4
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ Is your ring commutativev? If not then (1) doesn't make sense since the dual switches between left and right modules. $\endgroup$ Commented Feb 5, 2024 at 11:17
  • $\begingroup$ @BenjaminSteinberg yes it is commutative, I will edit the question accordingly $\endgroup$ Commented Feb 5, 2024 at 11:23
  • 2
    $\begingroup$ 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). $\endgroup$ Commented Feb 5, 2024 at 14:48
  • 1
    $\begingroup$ @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 $\endgroup$ Commented Feb 5, 2024 at 15:48
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Feb 6, 2024 at 13:20

0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.