# Depth of ideals in a commutative ring

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

$R = k[a,b,c,d]$, $I= (a, b)\cap (c,d)$ 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 $\mathrm{depth}R/J = 2$.
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.