Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have asked this question on MS however I did not receive any answer. Please help me to solve it. Thank you very much!

Let $R$ is a commutative Noetherian graded ring and $I$ is an ideal of $R$, $I'$ to be the ideal generated by all the homogeneous element contained in $I$. Prove that $\operatorname{ht}(I)-1\le\operatorname{ht}(I')\le\operatorname{ht}(I)$.

share|improve this question
3  
This was posted on math.SE: math.stackexchange.com/questions/148294/… @variete: You should not withhold this (on both sides). –  Martin Brandenburg May 23 '12 at 16:03

3 Answers 3

The answer follows from Theorem 1.5.8 of Bruns/Herzog "Cohen-Macaulay-Rings": The basic reason is that for any non-graded prime $p$ one has $\text{ht}(p/p') = 1$. To see this replace $R$ by $R/p'$ and assume that $p'=0$ and that $p$ does not contain a nonzero homogeneous element. Now invert all homogeneous elements, passing to $R_{(0)}$ and use the fact that in a graded ring $R$ all homogeneous elements are invertible if and only if $R$ is a field or $R_0 = k$ is a field and $R=k[t,t^{-1}]$ for some homogeneous element $t\in R$ of positive degree and transcendental over $k$ (BH, Lemma 1.5.7). Since $R_{(0)}$ has the non-zero prime $pR_{(0)}$ we get $\text{ht}(p) = 1$. That all said, choose $p'\supset I'$ with $\text{ht}(p') = \text{ht}(I')$. The first inequality follows since $\text{ht}(IR_{(0)})=1$ and in the original ring we can conclude $\text{ht}(I) = \text{ht}(p') + 1 = \text{ht}(I') + 1$. The second inequality is obvious since $I'\subset I$.

share|improve this answer
    
Sorry, but in your answer you has just dealt with only prime ideal. My question is about a general ideal. –  variete May 23 '12 at 15:31
    
I've updated my answer. –  Thomas Kahle May 24 '12 at 10:25
    
Dear Thomas Kahle, in the last line of your answer, you concluded that $\text{ht}(I)=\text{ht}(I')+1$, but we have to prove that $\text{ht}(I)\le\text{ht}(I')+1$ and why do we get $\text{ht}(IR_{(0)})=1$ ? I am so sorry if you feel my question is stupid. Thank you very much for your answer. –  variete May 24 '12 at 18:35
    
Hi variety, that's not a stupid question. The case only $ht(I) = ht(I')$ occurs when $I$ is already homogenous. I excluded that case in the beginning: I assumed that $p$ is non graded (and I'll also assume that $I$ is not graded, otherwise everything is trivial). We get $\text{ht}(IR_{(0)}=1$ since only the zero ideal has height zero and $IR_{(0)}$ is not the zero ideal since it contains a non-homogeneous element. –  Thomas Kahle May 24 '12 at 20:54

I think the above answer is enough to prove $ht \; I \le ht \; I' +1 $ if $I \neq I'$. Pick a prime ideal $q$ that contains $I'$ such that $ht\; q = ht\; I'$. Recall that $q$ is homogenous since it is a minimal prime over a homogeneous ideal. By going modulo $q$ we may assume that $q = 0$. Now use the fact stated above by localizing at (0) as above. The image of $I$ is not zero since $I$ in $R_{(0)}$ is not in $q$. And it is not the whole ring since we are only inverting homogeneous elements that are not in $q$ which contains $I'$. Then since $R_{(0)}$ is a PID in particular one diemsimonal domain, the height of the image of $I$ is 1. Now apply the above result again to conclude that $ht I \le ht I' + 1$.

-------- Added 5/27/2012 -----------------------------------------------------

I don't think $ht \;I = ht\; I' + 1$ if $I \neq I'$ in general. Here is an example. Let $R = k[x_1,\dots,x_n]_{(x_1,\dots,x_n)}$ and $m = (x_1,\dots,x_n)R$. Let $I = (f)+m^s$ where $s$ is large and $f \in m$ is an non-homogeneous polynomial whose terms have degree less than $s$. Then we have $m^s \subset I' \subset I \subset m$. Hence $ht \; I' = ht \; I \; (= d)$.

I needed to say "if $ht\; I \neq ht \;I'$" instead of "if $I \neq I'$" in my original answer. This happens because there may exist a homogeneous prime ideal that is minimal over both $I$ and $I'$. In other words, the converse of the statement "every minimal prime of a homogeneous ideal is again homogeneous" is not true.

share|improve this answer
1  
Dear @Youngsu : Could you please tell me when the equalities occur? –  variete May 24 '12 at 18:41

This is a comment to 23950's question.

To me, it seems like you are asking two different questions.
(1) How does height of an ideal in a ring related to height of it image in a factor ring?
(2) How does height behave under localizations?

For 1): I don't think it behaves well in general. If a ring is caterary and equidimensional than it behaves better. For example, Lemma 2, p. 250, Matsumura, Commutative Ring Theory, says

  • If an equidimensional local ring $(A,m)$ is catenary then ht $p_2$ = ht $p_1$ + ht $(p_2/p_1)$ for all $p_1,p_2 \in$ Spec $A$ with $p_1 \subset p_2$.

For 2): Consider, $R = k[x,y,z]$ and $I = (xz,yz)$. Notice that ht $I = 1$ since $I \subset (z)$. Localize at $z$. Then $R_z = k[x,y,z,z/1]$ and $IR_z = (xz,yz)R_z = (x,y)R_z$. Hence ht $IR_z = 2$. This is a famous example that R/I is not Cohen-Macaulay at $(x,y,z)$ since it is not equidimensional (a line passes though a plane).

--------------------------------------------------------------- added June 3rd 2012

This is to answer the question below Height of ideal in graded ring .

Hi. Assume ht $I \neq $ ht $I'$. We can find a homogeneous prime $q$ of the same height as $I'$. By going modulo $q$ we may assume the following

  • $R$ is graded,
  • $I$ is not zero, $I' = 0$,
  • $q = 0$.

Now we goto the localization $R_{(0)}$ as introduced by Thomas above. Since $I$ is not zero this ring is isomorphic to $k[t,t^{-1}]$ which is a PID. Hence $IR_{(0)}$ is of height $1$ and contained in a maximal ideal Q of height $1$. Here you can see $Q' = 0$. This shows that the preimage of $Q$ in $R$ contains $I$ and its $'$ is $q$. Now use 1.5.8b in Brunz-Herzog's book.

In other words, there exists a prime ideal $Q$ in $R$ such that $q = Q'$ and $I \subset Q$. Hence ht $I -1 \le$ ht $Q -1 = $ ht $q = $ ht $I'$ where the equality ht $Q - 1 = $ ht $q$ is from 1.5.8b.

share|improve this answer
    
A useful inequality for one is: In any Noetherian ring $R$ with proper ideal $I$ you have $\text{ht}(I) + \text{dim}(R/I) \leq \text{dim}(R)$. –  Thomas Kahle May 29 '12 at 9:32
    
@Youngsu If the height of an arbitrary ideal behaves so badly when localize and take factor rings, how can you conclude in your answer to the original problem that $\text{height }I\le\text{height }I'+ 1$? In fact you got the following: $\text{height }(I+q)/q\le 1$ in $R/q$. Is this enough to prove that $\text{height }I\le\text{height }I'+ 1$? Or maybe I didn't get your reasoning? –  user26857 Jul 20 at 22:29

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.