Derived Nakayama for complete modules I have encountered the following "Nakayama Lemma" recently:

Let $A$ be a ring and $I$ some finitely generated ideal. Let $\mathcal
 C_\bullet$ be a chain complex of $I$-(derived) complete $A$-modules,
  bounded above by zero. Then $\mathcal C$ is acyclic if 
  $C\otimes^{\mathbf L}_A A/I$ is acyclic.

I have tried to prove it first in the case where $C$ consists of classically $I$-complete $A$-modules, but I encountered some problems. Here is what I tried so far:
Suppose, $\mathcal C$ is not acyclic and let $n$ be the greatest integer such that $H_n(C)\neq 0$. We can apply the usual Nakayama Lemma (for separated modules, Theorem 8.4 in Matsumuras Commutative Ring THeory) to the $I$-adic completion $H_n(\mathcal C)^\wedge$ to see that $H_n(\mathcal C)^\wedge \otimes_A A/I\neq 0$. Now I somehow want to use that $H_n(\mathcal C\otimes_A^{\mathbf L} A/I)=H_n(\mathcal C) \otimes_A A/I$ by our choice of $n$ but I have a mental blockage when it comes to relating it to $H_n(\mathcal C)^\wedge \otimes_A A/I$. How do I proceed? How would I proceed in the derived case? Are there any good references to read up on derived completions?
 A: Let $A$ be a commutative ring and $I\subset A$ be a finitely generated ideal.  The basic facts are:


*

*For any complex of derived $I$-complete $A$-modules $C^\bullet$, the cohomology modules $H^*(C^\bullet)$ are derived $I$-complete.

*For any nonzero derived $I$-complete $A$-module $B$, the $A/I$-module $B/IB$ is nonzero.
Your desired assertion follows from 1. and 2. (apply assertion 2. to the highest cohomology module $B$ of your complex $C^\bullet$, taking into account that $B$ is derived complete by assertion 1.)
Concerning the proof of 2., you can proceed by induction in the number of generators $s_1,\dotsc,s_m$ of the ideal $I$: prove that $B\ne0$ implies $B/s_1B\ne0$ implies $B/(s_1B+s_2B)\ne0$ etc.  Here it helps to observe that:


*If $B$ is derived $I$-complete and $J\subset I$, then $B$ is derived $J$-complete.

*The kernel and cokernel of any morphism of derived $I$-complete modules is derived $I$-complete.  (This is just a restatement of 1.)
The observations 3. and 4. reduce your question to proving that $B/sB\ne0$ whenever $B\ne0$ is a derived $(s)$-complete $A$-module, for the principal ideal $(s)$ generated by a single element $s\in A$.  
At this point you recall that $B$ being derived $(s)$-complete means that $Hom_A(A[s^{-1}],B)=0=Ext^1_A(A[s^{-1}],B)$.  It remains to check that the equation $B=sB$ implies surjectivity of the natural map $Hom_A(A[s^{-1}],B)\to B$, which is a straightforward exercise.  So $Hom_A(A[s^{-1}],B)=0$ together with $sB=B$ imply $B=0$.
References:


*

*Bhatt's short paper https://doi.org/10.1016/j.jpaa.2018.08.008 .

*My long paper https://arxiv.org/abs/1605.03934 (in which "derived complete modules" are called "contramodules").

