(Remark: I first asked this question at math.stackexchange. As it received no answer, I'm posting it here).
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!