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Recall that a module M over some ring R is pseudo-coherent if it admits a resolution whose terms are finitely generated projective R-modules. Another notion is to ask whether M is reflexive when regarded as an object in the derived category $D(Mod_R)$, i.e., $$M \to RHom(RHom(M, R), R))$$ is a quasi-isomorphism, where RHom denotes the derived functor of the usual homomorphisms of R-modules.

  1. For what rings does pseudo-coherence imply reflexivity in this derived sense?
  2. When does the converse hold? (Maybe for nice rings, even for the integers?), i.e., are reflexive modules (where reflexivity is always understood in the derived sense) pseudo-coherent?
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  • $\begingroup$ Small comment: if $R$ is Noetherian, then pseudo-coherent just means finitely generated. In the non-derived sense, reflexive does not imply finitely generated; for example $\mathbf Z^{(\mathbf N)}$ is reflexive. This depends highly on the ring; for example for a DVR $R$ the analogous result is true if and only if $R$ is not complete; see e.g. this post. $\endgroup$ Commented May 15, 2020 at 15:53
  • $\begingroup$ Sorry for my ignorance: why are pseudocoherent modules "derived" reflexive? For example, I don't know whether $\operatorname{Hom}_R(K_*,R)$ is K-projective when $K_*$ is a homologically bounded below complex of free modules. Note that the dual is not ncessarily bounded below. $\endgroup$
    – user20948
    Commented Feb 8, 2021 at 11:28
  • $\begingroup$ @R.vanDobbendeBruyn Is it obvious that all higher Ext groups $\operatorname{Ext}^i(\prod_{\mathbb N}\mathbb Z,\mathbb Z)$ vanish? The OP mentioned that the reflexivity in question is derived. $\endgroup$
    – user20948
    Commented Feb 8, 2021 at 11:34
  • $\begingroup$ @R.vanDobbendeBruyn I found that this is false even when $i=1$. $\endgroup$
    – user20948
    Commented Feb 8, 2021 at 11:59
  • $\begingroup$ @Yai0Phah: Thanks - I think you are right, the assertion I made in the question is unclear. It might still be correct if R has finite global dimension. I will edit the question accordingly. $\endgroup$
    – Jakob
    Commented Feb 9, 2021 at 13:09

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