# Cohomological dimension and height of ideals

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.

-
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

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.
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