What is the unmixedness theorem?

Question

I want to show that $R$ is Noetherian local ring and $S$ is Gorestein local ring s.t. $R=S/J$ and $f$ is an $R$-regular element, if for every $p\in\operatorname{Ass}_{S}(R),\operatorname{ht}_{S}(p)=\operatorname{ht}_{S}(J)$, then for every $p\in\operatorname{Ass}_{S}(R/f),\operatorname{ht}_{S}(p)=\operatorname{ht}_{S}(J)+1$.

But I am confused.

Let $R$ be a Cohen-Macaulay ring. Then for every ideal $I \subset R$, $p\in\operatorname{Ass}_{R}(R/I)$, $\operatorname{ht}(p)=\operatorname{ht}(I)$. Is this the unmixedness theorem?

Manos · Accepted Answer · 2014-04-09T14:30:34.287

2

An ideal $I$ of a ring $R$ is called unmixed if it has not embedded associated primes, i.e. if $\operatorname{height} P = \operatorname{height}I$ for every $P \in \operatorname{Ass}_R (R/I)$.

We say that the unmixedness theorem is true, if for every ideal $I$ of height $r$ that is generated by $r$ elements, $I$ is unmixed.

Finally, we have the theorem that a Noetherian ring $R$ is Cohen-Macaulay if and only if the unmixedness theorem is true.

So i don't believe that what you wrote in your first paragraph is true. You need $I$ to be generated by $\operatorname{height}I$ elements.

edited Apr 09 '14 at 14:30

answered Apr 08 '14 at 23:39

Manos

26,949

Thank you! $\operatorname{Ass}{R}(I)$ means $\operatorname{Ass}{R}(R/I)$? – user141617 Apr 09 '14 at 11:54
Yes, that was a typo, thanks. I fixed it. – Manos Apr 09 '14 at 14:30
Than you very much . Maybe What I wrote in first paragraph is not true.Thanks – user141617 Apr 12 '14 at 12:07

What is the unmixedness theorem?

1 Answers1