Tensoring by locally free sheaves preserves quasi-isomorphism

Question

Let X be a scheme. Given a quasi-isomorphism of complexes of quasi-coherent sheaves $f:\mathcal{F}^.\to \mathcal{G}^.$ and a locally free sheaf $\mathcal{E}$ we could consider the induced map $f\otimes1: \mathcal{F}^.\otimes\mathcal{E}\to \mathcal{G}^.\otimes\mathcal{E}$ where you can just think of tensoring everything by $\mathcal{E}$ as a sheaf or of taking the tensor product of complexes with $0\to\mathcal{E}\to0$.

I believe that using a Cech cohomology argument one should be able to prove that the induced map $f\otimes1$ is also a quasi-isomorphism assuming for example that X is flat and separated.

Do you know of any reference for that?

Answer 1

I think this is a simpler argument. A morphism $f\colon \mathcal{F}^\bullet \to \mathcal{G}^\bullet$ is a quasi-isomorphism if and only if $$\mathcal{F}^\bullet \overset{f}{\longrightarrow} \mathcal{G}^\bullet \longrightarrow 0 \overset{[1]}{\longrightarrow}$$ is a distinguished triangle. Applying the tensor product $- \otimes^{\mathbf{L}} \mathcal{E}$ is exact, and coincides with the regular tensor product $- \otimes \mathcal{E}$ since $\mathcal{E}$ is flat, so we get the distinguished triangle $$\mathcal{F}^\bullet \otimes \mathcal{E} \overset{f \otimes 1}{\longrightarrow} \mathcal{G}^\bullet \otimes \mathcal{E} \longrightarrow 0 \overset{[1]}{\longrightarrow}$$ Thus, $\mathcal{F}^\bullet \otimes \mathcal{E}$ and $\mathcal{G}^\bullet \otimes \mathcal{E}$ are quasi-isomorphic. $\blacksquare$

I suppose that lurking here is how you would even define derived tensor products, though.

Answer 2

I guess I found a reference: Lemma 15.50.4, Stack project on derived tensor product.

http://stacks.math.columbia.edu/tag/06XY