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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: http://mathoverflow.net/questions/99828/lifting-isomorphisms-between-derived-categories)

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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

1 Answer 1

up vote 2 down vote accepted

Answered on mathoverflow at the following link:


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