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?

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    $\begingroup$ 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).

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