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 $\text{Tor}^T_n(A \otimes_R T,C)$?
Best Answer
$\mathcal{A}, \mathcal{B}, \mathcal{C}$ 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.