Height of ideal in graded ring

Question

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 $\operatorname{ht}(I)-1\le\operatorname{ht}(I')\le\operatorname{ht}(I)$.

Answer 1

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 $I'\subset I$.

Answer 2

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.

Answer 3

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.