Timeline for A Module with $Ext^i(M,R) = 0$ for all $i > 0$
Current License: CC BY-SA 3.0
11 events
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| S Jan 24, 2015 at 10:59 | history | suggested | user 1 |
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| Jan 24, 2015 at 10:48 | review | Suggested edits | |||
| S Jan 24, 2015 at 10:59 | |||||
| Dec 21, 2011 at 16:33 | comment | added | User3568 | @Mahdi, thanks. @Benjamin, thanks. I was not aware of this conjecture. | |
| Dec 21, 2011 at 10:53 | vote | accept | User3568 | ||
| Dec 20, 2011 at 16:25 | answer | added | Graham Leuschke | timeline score: 3 | |
| Dec 20, 2011 at 16:12 | comment | added | Benjamin Steinberg | Compare with the Auslander-Reiten conjecture that $Ext^i(M,M\oplus R)=0$ for all $i>0$ implies $M$ is projective. They were originally interested in Artin algebras but people have considered things like Gorenstein rings. | |
| Dec 20, 2011 at 15:34 | comment | added | Mahdi Majidi-Zolbanin | Sorry Graham, looks like I answered the exercise! | |
| Dec 20, 2011 at 15:32 | comment | added | Mahdi Majidi-Zolbanin | I think if you assume $M$ of finite projective dimension, then the answer is yes over any ring (you don't need CM), because if $M$ is not free and you take a minimal free resolution of $M$ and apply $\mathrm{Hom}_R(\:\cdot\:,R)$ to it, the last map on the left after you apply Hom cannot be surjective and will have a cokernel. It cannot be surjective because the resolution is minimal. | |
| Dec 20, 2011 at 15:32 | comment | added | Graham Leuschke | There are many counterexamples, mostly (but not all) having to do with the ring being Gorenstein. If that's not a familiar word, you might look it up; it fits between regular and CM. Also, if M has finite projective dimension and all the Ext's vanish, then M is free -- this is a relatively easy exercise. | |
| Dec 20, 2011 at 15:20 | history | edited | User3568 | CC BY-SA 3.0 |
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| Dec 20, 2011 at 14:04 | history | asked | User3568 | CC BY-SA 3.0 |