# Tensoring a flasque resolution by a line bundle give s a flasque resolution

I had an argument explained to me the other day and I didn't quite understand one of the steps. Here's my best reconstruction:

Let $\mathscr{F}$ be a quasi-coherent sheaf, and $\mathscr{L}$ a line bundle, on some scheme $X$. Consider an injective, thus flasque, resolution

$$0 \to \mathscr{F} \to \mathcal{J}_0 \to \mathcal{J}_1 \to \cdots$$

Tensoring by $\mathscr{L}$, we obtain a flasque resolution

$$0 \to \mathscr{F} \otimes \mathscr{L} \to \mathcal{J}_0 \otimes \mathscr{L} \to \mathcal{J}_1 \otimes \mathscr{L} \to \cdots$$

of $\mathscr{F} \otimes \mathscr{L}$.

Two questions:

1. Is this correct as stated? What is the proof?
2. Why is the hypothesis that $\mathscr{L}$ is a line bundle needed?
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Line bundles are flat. More generally, locally free modules are flat, because flatness is a local property and of course free modules are flat. This shows that the sequence is exact. The rest follows from the following Lemma (cf. Hartshorne, III.6.6):

If $F$ is injective and $\mathcal{L}$ is locally free, then $F \otimes \mathcal{L}$ is injective.

The reason is that $\hom(-,F \otimes \mathcal{L}) \cong \hom(- \otimes \mathcal{L}^*,F)$ is a composition of exact functors, hence exact.

By the way, we can derive from this $\mathrm{Ext}^p(F \otimes \mathcal{L},G) \cong \mathrm{Ext}^p(F,\mathcal{L}^* \otimes G)$.

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PS: I don't know if the tensor product of a flasque sheaf with a locally free sheaf is flasque. – Martin Brandenburg Sep 6 '13 at 9:37