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Lemma 31.12.8. Let $X$ be an integral locally Noetherian scheme. Let $\mathcal{F}$, $\mathcal{G}$ be coherent $\mathcal{O}_ X$-modules. If $\mathcal{G}$ is reflexive, then $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})$ is reflexive.

Proof. The statement makes sense because $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})$ is coherent by Cohomology of Schemes, Lemma 30.9.4. To see the statement is true, see More on Algebra, Lemma 15.23.8. Some details omitted. $\square$

Comments (2)

Comment #1587 by on

It might be a bit silly to assume and only quasicoherent, because you then take coherent and reflexive whose definition incorporates coherence.

Comment #1663 by on

OK, this is actually a rather important general remark on definitions. If you look at the definition for reflexive modules, then you will see that the definition assumes is coherent and then defines what it means for to be reflexive in that setting. So, in a lemma or in the text, we can never say that a module is reflexive without first saying that the module is coherent. The reason being that later on in the text we might define what it means for a module to be reflexive in a more general setting (this will probably never happen, but it can and does happen with other definitions).

Conclusion: I have changed the hypothesis into saying that and are coherent, and then assume in addition that is reflexive. See this commit

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