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.

  • 2
    $\begingroup$ For Noetherian commutative rings, this follows from Auslander-Buchsbaum formula. $\endgroup$
    – Mohan
    Dec 17, 2018 at 14:42
  • 1
    $\begingroup$ Probably you intend some finiteness condition? There are infinitely generated non-projective reflexive abelian groups. $\endgroup$ 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

1 Answer 1


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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. $\endgroup$ Dec 24, 2018 at 12:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.