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Suppose $X$ is a scheme, and $\mathcal{F}$ be a locally free sheaf (mayn't be of finite rank).

Definition of reflexive is given here: https://stacks.math.columbia.edu/tag/0AVU

Note that in the definition of reflexive, they stacks project assumes $\mathcal{F}$ is coherent, and hence finite type. But for the definition of reflexive, we don't need $\mathcal{F}$ to be coherent.

I know if $\mathcal{F}$ be a locally free sheaf of finite rank, then it would be reflexive. But I don't know if this is true for any locally free sheaf. Any help would in proving or disapproving this would be appreciated.

Ramandeep Singh Arora
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    If $X = \text{Spec } K$ then $F$ is just a $K$-vector space and you can check that (assuming the axiom of choice) $F$ is reflexive iff it's finite-dimensional. – Qiaochu Yuan Jul 16 '24 at 03:34
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    The linked duplicate gives a counterexample in the case when $X$ is the spectrum of a field, modulo some translation. – KReiser Jul 16 '24 at 03:35
  • @QiaochuYuan This resolves my question. Just have a follow through question. Are finite locally free sheaves reflexive? I think they should be. Same proof as locally free sheaves of finite rank should work. – Ramandeep Singh Arora Jul 16 '24 at 04:06
  • @RamandeepSinghArora "They're the same picture." – Alex Youcis Jul 16 '24 at 06:11

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