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Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $M,N$ be non-zero finitely generated $R$-modules.

Is it known that $M\otimes_R^{\mathbf L} N$ has finite projective dimension if and only if $M$ and $N$ have finite projective dimension?

[Since $M\otimes_R^{\mathbf L} N$ is represented by the chain complex $M \otimes_R F_{\bullet}$ where $F_{\bullet}$ is a resolution of $N$ by finite free modules, so let me recall here that a homologically bounded below chain complex $C$ of finitely generated modules is said to have finite projective dimension if $C $ is isomorphic, in the derived category, to a bounded complex of finitely generated projective modules (free in our case) ]

I would be happy even with an explicit reference.

Thanks

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    $\begingroup$ One direction here is of course clear (namely, if M and N have finite projective dimension, then so does their derived tensor product). Also, you probably want to assume that M,N are both non-zero, otherwise this is of course false. $\endgroup$
    – the L
    May 23, 2021 at 10:55
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    $\begingroup$ Take for example $M=0$ and $N$ some module of infinite projective dimension.. I think you need to add some hypothesis to your question (e.g. $\operatorname{supp}M=\operatorname{Spec}R$) $\endgroup$ May 23, 2021 at 10:56
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    $\begingroup$ @Denis Nardin: I have assumed $M,N$ are both non-zero $\endgroup$
    – strat
    May 23, 2021 at 12:13
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    $\begingroup$ By (A.7.4) and (A.5.7.2) of the book L.W. Christensen, Gorenstein Dimensions; a homologically bounded below complex $X$, with finitely generated homologies, has finite projective dimension if and only if the Poincare series has finite degree. Hence the result you are asking for follows by (A.7.6) of the same book. $\endgroup$
    – uno
    May 23, 2021 at 16:42
  • $\begingroup$ @uno Why not make it an answer? (Preferably also stating that (A.7.6)) $\endgroup$ May 23, 2021 at 19:17

1 Answer 1

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Let me preface this by saying that I don't know a reference - so if that's what you're really looking for, someone else will have to answer.

Let $k:=R/m$ denote the residue field (I'm assuming "commutative" was implicit in your question).

Lemma 1 : Suppose $X$ is a bounded below chain complex of finitely generated $R$-modules such that $X\otimes^L_R k = 0$. Then $X=0$.

This is a simple application of Nakayama's lemma and $H_0(X\otimes^L_R k) = H_0(X)\otimes_R k$ if $X$ is nonnegatively graded.

Proposition 1: Suppose $X$ is a bounded below chain complex of finitely generated $R$-modules. If $X\otimes^L_R k$ is perfect, then so is $X$.

"Perfect" here is what you call "finite homological dimension".

Proof : By induction on the number of nonzero homology groups of $X\otimes^L_Rk$.

The base case, where $X\otimes^L_Rk = 0$, is lemma 1.

Assuming $X$ is nonnegatively graded and the least nonzero homology group is $H_0$, let $k^n \to H_0(X\otimes_R^Lk)\cong H_0(X)\otimes_R k$ be an isomorphism, and lift it to a surjection (by Nakayama's lemma) $R^n\to H_0(X)$, which we can then lift to a morphism $R^n\to X$ in $D(R)$, and take its cofiber $Y$.

So we have a cofiber sequence $R^n\to X\to Y$. $Y$ is also equivalent to a bounded below chain complex of finitely generated $R$-modules and is perfect upon tensoring with $k$.

Furthermore if we tensor this cofiber sequence with $k$, the long exact sequence of homology groups shows that $Y\otimes^L_Rk$ has less nonzero homology groups that $X\otimes^L_Rk$ (it has the same homology groups except it has no $H_0$). By induction, it follows that $Y$ is perfect over $R$, and therefore so is $X$.

From there we can conclude about the non-straightforward direction in your question, namely:

Proposition 2: Suppose $M,N$ are bounded below complexes of finitely generated $R$-modules that aren't acyclic. If $M\otimes_R^L N$ is perfect, then so are $M,N$.

Proof: $-\otimes^L_Rk$ is symmetric monoidal and maps perfect $R$-complexes to perfect $k$-complexes, so by proposition 1 and lemma 1 it suffices to prove the result over $k$.

Over $k$ we have the Künneth formula and formality of chain complexes which make this into a straightforward result.

(note that lemma 1 is necessary here to ensure that $M\otimes_R^Lk$ and $N\otimes^L_Rk$ are nonzero)

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    $\begingroup$ @მამუკაჯიბლაძე The Stacks Project lemma is using cohomological notation (i.e., cochain complexes), but the OP and Z. M. are using homological notation (i.e., chain complexes), so they both involve the same sort of boundedness. $\endgroup$ May 23, 2021 at 21:37
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    $\begingroup$ @Z.M $R$ is local, so every perfect object is isomorphic to a bounded complex of finite free modules, isn’t it? $\endgroup$ May 23, 2021 at 23:55
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    $\begingroup$ @Z.M. : I'm not inducting on the homology groups of $X$, but those of $X\otimes_R^L k$, and there I can assume I have a bijection $\endgroup$ May 24, 2021 at 7:19
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    $\begingroup$ @მამუკაჯიბლაძე I think so, together with the Nakayama lemma. You need to replace Stacks' "bounded above" with bounded below (cohomological vs homological), and then if E is bounded in both directions (corresponding to perfectness, as E is a complex of finite stably free modules), then so must P be (because $P\otimes R/I\cong E$ - an honest isomorphism, if I'm reading Stacks correctly - with Nakayama's lemma). And therefore $P$ is perfect as well (it's a bounded complex of finite stably free modules). And Stacks says that $P$ is quasi-isomorphic to $K$ $\endgroup$ May 24, 2021 at 10:26
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    $\begingroup$ @მამუკაჯიბლაძე Sure, but I don't think this changes the fact that the Stacks lemma is applicable, since it's an existence statement, and a statement about representatives. Namely, if $X\otimes k$ is perfect, then it will have a representative of the form $E$ as in the Stacks lemma, and then you're rolling $\endgroup$ May 24, 2021 at 10:55

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