1
$\begingroup$

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 quasi-isomorphic as complexes of $B$-modules?

$\endgroup$
3
  • 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 quasi-isomorphic 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 quasi-isomorphic to $(A/I)\otimes_{A/I}^L(B/IB)$ in the derived category of $B$-modules. $\endgroup$ Jan 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 quasi-isomorphic to $(A/I) \otimes_A^L (B/IB)$ as complexes of $B$-modules? $\endgroup$
    – Yellow Pig
    Jan 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 Pig
    Jan 1, 2016 at 14:51

1 Answer 1

3
$\begingroup$

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 $M$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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