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Let $(R, \mathfrak m)$ be a one-dimensional commutative Noetherian local ring. Let $M$ be a finitely generated $R$-module with finite injective dimension $\operatorname{injdim}_R(M).$ One can prove that $R$ is Cohen-Macaulay by showing first that $\operatorname{injdim}_R(M) \geq 1$ (cf. this post) and subsequently appealing to the fact (proven by Bass) that $\operatorname{injdim}_R(M) = \operatorname{depth}(R)$; however, my aim is to prove the following.

Let $(R, \mathfrak m, k)$ be a one-dimensional commutative Noetherian local ring. If $R$ admits a finitely generated reflexive $R$-module $M$ with finite injective dimension, then $R$ must be Gorenstein.

Unfortunately, I fear that the line of reasoning used to prove that $R$ is Cohen-Macaulay is of little help in proving that $R$ is Gorenstein. But at the same time, once we know that $R$ is Cohen-Macaulay, we can invoke Grothendieck's Local Duality Theorem. Ultimately, we must show that $\operatorname{Ext}_R^1(k, R) \cong k$ and $\operatorname{Hom}_R(k, R) = 0.$

Edit: As @metalspringpro points out, there was an unresolvable error in my initial attempt, so I have removed it. Furthermore, it suffices to prove that if $R$ admits a reflexive canonical module $\omega_R,$ then $R$ is Gorenstein.

I would greatly appreciate any advice, comments, or suggestions.

Dylan C. Beck
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    Hint: Passing to the completion if necessary, you may assume $R$ has a canonical module $\omega_R$. Then as $M$ is reflexive and as $R$ is a Cohen-Macaulay ring of dimension $1$, $M$ is maximal Cohen-Macaulay. As it also has finite injective dimension, it follows that $M \cong \omega_R^{\oplus n}$ for some $n$. In particular, $\omega_R$ is reflexive. So your question reduces to why $\omega_R$ being reflexive forces $R$ to be Gorenstein in this setting. – metalspringpro Aug 13 '21 at 23:42
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    Also, I should point out your application of the functor $\operatorname{Hom}_R(-,R)$ is not correct. If $0 \to M \to Q \to Q' \to 0$ is the injective resolution you describe, one would have an exact sequence $$0 \to \operatorname{Hom}_R(Q',R) \to \operatorname{Hom}_R(Q,R) \to \operatorname{Hom}_R(M,R) \to \operatorname{Ext}^1_R(Q',R) \to \operatorname{Ext}^1_R(Q,R) \to \cdots$$ and one cannot conclude e.g. that $\operatorname{Ext}^1_R(Q,R)=0$ from injectivity of $Q$. We have $\operatorname{Ext}^1_R(-,Q)=0$ when $Q$ is injective, not that $\operatorname{Ext}^1_R(Q,-)=0$. – metalspringpro Aug 14 '21 at 02:54
  • @metalspringpro, your second comment is very useful. It makes sense; I had simply erroneously recalled the property that $\operatorname{Tor}_i^R(M, N) \cong \operatorname{Tor}_i^R(N, M)$ whenever $R$ is a commutative ring and conflated it with a property of Ext. – Dylan C. Beck Aug 14 '21 at 06:07
  • @metalspringpro, in fact, to show that the reflexivity of $\omega_R$ implies that $R$ is Gorenstein was my original problem; however, it was pointed out to me that the correct line of reasoning would show moreover that the existence of any finitely generated reflexive module of finite injective dimension would imply that $R$ is Gorenstein. I am still unsure how to solve the problem, but I appreciate your suggestions very much. – Dylan C. Beck Aug 14 '21 at 06:10
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    If you've already reduced to the case where $M=\omega_R$, then I'll add an additional hint that any reflexive module is a second syzygy, and that you should think about canonical duality. – metalspringpro Aug 14 '21 at 07:26
  • @metalspringpro, I have filled out most of the details you have suggested, but I am not sure why it is true that $M$ is a direct sum of copies of $\omega_R.$ – Dylan C. Beck Aug 18 '21 at 00:07
  • @metalspringpro, what are you calling "canonical duality?" I am not familiar with this. – Dylan C. Beck Aug 18 '21 at 00:18
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    If $R$ admits a canonical module, then any maximal Cohen-Macaulay module of finite injective dimension is isomorphic to $\omega_R^{\oplus n}$ for some $n$. A good reference for this fact as well as canonical duality as a whole is chapter 11 of "Cohen-Macaulay Representations" by Leuschke-Wiegand (cf. https://math.stackexchange.com/q/3766831). In particular see 11.7 there. Bruns and Herzog also has a nice chapter (Chapter 3) around this. The canonical dual of a Cohen-Macaulay $R$-module $M$ of dimension $t$ is $\operatorname{Ext}^{d-t}_R(M,\omega_R)$ where $d=\dim R$. – metalspringpro Aug 18 '21 at 01:46
  • @metalspringpro, I have thought about this a lot, and I have written down a lot of new information regarding the canonical module and canonical duality, but I am still not seeing where to go once we have that $0 \to \omega_R \to R^m \to R^n \to M \to 0$ is a short exact sequence for some finitely generated $R$-module $M$ and some integers $m, n \geq 1.$ – Dylan C. Beck Aug 26 '21 at 19:16
  • I have created a chat room for us here. – Dylan C. Beck Aug 26 '21 at 19:24

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