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In preparing for a talk, I am searching for an explicit example of Matlis duality. My attempt was to consider the Noetherian local ring $\mathbb{Z}_p$ which is the localization at the prime ideal $(p)$. Then take the injective hull of the residue field which is the Prüfer $p$-group $E=\mathbb{Z}(p^\infty)$. Now let $R$ be the completion of $\mathbb{Z}_p$ at its maximal ideal.

I would like to give an explicit description of $\operatorname{Hom}_R(\mathbb{Z}_p,E)$ and then show that taking the dual again recovers $\mathbb{Z}_p$. So far, I have been unsuccesful. (Edit: Comment shows this does not make sense).

So my question is how do I go about computing the explicit description I want above?

Or better yet, is there an explicit example where I take my Noetherian local ring to be complete as well and with a nicely chosen Artinian module to demonstrate?

Edit: From the comments, I guess an easy example where Matlis duality holds is:

Let $k$ be a field. Then its injective hull is itself and its completion is just itself. A finitely generated $k$-module is just finite dimension vector space. Then Matlis duality is an obvious consequence of properties of $\operatorname{Hom}_k$.

Edit: In this case, I would like to just give a module over the completion $R$ which is finitely generated and in which $\operatorname{Hom}_R(M,E)$ is clearly Artinian.

FShrike
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Shrugs
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  • How is $\mathbb{Z}_p$ a $R$-module? Regardless, as every prime different from $p$ acts invertibly on $E$, $Hom(\mathbb{Z}_p,E) \cong Hom(\mathbb{Z},E) =E$… no? – Aphelli Oct 07 '22 at 08:54
  • @Aphelli Oops. Yes you're correct. One example of Maltis duality is might be easy e.g. $Hom_{\mathbb{Z}_p}(\mathbb{Z}_p,E)=E$ and showing that $E$ is finitely generated over $R$. What I want is to go the other direction i.e. to give an $R$-module $M$ in which $Hom_R(M,E)$ is obviously Atinian. I want an example where $M\neq R$. – Shrugs Oct 07 '22 at 14:55
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    Hm… $E$ is not finitely generated over $R$, is it? Also, $R$ is a PID, so any finitely generated module over $R$ is a direct sum of cyclic torsion modules and of a free module… – Aphelli Oct 07 '22 at 16:09
  • Okay - so the example with $E$ won't work since Maltis duality needs finitely generated. I guess the only examples I'll have that are interested are $Z_p$ and quotients of $Z_p$. But that isn't so exciting.. Maybe I'd have to switch my rings. – Shrugs Oct 07 '22 at 18:00
  • But are there any non-trivial examples with finitely generated injective hulls? If $R$ is a local domain, and $M$ is injective, then $M$ is divisible then (by Nakayama) not finitely generated (unless $R$ is a field)? – Aphelli Oct 07 '22 at 18:20
  • A version of Matlis duality does not need that finiteness: https://stacks.math.columbia.edu/tag/0A6V – Yai0Phah Oct 07 '22 at 18:23
  • The version of Matlis duality I am presenting requires finiteness for the module over the completion, but not condition is imposed for the Artinian module over the ring itself. So I'd like to keep the finiteness condition in my example.

    Based on the comment on local domains, I might need a non-integral domain example.

     – Shrugs Oct 07 '22 at 18:30
  • But if $R$ is noetherian, $I \subset \mathfrak{p}$ an ideal with $\mathfrak{p}$ prime, and $M$ any injective $R$-module, then $Ext^1_{R_{\mathfrak{p}}}((R/I){\mathfrak{p}}, M{\mathfrak{p}})=Ext^1_R(R/I,M)_{\mathfrak{p}}=0$, so localizations of $M$ at prime ideals are injective. – Aphelli Oct 07 '22 at 21:25

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