Let $A$ be a Noetherian ring, and $I\subset A$ a proper ideal of height $r$. Is it true that there exist $a_1,\ldots,a_r\in I$ such that $$\operatorname{ht}(a_1,\ldots,a_j)=j$$ for all $j=1,\ldots,r$ ? It sounds easy but somehow I can't seem to prove it. If I can find an element $a\in I$ such that $\operatorname{ht}(a)=1$ then maybe I can use induction, but how do I find such $a$ ? The following (well-known) fact comes to my mind:

If $\mathfrak{p}\subset A$ is a prime ideal of height $r$, then there exist $a_1,\ldots,a_r\in\mathfrak{p}$ such that $\mathfrak{p}$ is minimal over $(a_1,\ldots,a_r)$.

but I don't see how it can be of help.

Edit: I also know that if $b\in I$ and $\operatorname{ht}(b)=1$, then $\operatorname{ht}(I/(b))=\operatorname{ht}I-1$.


1 Answer 1


Since $R$ is Noetherian, every ideal of $R$ has finitely many minimal primes. If $r = 0$ then the result holds vacuously. If $r \ge 1$, then by definition $I$ is not contained in any minimal prime of $R$, so by prime avoidance there exists $a_1 \in I \setminus \bigcup_{p \in \text{Min}(R)} p$. Then $\text{ht}(a_1)> 0$, so $\text{ht}(a_1) = 1$ by Krull's Altitude Theorem (which finishes the case $r = 1$).

If $r > 1$, then $I$ is not contained in any minimal prime of $(a_1)$, so there exists $a_2 \in I \setminus \bigcup_{p \in \text{Min}(a_1)} p$. Then $\text{ht}(a_1, a_2) > 1$ (otherwise $(a_1,a_2) \subseteq p$ for some height $1$ prime $p$, so necessarily $p \in \text{Min}(a_1)$, contradicting choice of $a_2$), so $\text{ht}(a_1, a_2) = 2$ by the altitude theorem.

This process can continue until a length $r$ sequence $a_1, \ldots, a_r \in I$ is reached: having constructed $a_1, \ldots, a_i \in I$ satisfying $\text{ht}(a_1, \ldots, a_i) = i$, pick $a_{i+1} \in I \setminus \bigcup_{p \in \text{Min}(a_1, \ldots, a_i)} p$, which will satisfy $\text{ht}(a_1, \ldots, a_{i+1}) = i+1$. By construction, $\text{ht}(a_1, \ldots, a_j) = j$ for every $j = 1, \ldots, r$.

  • $\begingroup$ My understanding is that Krull's altitude Theorem requires $a_1$ to be a non-zero-divisor. The condition $a_1\not\in\bigcup_{\mathfrak{p}\in\operatorname{Min}(R)}\mathfrak{p}$ is not enough for $a_1$ to be a non-zero-divisor, is it? $\endgroup$
    – ashpool
    Commented Sep 28, 2014 at 4:11
  • 2
    $\begingroup$ @ashpool: Krull's Altitude Theorem does not require any assumptions about being a nonzerodivisor - all that is required is that the ideal is proper. It is true that $a_1 \not \in \bigcup_{p \in \text{Min}(R)} p$ does not imply being a nonzerodivisor, if the ring is not reduced $\endgroup$
    – zcn
    Commented Sep 28, 2014 at 4:32

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