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?



Derived Nakayama for complete modules
Let $A$ be a commutative ring and $I\subset A$ be a finitely generated ideal. The basic facts are:


1. For any complex of derived $I$-complete $A$-modules $C^\bullet$, the cohomology modules $H^\*(C^\bullet)$ are derived $I$-complete.
2. 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:


3. If $B$ is derived $I$-complete and $J\subset I$, then $B$ is derived $J$-complete.
4. 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:


1. Bhatt's short paper <https://doi.org/10.1016/j.jpaa.2018.08.008> .
2. 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:


1. Dwyer, Greenlees "Complete modules and torsion modules", <https://doi.org/10.1353/ajm.2002.0001>
2. 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.


