Let R be a local noetherian ring and let M, N be two finitely generated modules.



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> Is it true that if $M \otimes\_R N$ has finite length, then $Tor\_i^R(M,N)$ also has finite length for all i?
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I know a reference for a special case of this in Serre's book on local algebra. But I really don't want to assume R to be regular.
My guess is that this is true as $supp(M \otimes^L N) = supp(M) \cap supp(N)$ and $supp(M \otimes^L N) = \bigcup\_i supp Tor\_i(M,N)$, but I can't make it precise.



If tensor product has finite length, do higher Tors also have finite length?
To expand on Mohan's comment:


1) If $S$ is a local noetherian ring and $A$ an fg $S$-module, then $A$ has finite length if and only if it is supported on the maximal ideal.


2) If $S$ is a local noetherian ring with residue field $k$ and $A,B$ are fg $S$-modules such that $A\otimes\_SB=0$, then
$$(A\otimes\_Sk)\otimes\_k(B\otimes\_Sk)=(A\otimes\_SB)\otimes\_Sk=0$$ so (because $A\otimes\_Sk$ and $B\otimes\_Sk$ are vector spaces) either $A\otimes\_Sk=0$ or $B\otimes\_Sk=0$, so (by Nakayama) either $A=0$ or $B=0$, so $Tor\_i^S(A,B)=0$.


3) Now in your situation (assuming $M\otimes\_RN$ of finite length) let $Q$ be any non-maximal prime, $S=R\_Q$, $A=M\_Q$, $B=N\_Q$. By 1), $A\otimes\_SB=(M\otimes\_RN)\_Q=0$. By 2), $0=Tor\_i^S(A,B)=Tor\_i^R(M,N)\_Q$. This is so for every non-maximal prime, so $Tor\_i^R(M,N)$ is supported on the maximal ideal, so by 1) (now in the opposite direction), $Tor\_i(M,N)$ has finite length.


