Proving Two Complexes are Not Quasi-Isomorphic

Question

In Richard Thomas' paper "Derived Categories for the Working Mathematician" he mentions (page 6) that the two complexes $$ \begin{align*} C^\bullet&= \mathbb{C}[x,y]^{\oplus 2}\xrightarrow{(x,y)}\mathbb{C}[x,y]\\ D^\bullet&= \mathbb{C}[x,y]\xrightarrow{0}\mathbb{C} \end{align*} $$ have the same cohomology but are not quasi-isomorphic. Proving that $C^\bullet$ and $D^\bullet$ have the same cohomology is straightforward, but I'm having trouble proving that they are not quasi-isomorphic. Any suggestions?

Thanks in advance!

Answer 1

As was pointed out by Andrea in the comments to the other answer, the other answer does not answer the question properly.

An argument that the two complexes cannot be quasi-isomorphic in the general sense is sketched on page 2 of:

https://www.math.upenn.edu/~tpantev/rtg09bc/secnotes/ainfnity.davidovich.pdf

Answer 2

I guess we are working with $\Bbb C[x,y]$-modules.

The degree-zero part of the quasi-isomorphism would have to be a morphism $\Bbb C[x,y] \oplus \Bbb C[x,y] \to \Bbb C[x,y]$ inducing an isomorphism between $K$ and $\Bbb C[x,y]$, where $K$ is the kernel of the differential of $C^*$, that is $$ K = \{ (y R, x R) \, | \, R \in \Bbb C[x,y]\}.$$

But such a morphism is necessary of the form $(P,Q) \mapsto PU + QV$, so it would map $K$ into the ideal $x \Bbb C[x,y] + y \Bbb C[x,y] \varsubsetneq \Bbb C[x,y]$, a contradiction.

Answer 3

As this is a (the?) standard example of a pair of complexes whose homologies agree despite their being non-isomorphic in the derived category, it is perhaps worth spelling out a proof here. Instead of $\mathbb{C}$, we'll work over a arbitrary commutative ring $R$ (in the category of $R\left[x,y\right]$-modules); the argument works in generality.

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First, remark that your latter complex admits an evident quasiisomorphism from the (bounded, objectwise projective) complex $$0\to R\left[x,y\right]\oplus 0\overset{\begin{pmatrix}y \\ -x\end{pmatrix}\ \oplus\ 0}{\to} R\left[x,y\right]^{2}\oplus R\left[x,y\right]\overset{\begin{pmatrix}x & y\end{pmatrix}\ \oplus\ 0}{\to} R\left[x,y\right]\oplus 0 \to 0$$ (by replacing the direct summand $R\left[x,y\right]$ with its Koszul resolution). We'd like to show that this is not isomorphic in the derived category to the (bounded, objectwise projective) complex $$0\to 0\to R\left[x,y\right]^{2}\overset{\begin{pmatrix}x & y\end{pmatrix}}{\to} R\left[x,y\right]^{1} \to 0\text{.}$$ But (e.g., as a corollary of Stacks Lemma 0649(1)), bounded, objectwise projective complexes are isomorphic in the derived category iff they're homotopy equivalent. Homotopy equivalence of chain complexes is in turn a purely diagrammatic condition, so is conserved by all additive functors. In particular, the tensoring functor $-\otimes_{R\left[x,y\right]}R\left[x,y\right]/\left(x,y\right)$ is an additive functor which sends the two complexes to $$0\to R\left[x,y\right]/\left(x,y\right)\overset{0}{\to} \left(R\left[x,y\right]/\left(x,y\right)\right)^{2}\oplus R\left[x,y\right]/\left(x,y\right)\overset{0}{\to} R\left[x,y\right]/\left(x,y\right) \to 0$$ and $$0\to 0\to \left(R\left[x,y\right]/\left(x,y\right)\right)^{2}\overset{0}{\to} R\left[x,y\right]/\left(x,y\right) \to 0$$ respectively. But the homologies disagree, so these latter two are not chain homotopy equivalent, and thus neither are the former two!

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Remark: When exhibiting examples of non-homotopy-equivalent CW complexes with the same homotopy groups (as is done here), the two are often distinguished by their (co)homology groups. Here we distinguish two complexes with the same homologies by constrasting the homologies of their $-\otimes_{R\left[x,y\right]}^{\mathbb{L}}R\left[x,y\right]/\left(x,y\right)$s.