Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $A$ and $B$ are commutative rings. Let $A\to B$ be a surjective ring homomorphism. I will denote by $D(A)$ and $D(B)$ the derived categories of unbounded complexes over $A$ and $B$.

Suppose $M,N \in D(B)$ are two complexes over $B$. Let $F:D(B)\to D(A)$ be the forgetfull functor.

Suppose that we know that $F(M) \cong F(N)$. Does it follows that $M\cong N$ in $D(B)$?

If we had a quasi-isomorphism $F(M) \to F(N)$, then it will of course lift to $D(B)$, because since $A\to B$ is surjective, an $A$-linear map of complexes over $B$ will automatically be $B$-linear.

However, isomorphisms in the derived category might pass through a third object $K$, which might not be defined over $B$. Thus, I suspect the answer to my question is no, but I have no idea how to find a counterexample.

Thank you for any idea!

(remark: Since I did not get any answer, I posted this question to mathoverflow:

share|cite|improve this question
Since this question got an answer on mathoverflow, it would be good to add an answer with a link thereto and accept it. – Julian Kuelshammer Feb 7 '13 at 13:32
@JulianKuelshammer, done. – the L Feb 10 '13 at 7:26
up vote 2 down vote accepted

Answered on mathoverflow at the following link:

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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