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(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!

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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 $\operatorname{Ext}^2_B(k,k)\to\operatorname{Ext}^2_A(k,k)$.

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