In [this question](https://mathoverflow.net/questions/4590/when-are-dual-modules-free) 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?



Criterion for being reflexive via Ext
**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.


