Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A$ be a Noetherian domain. Is the set $\{P\subset A \mid P \mbox{ prime ideal, } \dim A_P=1\}$ always finite?

I can prove for $f \neq 0, f\in A$, the set $\{P\subset A \mid \dim A_P=1, f\in P\}$ is finite (by using the primary decomposition of $\sqrt{(f)}$). The above statement is just the case when $f=0$.

share|cite|improve this question
If you think about this geometrically, the primes of height 1 correspond to the subvarieties of codimension 1 of $\text{Spec}(A)$ (at least if A is nice). There are lots of examples of noetherian domains A such that $\text{Spec}(A)$ has an infinite number of codimension 1 subvarieties, for example $A = \mathbb{C}[x,y]$ – Sebastian Sep 8 '11 at 15:21
up vote 13 down vote accepted

No. Look at the ring $A=\mathbb{Z}$.

share|cite|improve this answer
Yes, you are right! – Li Zhan Sep 8 '11 at 15:45
This example shows the value of looking at examples! – Mariano Suárez-Alvarez Sep 8 '11 at 17:36

If $\dim A=0$, then the answer is obviouly yes.

If $\dim A=1$, then the answer is yes if and only if $A$ is semi-local (i.e. has only finitely many maximal ideals). This is because height one prime ideals are then maximal.

If $\dim A>1$, then there are always infinitely many prime ideals of height 1. Actually, $A$ has a prime ideal $\mathfrak m$ of height $>1$. EDIT [After taking the quotient of $A$ by a minimal prime ideal appearing in a chain of prime ideals contained in $\mathfrak m$ of positive length, one can assume that $A$ is integral.]End of Edit. Let $a\in \mathfrak m$ be non-zero. Then the minimal prime ideals over $a$ have height $1$ by Krull's principal ideal theorem. Pick one of them $\mathfrak p_1$. Pick $a_1\in \mathfrak m\setminus \mathfrak p_1$ and let $\mathfrak p_2$ be a minimal prime ideal over $a_1$. Then $\mathfrak p_2\ne \mathfrak p_1$. Let $a_2\in \mathfrak m\setminus (\mathfrak p_1\cup \mathfrak p_2)$ and pick a minimal prime ideal $\mathfrak p_3$ over $a_2$ and so on. You get an infinite sequence of prime ideals of height $1$.

share|cite|improve this answer
Thank you for your comprehensive answer! – Li Zhan Oct 25 '11 at 20:15
A really unexpected result (for me), dear QiL! A little nitpick : a minimal prime ideal over $a$ might have height $0$, not $1$, if $a$ is nilpotent. But of course that would only strengthen your argument that there exists $a_1\in \mathfrak m\setminus \mathfrak p_1$ ! Et meilleurs voeux pour 2012. – Georges Elencwajg Jan 12 '12 at 23:13
@GeorgesElencwajg, you are right as usual. Thanks ! I will edit the answer. Très bonne année 2012 ! – user18119 Jan 13 '12 at 21:12
Dear @GeorgesElencwajg, for future reference this "unexpected" result is Theorem 144 in Kaplansky, Commutative Rings, 1974. – user26857 Apr 27 '14 at 7:01
Thanks for the reference, user26857. – Georges Elencwajg Apr 27 '14 at 7:47

Your Answer


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.