Let $A$ be a ring and $f \in A$ an element. If $M$ is an $A$-module on which multiplication by $f$ is an isomorphism, then $M$ is in fact an $A_f$-module.

Now suppose that $C \in D(A)$ is a complex in the derived category of $A$-modules and that the multiplication by $f$ map $C \xrightarrow{f} C$ is a (quasi-)isomorphism. Does $C$ necessarily come from $D(A_f)$? That is, is there some complex $C'$ of $A_f$-modules and a (quasi-)isomorphism $C' \rightarrow C$ of complexes of $A$-modules. What if $C$ is bounded or bounded above?

  • 2
    $\begingroup$ The two sides of your "that is" are different questions. The map $C\rightarrow C_f$ is a quasi-isomorphism, so $C$, as an object in $D(A)$, does (up to isomorphism in the derived category) "come from" $D(A_f)$. But that gives a map $C\rightarrow C'$, not $C'\rightarrow C$. Does this matter to you? Also, of course, this might not work if you're interested in the derived category of finitely generated modules. Is that what you intended? $\endgroup$ – Steven Landsburg Jun 21 '16 at 1:24
  • $\begingroup$ @nfdc23: I think you need your map to be surjective. Otherwise I can just add any direct summand I want to onto $M$, and that direct summand need not be an $A_f$ module. $\endgroup$ – Steven Landsburg Jun 21 '16 at 2:00
  • $\begingroup$ @StevenLandsburg: sorry, even the zero map shows what I was suggesting is wrong. $\endgroup$ – nfdc23 Jun 21 '16 at 2:09
  • $\begingroup$ @StevenLandsburg: Thank you, that answers my question. I meant to ask for an isomorphism in the derived category (as you point out my question is a little ill-posed), so the direction of the arrow does not matter to me. $\endgroup$ – Lisa S. Jun 21 '16 at 23:16
  • $\begingroup$ @LisaS : Thank you for the thank you. Since this was the answer you were looking for, I'll repost it as an answer. $\endgroup$ – Steven Landsburg Jun 21 '16 at 23:24

The two sides of your "that is" are different questions:

1) Does $C$ necessarily come from $D(A_f)$ ?

2) Is there some complex $C'$ of $A_f$ modules and a quasi-isomorphism $C'\rightarrow C$ of complexes of $A$-modules?

Regarding 1): By assumption, $f$ induces an isomorphism on homology. Therefore, by your first observation, the homology of $C$ is already an $A_f$ module. Therefore this homology won't change if we localize at $f$. Therefore the map $C\rightarrow C_f$ is a quasi-isomorphism, which shows that in the derived category, $C$ is quasi-isomorphic to an object that comes from $D(A_f$).

This, however, constructs a map $C\rightarrow C'$, not a map $C'\rightarrow C$, and therefore does not speak to question 2).

| cite | improve this answer | |

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.