A kind of ideals in Gorenstein local rings.

Asked Sep 30 '22 at 21:53

Active Sep 30 '22 at 21:59

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Let $(R,m)$ be a Gorenstein local ring. Then there exists an irreducible ideal $I\subset R$, such that $\sqrt I=m$, and $\operatorname{pd}_RI<\infty$.

I am looking for such $I$. Clearly, if such $I$ exists, $R/I$ has to be Gorenstein. And as an $R$-module, $R/I$ is of finite injective dimension.

I know such $I$ can't be $m$. Otherwise, $R$ has to be regular.

edited Sep 30 '22 at 21:59

asked Sep 30 '22 at 21:53

Bromelain

In any Cohen-Macaulay local ring. $R$, take $I$ to be an ideal generated by a maximal regular sequence. Then $R/I$ has finite projective dimension. – Mohan Oct 01 '22 at 22:00
@Mohan So the Gorenstein condition implies $R/I$ is also Gorenstein, which means $I$ is irreducible, right? – Bromelain Oct 01 '22 at 22:09
Yes. Does that answer your questions? – Mohan Oct 02 '22 at 14:53
@Mohan Yes, thanks, – Bromelain Oct 02 '22 at 17:05

A kind of ideals in Gorenstein local rings.

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