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.

A well-known fact:

Let $R$ be a Noetherian ring and $I$ an ideal. Then $${\rm ht}(I) \leq {\rm cd}(I,R).$$

I looked, but could not find the proof of this fact.

I want to see the proof of this fact.

share|improve this question
    
Could you include the definition of $cd(I,R)$? –  Martin Brandenburg Jan 2 '13 at 14:10
    
@MartinBrandenburg $cd(I,M)=\sup\{i:H_I^i(M)\neq 0\}$ –  user26857 Jan 2 '13 at 16:47
add comment

1 Answer

up vote 1 down vote accepted

Set $n=\operatorname{ht}I$ and take $\mathfrak p$ a prime ideal containing $I$ such that $\operatorname{ht}\mathfrak p=n$. Then we have $$H_I^n(R)_{\mathfrak p}\simeq H_{IR_{\mathfrak p}}^n(R_{\mathfrak p})\simeq H_{\mathfrak pR_{\mathfrak p}}^n(R_{\mathfrak p}),$$ where the last isomorphism holds because $\sqrt{IR_{\mathfrak p}}=\mathfrak pR_{\mathfrak p}$.

But $\dim R_{\mathfrak p}=n$ and then $H_{\mathfrak pR_{\mathfrak p}}^n(R_{\mathfrak p})\neq 0$. In particular, $H_I^n(R)\neq 0$. This shows the desired inequality.

share|improve this answer
    
Sorry, another question: What is $\mathrm{ara}(I)$? –  Martin Brandenburg Jan 2 '13 at 18:43
    
@MartinBrandenburg Arithmetical rank of $I$. –  user26857 Jan 2 '13 at 19:09
    
@YACP I know that ${\rm cd}(I,R) \leq {\rm ara}(I)$... –  Sang Cheol Lee Jan 3 '13 at 0:58
    
@YACP Thanks, YACP –  Sang Cheol Lee Jan 3 '13 at 6:23
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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