The Stacks project

15.92 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 15.86.10. 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 15.90.14). This “naive” derived completion construction does not agree with the adjoint of Lemma 15.90.10 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 15.90.10 gives $K^\wedge \cong \mathbf{Z}_ p[1] \oplus \mathbf{Z}_ p[0]$ (computation omitted). Lemma 15.91.2 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 15.92.1. In Situation 15.90.15. 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 13.41.4 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 13.41.3 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 $J^ t \subset (f_1^{m - n}, \ldots , f_ r^{m - n})$ for $m = n + tr$. Thus by Artin-Rees (Algebra, Lemma 10.51.2) for some $m$ large enough we see that 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 this $m$ we get the zero map. $\square$

Proposition 15.92.2. 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 15.90.10.

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 15.90.18 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 15.92.1 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 15.86.12. $\square$

Lemma 15.92.3. 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 15.92.2 and Lemma 15.86.4. $\square$

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

Lemma 15.92.4. 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 15.63.17. 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 15.92.2) and since the $R\mathop{\mathrm{lim}}\nolimits $ is just given by termwise $\mathop{\mathrm{lim}}\nolimits $ (Lemma 15.86.1) 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 15.90.20. $\square$

Lemma 15.92.5. 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 15.92.2 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 15.90.7. 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 15.92.6. 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 15.92.3 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 15.92.7. 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 15.92.2 and Lemma 15.92.6. 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 $K_ m^\bullet \to A/(f_1^ m, \ldots , f_ r^ m)$.

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