Length of dual module

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.

• Is your ring commutativev? If not then (1) doesn't make sense since the dual switches between left and right modules. Feb 5 at 11:17
• @BenjaminSteinberg yes it is commutative, I will edit the question accordingly Feb 5 at 11:23
• 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).
– YCor
Feb 5 at 14:48
• @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
– YCor
Feb 5 at 15:48
• @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.
– YCor
Feb 6 at 13:20