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

Let $I, J \subset S = k[x_1,\dots,x_n]$ be two monomial ideals and $k$ a field. If every element of $S$ which is $S/J$-regular is also $S/I$-regular is it true that depth$_S S/I \geq$ depth$_S S/J$ ?

share|improve this question
    
What if we add $I \subset J$ ? –  Andrei Jul 7 '13 at 21:24

2 Answers 2

up vote 5 down vote accepted

$R = k[a,b,c,d]$, $I= (a, b)\cap (c,d)$ and $J = (ac, bd) = (a, b) \cap (a, d)\cap (c, b)\cap (c, d)$. Then $\mathrm{Ass}R/I \subseteq \mathrm{Ass}R/J$. Hence if $x$ is a regular element of $R/J$ then $x$ is a regular element of $R/I$ but $\mathrm{depth}R/I = 1$ and $\mathrm{depth}R/J = 2$.

share|improve this answer
    
Thanks, but what if we add, as I said above, the condition $I \subset J$ ? –  Andrei Jul 8 '13 at 20:10
    
If your question has an affirmative amswer then $R/I^{(n)}$ is Cohen-Macaulay iff $R/I^{(n)}$ is Cohen-Macaulay for all $m, n$, where $I^{(n)}$ denotes the $n$th-symbolic power of $I$. This is not true in general (See arxiv.org/pdf/1003.2152.pdf). –  Pham Hung Quy Jul 10 '13 at 2:05

The answer is no without the monomial ideal hypothesis. I do not know whether it is true with that hypothesis. There exists prime ideals $P\subset S$ such that $S/P$ is not Cohen-Macaulay but, and ideal $J\subset P$ with radical of $J$ equals $P$ and $S/J$ Cohen-Macaulay. Taking $P=I,J$ in your question, gives examples where the above inequality does not hold. It is easy to construct characteristic $p>0$ examples. For a characteristic zero example, see the review of a paper by Cowsik and Nori, MR0393004.

share|improve this answer
    
Can u please give me the name of the paper? –  Andrei Jul 6 '13 at 17:07
    
Cowsik-Nori, On Cohen-Macaulay rings, J. Alg. (38), 1976 –  Mohan Jul 7 '13 at 16:25

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.