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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 $f=0$.

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    $\begingroup$ 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 $A = \mathbb{C}[x,y]$ $\endgroup$ – Sebastian Sep 8 '11 at 15:21
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No. Look at the ring $A=\mathbb{Z}$.

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  • $\begingroup$ Yes, you are right! $\endgroup$ – Li Zhan Sep 8 '11 at 15:45
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    $\begingroup$ This example shows the value of looking at examples! $\endgroup$ – Mariano Suárez-Álvarez Sep 8 '11 at 17:36
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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 $1$.

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  • $\begingroup$ Thank you for your comprehensive answer! $\endgroup$ – Li Zhan Oct 25 '11 at 20:15
  • $\begingroup$ 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. $\endgroup$ – Georges Elencwajg Jan 12 '12 at 23:13
  • $\begingroup$ @GeorgesElencwajg, you are right as usual. Thanks ! I will edit the answer. Très bonne année 2012 ! $\endgroup$ – user18119 Jan 13 '12 at 21:12
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    $\begingroup$ Dear @GeorgesElencwajg, for future reference this "unexpected" result is Theorem 144 in Kaplansky, Commutative Rings, 1974. $\endgroup$ – user26857 Apr 27 '14 at 7:01
  • $\begingroup$ Thanks for the reference, user26857. $\endgroup$ – Georges Elencwajg Apr 27 '14 at 7:47

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