# $\operatorname{depth} R-\dim(R/I)\le \operatorname{grade} I$

Could any one help me to solve this one? If $R$ be a Noetherian Local ring then $\operatorname{depth} (R)-\dim(R/I)\le \textrm{grade} (I)$ for any ideal $I$ of $R$?

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How is $grade(I)$ defined? –  Giovanni De Gaetano May 14 '12 at 14:50
Let R be a Noetherian, M is f.g R-module,let $I\subseteq R$ be an ideal such that $IM\neq M$, then all maximal $M$ regular sequences in $I$ have the equal length and is denoted by grade(I,M). –  Une Femme Douce May 14 '12 at 15:09
So, by $grade(I)$ you mean $grade(I,I)$? Is it clear $I^2\neq I$? –  Giovanni De Gaetano May 14 '12 at 15:14

Let $(R,m)$ be our Noetherian local ring, by definition $\text{depth}(R)=gr(m).$ In general, for any proper ideal $J$ of a Noetherian ring ( as I noted, in your recent question) $gr(J)\leq ht(J).$ It is now, enough to show that $gr(m) \leq gr(m/I)+gr(I),$ which you need nothing more that the definition of a regular sequence and the fact that for an $R$-module ($R$ is not necessarily local) $M$, giving an $M$-sequence $x_1,\cdots ,x_n$ is equivalent to giving an $M$-sequence $x_1,\cdots, x_i$ (for a fixed $1 \leq i \leq n$) and an $M/(x_1,\cdots, x_i)M$-sequence $x_{i+1}, \cdots, x_n.$ I hope, it will be helpful.

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I think there is something wrong. The equality between the depth and the height holds only when $R$ is Cohen-Macaulay. –  user18119 May 14 '12 at 19:36
@Qil: Am I using the equality? –  Ehsan M. Kermani May 14 '12 at 19:38
you said depth(R)=ht(m) in the first line, unless I misunderstood. –  user18119 May 14 '12 at 19:42
dear @Qil: oh, yes, you're right! I meant $gr(m),$ thanks! –  Ehsan M. Kermani May 14 '12 at 19:49

This is exercise 1.2.23 from Bruns & Herzog, Cohen-Macaulay Rings. Use induction on $\textrm{grade}(I)$ and reduce the problem to the case $\textrm{grade}(I)=0$. Then choose a prime $p\in\text{Ass}(R)$ such that $I\subset p$ and apply Proposition 1.2.13 from the same book.

Remark. ehsanmo "proved" that $\operatorname{depth} (R)-\operatorname{depth}(R/I)\le \textrm{grade} (I)$ which is obviously wrong: consider the case $\textrm{grade}(I)=0$ and find a counterexample or look at http://mathoverflow.net/questions/21484/two-questions-about-cohen-macaulay-rings, question #2.

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