Counterexample in the derived category

Question

Suppose we have the derived category $\mathcal{D}(A)$ of some algebra $A$ over some commutative ring $R$.

I would like an example of two objects $X,Y$ in $\mathcal{D}(A)$ such that for all $n \in \mathbb{Z}$, $H^n(X) \cong H^n(Y)$, but they are not isomorphic in $\mathcal{D}(A)$.

Though I am not sure such an example exists, the converse statement seems even more untrue to me.

Consider $A$, the algebra of dual numbers, or $A = k \oplus k[\epsilon]$, with $\epsilon^2 =0$. Then the following complexes have isomorphic homology, but there is no quasi-isomorphism between them, since $k$ as an $A$-module is isomorphic to the direct summand $k[\epsilon]$ of $A$. $$ 0 \xrightarrow{} A \xrightarrow{\cdot \epsilon} A \rightarrow 0 \rightarrow k \xrightarrow{0} k \rightarrow0 $$

$$ 0\rightarrow k \xrightarrow{0} k\rightarrow 0\rightarrow A \xrightarrow{\cdot \epsilon} A \rightarrow0 $$

However I am not sure how to prove that there is no zigzag of quasi-isomorphisms connecting the two.

Answer 1

In this case, a way to see that the complexes $0\to k\stackrel{0}{\to} k\to 0$ and $0\to A\stackrel{\epsilon}{\to} A\to 0$ are not isomorphic is to prove that the former is not indecomposable, while the latter is.

That $0\to k\stackrel{0}{\to} k\to 0$ is not indecomposable is clear: it is the direct sum of two complexes, each with $k$ in a certain degree and $0$ in all others.

To see that $0\to A\stackrel{\epsilon}{\to} A\to 0$ is indecomposable, we compute its endomorphism algebra and show that it is local. Luckily, it is a bounded complex of projective $A$-modules, so its endomorphism algebra in $\mathcal{D}(A)$ is isomorphic to its endomorphism algebra in the homotopy category $K(A)$.

Now, the endomorphism algebra of $0\to A\stackrel{\epsilon}{\to} A\to 0$ in the category of complexes $C(A)$ is $E=\{(a+b\epsilon, a+c\epsilon) | \ a,b,c\in k\}$, with coordinatewise multiplication. The ideal of null-homotopic morphisms is $I=\{(b\epsilon, b\epsilon) \ | \ b\in k\}$. Their quotient, $E/I$, is the endomorphism algebra of $0\to A\stackrel{\epsilon}{\to} A\to 0$ in $K(A)$. Moreover, $E/I$ is isomorphic to $A$, which is a local ring.

Thus $0\to A\stackrel{\epsilon}{\to} A\to 0$ is indecomposable in $\mathcal{D}(A)$. Since $0\to k\stackrel{0}{\to} k\to 0$ is not, the two cannot be isomorphic.