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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jul 21, 2016 at 15:50 answer added Hailong Dao timeline score: 6
Jul 18, 2016 at 16:20 vote accept Hans
Jul 17, 2016 at 17:58 answer added Jason Starr timeline score: 10
Jul 17, 2016 at 16:34 comment added Jason Starr 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.
Jul 17, 2016 at 16:26 comment added Jason Starr 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.
Jul 17, 2016 at 16:19 comment added Jason Starr 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.
Jul 17, 2016 at 15:46 history asked Hans CC BY-SA 3.0