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



derived tensor product and finite projective dimension
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).



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> **Lemma 1** : Suppose $X$ is a bounded below chain complex of finitely generated $R$-modules such that $X\otimes^L\_R k = 0$. Then $X=0$.
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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 $X$.
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"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 $M,N$.
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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)


