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Recall that a module $M$ over a ring $R$ is reflexive in case the natural evaluation map $f_M:M \rightarrow M^{**}$ (where $M^{*}=Hom_R(M,R)$) is an isomorphism, where $f_M(m)=g$ with $g(h)=h(m)$, when $h \in Hom_R(M,R)$. For example every projective module is reflexive. Assume that all modules in the following are finitely generated. Call a reflexive module non-trivial in case it is not projective.

Question: Does an arbitrary ring of finite global dimension have global dimension at most two if and only if there is no non-trivial reflexive module?

Here two similar questions restricted to finite dimensional algebras:

Does a finite dimensional algebra have a non-trivial reflexive module in case it has finitistic dimension at least 3?

Does a finite dimensional algebra of finite global dimension have global dimension at most two if and only if it has no non-trivial reflexive modules?

In general a non-trivial reflexive module implies that the global dimension of a ring with finite global dimension is at least three but I do not know about the other direction. The answer should be positive for QF-3 algebras.

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    $\begingroup$ For Noetherian commutative rings, this follows from Auslander-Buchsbaum formula. $\endgroup$
    – Mohan
    Dec 17, 2018 at 14:42
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    $\begingroup$ Probably you intend some finiteness condition? There are infinitely generated non-projective reflexive abelian groups. $\endgroup$
    – Jeremy Rickard
    Dec 18, 2018 at 9:13
  • $\begingroup$ @JeremyRickard Thanks, I was actually only thinking about finitely generated modules. I edited it. $\endgroup$
    – Mare
    Dec 18, 2018 at 9:32

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The algebra with quiver

$\require{AMScd}$ \begin{CD} \bullet @>>>\bullet@>>>\bullet\\ @AAA&@AAA\\ \bullet@>>>\bullet \end{CD}

and radical square zero has global dimension three and (according to my calculations) no non-projective reflexive modules.

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  • $\begingroup$ QPA says you are right. It must have been some work to come up with this example (or do you have a trick?). Thank you very much! $\endgroup$
    – Mare
    Dec 24, 2018 at 12:01
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    $\begingroup$ @Mare Well, for a radical square zero quiver algebra, it's not hard to see that having no non-projective simple reflexive modules is equivalent to having no arrows $\alpha:i\to j$ such that $j$ is not a sink and $\alpha$ is the only arrow starting at $i$ and the only arrow ending at $j$. So then you only have to worry about indecomposable non-simple, non-projective modules; and this algebra only has three of those. $\endgroup$
    – Jeremy Rickard
    Dec 24, 2018 at 12:46

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