Lemma 20.49.1. Let $X$ be a ringed space. Let $\mathcal{I}$ be an injective $\mathcal{O}_ X$-module. Let $\mathcal{E}$ be an $\mathcal{O}_ X$-module. Assume $\mathcal{E}$ is finite locally free on $X$, see Modules, Definition 17.14.1. Then $\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{I}$ is an injective $\mathcal{O}_ X$-module.

Proof. This is true because under the assumptions of the lemma we have

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{I}) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}( \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{E}^\vee , \mathcal{I})$

where $\mathcal{E}^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}, \mathcal{O}_ X)$ is the dual of $\mathcal{E}$ which is finite locally free also. Since tensoring with a finite locally free sheaf is an exact functor we win by Homology, Lemma 12.27.2. $\square$

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