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. $\square$

Comment #1587 by on

It might be a bit silly to assume $\mathcal{F}$ and $\mathcal{G}$ only quasicoherent, because you then take $\mathcal{F}$ coherent and $\mathcal{G}$ 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 $\mathcal{G}$ is coherent and then defines what it means for $\mathcal{G}$ 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 $\mathcal{F}$ and $\mathcal{G}$ are coherent, and then assume in addition that $\mathcal{G}$ is reflexive. See this commit

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