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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?

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2 Answers 2

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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:

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

  2. 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.

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  • $\begingroup$ What about unbounded case? In topology, if I am not mistaken, given a $p$-complete spectrum without boundedness conditions, if $X/p\simeq0$, then $X\simeq0$. $\endgroup$
    – user20948
    Jun 7, 2020 at 9:11
  • $\begingroup$ @Yai0Phah I've edited the answer above, adding an answer to your question. $\endgroup$ Jun 7, 2020 at 13:49
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    $\begingroup$ It is not necessary to use MGM duality. Say $C \otimes_A^L A/I = 0$ with $C$ derived $I$-complete. Choose generators $x_1,...,x_r \in I$. Then $C \otimes_A^L Kos(A; x_1,...,x_r) = 0$ as the Koszul complex has finitely many homology groups and all of those are $A/I$-modules. Now note that $C \otimes_A^L Kos(A;f) = 0$ iff $f$ acts invertibly on $C$; thus, if $C$ is also derived $f$-complete (i.e., $Rlim_f C = 0$), this forces $C = 0$. Using this observation, proceed by descending induction to show that $C \otimes_A Kos(A;x_1,...,x_i) = 0$ for $i \geq 0$. The $i=0$ case implies the lemma. $\endgroup$
    – Anonymous
    Jun 7, 2020 at 15:43
  • $\begingroup$ @Anonymous Yes, this simplified version of the argument also works. $\endgroup$ Jun 7, 2020 at 18:56
  • $\begingroup$ Thanks a lot to you along with @Anonymous. If I am not mistaken, the Čech complex in question is closely related to the Koszul complex, as the second proof shows. $\endgroup$
    – user20948
    Jun 8, 2020 at 9:29
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There is a version of Nakayama with finiteness of cohomology when the ring is noetherian. See Theorem 2.2 of the paper [PSY2].

taken from the paper [PSY]

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).

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