2
$\begingroup$

Let $(R,\mathfrak m)$ be a Noetherian local ring.

Definition: $I$ is called locally complete intersection ideal if $I_p$ is a complete intersection for all $p\in V(I)$.

I want an example of an ideal $I$ satisfying the following three properties:

1) grade$(I)\geq 1$,

2) $I$ locally complete intersection ideal but not an $\mathfrak m$-primary and complete intersection ideal,

3) $I$ is not integrally closed.

Any suggestion or reference will be extremely helpful. Thank you in advance.

$\endgroup$
5
  • $\begingroup$ What is a locally complete intersection and non-complete intersection ideal of a local ring? $\endgroup$
    – A.G
    Commented Feb 9, 2017 at 17:19
  • $\begingroup$ An ideal I is called a locally complete intersection if its localization at every prime ideal containing I is complete intersection. $\endgroup$ Commented Feb 11, 2017 at 4:28
  • $\begingroup$ An ideal I is called a complete intersection if it is generated by regular sequence. $\endgroup$ Commented Feb 11, 2017 at 4:28
  • $\begingroup$ In a noetherian local ring both definitions are obviously the same, so you cannot find an ideal satisfying one and not the other. $\endgroup$
    – A.G
    Commented Feb 11, 2017 at 22:27
  • $\begingroup$ Do you know how to construct a variety from the embedding of $\mathbb{P}^1$ in $\mathbb{P}^3$ using $\mathcal{O}(3)$? $\endgroup$
    – 54321user
    Commented Jun 9, 2017 at 20:24

1 Answer 1

1
$\begingroup$

Let ,$R=\mathbb{Z}/3\mathbb{Z}[x,y,z,w]/(x^4+y^3+z^4)$. Then, $y\in (x,z)^F\subseteq (x,z)^*\subseteq \overline{(x,z)}$, and thence $(x,z)$ is not integrally closed, however $x,z$ is a regular sequence because $R$ is a complete intersection and $x,z,w$ is a system of parameters for $R$.

$\endgroup$
1
  • $\begingroup$ I am mainly interested in the case where the ideal itself is not complete intersection. I have examples when the ideal itself is complete intersection. Sorry for not mentioning that in the question. $\endgroup$ Commented Feb 9, 2017 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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