EDIT: A question was asked in the comments whether the assertion remains true for unbounded complexes.  The answer is positive, but the proof below is more complicated.
Let $s_1,\dotsc,s_m$ be a finite set of generators of the ideal $I\subset A$. Denote the sequence $s_1,\dotsc,s_m$ by the single letter $\mathbf s$ for brevity. The augmented Cech complex $\check C(A;\mathbf s)$ is defined as the tensor product
$$
 \check C(A;\mathbf s)=(A\to A[s_1^{-1}])\otimes_A\dotsb
 \otimes_A(A\to A[s_m^{-1}]).
$$
Consider the unbounded derived category of $A$-modules $D(A{-}Mod)$, and denote by $D_{I-tors}(A{-}Mod)\subset D(A{-}Mod)$ the full triangulated subcategory of complexes of $A$-modules with $I$-torsion cohomology modules.  Denote by $D_{I-com}(A{-}Mod)\subset D(A{-}Mod)$ the full triangulated subcategory of complexes of $A$-modules with derived $I$-adically complete cohomology modules.  The full subcategory $D_{I-com}(A{-}Mod)$ coincides with what is usually called the full subcategory of derived $I$-adically complete complexes in $D(A{-}Mod)$.
The key fact for the argument below is that there is an equivalence of categories
$$
 D_{I-tors}(A{-}Mod)\simeq D_{I-com}(A{-}Mod)
$$
provided by the functor of tensor product with the augmented Cech complex
$$
 \check C(A;\mathbf s)\otimes_A{-}\colon D_{I-com}(A{-}Mod)
 \longrightarrow D_{I-tors}(A{-}Mod),
$$
whose quasi-inverse is the derived Hom functor
$$
 \mathbf R Hom_A(\check C(A;\mathbf s),{-})\colon D_{I-tors}(A{-}Mod)
 \longrightarrow D_{I-com}(A{-}Mod).
$$
This result is known as the Matlis-Greenlees-May (MGM) duality or Matlis-Greenlees-May equivalence.
Specifically, it is important for us that, for any complex of $A$-modules $B^\bullet$ with derived $I$-adically complete cohomology modules (or in other words, for any derived $I$-adically complete complex $B^\bullet$), the complex $\check C(A;\mathbf s)\otimes_A B^\bullet$ is not acyclic if the complex $B^\bullet$ is not acyclic.
Let us prove that the complex $B^\bullet$ is acyclic if the complex $A/I\otimes_A^{\mathbf L}B^\bullet$ is.  First of all, we can replace $B^\bullet$ by its homotopy flat ("K-flat") resolution.  This allows us to assume that $B^\bullet$ is a homotopy flat complex (to simplify matters, one can additionally assume $B^\bullet$ to be a complex of flat $A$-modules).
Then we have $A/I\otimes_A^{\mathbb L} B^\bullet=A/I\otimes_A B^\bullet$, that is the derived tensor product coincides with the underived one.  Assume that the complex $A/I\otimes_A B^\bullet$ is acyclic.
Since $B^\bullet$ is a homotopy flat complex of $A$-modules, $A/I\otimes_A B^\bullet$ is also a homotopy flat complex of $A/I$-modules.  It follows that the complex $N\otimes_A B^\bullet = N\otimes_{A/I}(A/I\otimes_A B^\bullet)$ is acyclic for any $A/I$-module $N$.  Hence the complex $N\otimes_A B^\bullet$ is also acyclic for any $A/I^k$-module $N$, where $k\ge1$ is any integer.  Passing to the direct limit, we can conclude that the complex $N\otimes_A B^\bullet$ is acyclic for any $I$-torsion $A$-module $N$.
Finally, it follows that the complex $M^\bullet\otimes_A B^\bullet$ is acyclic for any finite complex of $A$-modules $M^\bullet$ with $I$-torsion cohomology modules.  For example, $M^\bullet = \check C(A;\mathbf s)$ is such a complex.  We have shown that the complex $\check C(A;\mathbf s)\otimes_A B^\bullet$ is acyclic.  According to the above MGM equivalence theorem, it follows that the complex $B^\bullet$ is acyclic.
References to MGM duality/equivalence:


*

*Dwyer, Greenlees "Complete modules and torsion modules", https://doi.org/10.1353/ajm.2002.0001

*My paper "Dedualizing complexes and MGM duality", https://arxiv.org/abs/1503.05523 , https://doi.org/10.1016/j.jpaa.2016.05.019 , Section 3.
A: There is a version of Nakayama with finiteness of cohomology when the ring is noetherian. See Theorem 2.2 of the paper [PSY2].

For MGM Equivalance see [PSY1].
References:
[PSY1]: M. Porta, L. Shaul and A. Yekutieli, On the Homology of Completion and Torsion, Algebras and Repesentation Theory 17 (2014), 31–67.
http://dx.doi.org/10.1007/s10468-012-9385-8.
Erratum to: On the Homology of Completion and Torsion
Algebras and Representation Theory: Volume 18, Issue 5 (2015), Page 1401-1405
http://dx.doi.org/10.1007/s10468-015-9557-4
[PSY2]: M. Porta, L. Shaul and A. Yekutieli, Cohomologically Cofinite Complexes, Comm. Algebra 43 2015, 597–615,
(https://www.tandfonline.com/doi/full/10.1080/00927872.2013.822506).
