Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $R \to T$ is a ring map such that $T$ is flat as an $R$-module. Then for $A$ an $R$-module, $C$ a $T$-module there is an isomorphism

$$\text{Tor}^R_n(A,C) \simeq \text{Tor}^T_n(A \otimes_R T,C)$$

This can be proved directly by choosing an $R$-projective resolution $P^\bullet \to A$ and following through with the homological algebra. A more 'sledgehammer' approach us to use the Grothendieck spectral sequence $$E_2^{s,t} = \text{Tor}^{T}_s(\text{Tor}_t^R(A,T),C) \Rightarrow \text{Tor}^R_{s+t}(A,C),$$

which collapses under the assumptions above to give the required isomorphism.

In the case that $T$ is not flat as an $R$-module is it possible to build a spectral sequence that abuts to $\text{Tor}^T_n(A \otimes_R T,C)$?

share|cite|improve this question
And I think you meant that $A$ is an $R$-module? Regards. – awllower Apr 7 '13 at 3:31

$\mathcal{A}, \mathcal{B}, \mathcal{C}$ are abelian categories, $F:\mathcal{A} \rightarrow \mathcal{B}$ and $G: \mathcal{B} \rightarrow \mathcal{C}$ are right exact functors. You want to compute right derived functors of $G\circ F$.

To apply's Grothendieck's spectral sequence you have to ensure that $F$ maps acyclic complexes in $\mathcal{A}$ to acyclic complexes in $\mathcal{B}$.

In your example $F(A) = A \otimes T$ and $G(B)= B \otimes C$, if $T$ is flat then $F$ takes acyclic complexes to acyclic ones but if $T$ is not flat then I do not know of a general characterization in that case.

a way around would be to extend your underlying category to complexes of modules. Then you can replace $T$ by a flat/free resolution. In that case Grothendieck's machinery will work but it will give hyper-tor instead of tor.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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