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It is well known that, given a commutative ring $R$ and an $R$-module $M$, the dual module $M^\vee = \operatorname{Hom}_R(M, R)$ does not always satisfy $M^\vee \cong M \ (1)$, and not even $M^{\vee \vee} \cong M$ (the latter is satisfied, for example, when $M$ is finitely generated projective). My question is, are there any known conditions for $(1)$ to be satisfied? I am primarily investigating finitely generated $M$ over a commutative finite principal ideal ring (not necessarily local or a domain), but would welcome a more general result, if it exists. In the case that conditions for an isomorphism are nonexistent or too deep, is there a known way to at least compare the lengths of $M^\vee$ and $M$ as $R$-modules? In the case that mainly interests me (finitely generated over a finite PIR), I am thinking that they should have equal length, but could not find a proper proof.

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  • $\begingroup$ Is your ring commutativev? If not then (1) doesn't make sense since the dual switches between left and right modules. $\endgroup$ Feb 5 at 11:17
  • $\begingroup$ @BenjaminSteinberg yes it is commutative, I will edit the question accordingly $\endgroup$
    – JBuck
    Feb 5 at 11:23
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    $\begingroup$ If $R$ is artinian, say local, there is a useful duality, given by $\mathrm{Hom}(-,E)$, where $E$ is the injective hull of the residual field. In some cases, $E\simeq R$ ($R$ is called self-injective — I guess it holds in the PIR case). $\endgroup$
    – YCor
    Feb 5 at 14:48
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    $\begingroup$ @JBuck I forgot to say this is called Matlis duality. A good reference: W. Bruns, J. Herzog. “Cohen–Macaulay rings”, rev. ed., Cambridge Stud. Adv. Math., 39, Cambridge Univ. Press, Cambridge, 1998 $\endgroup$
    – YCor
    Feb 5 at 15:48
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    $\begingroup$ @JBuck Well, passing from local to semilocal is somewhat trivial. The main purpose (as far as I view it) of Matlis duality is to describe artinian modules over noetherian rings, and this always gives modules over some semilocal ring. $\endgroup$
    – YCor
    Feb 6 at 13:20

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