10
$\begingroup$

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 $pd(R/\mathfrak{m}^2)< \infty$?

$\endgroup$
  • $\begingroup$ If $R$ is regular, every $R$-module has finite projective dimension. $\endgroup$ – user91132 Sep 10 '12 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 $\dim R > 0$. $\endgroup$ – Pham Hung Quy Sep 10 '12 at 10:08
18
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
6
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.