Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $R$.

share|improve this question

closed as off-topic by user26857, Ivo Terek, Alizter, drhab, Joonas Ilmavirta Dec 8 at 19:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Ivo Terek, Alizter, drhab, Joonas Ilmavirta
If this question can be reworded to fit the rules in the help center, please edit the question.

    
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). –  El Angel Exterminador 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

2 Answers 2

up vote 0 down vote accepted

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.

share|improve this answer
    
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 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.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.