Let $(R, \mathfrak{m})$ be a Noetherian local ring. It is well know that

$R$ is regular iff $pd(R/\mathfrak{m}) < \infty$ (i.e. $R/\mathfrak{m}$ has finite projective dimension).

Assume that $\dim R > 0$. Is $R$ regular, if $pd(R/\mathfrak{m}^2)< \infty$?

  • $\begingroup$ If $R$ is regular, every $R$-module has finite projective dimension. $\endgroup$
    – user91132
    Sep 10, 2012 at 9:14
  • 1
    $\begingroup$ I know it. But i ask: Is it true that if $pd(R/mathfrak{m}^2)<\infty$, then $R$ is regular? provided $\dim R > 0$. $\endgroup$ Sep 10, 2012 at 10:08

2 Answers 2


Levin-Vasconcelos (journal link): 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.

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.


If $R$ has positive dimension, then for any $t\ge 1$, the ideal $I=\mathfrak m^t$ is a so called Burch ideal (as defined by Dao, Kobayashi, Takahashi; Burch ideals and Burch rings ) i.e. $\mathfrak m(I:\mathfrak m)\ne \mathfrak mI$.

It is a Theorem of Burch that if $I$ is a Burch ideal and $R/I$ has finite projective dimension, then $R$ is regular.


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