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In this question it was claimed that if a module $M$ over a noetherian domain $R$ satisfies $\rm{Ext}^i(M,R)=0$ for $i=1,2$, then $M$ is reflexive. Is this true? Does someone know a reference or a proof for this? If it is not correct, is there another criterion for being reflexive via some $\rm Ext$-goups vanishing?

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  • $\begingroup$ For $R=k[x,y,z,w]$, for the torsion-free, pure $R$-module $M=\langle x,y,z,w\rangle$, i.e., $M$ equals the maximal ideal of the origin, both $\text{Ext}^1_R(M,R)$ and $\text{Ext}^2_R(M,R)$ are zero. This follows from the self-duality of the Koszul complex of the regular sequence $(x,y,z,w)$. The module $M$ is not reflexive. $\endgroup$ Jul 17, 2016 at 16:19
  • $\begingroup$ I am sure the poster of that answer just wrote the wrong pair of integers. There is a characterization of reflexive modules in terms of the cohomological dimension for local cohomology. If you translate that into Ext groups, there is probably a shift of degrees that got lost in the post. $\endgroup$ Jul 17, 2016 at 16:26
  • $\begingroup$ If $R$ is a regular local ring of dimension $d$ so that we have local duality. Assuming that my computations are correct, a finitely generated $R$-module $M$ has depth $\geq 2$ (i.e., it is reflexive) if both $\text{Ext}^{d-1}_R(M,R)$ and $\text{Ext}^d_R(M,R)$ are zero. $\endgroup$ Jul 17, 2016 at 16:34

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Edit. Hailong Dao points out a serious error in what I originally wrote. I have edited the statement below. Unfortunately, the corrected condition on the Ext modules is now rather complicated: a dimension condition on the support of every Ext module, not just vanishing of the final two Ext modules.

I am just writing my comments as an answer. Edit. For a normal Noetherian integral domain $A$ and for a torsion-free, finitely generated $A$-module $N$, $N$ is reflexive if and only if, for every prime ideal $\mathfrak{p}\subset A$ of height $\geq 2$, for the local Noetherian ring $R=A_{\mathfrak{p}}, \mathfrak{m}= \mathfrak{p}A_{\mathfrak{p}}$ and for the localized module $M=N\otimes_A A_{\mathfrak{p}}$, the depth of $M$ is $\geq 2$ as an $R$-module. For a local Noetherian ring $(R,\mathfrak{m})$ of dimension $\geq 2$ (Edit: that is normal), for a finitely generated $R$-module $M$ (Edit: that is torsion-free), $M$ is "reflexive" in the usual sense if and only if the depth of $M$ is $\geq 2$. Also, the depth is $\geq 2$ if and only if the local cohomology groups $H_{\mathfrak{m}}^i(M)$ are zero for $i=0,1$. One reference is the exercises part of Section II.3, pp. 216-218, of Hartshorne's "Algebraic Geometry". If $(R,\mathfrak{m})$ is a regular local ring of dimension $d$, then those local cohomology groups are dual (in the sense of local duality theory) to the Ext groups $\text{Ext}^{d-i}_R(M,R).$ This is Corollary V.6.8, p. 276, and the comments at the end of the section, p. 281, of Hartshorne's "Residues and Duality", cf. also 6.3 of "Local Cohomology". Thus, in the regular case, $M$ is reflexive if and only if both $\text{Ext}^{d-1}_R(M,R)$ and $\text{Ext}^d_R(M,R)$ are zero.

Edit. Corrected Necessary and Sufficient Ext Condition. Assuming that $A$ is a regular Noetherian integral domain (so that also each localization $R$ is regular by Serre), this becomes a condition in terms of the Ext modules of the original module $N$ over the original ring $A$. A torsion-free, finitely generated module $N$ is reflexive if and only if, for every integer $e\geq 1$, every minimal associated prime of $\text{Ext}^e_A(N,A)$ has height $\geq e+2$, i.e., the support of the module has codimension $\geq e+2$. For example, when $A$ is regular of dimension $2$, this is equivalent to saying that $N$ is projective. When $A$ is regular of dimension $d \geq 2$, this condition implies that both $\text{Ext}^d_A(N,A)$ and $\text{Ext}^{d-1}_A(N,A)$ are zero, but vanishing of these Ext modules is not sufficient.

Having explained that the wrong statement in that previous post seems to have just been a transcription problem (Edit. Hailong Dao points out that it is more complicated than I thought), it is straightforward to construct examples of modules $M$ that are not reflexive where $\text{Ext}^1_R(M,R)$ and $\text{Ext}^2_R(M,R)$ are zero. Of course we need to choose $d>2$. For $d\geq 4$, the maximal ideal $M=\langle x_1,\dots,x_d \rangle$ in $k[x_1,\dots,x_d]$ has vanishing $\text{Ext}^1_R(M,R)$ and $\text{Ext}^2_R(M,R)$, yet $M$ is not reflexive.

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    $\begingroup$ Hi Jason, thanks for this answer! Just two quick question: What happens for rings of dimension 1? Is it true that reflexive is equivalent to projective? And do you have a reference for your first statement, namely that the module is reflexive iff its depth is at least two? $\endgroup$
    – Hans
    Jul 18, 2016 at 15:18
  • $\begingroup$ The definition that I wrote is correct if $R$ is normal. If $R$ is not normal, then you will find commonly used at least two inequivalent definitions, cf. stacks.math.columbia.edu/tag/0AY1. If $R$ is normal, then every finitely generated, torsion-free $R$-module is reflexive at all points of codimension $0$ and codimension $1$. Then one reference relating reflexive to depth $\geq 2$ is stacks.math.columbia.edu/tag/0AY5. $\endgroup$ Jul 18, 2016 at 15:30
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    $\begingroup$ It is not true that $M$ is reflexive if depth is at least $2$. Over a normal domain you need it to be $S_2$, a more subtle condition. For example, take $M=(x,y,z)$ in $k[x,y,z,t]$. It has depth 2 but not reflexive, since a reflexive ideal would have to have height one. $\endgroup$ Jul 21, 2016 at 13:42
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    $\begingroup$ @HailongDao: You are correct, and I will edit my answer. For $M=\mathfrak{p}=\langle x,y,z \rangle \subset k[x,y,z,t]$, the module does have depth $2$ and one regular sequence of length $2$ is $(t,x)$. Of course the localization $M_{\mathfrak{p}}$ as a $k[x,y,z,t]_{\mathfrak{p}} = k(t)[x,y,z]_{\langle x,y,z \rangle}$ does not have depth $2$. $\endgroup$ Jul 21, 2016 at 13:54
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About the last question, I should point out that there is indeed a Ext-vanishing criterion for reflexivity. Namely, $M$ is reflexive iff $Ext^{1,2}(Tr(M),R)=0$ where $Tr(M)$ is the Auslander-Bridger transpose of $M$. For details see for instance Lemma 2.2 in this paper: https://arxiv.org/pdf/0809.1958v3.pdf

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