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

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?

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.

  • $\begingroup$ Can you show that if $M\otimes_R N=0$, then $M=0$ or $N=0$? If you can, reduce to this case by localizing. $\endgroup$
    – Mohan
    Oct 4 '16 at 18:31
  • $\begingroup$ @Mohan could you please elaborate? $\endgroup$ Oct 4 '16 at 18:36
  • 2
    $\begingroup$ To show that $T_i=Tor_i^R(M,N)$ is finite length, suffices to show that $T_{i_Q}=0$ for all non-maximal primes $Q$. But, this is just $Tor_i^{R_Q}(M_Q, N_Q)$. Since $M_Q\otimes_{R_Q}N_Q=0$, we have one of them is zero and then so are these tors. $\endgroup$
    – Mohan
    Oct 4 '16 at 19:05
  • 1
    $\begingroup$ You don't need to go to the derived category. An fg module has finite length if and only if it is supported on the maximal ideal. So after localizing away from the maximai ideal, your statement is equivalent to $M\otimes_RN=0\Rightarrow Tor_i^R(M,N)=0$. $\endgroup$ Oct 4 '16 at 20:11

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.

  • $\begingroup$ thanks. In 3), I don't want to assume M and N to be of finite length. I only know that $M \otimes N$ is of finite length. $\endgroup$ Oct 5 '16 at 17:11
  • $\begingroup$ FIxed. $\phantom{.............}$ $\endgroup$ Oct 5 '16 at 17:18
  • $\begingroup$ I finally had time to have a look at it, thanks for the clear explanation. $\endgroup$ Oct 6 '16 at 20:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.