Let $A$ be a finitely generated $k$algebra, where $k$ is a field, let $I$ be an ideal in $A$, let $M$ be a finitely generated $A/I$module, and let $M^{\prime}$ denote $M$ considered as an $A$module. Let $B$ be a finitely generated $A$algebra. Is it true (perhaps under some additional conditions on $A$ and $I$, though I need the case when $B$ is not flat over $A$) that $M^{\prime} \otimes^L_A B$ and $M \otimes^L_{A/I} B/IB$ are quasiisomorphic as complexes of $B$modules?

2$\begingroup$ What about the case when $M$ equals $B$ equals $A/I$? Then you would be asking that $(A/I)\otimes_A^L (A/I)$ is quasiisomorphic to $(A/I)\otimes_{A/I}^L(A/IA)$. This fails already for $A=\mathbb{Z}$ and $I = n\mathbb{Z}$ (or $A=\mathbb{C}[t]$ and $I=t\mathbb{C}[t]$). Perhaps you should add a hypothesis that $(A/I)\otimes_A^LB$ is quasiisomorphic to $(A/I)\otimes_{A/I}^L(B/IB)$ in the derived category of $B$modules. $\endgroup$– Jason StarrJan 1, 2016 at 14:29

$\begingroup$ @JasonStarr Thank you very much! Sorry, did you mean that I should add the hypothesis that $(A/I) \otimes_A^L B$ is quasiisomorphic to $(A/I) \otimes_A^L (B/IB)$ as complexes of $B$modules? $\endgroup$– Yellow PigJan 1, 2016 at 14:37

$\begingroup$ @JasonStarr Sorry, never mind, my previous comment was stupid. I guess what you meant is that I should add the hypothesis that $Tor^i_A(A/I,B)=0$ for all $i \neq 0$, right? Do you by any chance also think that assuming that $Tor^i_A(A/I,M)=0$ for $i \neq 0$ would work instead? $\endgroup$– Yellow PigJan 1, 2016 at 14:51
1 Answer
There is an associativity identity for total derived tensor products, cf. Stacks Project Tag 08YU. In your case, for the triple of rings, $$A\twoheadrightarrow A/I \xrightarrow{\text{Id}}A/I,$$ this gives an equivalence in the derived category, $$ M\otimes_{A/I}^{\textbf{L}}(A/I\otimes_A^{\textbf{L}} B) \cong M\otimes_A^{\textbf{L}} B. $$ Thus, if the natural truncation morphism, $$A/I\otimes_A^{\textbf{L}}B \to h_0(A/I\otimes_A^{\textbf{L}}B), \text{ i.e., }\ A/I\otimes_A^{\textbf{L}}B \to B/IB, $$ is an equivalence in the derived category, then that gives the equivalence that you are asking about, $$M\otimes_{A/I}^{\textbf{L}} B/IB \cong M\otimes_A^{\textbf{L}}B.$$
As the OP points out, the natural truncation morphism is an equivalence in the derived category if and only if $h_i(A/I\otimes_A^{\textbf{L}} B)$ vanishes for every $i>0$, i.e., if and only if $\text{Tor}_i^A(A/I,B)$ vanishes for every $i>0$. The OP asks whether it might suffice to have vanishing of $\text{Tor}_i^A(A/I,M)$ for $i>0$? It is hard for me to imagine any situation where this vanishing would hold, since $\text{Tor}_1^A(A/I,M)$ is naturally isomorphic to $I\otimes_A M$ for every $A/I$module $M$.