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$_S S/J$ ?
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$\begingroup$ What if we add $I \subset J$ ? $\endgroup$– AndreiJul 7, 2013 at 21:24
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2 Answers
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$R = k[a,b,c,d]$, $I= (a, b)\cap (c,d)$ 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 $\mathrm{depth}R/J = 2$.
answered Jul 8, 2013 at 14:50
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$\begingroup$ Thanks, but what if we add, as I said above, the condition $I \subset J$ ? $\endgroup$– AndreiJul 8, 2013 at 20:10
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$\begingroup$ 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). $\endgroup$ Jul 10, 2013 at 2:05
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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.
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$\begingroup$ Cowsik-Nori, On Cohen-Macaulay rings, J. Alg. (38), 1976 $\endgroup$– MohanJul 7, 2013 at 16:25