The Stacks project

Proof. We have an inverse system of distinguished triangles

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

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. $\square$

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

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. $\square$

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

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. $\square$

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). $\square$

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 $M_ f = M \otimes _ A A_ f$. $\square$

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

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). $\square$