Let $R$ be a local artinian Gorenstein ring and $M$ a finitely generated $R$-module, then $\mathrm{Ext}_R^1(M,M) = 0$ if only if $M$ is projective?
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$\begingroup$ Dear TmobiusX, would you care to explain any motivation for your question? $\endgroup$– Hailong DaoCommented May 26, 2010 at 16:17
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This would be a very strong version of the Auslander-Reiten Conjecture (see here, for example) in the Gorenstein case. The Conjecture is still open, though many partial results are known.
By the way, this paper (ScienceDirect link, may not be visible to everyone) claims a proof of the Conjecture in the Gorenstein case, which would give an affirmative answer to your question, but I --- and several other people I've talked to --- believe there is a gap in the proof. Their assumption is that $\mathrm{Ext}_R^{i}(M,M)=0$ for $i =1,2$. The questionable step is at the top of page 2163, the second line, where they say "therefore ... is exact".
Edit 2023-05-15: The paper I referred to above is: "When are Torsionless modules projective?", by Rong Luo and Zhaoyong Huang, Journal of Algebra 320 (2008) 2156-2164, MR2437647.
answered May 26, 2010 at 13:53
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2$\begingroup$ Fixed the leuschke.org link, thanks. Included citation information for the other paper, which I should have done at the time. $\endgroup$ Commented May 15, 2023 at 14:14
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$\begingroup$ Is this result now accepted by the community, or is there still a belief that a gap may exist? $\endgroup$ Commented May 15, 2023 at 16:53
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1$\begingroup$ As far as I know the situation is still the same as in 2010. I still doubt the result. It has been cited in half-a-dozen papers since then. $\endgroup$ Commented May 15, 2023 at 18:22
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$\begingroup$ I see. Thank you for your insight. $\endgroup$ Commented May 16, 2023 at 12:46