If $R$ is a Noetherian commutative ring with unity having finitely many height one prime ideals, one could derive from the "Principal Ideal Theorem", due to Krull, that $R$ has finitely many prime ideals (all of height less than or equal to $1$). It may comes to mind that, in general,

if $R$ is a commutative Noetherian ring with unity of finite Krull dimension $d$ has finitely many prime ideals of height $n$ with $0<n≤d$, then $n=d$.

My question: is this generalization valid?

I shall appreciate any help.


Suppose that $0<n<d=\dim R$. Let $\mathfrak p$ be a prime ideal of height $d$. There is a chain of prime ideals $\mathfrak p_0\subset\cdots\subset\mathfrak p_d=\mathfrak p$. Somewhere in this chain there is a prime ideal $\mathfrak q$ of height $n$ which is contained in a prime ideal $\mathfrak q'$ of height $n+1$ and contains a prime ideal $\mathfrak q''$ of height $n-1$. By Theorem 144 from Kaplansky, Commutative Rings, it follows that there are infinitely many prime ideals between $\mathfrak q''$ and $\mathfrak q'$ and all have height $n$.

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    $\begingroup$ Theorem 144 from Kaplansky, Commutative Rings, it is proved in the answer of zcn. $\endgroup$ – user26857 May 16 '14 at 20:11

This is true - in a Noetherian ring, the only possible heights for which there are finitely many primes of that height are either $0$, or maximal ideals. This follows from the following fact:

Proposition: If $R$ is Noetherian and $\mathfrak p_0 \subsetneq \mathfrak p_1 \subsetneq \mathfrak p_2$ is a chain of distinct prime ideals in $R$, then there are infinitely many primes $\mathfrak q$ such that $\mathfrak p_0 \subsetneq \mathfrak q \subsetneq \mathfrak p_2$.

Proof: Localizing at $\mathfrak p_2$ and quotienting by $\mathfrak p_0$, it suffices to show that any Noetherian local domain of dimension $2$ has infinitely many primes. If there were only finitely many height $1$ primes, then since $\mathfrak p_2$ is not contained in any of them, by prime avoidance it is not contained in their union, so there exists $x \in \mathfrak p_2$, but not in any height $1$ prime. But this contradicts the Principal Ideal Theorem, since $\text{ht}(x) \le 1$.

  • $\begingroup$ "the only possible heights... are either $0$, or maximal ideals". The second part means height equal to $\dim R$? $\endgroup$ – user26857 May 16 '14 at 20:17
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    $\begingroup$ @user26857: Yes, modulo issues of equidimensionality. Such a statement is tricky to make - perhaps a clearer but equivalent version is: if $p$ is a nonminimal and nonmaximal prime of height $n$, then there are necessarily infinitely many primes of height $n$ $\endgroup$ – zcn May 16 '14 at 20:24

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