most recent 30 from http://mathoverflow.net 2013-05-19T03:08:37Z http://mathoverflow.net/feeds/question/106793 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106793/regularity-of-local-ring Pham Hung Quy 2012-09-10T09:07:26Z 2012-09-10T11:14:01Z

<p>Let $(R, \mathfrak{m})$ be a Noetherian local ring. It is well know that</p> <p>$R$ is regular iff $pd(R/\mathfrak{m}) &lt; \infty$ (i.e. $R/\mathfrak{m}$ has finite projective dimension).</p> <p>Assume that $\dim R > 0$. Is $R$ regular, if $pd(R/\mathfrak{m}^2)&lt; \infty$?</p>

http://mathoverflow.net/questions/106793/regularity-of-local-ring/106802#106802 Graham Leuschke 2012-09-10T10:53:21Z 2012-09-10T11:14:01Z

<p><a href="http://www.ams.org/mathscinet-getitem?mr=230715" rel="nofollow">Levin-Vasconcelos</a> (<a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.pjm/1102986272" rel="nofollow">journal link</a>): for $R$ a local ring with maximal ideal $\mathfrak{m}$, the existence of a finitely generated $R$-module $M$ such that $\mathfrak{m}M$ has finite projective dimension and $\mathfrak{m}M\neq 0$ implies R is regular.</p> <p>Applied to $M=\mathfrak{m}^{n-1}$, this implies that if any nonzero power of the maximal ideal has finite projective dimension, then $R$ is regular.</p>