# $\operatorname{depth} R-\dim(R/I)\le \operatorname{grade} I$; Bruns and Herzog, Cohen-Macaulay Rings, Exercise 1.2.23 [closed]

Could any one help me to solve this one?

If $R$ is 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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## closed as off-topic by user26857, Ivo Terek, Alizter, drhab, Joonas IlmavirtaDec 8 '14 at 19:05

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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). –  La Belle Noiseuse 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 grade(I)=n. Then there exists a regular $R$-sequence $x_1,...,x_n$ contained in $I$. Thus $x_1,...,x_n$ is contained in $\def\fm{\mathfrak m}\fm$. Therefore depth$(R/(x_1,...,x_n))\leq\dim R/\def\fp{\mathfrak p}\fp$ for all $\fp\in$ Ass$(R/(x_1,...,x_n))$. Thus depth $R-n\leq\dim R/{\fp}\leq\dim R/I$ and so the result follows.
depth$(R/(x_1,...,x_n))\leq\dim R/\def\fp{\mathfrak p}\fp$ follows from Proposition 1.2.13 in Bruns and Herzog, so I can't see any novelty in your answer compared to the older one. –  user26857 Nov 18 '14 at 19:01
Use induction on $\textrm{grade}(I)$ and reduce the problem to the case $\textrm{grade}(I)=0$. Then choose a prime $\mathfrak p\in\text{Ass}(R)$ such that $I\subset\mathfrak p$ and apply Proposition 1.2.13 from the same book.