Proposition 5

Let A be a noetherian ring and $p \subset q$ a non-saturated chain of prime ideals of A. The set E of prime ideals r of A such that $p \subset r \subset q$ is a chain is infinite. One has $\bigcup_{r \in E} r = q$ and $\bigcap_{r \in E} r = p$.

Replacing A by $A/p$ if necessary, one is reduced to the case where $p = \{0\}$.

By Lemma 1 of No. 1, one has $q = \bigcup_{r \in E} r$, and Prop. 2 of II, $§ 1$, No. 1 shows that E is infinite.

Let $y \neq 0$ be an element of $\bigcap_{r \in E} r$. The height of q is finite (No. 1, Cor. 1 of Prop. 2), and one has $\mathrm{ht}(q) \geq 2$ by hypothesis. There therefore exists a prime ideal $q' \subset q$ of height 2. The first part of the proof applied to $q'$ shows that the set E’ of prime ideals of height 1 contained in $q'$ is infinite; each of these ideals contains y by hypothesis. Now the local noetherian ring $B = A_{q'}/yA_{q'}$ is of dimension 1 by Corollary 2 of Prop. 2 of No. 1. For every $r \in E'$, the prime ideal $r/yA_{q'}$ of B is therefore minimal; consequently, the noetherian ring B has an infinity of minimal prime ideals, which is absurd (II, $§ 4$, No. 3, Cor. 3 of Prop. 14). Hence $\bigcap_{r \in E} r = \{0\}$.