$ there exists an $m \geq n$ such that $K_ m^\bullet \to K_ n^\bullet$ factors through the map $K_ m^\bullet \to A/(f_1^ m, \ldots , f_ r^ m)$.

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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BKH. Beware of the difference between the letter 'O' and the digit '0'.