Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
    
If $R$ is regular, every $R$-module has finite projective dimension. –  Konstantin Ardakov Sep 10 '12 at 9:14
    
I know it. But i ask: Is it true that if $pd(R/mathfrak{m}^2)<\infty$, then $R$ is regular? provided $\dim R > 0$. –  Pham Hung Quy Sep 10 '12 at 10:08

1 Answer 1

up vote 12 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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