MathOverflow is a question and answer site for professional mathematicians. 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

(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!

share|cite|improve this question
up vote 13 down vote accepted

Let $A=k[x]$ and $B=k[x]/(x^2)$, let $X$ be the complex $\hskip{.1in}\dots\to 0 \to B\stackrel{x}{\to} B\to 0\to \dots$, and let $Y$ be $\hskip{.1in}\dots\to 0\to k\stackrel{0}{\to}k\to 0\to\dots$. Then $X$ and $Y$ are isomorphic in $D(A)$, but not in $D(B)$.

The point is that $X$ is isomorphic to the third object in a triangle containing a map $\zeta:k\to k[2]$, where $\zeta$ represents an element in the kernel of $\operatorname{Ext}^2_B(k,k)\to\operatorname{Ext}^2_A(k,k)$.

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.