Below is a very simple direct proof for any ring $\,R\,$ of algebraic integers.
Lemma $\ $ If $\ I \supsetneq P\ $ are ideals of $\,R,\,$ with $\,P\,$ prime then there is an integer $\,f_k\in I\,$ but $\,f_k\not\in P\,.$
Proof $\ $ Choose $\,\alpha\in I,\ \alpha\not\in P\,.\,$ Being an algebraic integer, $\,f(\alpha) = 0\,$ for a monic $\,f(x)\in \mathbb Z[x],\ $ $\,f(x) \, =\, x^n +\cdots + f_1\ x +\, f_0\,.\,$ Note $\,f_n = 1\not\in P\,.\,$ Let $\,k\,$ be least with $\,f_k\,\not\in P.\,$ $\ f(\alpha) = 0\in P\,$ $\, \Rightarrow\,$ $\,(\alpha^{n-k}+\cdots+f_k)\ \alpha^k\in P,\ \alpha\not\in P$ $\,\Rightarrow\,$ $\,\alpha^{n-k}+\cdots+f_k\in P\subset I\,.\,$ So $\ \alpha\in I$ $\,\Rightarrow\,$ $\,f_k \in I\,.\ $ QED
Corollary $\, $ A proper chain of prime ideals in $\,R\,$ cannot contract to a shorter such chain in $\,\mathbb Z\,.$
Remark $\ $ Alternatively, reduce to the simpler case $\,P = 0\,$ by way of factoring out the prime $\,P\,.\,$ Then $\,f_k\,$ is the constant term of a minimal polynomial for $\,\alpha\,$ over a domain, so $\,f_k\ne 0,\,$ i.e. $\,f_k\not\in P,\,$ which is explained in much detail in this post, as a generalization of rationalizing denominators.