Proposition 1

Let $A$ be a noetherian integral domain, $x$ a nonzero element of $A$ and $p$ an element minimal in the set of prime ideals of $A$ containing $x$. Then $p$ is of height 1.

Let $q \subset p$ be a prime ideal distinct from $p$. One has $x \notin q$ in view of the minimal character of $p$. Since $A$ is an integral domain, $A_p$ can be identified with a subring of $A_q$; for every integer $n \geq 0$, let $q_n$ denote the ideal $q^n A_q \cap A_p$ of $A_p$. The minimal character of $p$ means that the local ring $A_p / xA_p$ is of dimension 0; hence it is of finite length ($§ 1$, no 3, example 1), and there exists an integer $n_0 \geq 0$ such that one has

Fix the integer $n \geq n_0$. Given $y \in q_n$, there exists $a \in A_p$ such that $y - ax \in q_{n+1}$; one then has $ax \in q_n$, whence $a \in q_n$ since $x \notin q$, and finally one has $y \in q_{n+1} + xq_n$. Hence one has

Since $x$ belongs to the maximal ideal of the noetherian local ring $A_p$, Nakayama's lemma shows that one has $q_n = q_{n+1}$. Since one has $(qA_q)^n = q_n A_q$, one concludes

Since the ring $A_q$ is local and noetherian, one has $\bigcap_{n \geq 0} (qA_q)^n = \{0\}$ (III, § 3, no 2, corollary of the prop. 5) whence $(qA_q)^{n_0} = \{0\}$ and finally the prime ideal $qA_q$ of $A_q$ is reduced to 0. Hence $q = \{0\}$, which proves that $p$ is of height 1.